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ELEMENTS OF 
PEACTICAL AERODYNAMICS 



ELEMENTS OF 

PRACTICAL AERODYNAMICS 



BY 

BRADLEY JOJMES, M.S. 

Professor of Aeronautics^ University of Cincinnati 
M.LAe.S. 



SECOND EDITION 



NEW YORK 

JOHN WILEY & SONS, Inc. 

London: CHAPMAN & HALL, Limited 
1939 



4 ^'iJ3;is 

J76 ^5. 



sciENced 

UBRARt 



Copyright, 1936, 1939 
By BRADLEY JONES 



All Rights Reserved 

This book or any part thereof must not 
be reproduced in any form without 
the written permission of the publisher. 



Printed in U. S. A. 8-41 



Printing Composition Binding 

I. GILSON CO. TECHNICAL COMPOSITION CO. STANHOPE BINDERY 

BOSTON BOSTON BOSTON 



K; 



PREFACE TO SECOND EDITION 

Since the first presentation of this book, it has seemed desirable 
to include additional material which will be of value to a student 
and which properly merits place in an elementary textbook. For 
this reason in the present edition there has been added a descrip- 
tion of N.A.C.A. airfoils, the standard taper of a monoplane wing, 
a graphical method of developing a streamline contour, a mathe- 
matical treatment of range and take-off distance, and an abridge- 
ment of Oswald's method of rapid prediction of performance. In 
various places throughout the book, especially in the chapter on 
longitudinal balance, the treatment of the subject has been 
enlarged in an attempt to clarify it for the reader. 

In revising the original edition, the author wishes to acknowl- 
edge the valuable criticisms and suggestions of Professors 
A. Lavrow of the Detroit Institute of Technology, W. A. Bevan 
of the Iowa State College, V. W. Young of the Oklahoma A. and M. 
College, and T. E. Butterfield of Lehigh University. The author 
is especially grateful to his colleague at the University of 
Cincinnati, Mr. Mortimer Powell, for his valuable editorial 
assistance. 

Bradley Jones 

University of Cincinnati 
January, 1939 



138159 



PREFACE TO FIRST EDITION 

The existence of several excellent works on the subject of 
aerodynamics demands that an explanation be offered for the ap- 
pearance of a new one. 

My experience in teaching has convinced me that a book in- 
tended primarily for the classroom must be, in general, different 
in character and scope from books intended for reference by prac- 
ticing engineers. It is attempted in this textbook to present the 
subject matter as simply as possible. In two isolated cases of 
summation of forces, it is necessary to introduce the calculus to 
prove the correctness of certain statements. If these statements 
are accepted as being true, the proofs may be dispensed with, and 
no mathematical training higher than elementary algebra is needed 
for complete understanding of the text. 

Although aerodjTQamics deals with air in motion and aerostatics 
with air at rest, a short chapter on aerostatics has been included 
because it is believed that students of aerodynamics should be 
acquainted with the elements of the sister subject. 

Use has been made of data from various publications of the 
National Advisory Committee for Aeronautics and the Army and 
Navy Designers Handbook. I desire to acknowledge useful 
criticisms and suggestions from Professor Harold W. Sibert of 
this University. 

Bradley Jones 

University of Cincinnati 
April, 1936 



CONTENTS 

CHAPTER PAGE 

I. PHYSICAL PROPERTIES OF AIR 1 

II. EFFECTS OF DEFLECTING AIR STREAMS 11 

III. AIR FLOW 22 

IV. AIRFOILS 28 

V. INDUCED DRAG OF MONOPLANES 75 

VI. INDUCED DRAG OF BIPLANES 104 

VII. PARASITE DRAG 125 

VIII. ENGINES 138 

IX. PROPELLERS 148 

X. AIRPLANE PERFORMANCE 183 

XI. PERFORMANCE AT ALTITUDE 230 

XII. TURNS 251 

XIII. THE CONTROL SURFACES 265 

XIV. STABILITY 275 

XV. AUXILIARY LIFT DEVICES 299 

XVI. UNCONVENTIONAL TYPES OF AIRCRAFT 303 

XVII. MATERIALS AND CONSTRUCTION 311 

XVIII. INSTRUMENTS 322 

XIX. METEOROLOGY 357 

XX. AVIGATION 366 

XXL AEROSTATICS 384 

APPENDIX A. NOMENCLATURE 403 

APPENDIX B. ANSWERS TO PROBLEMS 419 

INDEX 427 



ELEMENTS OF 
PRACTICAL AERODYNAMICS 



CHAPTER I 
PHYSICAL PROPERTIES OF AIR 

Aerodynamics is the study of the motion of air and of the forces 
on soHds in motion relative to air. 

In a fluid the particles cohere so slightly that they may be easily 
made to change their relative positions by the application of small 
forces; i.e., the modulus of shear is small. The amount of in- 
ternal friction of a fluid may be very small, and one may conceive 
of a perfect fluid as a fluid wherein there is no internal friction, so 
that between two particles the action of any force must be normal 
to the contact surfaces, and cannot have any tangential component. 

Air is often referred to as a perfect fluid. Air is a gas. It 
is a physical mixture, not a chemical compound. 

The earth's atmosphere at sea-level has the following percentages 
by volume of these gases: 

Nitrogen 78.08 

Oxygen 20.94 

Argon 0.94 

Hydrogen 0.01 

Neon 0.0012 

Helium 0.0004 

Carbon dioxide 0.03 

Water vapor also is always present, the amount varying with the 
temperature and other factors, but averaging about 1.2 per cent 
at the earth's surface. 

Up to altitudes encountered by aircraft (40,000 ft.) there are 
always winds and vertical air currents to keep the various con- 
stituents commingled in approximately the same proportion, as 
listed. At high altitudes, undoubtedly the different gases com- 
posing the atmosphere separate according to their respective 

1 



2 PHYSICAL PROPERTIES OF AIR 

densities, hydrogen forming the outermost layer, helium the next, 
and so on. 

Because air is a gas, its density varies with temperature, 
pressure, etc. The standard conditions most commonly used 
(N.A.C.A.) are a barometric pressure of 29.92 in. (or 760 mm.) of 
mercury, and a temperature of 59° F. (or 15° C). 

Under these conditions, the mass density (p) of dry air is 
0.002378 slug per cubic foot. 

Using for standard acceleration of gravity (g) equal to 32.1740 
ft. per sec. per sec, the specific weight of " standard " dry air is 
0.07651 lb. per cu. ft. "* 

Up till 1926 the National Advisory Committee for Aeronautics 
used for standard conditions a barometric pressure of 29.92 in. of 
mercury and a temperature of 60° F. (15.6° C). The mass 
density of dry air under these conditions is 0.002372 slug and the 
weight density (g = 32.172 ft. per sec.^) is 0.07635 lb. per cu. ft. 
In using data from N.A.C.A. reports or other sources of informa- 
tion, care should be taken to note the standard conditions of the 
test. 

Air conforms to Boyle's law that at constant temperatures 
and up to certain Hmits the volume varies inversely as the pressure 
and to Charles's law that at constant pressure the volume varies 
directly as the absolute temperature. These two laws are usually 
expressed as the formula: 

PoVo PiVi 



where Po, Vq, and To are pressure, volume, and absolute tempera- 
ture under standard conditions, and Pi, Fi, and Ti are under 
other than standard conditions. 

As weight density is the weight divided by the volume, the 
density of a gas is increased by an increase in pressure, a decrease 
in volume, or a decrease in temperature. In a given volume, the 
density, as well as specific weight, varies directly as the pressure 
and inversely with absolute temperature. 

PO Po T 

Example. What is the density of dry air if the pressure is 25.93 
in. of mercury and the temperature is 45° F.? 



PHYSICAL PROPERTIES OF AIR 



3 



The absolute temperature (T) =45 -t^4"59.4 = 504.4. 



P_ To _ 25.93 

Po T ^° 29.92 "" 504.4 



5_1m/4 



002378 = 0.00212 



Example. What is the specific weight of dry air if the pressure is 
16.38 in. of mercury and the temperature is —10° F.? 
First method: 



^P_ To 

^ Po T^' 29.92 



16.38 ^ 518^ ^ 0.002378 = 0.001502 



449.4 



Specific weight pXg = 0.001502 X 32.1740 = 0.04832 lb. per cu. ft. 

Second method: 

Specific weight = 0.07651 X j^ X |^ = 0.04832 lb. per cu. ft. 
^ 29.92 449.4 



Problems 

1. Find the density of dry air at 20 in. pressure and 10° F. 

2. Find the density of dry air at 18.52 in. pressure and 0° F. 

3. Find the specific weight of dry air at 24 in. pressure and 25° F. 

TABLE I 

Altitude-Pressure-Density Relation 
Based on N.A.C.A. No. 218 



Altitude, 


Temperature, 


Pressure, 


p_ 


i/^ 




ft. 


°F. 


in. Hg 


PO 


Vp 


p 





59.0 


29.92 


1.000 


1.000 


0.002 378 


1000 


55.4 


28.86 


0.9710 


1.0148 


. oof 309 


2 000 


51.8 


27.82 


.9428 


1 . 0299 


.0#242 


3 000 


48.4 


26.81 


.9151 


1 . 0454 


002 176 


4 000 


44.8 


25.84 


.8881 


1.0611 


: 002 112 


5 000 


41.2 


24.89 


.8616 


1.0773 


.002 049 


6 000 


37.6 


23.98 


.8358 


1.0938 


.001987 


7 000 


34.0 


23.09 


.8106 


1.1107 


.mi 928 


8 000 


30.6 


22.22 


.7859 


1.1280 


.001869 


9 000 


27.0 


21.38 


.7619 


1.1456 


.001812 


10 000 


23.4 


20.58 


.7384 


1.1637 


.001756 


11000 


19.8 


19.79 


.7154 


1 . 1822 


.001701 


12 000 


16.2 


19.03 


.6931 


1.2012 


.001648 


13 000 


12.6 


18.29 


.6712 


1 . 2206 


. 001 596 


14 000 


9.2 


17.57 


.6499 


1 . 2404 


. 001 545 


15 000 


5.5 


16.88 


.6291 


1 . 2608 


.001496 


20 000 


-12.3 


13.75 


.5327 


1.3701 


. 001 267 


25 000 


-30.1 


11.10 


.4480 


1 . 4940 


. 001 065 


30 000 


-48.1 


8.88 


.3740 


1.6352 


. 000 889 


35 000 


-65.8 


7.04 


.3098 


1.7961 


.000 736 


40 000 


-67.0 


5.54 


.2447 


2.0215 


.000 582 


45 000 


-67.0 


4.36 


.1926 


2.2786 


.000 459 


50 000 


-67.0 


3.44 


.1517 


2.5674 


.000 361 



4 PHYSICAL PROPERTIES OF AIR 

Atmosphere. The air in the atmosphere close to the earth is 
compressed by the weight of the air above it. At higher altitudes, 
the air is under less pressure because there is less air above to cause 
pressure. The evidence from observations of meteors appears to 
indicate that our atmosphere extends upward at least 500 miles. 
In the upper limits the air is greatly rarified so that there is no 
exact demarcation at the upper edge. 

Heat is radiated from the sun, and this radiation passes through 
our atmosphere without any appreciable heating effect. The 
sun's radiation heats the earth. The layer of air resting immedi- 
ately on the earth is heated by conduction. This air in turn 
warms air superadj acent to it. Also the warm air will rise, and, 
in ascending to a region where the pressure is less, it will expand. 
If no heat is added or subtracted, which will be true if the upward 
movement is fairly rapid, when the air expands, its temperature 
drops. With increase in altitude above the earth, the temperature 
decreases. This decrease in temperature with altitude continues 
until at high altitudes the temperature has decreased to —72° F. 
(at latitude 45°); then no further decrease in temperature takes 
place with increased altitude. 

The lower region of the earth's atmosphere is called the tropo- 
sphere. In the troposphere, the temperature decreases with alti- 
tude, winds may blow from any direction, and there is moisture in 
the air so that there may be clouds. The upper region of the 
earth's atmosphere is called the stratosphere. In the strato- 
sphere, the temperature is constant, not varying with altitude; 
any winds that may blow are from a westerly direction and there 
is no moisture, consequently no clouds. 

The dividing surface between the troposphere and the strato- 
sphere is called the tropopause. The altitude of the tropopause is 
56,000 ft. in the tropics, about 38,000 ft. over the United States, 
and about 28,000 ft. in the polar regions. 

For convenience in aeronautics, a " standard " atmosphere has 
been adopted; that is, it has been agreed to assume that the 
temperature, pressure, and, consequently, density are fixed and 
constant for any altitude. This hypothetical '^standard " atmos- 
phere assumes that there is no moisture present in the air. This 
standard atmosphere represents average conditions at 40° latitude, 
but at any one time there may actually be considerable diver- 
gence from this standard, especially at low altitudes. The 



ATMOSPHERE 6 

standard atmosphere has been approved by the International 
Commission for Air Navigation and is therefore frequently referred 
to as the I. C.A.N, atmosphere. The standard atmosphere is 
given in Table I. 

The standard atmosphere assumes that at sea-level the baromet- 
ric pressure is 29.92 in. and the temperature is 59° F. (15° C). 
It assumes further that the temperature decreases 1° F. for every 
280 ft. increase in altitude up to a height of 35,332 ft., where the 
temperature is —67° F. ( — 55° C). From this altitude upward 
the temperature is assumed to be constant. It is assumed that 
the air is dry, that the change in acceleration due to gravity 
is negligible, and that air conforms to Boyle's and Charles' 
laws. 

The temperature at any altitude is found by subtracting the 
temperature drop with altitude from the standard sea-level 
temperature. 

r = 518.4 — aZ where T = absolute tempera- 

ture at altitude 
of Z feet 
a = standard tem- 
perature gradi- 
ent, 0.003566° F./ft. 

The difference in barometric pressure at two different altitudes 
is due merely to the weight of the column of air of unit cross- 
section between the two heights. In order to ascertain this 
weight, it is necessary first to know the average temperature of 
the column. Because the density varies with the altitude, the 
harmonic mean temperature is used. For altitudes below 35,332 
ft., the mean temperature is found by the following formula. 

^ aZ where Tm = mean temperature (abs.) 

"" , To To = standard temperature (abs.) 

^^'To - aZ Z = altitude in feet 

If the mean temperature (Tm) is known, the barometric pressure 
(p) at any altitude (Z) maybe found by the formula, due to Laplace; 
, po _ ppgTo y where po = standard pressure 
p poTm Po = standard density 

To = standard temperature 
g = standard acceleration of gravity 



6 PHYSICAL PROPERTIES OF AIR 

By substituting the values of standard conditions and using 
70.726 lb., the weight of one square foot of mercury, one inch high, 
to transform pressures from pounds per square foot to inches of 
mercury, the above formula is modified to 

o ono« 1 29.921 _ 0.002378 X 32.174 X 518.4 X Z 
2.3026 logio^ 29.921 X 70.73 X T. 

0.002378 X 32.174 X 518.4 X Z 



logiop = logio 29.921 - 



2.3026 X 29.921 X 70.73 X T, 
1.47597 - 0.0081398^ 

i- m 



m 



Example. Find the temperature, pressure, and density at 18,000 ft. 
altitude in standard atmosphere. 
Solution. 

T = 518.4 - aZ 

= 518.4 - 0.003566 X 18,000 

= 518.4 - 64.2 

= 454.2° F. absolute 

= -5.2° F. 

J- m — rp 



To - aZ 
64.2 



2.3026 logio^^ 
= 485.6° F. absolute 

logiop = 1.47597 - 0.0081398^ 

J- in 

= 1.47597 - 0.0081398^^ 

= 1.17423 
p = 14.936 in. of Hg. 



= 4^)(5) 



_ 0002378 yi^X^^ 
- 0.002378 X 29 92 X 454.2 

= 0.001355 slug per cu. ft. 



Problems 

1. Find the density of air in the standard atmosphere at 21,000 ft. 
altitude. 



VISCOSITY 7 

2. Find the density of air in the standard atmosphere at 27,000 ft. 
altitude. 

3. Find the density of air in the standard atmosphere at 32,500 ft. 
altitude. 

For altitudes above 35,332 ft., the formula for mean tempera- 
ture becomes 

392.4 Z 
^^- Z - 4704.9 

Example. Find the temperature, pressure, and density at 40,000 ft. 
in standard atmosphere. 
Solution. 

T = 392.4° F. absolute (isothermal temperature) 
= -67.0° F. 
_ 392.4 Z 
Tm - z - 4704.9 

392.4 X 40,000 



40,000 - 4704.9 
= 444.7° F. absolute 

logio P = 1.47597 - 0.0081398 |- 

= 1.47597 - 0.0081398^5^ 
P = 5.54 in. of Hg. 



' = '{fX^) 



-0 002S78V-5^V^^ 
- U.UU2d78 X 29 92 X 392 4 

= 0.000582 slug per cu. ft. 

Problems 

1. Find the density of air in the standard atmosphere at 38,200 ft. 
altitude. 

2. Find the density of air in the standard atmosphere at 44,400 ft. 
altitude. 

3. Find the density of air in the standard atmosphere at 49,200 ft. 
altitude. 

Viscosity. The viscosity of air is much less than for water or 
other liquids. It is not zero, nor in some calculations is it even 
negligible. In liquids, viscosity is due to the internal friction of 
the molecules. In gases, viscosity is due not so much to internal 
friction as to molecular vibration. Consider a rapidly moving 
stratum of gas immediately over another stratum of gas moving 



I 



8 



PHYSICAL PROPERTIES OF AIR 



more slowly in the same direction. Some molecules from the 
upper layer, owing to their vibratory motion, will wander into the 
lower layer and will accelerate the motion of the lower strata. 
Molecules passing from the lower to the upper stratum will retard 
the latter. Any mingling of the molecules between the two strata 
results in the two velocities becoming more nearly equal. 

In Uquids, an increase in temperature causes a decrease in vis- 
cosity, because the intermolecular friction is less. In gases, an 
increase in temperature causes an increase in viscosity, because 
there is an increase in molecular vibration and therefore an 
increase in molecular interchange. 

The coefficient of viscosity (ijl) of air varies approximately as 
the I power of the absolute temperature. 

TABLE II 

Coefficient of Viscosity 



Temperature, 


C.g.s. units, 


Slug-feet-second 


°C. 


poises 


units 


-30 


1 554 X 10-7 


3.25 X 10-7 


-20 


1 605 X 10-7 


3.35 X 10-7 


-10 


1 657 X 10-7 


3.46 X 10-7 





1 709 X 10-7 


3.57 X 10-7 


10 


1 759 X 10-7 


3.67 X 10-7 


20 


1 808 X 10-7 


3.78 X 10-7 


30 


1 856 X 10-7 


3.88 X 10-7 


40 


1 904 X 10-7 


3.98 X 10-7 


50 


1 951 X 10-7 


4.08 X 10-7 


60 


1 997 X 10-7 


4.17 X 10-7 


70 


2 043 X 10-7 


4.27 X 10-7 


80 


2 088 X 10-7 


4.36 X 10-7 


90 


2 132 X 10-7 


4.45 X 10-7 


100 


2 175 X 10-7 


4.54 X 10-7 



The N.A.C.A. uses a value of 3.73 X lO"^ for coefficient of vis- 
cosity at 15° C. 

Except for very high or very low pressure, the coefficient of 
viscosity, ju, is independent of pressure. 

Kinematic Viscosity. The coefficient of kinematic viscosity 
(v) is the ratio of the coefficient of viscosity (ju) to the density. 



V = 



Because the density of air is affected by both pressure and 



MODULUS OF ELASTICITY 9 

temperature, whereas the coefficient of viscosity is affected only 
by temperature, the coefficient of kinematic viscosity of air is less 
at high altitudes than at the ground. For example, at 15° C, the 
coefficient of kinematic viscosity (y) is 1.50 X 10""^ square feet per 
second; whereas, at —30° C, the coefficient of kinematic viscosity 
(v) is 1.21 X 10~4 square feet per second. 

The kinematic viscosity is a factor in Reynolds' number, which 
will be discussed in a later chapter. 

Modulus of Elasticity. For solids, the modulus of elasticity 
(E) is simply the ratio of stress to strain. In the case of gases 
under tension or compression, the temperature changes during 
stress should properly be considered. With rapid changes in the 
stress, such as would occur in compression waves, because of the 
low thermal conductivity of air, no appreciable heat is lost or 
gained so that the adiabatic law is assumed to be correct, i.e., that 

P7« = a constant where n = ratio of specific heats; for air, 

n = 1.405 

Since density varies inversely as volume, the adiabatic law may 
be written as 

P = kp"" 

By differentiating w4th respect to p 

Ctp 

The modulus of elasticity (E) being the ratio of stress to strain, 
if an increase of pressure from P to P + AP causes a change in 
volume from F to 7 — AV, so that the strain is — AV/V. 









V 


but 




' 


AV Ap 

V - p 


Therefore 






E^pf 
Ap 


Expressed as 


a derivative 




F-JP 
^-"dp 



t 






10 PHYSICAL PROPERTIES OF AIR 



but from the adiabatic law 

dP 



nkp^~^ 



dp 

E = pX nkp""-^ 
= nkp^ 
= nP 
Velocity of Sound. Sound is propagated through air by means 
of compression waves. Newton proved that the velocity of sound 
through a fluid varies directly as the square root of the modulus of 
elasticity and inversely as the square root of the density. 



IE 
Velocity of sound = V - 



For gases, however, 
P 
P 



p 
PV = — = RT where 72 = a constant for each gas (for air 



R = 3075) 
T = absolute temperature Centi- 
grade 
If this is substituted in the previous equation, 

velocity of sound = V nRT 

= Vl.405 X 3075 X T 
= Qd.WT 

For 15° C. (T = 288°), the velocity of sound is therefore 1,118 
ft. per sec. 

In designing propellers it is not considered good practice to have 
the tips rotating at a speed close to the velocity of sound. If the 
tip-speed corresponds to the velocity of sound, violent compression 
waves are initiated and there is great loss of power. At altitudes, 
the velocity of sound is greatly reduced and this must be reckoned 
with in designing propellers for sub-stratospheric flight. For 
example, at 40,000 ft. altitude (temperature, —55° C.) the velocity 
of sound is only 973 ft. per sec. 

It is to be noted that changes in the barometric pressure have no 
effect on the velocity of sound. 

BIBLIOGRAPHY 

Humphrey, Physics of the Air. 
Gregg, Aeronautical Meteorology. 
International Critical Tables. 



CHAPTER II 
EFFECTS OF DEFLECTING AIR STREAMS 

Isaac Newton formulated three well-known laws of motion. 

1. Every body persists in its state of rest or uniform motion 
in a straight line, unless it be compelled by some force to change 
that state. 

2. The rate of change of momentum of a body is proportional to 
the force acting on the body and is in the direction of the force. 

3. Action and reaction are equal and opposite. 

Air, though of less density than fluids or solids, does possess 
mass. It conforms to the laws of motion. If a mass of air is at 
rest, a force is required to put it in motion. If the mass of air 
is in steady rectilinear motion, a force is required to slow it, stop 
it, or change its direction. Air in motion has momentum. 

When a quantity of air is put in motion work has been done on 
the air and it has acquired momentum. Momentum is the prod- 
uct of the mass and the velocity. 

A cubic foot of air moving at a speed of V feet per second will 
have a momentum of pV. 

Kinetic Energy. A cubic foot of air at rest to be put in motion 
must have a force applied. Let F be the force applied to increase 
the velocity from zero to V feet per second. If this change in 
velocity takes place uniformly in t seconds, the acceleration is 
V/t feet per second per second. Since a force is measured by the 
mass times the acceleration, F = pV/t. In the t seconds during 
which the force is applied, the average velocity is J "T feet per sec- 
ond and the distance covered is this average velocity multiplied 
by the time in seconds, that is, ^Vt. The work done on the 
cubic foot of air in imparting to it a velocity of V feet per second 
is the product of the force and the distance through which the 
force acts. 

^ TF in foot-pounds 

Work = ^ X iVt = ipV^ p in slugs per cubic foot 

V in feet per second 
11 



12 EFFECTS OF DEFLECTING AIR STREAMS 

This moving mass of air is capable of doing work and is therefore 
said to possess energy of motion or kinetic energy. The kinetic 
energy is numerically equal to the work which was required to 
put the air in motion, that is, Jp V^ foot-pounds. 

Momentum Pressure. If a horizontal stream of moving air were 
to strike perpendicularly against a wall so as to lose all its momen- 
tum, the amount of momentum lost per second is equal to the mo- 
mentum of the quantity of air arriving at the wall per second, which, 
by Newton's second law, is numerically equal to the force exerted 
on the wall in pounds. If the air stream is A square feet in cross- 
section and has a velocity of V feet per second, the volume of air 
arriving at the wall each second is A F cubic feet. Since the mo- 
mentum of each cubic foot is the mass density times the velocity, 
pV, the total momentum oi AV cubic feet is pF multiphed by 
AV or pA V^. The force acting on the wall will be equal to this, or 

F in pounds 
p _ A-rr2 P^^ slugs per cubic foot 

A in square feet 
V in feet per second 

The pressure in pounds per square foot will be 

P in pounds per square foot 
P = pV^ p in slugs per cubic foot 

F in feet per second 

If conditions are altered so that, instead of a small stream strik- 
ing against a large flat surface, a large stream strikes against a 
small flat surface, the same reasoning apphes only that the area in 
the formula will be the area of flat plate instead of the cross-sec- 
tional area of the air stream. Again the force on the flat plate is 
exactly the same, whether the small plate is held stationary against 
a large stream of air which moves against the plate with a velocity 
of V feet per second, or whether the air is still and the plate is 
moved through the air with a velocity of V feet per second in a 
direction perpendicular to its flat surface. 

The foregoing presupposes that the air disappears or vanishes on 
arriving at the flat surfaces. Actually, when particles of air are 
stopped at the flat surfaces, they cannot get out of the way of the 
following particles of air ; some of the air is pocketed or trapped in 
the central portion of the plate; eddy currents and whirlpools of 



DYNAMIC PRESSURE 13 

air are created. The amount of eddying depends on the size and 
shape of the fiat surface. Because of these eddies, etc., the actual 
force against the flat surface is not exactly the same as given by the 
formula. Numerous tests have been made in which have been 
measured exactly the forces on flat plates of various sizes and 
shapes when in air streams of various velocities. The results of 
these tests have been quite consistent, so that, if the size and shape 
of the flat surface are known, the amount by which the true force 
differs from the theoretical force can be predicted. 

It is usual to multiply the theoretical force by a correction factor 
K to obtain the actual force on the flat surface. The magnitude 
of K varies slightly with size if the area is only a few square feet; 
it also varies with the shape, i.e., whether the flat surface is circular, 
square, or rectangular. Except for very precise work, in aero- 
nautics, it is customary to neglect these variations of K and as- 
sume that it has a constant value of 0.64, so that the equation for 
force on a flat plate normal to an air stream becomes 

p in slugs per cubic foot 
F = 0.64 pA 72 A in square feet 

V in feet per second 

For standard atmosphere this becomes 

F = 0.64 X 0.002378^72 
= 0.00152 A 72 

Since it is customary to measure airspeed in miles per hour, it 
is sometimes convenient to have a formula wherein velocity is in 
miles per hour. 

F = 0.64 X 0.002378 X ^ X (||7)2 I. ^^P^.^^^^ 

= 0.00327 A 72 '7m miles per hour 

A in square feet 

Example. A 40-mile-per-hour wind is blowing against a sign- 
board 8 ft. by 10 ft. in size. Atmosphere is normal density. What 
is the force acting against the signboard? 
Solution: F = 0.00327 A V^ 

= 0.00327 X 8 X 10 X (40)2 
= 419 lb. 

The force on the flat plate varies as the square of the airspeed. 
If the force acting when the relative speed is 1 mile per hour is known, 



14 EFFECTS OF DEFLECTING AIR STREAMS 

the force at any other speed can be found by multiplying by the square 
of the airspeed. 

Example. The force against an automobile windshield is 0.012 lb. 
when the car is moving forward at 1 mile per hour. What is the 
force when the car is traveling at 35 miles per hour? 

Solution: F (pounds) = 0.012 X 35 X 35 = 14.7 lb. 

Work and Power. The accomplishment of motion against the 
action of force tending to resist it is defined as work. Work is 
expressed in units of force times distance, for example, foot- 
pounds or mile-pounds. No time is involved. In overcoming a 
force of X pounds through a distance of y feet, the same work of 
xy foot-pounds is done whether accomplished in a short or long 
time. 

Power involves the element of time. To do the same work in 
half the time means that twice the power is required. Power is 
expressed in units of work divided by time. Work is force times 
distance. Velocity is expressed in units of distance divided by 
time. Then power, which is force times distance divided by time, 
is also force times velocity. It is usual to express power in terms 
of an arbitrary unit, a horsepower, which is 550 ft-lb. per sec. 
1 hp. =[550 ft-lb. per sec. 

= 33,000 ft-lb. per min. 

= 1,980,000 ft. lb. per hr. 

= 375 mile-lb. per hr. 

That is, overcoming a force of 1 lb. at a speed of 375 miles per 
hour, or of 375 lb. at a speed of 1 mile per hour, or 15 lb. at a speed 
of 25 miles per hour, requires 1 hp. 

Except for minor corrections, if the plate is smaller than 12 
sq. ft. in area, the force on a flat plate due to meeting air perpen- 
dicular to its surface is 0.64 pA7^ If this is multiplied by the 
velocity to get power, then power is 0.64 pA V^. 

1.28 1 A 73 H.P. in horsepower 

H.P. = rrT^ p in slugs per cubic foot 

A in square feet 
= 0.00233 1 ^ 73 y in feet per second 

for standard density 

H.P. = 0.00000276 AV^ F in feet per second 

H.P. = 0.0000087 A 73 7 in miles per hour 



INCLINED FLAT PLATES 15 

Example. A flat surface 5 ft. square is moved through the air at 
a speed of 30 miles per hour in a direction perpendicular to its surface; 
what horsepower is required to do this? 

Solution. The force on the plate will be 

F = 0.00327 X (5)2 X (30)2 
= 73.6 lb. 

the horsepower required will be 

73.6 X 30 



H.P. = 
or directly 



375 

= 5.88 hp. 



H.P. = 0.0000087 X 25 X (30)^ 

= 5.88 

Example. Driving at 1 mile per hour, the force on a certain auto- 
mobile windshield is 0.1 lb.; what horsepower is used up by the wind- 
shield at 70 miles per hour? 

F = 0.1 X 70 X 70 
= 490 lb. 
490 X 70 



H.P. = 



375 
91.5 hp. 



Problems 

1. What is the total force of a 40-mile-per-hour wind on a hangar 
door 40 ft. by 20 ft.? 

2. An auto windshield is 42 in. wide by 12 in. high. Neglect cor- 
rection for size and shape. What horsepower is used up by the wind- 
shield at 50 miles per hour? 

3. The radiator of an automobile is 18 in. wide by 24 in. high. 
With shutters closed, what horsepower is expended in overcoming its 
head resistance at 45 miles per hour? 

4. What horsepower is required for radiator in problem 3 if speed is 
60 ft. per sec? 

Inclined Flat Plates. If the stream of air does not strike the 
flat surface perpendicularly, as considered in the last few para- 
graphs, but at an angle so that the air on hitting the plate moves 
off parallel to the surface, the problem is not the same. Let a 
be the angle between the direction of the air and the plane surface. 
If A is the area of the flat surface, the cross-section of the air stream 
which comes in contact with the surface will have an area of 



^ \>t^-SurfaceofA 


_^ \ sq. tt. 


^ N. 




Fig. 1. Air striking an 
inclined plate. 



16 EFFECTS OF DEFLECTING AIR STREAMS 

A sin a square feet. If the air stream is moving at a speed of V 
feet per second, then A F sin o; cubic feet of air will approach the 
flat surface every second. This volume of AF sin a cubic feet 
will have a mass of pAV sin a slugs and a momentum of pAV^ 
sin a pounds. If all this momentum were given up and lost by the 
moving air, then the force on the flat sur- 
face would be equal to the momentum or 
pAV^ sin a pounds. Since the air on 
meeting the plate is diverted and slides 
off in a direction more or less parallel to 
the surface it still retains some momen- 
tum. Also there are bound to be some 
eddy currents formed when the air stream meets the flat surface. 
The amount of turbulence will depend on the angle at which 
the air meets the surface. At small angles the turbulence will 
be relatively little, gradually increasing as the angle increases 
to 90°. 

Energy is required to put air in motion, and energy consumed in 
moving air about in whirlpools or eddy currents will subtract 
from the energy of the moving air. 

The value of pA V^ sin a pounds can be considered as a maximum 
value of the force acting on the plate, provided that the effect of 
friction is neglected. 

Necessarily when air flows along a surface there is bound to be 
friction. The moving air tends to drag the plate along in the 
direction parallel to the plane of the plate. If there were no fric- 
tion, the only force on the flat plate due to the moving air would be 
perpendicular to the surface of the plate. With friction, there 
would be additional force parallel to the surface of the plate. The 
combination of these two forces is a resultant which is not perpen- 
dicular to the plate but is away from the perpendicular in the 
direction of flow of the air. 

In aerodynamic work, the force resulting from air meeting a sur- 
face is not as important as the two components of this force per- 
pendicular and parallel to the direction of the air stream. The 
component force perpendicular to the direction of the air stream is 
called the lift; the component parallel to the air stream is called 
the drag. 

The components of lift and drag will vary with the density of 
the air, the area of the flat surface, and the square of the velocity. 



INCLINED FLAT PLATES 



17 



.24 



.20 



.16 



.12 



.60 



The forces of lift and drag can therefore be expressed as a coefficient 

multipUed hy^AV^ (A in square feet, V in feet per second). The 

coefficient will be different for every different angle at which the 
flat plate is set with respect to the air stream. Figure 2 gives the 

value of the lift coeffi- 
cient {Cl) and drag co- 
efficient ((?£>) plotted 
against angle of attack, 
the use of these coeffi- 
cients being explained 
later in this text. 

The coefficients given 
in Fig. 2 were obtained 
by actual tests on a rec- 
tangular plane 90 cm. by 
15 cm, in size, the plate 
being set so that the air 
stream first meets one of 
the long edges. These 
coefficients are correction 
factors to allow for losses 
or gains due to turbu- 
lence. The amount of 
turbulence would be dif- 
ferent if the plate were 
turned so that a short 
side were first to meet the air; consequently the coefficients in this 
case would not be the same as in Fig. 2. If the ratio of long 
side to short side differs from six the coefficients will also be 
different. 

Problems 

1. A signboard is 18 ft. long by 3 ft. wide. A 30-mile-per-hour 
wind is blowing at an angle of 10° to the plane of the signboard, 
(a) What is the force in pounds on the signboard at right angles to 
the wind direction? (6) What is the force parallel to the wind di- 
rection? (c) What is the resultant of these two forces? {d) What 
is the component perpendicular to the face of the signboard? 

2. A kite having 6 sq. ft. area is balanced by its tail so that it 
slants 15° to the horizontal. What is the lifting force in a 10-mile- 
per-hour wind? 



,08 



.04 

















J 
















/ 
















/ 












Cd- 


V 
















/ 














/ 


^ 














/ 


r 




^Cl 








/ 


/ 














/^ 


/ 












/ 


/ 












/. 


y y 












/ 


/ 














/ 

















Fig. 2. 



.2 



2 4 6 8 10 12 14 16 
Angle of Attack in Degrees 

Graph of Cl and Cd for flat plate. 



18 



EFFECTS OF DEFLECTING AIR STREAMS 



3. A flat surface 6 ft. by 2 ft. is subject to a 25-mile-per-hour wind 
which comes in a direction 6° to the surface, (a) What is the force 
in pounds on the surface perpendicular to the wind? (6) What is 
the force parallel to the wind? (c) What angle does the resultant of 
these two forces make with the surface? 

4. The side of a freight car is 60 ft. long and 10 ft. high. What is 
the force against the side of the car due to an 8-mile-per-hour wind 
from a direction perpendicular to the side (a) when freight car is 
stationary; (b) when car is moving forward at 30 miles per hour? 

5. A flat plate is moving in a direction 15° from the plane at a speed 
of 250 ft. per sec. If plate has area of 50 sq. ft., what force applied 
in the direction of movement is required? 




Air striking curved plate. 



Curved Plates. A curved surface may be placed in a stream 
of air, so that the air meets the surface tangentially but is gradually 
deflected so that the air leaves 
the surface moving in a different 
direction from its original path 
(see Fig. 3). If the surface is 
smooth and there are no abrupt 
changes in direction, the air 
stream should suffer no dimin- 
ishment in speed. 

Velocity has both speed and 
direction. In deflecting the air 
stream the velocity has been 
changed. By the first law of motion, any matter in motion in a 
straight line continues in that straight line unless acted upon by 
an (outside) force. A restatement of this is that if the direction 
_d of motion is changed there must exist a 
force which produced the change. 

In Fig. 4, let ah represent in magnitude 
and direction the velocity (F) of the air 
stream before it meets the deflecting sur- 
face, ac the velocity as it leaves the 
surface, and e the angle between. Con- 
structing a parallelogram of velocity, ahcd, the diagonal he repre- 
sents the only velocity which combined with velocity ah will give 
a resultant velocity of ac. 

If the final velocity ac is equal in magnitude to entrance velocity 
ah, length ac equals length ah, and since ah = hd = ac = cd, 




Fig. 4. 



Forces on a curved 
plate. 



CURVED PLATES 19 

triangles ahc and dbc are equal, and angle abc equals angle chd. 
Then the direction of the velocity he bisects the angle ahd, and since 
angle ahd equals 180° — angle hac, angle ahc equals J (180° — e). 
By trigonometry, | side he = side ah X cos ahc. 

Side hc = 2V cos i(180 - e) 



2yJ ^ + cos (180 -6) 



2Y^ • - cos ^ 

.= 7\/2(l - cos e) 

To change the direction of the air stream, a velocity of the 
magnitude of 7x^2(1 — cos e) must be impressed on the moving 
air. The force required to produce this change is equal to the 
change in momentum, which is equal to the mass times the change 
in velocity. If it requires i seconds for a particle of air to traverse 
the curved surface and undergo deflection, the mass of air de- 
flected in i seconds will be pAVt. The change in velocity takes 
place during the time that the air is traversing the surface (t sec- 
onds) ; therefore the acceleration is -, as was shown 

previously. The deflecting force is therefore pA F2V2(1 — cos e). 

Assuming that there is no burbling, and neglecting friction, the 
direction of the deflecting force will bisect the angle between the 
directions of the air coming towards and leaving the surfaces. 
For every action there is an equal and opposite reaction, so if the 
plate acts on the air stream as explained, the air stream will react 
on the plate in an equal manner but opposite in direction as 
shown by F in Fig. 3. If there is friction, the moving air will 
tend to drag the plate along with it, so that there will be a sHght 
tangential force and the reaction will be more as shown by the 
dotted line F' in Fig. 3. 

If the curved surface be the part of a circular cyHnder, the air 
stream will follow a circular path while in contact with the surface. 
Let be the center and R the radius of the arc which is the path of 
the particles in the center of the air stream, as shown in Fig. 5. 
Let a be the position of a particle on the circular path and h be its 
position a brief interval of time, t seconds later. Then, since the 
velocity is V, the length of arc ah is Vt. Let ac and hd be tangents 



20 



EFFECTS OF DEFLECTING AIR STREAMS 



to the circular path at a and h respectively, and call the inter- 
section of these two lines point 
e. Draw lines from a, h, and 
e to 0, the center of the circle. 
Then the central angle aoh is 
equal to the exterior angle hec, 
and oe bisects angle aoh. If 
angle hec is called 6, angle aoe 
is 6/2. Draw ah intersecting 
oe at /. Provided angle aoh 
is small, line ah is very nearly 
equal in length to arc ah. As 
ah is Vt feet in length, af will 
be Vt/2 feet long. The sine 




Fig. 5. Forces on a circular 
cylindrical plate. 



of angle aoe will be af divided by oa, or 

. d Vt/2 



then 



«m 2 = 
2 sm - = 


R 

Vt 
R 


2^1 -^cos.^ 
FV2(1 - cos d) 


Vt 
R 

72 


t 


R 



In a previous paragraph, it has been shown that, if a curved 
surface deflects an air stream through a total angle of e, the ac- 



celeration or change in velocity per second is 



FV2(1 - cos e) 



If the surface is such that the air stream takes a circular path, and 

the radius is known, the acceleration 
can be expressed as V^/R. This is 
the usual way of expressing the ac- 
celeration due to centrifugal force. 

As an illustration of the application 
of this principle, let the fohowing 
example be considered. A rec- 
tangular metal sheet, 36 ft. by 6 ft., 
is bent as shown in Fig. 6 so that a 

horizontal stream of air striking the sheet tangentially is gently 




Fig. 6. Direction of reaction on 
circular cylindrical plate. 



CURVED PLATES 21 

deflected and leaves in a direction 10° below the horizontal. The 
air first meets the long edge of the deflecting surface. The air 
stream is 1 ft. thick; its velocity of 40 ft. per sec. is unchanged. 
What is the force of the air against the metal sheet, and in what 
direction does it act? 

In 1 sec, a volume of air 36 ft. wide, 1 ft. deep, and 40 ft. long 
meets the surface and, in having its direction changed, reacts 
against the plate. Since, in passing from front to rear, the air 
traverses a distance of 6 ft., the time required will be 6/40 or 0.15 
sec. In 0.15 sec, a volume of 0.15 X 36 X 1 X 40, or 216 cu. 
ft., will meet the surface. The mass of this volume of air will 
be 0.002378 X 216, or .514 slugs. The acceleration will be 

40V2(1 - cos 10°) _ ^ ,^ „, , .„ , 
^r-zr= = 46.5 it. per sec. per sec. Ihe force will be 

U. J.0 

0.514 X 46.5, or 23.8 lb. Its direction, neglecting friction, wifl 
be upward and backward, 5° from the vertical. 

Problems 

1. A curved surface, 36 ft. by 6 ft., deflects an air stream through 
an angle of 5°. The air stream is 1 ft. thick and first meets a longer 
side of the surface. The airspeed is 40 ft. per sec. before and after 
deflection, and there is no turbulence. What is the force against 
the surface? 

2. What is the force in problem 1 if the airspeed is 80 ft. per sec? 

3. What is the force in problem 1 if the angle of deflection is 2j°? 



CHAPTER III 
AIR FLOW 

If a body is moved through air, or if air flows around a body, 
the motion of the air may be either smooth or turbulent. Smooth 
flow is called continuous or streamline flow. Turbulent flow is 
discontinuous or sinuous. 

A stream of air may be conceived of as consisting of a number of 
particles moving in the same direction. On meeting an obstruc- 
tion, the path of some of these particles will be diverted. If the 
obstructing body is so shaped as to not cause a sudden and 
abrupt change in direction, the particles will move in new paths 
which are usually gentle curves. In this case, a particle, which 
was directly behind another particle in the original air stream, 
will follow exactly the new path taken by the leading particle in 
flowing around the obstruction. Succeeding particles will also 
foUow this path. A con- 
tinuous line describes this ^ ^ ~ 
path, and the flow is cafled — >- — ^ — 
streamline. Figure 7 il- ^ ^^ 
lustrates this flow pattern. 
Particle A follows the path Fig. 7. Streamline flow. 
AA'A", Particle B, im- 
mediately to the rear of particle A, follows the same path as A, as 
does particle C and all other particles behind C. 

In turbulent flow, the flow is discontinuous; that is, particles 
that follow each other in the original air stream do not follow in 
all cases the same path as preceding particles in flowing around 
obstructions. In Fig. 8, the air stream is shown meeting a flat 
plate. In Fig. 8a, particle A and a small number of succeeding 
particles will follow path AA' A" . Reaching the position A" ^ 
in order to continue in motion, particle A must break across path 
AA' A" , so that momentarily the flow is as shown in Fig. 86. 
Particle A moves onward as shown by path A" A'" A"". This 
particular path is followed only till most of the particles which 
moved into the vortex behind particle A have escaped onward, 

22 




SKIN FRICTION AND VISCOSITY 23 

when the original path (Fig. 8a) is resumed. The flow is thus 
intermittent, particles first following one path then another. 









(a) 




Fig. 8. Turbulent flow. 



Skin Friction and Viscosity. Even in so-called streamline flow 
around an object, some turbulence is present. The layer of air 
immediately adjacent to the body is retarded in its forward move- 
ment by friction with the surface of the body. No matter how 
smooth the surface may be, there is bound to be some retardation. 
This friction between moving air and a stationary body will 
always be greater than between air particles themselves. 

It is probably correct to assume that the layer of air in contact 
with the body has Uttle, if any, forward motion. The adjoining 
layer of air is slowed up by friction with the motionless layer. 
Contiguous layers affect each other, in that layers nearer the body 
act to retard layers farther away. This interaction continues 
through successive layers, the retarding effect diminishing with 
increasing distance from the surface. 

The viscosity and friction effects are very small, and it is only 
in a narrow zone clinging to the surface that this effect is notice- 
able. This region, in streamline flow, is only a few thousandths 
of an inch thick, and it is in this boundary layer that the velocity 
rises from zero at the immediate surface to its full value in the 
air stream. The velocity gradient, which is the ratio of velocity 
difference to distance between layers, is very high in the boundary 
layer, while very small in the main air stream. 

The effect of the different speeds is to destroy the streamUnes 

^ and break them up into tiny eddies 

as shown in Fig. 9, which is a greatly 

enlarged view of a portion of Fig. 7. 

Fig. 9. Turbulence in Boundary Consider a particle in a streamline 

Layer. close to but not touching the body. 

Owing to the velocity gradient, on the 

outer side it is being rubbed by particles moving forward with 

greater velocity, on the inner side it is being rubbed by particles 



24 AIR FLOW 

having a slower velocity. This tendency towards rotation of the 
individual particles serves to turn its path downward and causes 
minute vortices. 

In the thin boundary layer, the viscosity of air plays the pre- 
dominating part in determining the motion of the particles. Out- 
side of the boundary layer, viscosity has little effect and air may be 
considered as a non- viscous fluid. With fluids of greater viscosity 
than air, the boundary layer will be thicker. 

Reynolds' Number. The English physicist, Osborne Reynolds, 
made some interesting experiments with the flow of liquids in 
tubes. By using a glass tube, and injecting small streams of an 
insoluble colored fluid, he was able to study the form of flow. 

Using water as a fluid, and starting with one size of tube, he 
found that at low velocity the flow of the colored liquid in the water 
was a continuous line; that is, the flow was streamline or laminar 
except for an infinitesimal layer touching the side walls of the tube, 
where there were slight eddies. As he increased the velocity the 
flow remained streamline, though the outside layer in which 
burbling was taking place became slightly thicker. 

Increasing the rate of flow still more, he found that at a certain 
speed the main flow changed from streamline to turbulent. 
Above this critical speed the flow was turbulent; below, it was 
streamline. Increasing the diameter of the tube, he found that 
the critical speed, at which flow changed from streamline to 
turbulent, was lessened. Using fluid of different densities and 
different viscosities also varied the critical speed. 

He evolved an expression which is called Reynolds' number and 
commonly abbreviated as R.N. Reynolds' number is dimension- 
less. 

R.N. = '-^ 

V = average axial velocity (feet per second) 
R = inner radius of tube (feet) 
p = mass density of fluid (slugs per cubic foot) 
fx = coefficient of viscosity of fluid (pound-second 
per square foot) 
Reynolds found that for fluid flow through pipes or tubes if the 
R.N. was less than 1,100 the flow was laminar or streamline. 
Although later experimenters found a somewhat higher number 
as the critical condition, the value of 1,100 is the one commonly 



SIMILAR FLOWS 25 

used as being perfectly safe. Below that value, the flow is certain 
to be laminar. 

Reynolds' number is important to hydrauHc engineers in decid- 
ing the proper size of piping. With a given size of pipe, the 
quantity of fluid that can be conducted through the pipe increases 
with the velocity until the critical Reynolds' number is reached. 
With turbulent flow, the quantity of fluid is decreased greatly. 

It is to be noted that Reynolds' work was entirely with tubes 
and pipes. The critical value of Reynolds' number for flow inside 
of circular pipes is meaningless when the problem deals with flow 
of unconfined Siir. around objects, such as airplanes. It is con- 
jecturable, however, that even free air moving around an obstruc- 
tion may have a critical factor, corresponding to the Reynolds' 
number for pipes, which will differentiate whether the flow wil^)^ 
streamline or turbulent. ' ^0^ ' 

Similar Flows. The important appHcation of Reynolc^'wimber 
for the aeronautic engineer is its use in comparing the*iftow of air 7 
at different speeds around objects of varying size^ T|ie manner 
of air flow around an object, as shown in Fig. ^Oa, wouldl bfj 

(a) (6) 





(c) 
Fig. 10. Flow around similar figures. 

changed to a flow more like that shown in Fig. 106 if the speed 
of the air were increased. If the speed of flow was the same as 
for the flow shown in Fig. 10a, but the size of the object were 
increased, the flow would be as shown in Fig. 10c. By decreasing 
the velocity in an inverse ratio to the increase in size, i.e., keeping 



26 AIR FLOW 

VL the same, where L is a representative dimension of the 
object, the flow would be as shown in Fig. lOd. With the same 
VL the flows are geometrically similar. 

Again if the size of the object and speed of flow were the same 
as in Fig. 10a but the density of the air were increased, the flow 
would be changed to a flow, resembling that in Fig. 106. A 
decrease in coefficient of viscosity would have the same effect on 
the shape of the air flow as an increase of density or velocity. 

To summarize : If the Reynolds' number is the same, the flows 
are geometrically similar. With geometrically similar flows about 
two bodies of different sizes, at corresponding points in the two 
flows the direction of the streamlines will be the same and the 
magnitude of the forces will always have the same ratio to each 
other. 

This relation is important in the design of airplanes. Models of 
wings or airplanes may be tested in wind tunnels, and the results 
of these tests may be used in the computation of the performance 
of the full-size airplane, provided that the Reynolds' number of 
the model is the same as the Reynolds' number of the airplane. 
Since the models are small in size in comparison with the actual 
wings, in order to have the Rejmolds' number of the same magni- 
tude as the Reynolds' number of the wing the velocity or the 
density of the air in the wind-tunnel test must be much greater 
than in the actual flight. 

Where the Reynolds' number of the wind-tunnel test of a model 
wing is not the same as for the full-size wing in flight, the data 
obtained from the wind-tunnel tests cannot be expected to give 
exactly correct results when used in calculating forces on the 
full-size wings. At Langley Field, Virginia, the National Advisory 
Committee for Aeronautics has a variable-density wind tunnel. 
The tunnel is entirely enclosed in a steel shell. The air, after 
being drawn through the tunnel, past the model on test, flows 
around the outside of the tunnel proper to re-enter the throat 
once more. Since the air is at all times imprisoned inside the steel 
chamber it can be put under pressure. The Langley Field 
variable-density tunnel was designed for pressures as high as 21 
atmospheres and air velocities past the model wings up to 75 
ft. per sec. 

In computing Reynolds' number, velocity (V) must be in feet 
per second; a linear dimension of the object (L) must be in feet. 



SIMILAR FLOWS 27 

For wings, the length of the chord is commonly used for this 
dimension (L). If the test is conducted under standard conditions 
(15° C. and 760 mm. pressure) the density of the air (p) is 0.002378 
slug per cubic foot and the coefficient of viscosity (m) is 0.000000373 
slug per foot per second. If the temperature or pressure is not 
standard, corrections must be made to density (p) and coefficient 
of viscosity (jjl). 

Example. Find the Reynolds' number for a model wing of 3-in. 
chord, tests run at 100 miles per hour with standard air. 

3 in. = 0.25 ft. 
100 miles per hour = 146.7 ft. per sec. 

^ ^ _ 0.002378 X 146.7 X 0.25 _ r,oA nno 

^'^- 0.000000373 ^^^'^^^ 

Example. Find the R.N. for model wing of 3-in. chord. Tests 
run at 100 miles per hour. Air at normal pressure but at 100° C. 
temperature. 

r, ^ _ 0.002378 X m X 146.7 X 0.25 _ ..^ ^^^ 
^'^' 0.000000454 ^^^'^^^ 

Example. Find R.N. for model wing of 3-in. chord. Tests run at 
100 miles per hour. Air at standard temperature but 21 atmospheres 
pressure. Note that /x is independent of pressure (except for ex- 
tremely high or extremely low pressures). 

_ 0.002378 X 21 X 146.7 X .25 _ ^ 

^•^- 0.000000373 " 4,yiu,uuu 

Problems 

1. Find R.N. for an airplane wing, 4-ft. chord moving at 120 miles 
per hour through standard atmosphere. 

2. Find R.N. for an airplane wing with a 3-ft. 6-in. chord moving at 
180 miles per hour through standard air. 

3. Find R.N. for an airplane wing, 4-ft. chord moving at 150 miles 
per hour. Air is -f-40° C. and 21 in. barometer. 

4. Find the velocity at which tests should be run in a wind tunnel 
on a model wing of 4-in. chord in order that the R.N. shall be the same 
as for a wing with a 4-ft. chord at 100 miles per hour. Air under 
standard conditions in both cases. 

5. In a variable-density wind tunnel, under what pressure should 
tests be run on a model with a 3-in. chord, air velocity being 60 miles 
per hour, in order that the R.N. shall be the same as for a full-size 
wing of 4-ft. chord, moving at 100 miles per hour through the air? 
Air temperature is the same in each case. 

do'" 



CHAPTER IV 
AIRFOILS 

Flat plates are not suitable for airplane wings. To be struc- 
turally strong, there would have to be a considerable amount of 
bracing. This bracing, if located outside the plate, would cause 
much friction in moving through the air; if inside the plate, it 
would increase the thickness of the plate. By curving the plate 
even slightly it is possible to increase the weight-sustaining 
property greatly. By closing in the under side and making it 
more nearly streamline in shape, the resistance to forward move- 
ment through the air can be greatly decreased. 

The wings of the first Wright glider were simply curved strips 
of wood covered only on the upper side with cloth (see Fig. 11a). 
The Wrights soon re- 
alized that when the 
wing was horizontal 

there would be a (^) 

great amount of 

burbling on the un- ^ _y^5^^-^ 

der side (see Fig. ^ 

116). They conse- (5) 

quently covered up ^^^ ^^ Early Wright Wing. 

this space, giving the 

wing appreciable thickness and reducing its resistance to forward 

motion. 

Changing the amount of curvature of a wing or its thickness will 
change the lifting power and resistance. There is no ideal wing 
which is suitable for all types of airplanes. The proper shape 
must be chosen for each individual requirement. 

Airfoils. Although the word airfoil may be used interchange- 
ably with the word wing, common usage dictates that the word 
wing be used in referring to the actual wing of an airplane, and that 
the word airfoil be applied in describing the contour or shape of 
the vertical cross-section of a wing. Many differently shaped 
airfoils have been proposed and their properties investigated. 
Only a few of these airfoils will be discussed in this textbook. 

28 





CHORD AND CAMBER 29 

For descriptions of other airfoils, the reader is referred to reports 
of the National Advisory Committee for Aeronautics and other 
aeronautical testing laboratories. 

Chord and Camber. To describe the curvature of the upper 
and lower surfaces, a base Une is used as a-reference. Coordinates 
are given with respect to a point on this reference line to locate 
sufficient points on the airfoil to enable the curve to be drawn. 
This reference line is the chord of the airfoil. 

If the airfoil has convex curvature on both upper and lower 

surfaces, as in Fig. 12a, the chord 

is the line joining the most forward 

point on the front edge (leading 

edge) to the rearmost point (trailing 

edge). If the lower surface of the 

airfoil is predominantly flat, as in 

Fig. 126, the chord is the straight 

line that coincides for most of its 

(b) length with the lower surface. With 

Fig. 12. Chords, (a) Double other shapes of airfoils the datum 

camber, (6) Single camber. line is more or less arbitrary. This 

causes no confusion as the chord 
and the airfoil are never dissociated. Whenever an airfoil is 
described, the chord is known. 

The chord length is the length of the projection of the airfoil 
section on its chord. Its S5mibol is c. 

Camber is the length of the ordinate perpendicular to the chord. 
Top camber is the distance of a point in the upper surface of the 
airfoil from the chord line; bottom or lower camber is the distance 
of a point in the lower surface of the airfoil from the chord line. 
Because airfoil sections, which are geometrically similar but of 
different size, would be designated by the same name, amount of 
camber is always given as a percentage of the chord length. The 
abscissa, or length along the chord, of a point in the airfoil surface 
is also given in percentage of chord. 

In Table III are given the dimensions of a Clark Y airfoil, a 
typical medium-thickness wing. When once the chord length is 
decided upon, the actual camber can be found from the above 
table by multiplying the chord length by the suitable percentages. 
The use of this table is shown in Fig. 13, the points laid out by 
means of these coordinates defining a curve. 



% 



30 



AIRFOILS 

TABLE III 

Shape of Clark Y Airfoil 

Percentage of Chord 



Distance from 


Upper 


Lower 


Leading Edge 


Camber 


Camber 





3.50 


3.50 


1.25 


5.45 


1.93 


2.5 


6.50 


1.47 


5 


7.90 


0.93 


7.5 


8.85 


0.63 


10 


9.60 


0.42 


15 


10.69 


0.15 


20 


11.36 


0.03 


30 


11.70 


• 


40 


11.40 





50 


10.52 





60 


9.15 





70 


7.35 





80 


5.22 





90 


2.80 





95 


1.49 





100 


0.12 






The maximum camber is an important dimension and refers to 
the greatest departure of the curve of the airfoil from the chord 
line. Mean camber is the camber of a point equidistant from the 
upper and lower surfaces. 





Fig. 13. Clark Y wing contour. 

Span and Aspect Ratio. Span is the distance from wing tip to 
wing tip, inclusive of ailerons. It may be considered as the least 
width of hangar doorway through which the airplane can be pushed 
straight. The symbol for span is h. 

The area of a wing is the area of the projection of the actual 
outline on the plane of the chord. The sjmibol for area is S. 



FORCES ON AN AIRFOIL 31 

The aspect ratio of a wing is the ratio of the span to the chord. 
This is for rectangular wings. For wings that are not rectangular 
in shape when viewed from above, the aspect ratio is the ratio of 
the square of the span to the area. 

c S 

It is customary in wind-tunnel tests to use model airfoils whose 
aspect ratio is 6. Force measurements from these tests need to 
be corrected when applying them to airfoils of differing aspect 
ratio. 

Relative Wind. Relative wind is the motion of the air with 
reference to the wing. If the wing is moving horizontally forward, 
then the relative wind is horizontally backward. If the wing is 
moving both forward and downward, as when the plane is settling, 
the relative wind is upward as well as backward. 

The angle of attack of a wing is the acute angle between the 
chord and the direction of the relative wind. 

Forces on an Airfoil. Air flowing around an airfoil exerts a 
pressure on each little portion of the airfoil surface. This pressure 
is considered positive if it is greater than atmospheric, negative 
if it is less than atmospheric. 

At small positive angles of attack the air flows smoothly over 
the upper surface. Each particle of air sweeping along a surface 
contributes to giving a small negative pressure as long as its 
motion is parallel to the surface. If its motion is not parallel but 
towards the surface it will contribute its component of positive 
Impact pressure. The direction of flow is shown in Fig. 14a, and 
the pressures, to scale, are shown in Fig. 146. 

At larger angles of attack the shape of the stream lines and the 
magnitude of the pressures change. This is shown in Fig. 15. 

As the angle of attack is increased still more, the air cannot 
follow the upper surface of the wing as that would entail too great 
a change in direction. The streamlines no longer conform to the 
contour of the airfoil (Fig. 16). Burbling starts at the trailing 
edge. If the angle of attack is made greater, the burbling will 
extend farther forward. 

Wherever burbling takes place, that portion of the wing has 
lost its weight-sustaining property to a large extent. Burbling 
also increases the resistance of the wing to forT\^ard motion. 



32 



AIRFOILS 





(b) 
Fig. 14. Flow at low angle of attack. 




Fig. 15. Flow at medium angle of attack. 




Fig. 16. Flow at high angle of attack. 



It will be noted in Fig. 156 that the forces on the upper side of 
the airfoil are predominantly upward, and the magnitude of these 



LIFT AND DRAG COMPONENTS 33 

forces is greater than that of the upward forces on the under side 
of the airfoil. Tests have shown that, for typical wings at 0° 
angle of attack, 100 per cent of the total upward force on a wing 
is derived from the upper surface; at 5° angle of attack, 74 per 
cent is due to forces on the upper surface; and at 10°, 68 per 
cent. 

All the small forces acting on both upper and lower surfaces of 
the airfoil may be added together vectorially, i.e., taking into 
account both magnitude and direction, and this summation is 
called the resultant force. At small angle of attack, this resultant 
is small in magnitude and it acts near the trailing edge; as the 
angle of attack is increased, the resultant becomes larger and its 
point of action, called the center of pressure, moves forward. 

Lift and Drag Components. The resultant can, of course, be 
described by giving its magnitude and direction. It can also be 
determined if its two components are given about known axes. 
It is customary and most useful to give a resultant in terms of 
lift and drag components.. 

Lift is the component of the resultant on a wing which is per- 
pendicular to the relative wind. 
*"' Drag is the component of the resultant which is parallel to the 
relative wind. 

The resultant force on a wing varies directly with the air 
density, the area of the wing, and the square of the velocity. It 
also depends on the angle of attack. The lift and drag compo- 
nents, also, vary in the same way. These relations could be 
expressed in a formula by stating that the component was a factor 
times air density times wing area times velocity squared, but it is 
more desirable to use a different form of the formula containing 
one-half the density instead of the density itself. The standard 
formulas are 

j'fi- — n P QV2 -^^^^ ^^^ ^^^S are in pounds 

^2 p is in slugs per cubic foot 

T^ n P arro aS is in square feet 

Drag = Cd^SV^ jr • - f\. j 

2 V IS m feet per second 

Cl is called the lift coefficient; Cd, the drag coefficient. They 
are " absolute " or dimensionless coefficients and therefore can be 
used with metric units of measurement. That is, the formulas 
for lift and drag are perfectly valid if air density is given in metric 



34 AIRFOILS 

slugs, area in square meters, and velocity in meters per second. 
Lift and drag will then be in kilograms. 
The equations may be written in the form 



P 
y = 

and Drag = CoqS 



Lift = CLqS ^ = I ^' 



Characteristic Curves. The lift coefficient Cl, the drag coeffi- 
cient Cd, and the center-of-pressure location, all for different 
angles of attack, are considered the characteristics of an airfoil. 
This information may be given in tables, but it is more usual to 
plot these data in the form of a curve. The characteristic curves 
for the Clark Y airfoil, for an aspect ratio of 6, are given in Fig. 17. 
Since lift coefficients are at most angles of attack much larger than 
drag coefficients, the drag coefficients are plotted on a larger scale 
in order to be legible. 

Careful study should be given to Fig. 17. It should be noted 
that the lift coefficient {Cl) curve crosses the zero ordinate at 
— 5° angle of attack. This does not mean that there are zero 
forces acting on the wing at this angle of attack, but that the sum 
of the upward or positive lift forces equals the sum of the down- 
ward or negative lift forces. The angle of attack, where the 
upward and downward lift are equal, is called the angle of zero 
lift. For symmetrical sections, that is, airfoils that have the same 
camber on both upper and lower sides, the angle of zero lift is at 
0° angle of attack. 

From the angle of zero lift, the curve of lift coefficient is practi- 
cally a straight line for a considerable distance. The slope is 
constant. The lift coefficient {Cl) is proportional to the angle of 
attack provided angle of attack is measured from the angle of 
zero lift. 

At the larger angles of attack, the lift coefficient curve begins to 
deviate from a straight line. The lift increase is no longer propor- 
tional to the increase in angle of attack. At some angle of attack, 
for the Clark Y airfoil, it is 18^°; the lift coefficient is a maximum. 
This angle of maximum lift is also called the critical angle, the 
burble point, or the stalfing angle. 

At angles above the angle of maximum lift, the lift coefficient 
decreases with increasing angle of attack till lift coefficient becomes 
zero at some angle slightly greater than 90°. 



CHARACTERISTIC CURVES 



35 



The drag coefficient curve resembles a parabola in part. At 
some small angle of attack, for the Clark Y it is — 3i°, the drag 
coefficient has a minimum value. Whether the angle of attack is 
increased or decreased from this angle of minimum drag, the drag 
increases. For a few degrees above or below the angle of minimum 
drag, there is very little change in the value of the drag coefficient. 



.30 



d .50 



22 


r .22 


20 


- .20 


18 


- 18 


^Ql6 


-<?.16 


14 


- 14 


12 


- .12 


10 


- .10 


8 


.08 


6 


.06 


4 


- .04 


2 


- .02 























1 1 1 

r P in oercent chord 














_,^ 




--' 






r 






/ 


— 


^ 






/ 


y 














/ 


/ 








i 


1 
















/ 


























/ 


























/ 










/ 






/ 


' — 


\^ 




/ 


^ 








i 


\ 




\ 


/ 




N 


\, 


/ 


/ 
















/ 








s 


><. 












/ 






/ 








/ 




\ 








/ 














> 


/ 






\ 


N 


/ 
















/ 










/ 


\ 














/ 










/ 






\ 


-^ 




i 




/ 


/ 








'/^ 
















/ 






y 


y 


















/ 




' 




y 


















-1 


r 


eS 


r 























































1.5 

14 
1.3 
12 
1.1 
1.0 
9 

.7 
6 
5 
.4 
.3 
.2 



-6-4-2 2 4 6 8 10 12 14 16 18 20 22 
Angle of Attack 

Fig. 17. Characteristics of Clark Y airfoil, aspect ratio 6. 



With bigger increases in the angle of attack, the drag coefficient 
increases greatly. 

With a sjmametrical airfoil, the angle of minimum drag is at 0° 
angle of attack and the curve is symmetrical about a vertical axis 
through this point. With the more common airfoils, having more 



36 ^^==0. AIRFOILS 

camber on the upper than the lower side, the angle of minimum 
drag is a small negative angle of attack and the part of the curve 
for more negative angles of attack has a steeper slope than the 
other side of the curve. 

The center-of-pressure curve usually shows that at the angle of 
zero lift the center of pressure is at the trailing edge. With a 
sHght increase in angle of attack, the center of pressure moves 
forward. It has its most forward position when the angle of 
attack is a few degrees below the angle of maximum lift. For 
symmetrical airfoils, there is practically no movement of the center 
of pressure; for this reason symmetrical airfoils are referred to as 
stable airfoils. 

Lift Equation. For horizontal flight, except for a small vertical 
force on the tail and a small component of thrust which need not 
be considered at this time, the weight of the airplane equals the 
lift of the wing. If the Hft is greater than the weight, the airplane 
will rise; if the weight is greater, the airplane will lose altitude. 
Then for horizontal flight 

W in pounds 

S in square 1 

V in feet per second 

The above equation may be used in other forms, viz. 

W 



rr XXX ^Kj\j.xx\jio 

W = Cl~SV' Sin square feet 



>S = 



w 



'-^^Ji-i 



The ratio of total weight to the area of the wing (W/S) is the 
wing loading, expressed in pounds per square foot. From the 
above formulas it will be seen that, for a given angle of attack, 
the proper velocity depends on the square root of the wing loading. 

Example. What weight can an airplane have to fly level with a 
Clark Y wing 250 sq. ft. in area, at 4° angle of attack and airspeed of 
100 miles per hour at sea-level? ^ 



LIFT EQUATION 37 

From Fig. 17, Cl at 4° angle of attack = 0.649; 100 miles per 
hour = 146.7 ft. per sec. 

.(10237 



W « 0.649 X ^ X 250 X 146.7' = 4,12(1) lb. 

Example. At what angle of attack should an airplane weighing 
3,000 lb. be flown, if the wing is 300 sq. ft. in area, Clark Y section, 
airspeed 90 miles per hour? 

n _ ^)000 _ Q ^oo 

^ %.m\\% X 300 X (90 X 1.47)2 

Angle of attack (from Fig. 17) = +1.7° 

Example. An airplane has a wing loading of 9 lb. per sq. ft.; the 
angle of attack of the Clark Y wing is 6°. What should be the air- 
speed? 

F = a/q 4 / ^ 

^ V 0.791 X 0.00118 

= 97.9 ft. per sec. 

= 66.8 miles per hour 

Problems 

(Standard air density unless otherwise specified) 

1. What is the lift on a Clark Y wing 300 sq. ft. in area at 8° angle 
of attack and airspeed of 80 miles per hour? 

2. What is the lift on a Clark Y wing' 450 sq. ft. in area at 6° angle 
of attack, airspeed 88 ft. per sec? 

3. A monoplane weighing 4,460 lb. has a rectangular Clark Y wing, 
353 sq. ft. in area; at an airspeed of 100 miles per hour, what should 
be the angle of attack? 

4. At what angle of attack should the airplane in problem 3 fly- 
when airspeed is 85 miles per hour? 

5. With a wing loading of 12 lb. per sq. ft., at what angle of attack 
should an airplane with a Clark Y wing fly, if airspeed is 70 miles per 
hour? 

6. What should be the area of a Clark Y wing to support a total 
weight of 5,000 lb. when flying at 7° angle of attack and a velocity of 
90 ft. per sec? 

7. What weight will be supported by a Clark Y wing 525 sq. ft. in 
area, at 5° angle of attack, and an airspeed of 120 ft. per sec. at sea- 
level? 

8. What weight will be supported by the wing in problem 7, at the 
same angle of attack and same airspeed, but flj^ing at 10,000 ft. alti- 
tude (air density at 10,000-ft. altitude is 0.00176 slug per cu. ft.; 
see Table I)? 



38 AIRFOILS 

9. At what airspeed should a 6,000-lb. airplane be flown, having an 
area of 700 sq. ft., Clark Y wing, at 10° angle of attack? 

10. At what airspeed should the airplane described in problem 9 
fly at an altitude of 10,000 ft.? 

11. An airplane having a Clark Y wing at 6° angle of attack is to be 
flown at 150 miles per hour. What should be the wing loading? 

12. What should be the wing loading of a plane with a Clark Y 
wing, if it is desired to fly at 6° angle of attack and 150 miles per hour 
airspeed at an altitude of 10,000 ft.? 

Pw I i 

Minimum Speed. Examination of formula F = 1/ -^ / tt 

shows that, with a fixed weight and a fixed wing area (fixed wing 
loading), the lift coefl&cient must vary inversely as the square of 
the velocity. It is axiomatic that small angles of attack mean 
highspeed; large angles of attack mean slow speed. The smallest 
velocity will be when the lift coeflicient is maximum. This 
slowest velocity is the stalling speed. Some airplanes (low-wing 
monoplanes), when gliding down to land, pocket air under their 
wing, which has a cushioning or buoyant effect, helping to support 
the airplane slightly. This added support enables the airplane 
to fly at slightly less speed than it could if the ground effect were 
not present. This effect is ordinarily negligible, and the minimum 
airspeed is considered to be the landing speed or take-off speed. 



F. 



v^/— 



t / ^Zmax. X c\ 



2 

Y in feet per second 

Example. What is the landing speed of an airplane weighing 2,500 
lb., having a Clark Y wing 300 sq. ft. in area? 



V 300 V 1.56 X 0.00118 
= 67.0 ft. per sec. 
= 45.7 miles per hour 



Problems 

1. What is the landing speed of an airplane weighing 4,500 lb., 
with a Clark Y wing 350 sq. ft. in area? 



DRAG 39 

2. What is the landing speed of an airplane with a Clark Y wing and 
a wing loading of 12 lb. per sq. ft.? • 

3. What area should a Clark Y wing have in order that an airplane 
weighing 1,500 lb. shall not land faster than 30 miles per hour? 

4. An airplane has a Clark Y wing. What would be the greatest 
wing loading in order that the landing speed should not exceed 35 
miles per hour? 

5. An airplane has a Clark Y wing 425 sq. ft. in area. What is the 
greatest weight this airplane can have with a landing speed not more 
than 40 miles per hour? 

6. An airplane with a Clark Y wing has a wing loading of 14 lb. per 
sq. ft. What is the minimum speed at an altitude of 10,000 ft.? 

7. It is desired that a pursuit airplane be able to fly at 50 miles per 
hour at an altitude of 10,000 ft. What should be the wing loading 
of its Clark Y wing? 

8. An airplane weighing 5,000 lb. has a Clark Y wing 625 sq. ft. in 
area. What is its minimum speed (a) at sea-level, (b) at 10,000-ft. 
altitude? 

Drag. Drag is the force, in pounds, with which a wing resists 
forward motion through the air. The drag multiplied by the 
velocity (in feet per second units) gives the power, in foot-pounds 
per second, required to move the wing forward. One horsepower 
is 550 ft-lb. per sec. Hence the drag times the velocity, divided 
by 550, gives the horsepower required to move the wing forward. 

-rrp _ D X V Din pounds 

'^^^' 550 F in feet per second 

But D = Cd^SV' 

Therefore H.P.req. = — ^ — 

Example. A Clark Y wing 350 sq. ft. in area is moving through the 
air at 80 ft. per sec. at 6° angle of attack. What is the drag? What 
horsepower is required? 

Prom Fig. 17, at a = 6°, Cd = 0.0452 

Drag = 0.0452 X 5:5^^ x 350 X (80)2 
= 120 lb. 
120 X 80 



H.P.rea. = 



req. 



550 

= 17.5 hp. 



40 



AIRFOILS 



Problems 

1. What is the drag of a Clark Y wing, 300 sq. ft. in area, at 8° 
angle of attack, at 120 ft. per sec. airspeed? 

2. What is the drag of a Clark Y wing, 300 sq. ft. in area, at 8° 
angle of attack, at 120 miles per hour airspeed? 

3. What is the horsepower required to move a Clark Y wing 430 
sq. ft. in area at 2° angle of attack and airspeed of 150 miles per hour? 

4. What is the horsepower required to move a Clark Y wing 270 
sq. ft. in area at 10° angle of attack and an airspeed of 70 ft. per sec? 

5. At an altitude of 10,000 ft., what is the drag of a Clark Y wing, 
480 sq. ft. in area, angle of attack 4°, airspeed 95 ft. per sec? What 
horsepower is required? 

























/ 




\ 
























/ 




























/ 




























/ 










1 


















/ 


/ 








1 




















/ 




























/ 












/ 
















V 


/ 










/ 
















/ 


/ 










/ 


/ 
















/ 










/ 


/ 
















/ 










/ 


/ 
















/ 


/ 








y 


v 


























y 


/ 


















/ 








y 


/ 


















/ 


/ 




^ 


r"' 






















± 


— 


"^ 



























1.5 
1.4 
1.3 
1.2 
1.1 
1.0 

.9 

.8 6' 

.7 

.6 

.5 

.4 



-6 -4 



2 4 6 8 10 12 14 16 18 20 

Angle of Attack 



Fig. 18. Characteristics of Gottingen 398 airfoil, aspect ratio 6. 



Difference in Airfoils. Many different shapes have been pro- 
posed and tested. Airfoils with contours radically different from 
those in general use do not ordinarily prove to be satisfactory. 



DIFFERENCE IN AIRFOILS 



41 



This indicates that certain general rules must be followed in 
designing airfoils. 

The leading edge should be slightly rounded. The camber of 
the upper surface should be such that the highest point or maxi- 
mum ordinate is between one-quarter and one-third the chord 
length back from the leading edge. 



^4 

.22 






















/ 




l.l 
















/ 






^ 




.20 
.18 
.16 
















/ 




i 






1.0 
.9 
.8 














y 


r 




/ 
















J 


/ 






/ 
















/ 








/ 






.14 












/ 








/ 






.7 












f 






/ 








.12 

a 
o 

.10 
.08 
.06 
.04 
.02 











/ 








/ 








.6 

< 

.5 

.4 

.3 

.2 

.1 










/ 


/ 








/ 














/ 








y 


f 












/ 


/ 






J 


/ 












/ 


/ 






y 


/ 














/ 




^^ 




^ 














/ 






.--'^ 










































-i 


> ( 


) : 


> <t 


\ < 


3 i 


\ 1 

^ngle 


1 
f Attac 


2 1 

k 


4 1 


e 1 


8 







Fig. 19. Characteristics of C-80 airfoil, aspect ratio 6. 



Increasing the camber increases the lift at any angle of attack 
but it also increases the drag. The maximum Kft coefficient is 
increased, but the minimum drag coefficient is increased also. 
No wing should have an upper camber greater than one-quarter 
of the chord length. 

An airfoil having a symmetrical section, the upper and lower 
surfaces of equal camber, is streamline in appearance and conse- 
quently will have less minimum drag than a non-symmetrical 
airfoil of the same thickness. Symmetrical airfoils have zero 
lift at zero angle of attack, and this is also the angle of minimum 



42 



AIRFOILS 



drag. Non-symmetrical airfoils have zero lift at a slightly- 
negative angle of attack. 

Concave lower camber increases the Uft coefficient at any angle 
of attack but adds to the drag coefficient, especially at smaller 
angles of attack. 



22 


.22 


20 


.20 


18 


.18 


i6 


:i6 


14 


M 


12 
10 


o 

.10 


8 


.08 


6 


.06 


4 


.04 


2 


.02 



















































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/ 






















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\ 




/ 












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y 


/ s 


\ 








/ 




/ 






/ 


/ 






N 


N, 


/ 






/ 






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/ 










/ 


r 




N 




( 


/ 


/ 








Cd/ 










] 




/ 






y- 


y 














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y 














i 


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1.3" 
1.2 
1.1 
1.0 
.9 
.8 

o 

.6 

.5 
.4 
.3 
.2 
.1 

u z ^ o 8 10 12 14 16 18 20 22 ° 
Angle of Attack 

Fig. 20. Characteristics of M-6 airfoil, aspect ratio 6. 

Figures 18, 19, 20, and 21 give characteristics of airfoils in 
common use. Thick wing sections are commonly called " high 
lift " sections because the maximum lift coeflficients are large. 
With a predetermined landing speed and wing area, more load 
can be carried by a thick than a thin wing section. 

Problems 

1. (a) What is the lift on a Gottingen 398 wing 400 sq. ft. in area, 
at 2° angle of attack and airspeed of 90 ft. per sec? (6) What is the 
drag? 



DIFFERENCE IN AIRFOILS 



43 



2. What is the lift on a C-80 wing 400 sq. ft. in area at 2° angle of 
attack and airspeed of 90 ft. per sec? 

3. What is the lift on a C-80 wing 400 sq. ft. in area at 7° angle of 
attack and airspeed of 90 ft. per sec? 

4. What is the landing speed of an airplane weighing 5,000 lb. with 
a Gottingen 398 wing 400 sq. ft. in area? 



.30 

C.40 
o 

'Si 

£.50 

ol 

".60 
.70 .24 
.22 
.20 
.18 
.16 
.14 



o 
o .12 













c_^ 


















/ 
























/ 






















/ 












































1 


















/ 


^ 


\ 


/ 
















/ 


/ 




\ 


/ 
















/ 








\ 














/ 






















Cl/ 


/ 






















/ 








1 














/ 










J 












/ 


/ 








y 


/ 












/ 








/ 


/ 












/ 








y 














/ 


/ 






y' 














' 


y 






















/ 





-2" 0° 2° 4° 

Fig. 21. Characteristics of RAF-15 airfoil, aspect ratio 6 



go go 10« 12° 14° 160 180 20° 
Angle of Attack 



1.2 
1.1 
1.0 

.9 

.8 

.7 

.6C5' 

.5 

.4 

.3 

.2 

.1 





5. What is the landing speed of an airplane weighing 5,000 lb., 
with a C-80 wing 400 sq. ft. in area? 

6. What total weight can a Gottingen 398 wing 400 sq. ft. in area 
sustain if landing speed must not exceed 45 miles per hour? 

7. What total weight can a C-80 wing 400 sq. ft. in area sustain if 
landing speed must not exceed 45 miles per hour? 



44 AIRFOILS 

N.A.C.A. Four-Digit Airfoils. The National Advisory Com- 
mittee for Aeronautics, in making a study of a large number of air- 
foils that had proved to be good in practice, discovered a certain 
relationship between dimensions of these airfoils. If a line is 
drawn equidistant between the upper and lower contours of an 
airfoil, this hne is called the median line. Various airfoils may 
each have a different thickness; but, if the maximum thickness of 
each of several airfoils be made the same, the thickness at other 
points being reduced (or increased) in the same proportion as the 
maximum thickness was reduced (or increased), the various con- 
tours will coincide if the median line is straightened. 

With the median line straight, the airfoil is symmetrical. The 
contour is a curve fitting the following equation 

±2/ = 0.2969 v^ - 0.1260 x - 0.3516 x^ + 0.2843 x^ - 0.1015 x^ 
where y and x are the vertical and horizontal coordinates in percent- 
age of chord. The above equation gives a maximum thickness 
which is 20 per cent of the chord, and this maximum thickness is 
located 30 per cent of the chord back from the leading edge. To 
obtain a wing of any desired thickness t (expressed in fraction of 
chord), each value of y found by the above equation is multiplied 
by the constant ^/0.20. 

For non-symmetrical airfoils, the median line is made curved, 
the thickness at the various points along the chord being kept the 
same. The amount and shape of the curvature of the median line 
is defined by the distance of the maximum ordinate back of the 
leading edge (p), and the amount of the maximum ordinate. For- 
ward of the maximum ordinate, the curvature of the median line 
fits the parabolic equation 

Vc = -2 (2 px - a;2) 
Aft of the maximum ordinate the following equation applies 

The National Advisory Committee for Aeronautics has made an 
extensive and systematic study of airfoils conforming to this 
description. To identify each airfoil, a number consisting of four 
digits was assigned to each ; the first digit representing the amount 
of camber of the median line in percentage of chord, the second 
digit representing the position of the point of maximum camber in 



N.A.C.A. FIVE-DIGIT AIRFOILS 



45 



tenths of chord from the leading edge, and the last two digits 
representing the maximum thickness in percentage of chord. For 
example, in airfoil N.A.C.A. 2421, the maximum ordinate of the 
median line is 2 per cent of the chord, this ordinate is 40 per cent 
of the chord back from the leading edge, and the greatest thick- 
ness of the airfoil is 21 per cent of the chord. 




-lOL 



'Radius through end of chord 



1.00 



(b) 



Fig. 22. N.A.C.A. airfoils. 

N.A.C.A. Five-Digit Airfoils. Continuing their systematic 
study of airfoils, the N.A.C.A. wished to find the effect of changing 
the position of the point of maximum camber of the median line. 
They had found by investigation that moving the position of the 
maximum camber of the median line forward of 30 per cent of the 
chord had a pronounced effect on the maximum lift. Two groups 
have been investigated, one group having a simple curve for a 
median line, the other having a median line with reverse or reflex 
curvature. This reverse curvature was introduced to improve 
the pitching moment characteristics. The five-digit designation 
identifies the airfoils as follows. The first digit gives the amount 
of maximum camber of the median line in percentage of chord. 
The second digit gives the position of the point of maximum 
camber in twentieths of the chord. The third digit describes the 
median line; if it is the simple curve,- 1 if it is the reflexed curve. 
The last two digits are the thickness in percentage of chord. For 
example, in airfoil N.A.C.A. 24112, the maximum camber of the 
median line is 2 per cent of the chord; this ordinate is four- 



46 



AIRFOILS 



twentieths (20 per cent) of the chord back from the leading edge; 
the median Hne has a reverse curvature and the maximum thick- 
ness of the airfoil is 12 per cent of the chord. 

Tapered Wings. The simplest wing shape is rectangular in 
plan. In tapering it is usual to leave the center section unchanged 
but, starting at the root (where the wing proper joins the center 
section), the chord is progressively decreased out to the tip. The 
decrease in chord length at any section is to the total decrease as 



168 c 




A.R. = 6 

Area = 3.719 C^ 

Length of MAC. = 0.843 C 

LE. MAC. Aft L.E. Root Chord = 0.039 C 



Root Chord Thickness - 
Tip Chord Thickness 
(Section C-C unfaired) 



■30% Stations 



Fig. 23. Standard wing taper. 



the distance from the root out to that section is to the total dis- 
tance from root to tip. If the same airfoil is used throughout, 
because the chord is less towards the tip, the thickness is also 
less in the same ratio. It is usual to make the wing in such a 
manner that lines, joining the points of maximum upper camber in 
each section, will be horizontal; this will make the lower surfaces 
of a tapered wing slope upward to the tip. 

It is becoming more customary to change arbitrarily the airfoil 
section from the root towards the tip. By doing this, it is possible 
to reduce the twisting action of the air. The standard taper 
adopted by the Army is illustrated in Fig. 23. 

When a wing is tapered by making the chord progressively 
smaller from root to tip, while the same airfoil section is used 
throughout, it is termed '' taper in planform only," even though 
the thickness does decrease from root to tip. When the thickness 
decreases at a faster rate than the chord, by using thinner airfoil 
sections towards the tip, the wing is said to " taper in planform 
and thickness." 



VELOCITY VERSUS ANGLE OF ATTACK 



47 



Velocity versus Angle of Attack. An airplane, if it flies at a 
low angle of attack, must fly faster than if it flies at a high angle of 
attack. For each speed there is one angle of attack for level 
flight to be maintained. If the angle of attack is greater, the 
airplane will gain altitude ; if the angle of attack is less, the airplane 
will descend. 

340 



320 

300 

280 

260 

240 

I 220 

J 200 

^180 

^ 160 

140 

120 

100 

80 

60 

40 





































































































































































































V 
























V 


s 






















v^ 


s^ 






















\ 


\ 




^^ 


^i5 


















\ 




^^ 






















-<i 


f^ 


^^ 














































6 8 10 12 14 16 18 20 
Angle of Attack * 



Fig. 24. Angle of attack versus velocity. 



Conversely, for each angle of attack there is only one speed for 
level flight. This depends on the relation 

T7 1 
Ol - -o - — 

2 

If the wing loading (W/S) is changed, the relation between 
angle of attack and speed is changed. A partly loaded airplane, 



48 AIRFOILS 

fljdng level, will either have to increase its speed or angle of attack 
if additional load is put on. Figure 24 shows the relation between 
airspeed and angle of attack for a Clark Y airfoil. 

Problems 

1. Plot airspeed versus angle of attack for a Gottingen 398 airfoil 
with a wing loading of 14 lb. per sq. ft. 

2. Plot airspeed versus angle of attack for a R.A.F. 15 wing with a 
wing loading of 18 lb. per sq. ft. 

3. For an M-6 airfoil, plot wing loading versus angle of attack for 
an airspeed of 100 miles per hour. 

4. For a C-80 airfoil, plot wing loading versus airspeed, in feet per 
second, for an angle of attack of 4°. 

Flying Level at Altitude. At altitudes above sea-level, the air 
density is less than its standard value. Since both lift and drag 
coefficients are multiplied by the air density, as well as wing area 
and the square of the velocity, to give hft and drag; at the same 
angle of attack or at the same airspeed, both lift and drag will be 
less at altitude. In order to fly level no matter what altitude, the 
lift must be equal to the weight. 

As the wing area remains constant, when an airplane ascends 
to higher altitudes either lift coefficient or velocity, or both, must 
be increased to make up for the decrease in density. The case of 
flying at the same angle of attack as at the ground will be considered 
first. 

The angle of attack being unchanged, Cl and Cd will be the same 
at all altitudes. Let po and Vo be the air density and velocity at 
zero altitude and px and Vx the density and velocity at any 
altitude, x feet. 

W = Cl^SVo' 
W = Cl^SVx' 
Then poVo' = PxVx' and ^ Vo' = VJ" 

Px 

Drag at zero altitude = Do = Cdi^SVo' 



Drag at x feet altitude = Dx = Cdi-SYx" 



FLYING LEVEL AT ALTITUDE 49 

But V,' = ^ Vo' 

Px 

Therefore D, = Cd^S^Vo' 

^ Px 

= Cd^SVo' 

Therefore, whatever the altitude, the drag of the wing is the 
same, provided the angle of attack is the same. This may seem 
strange, but it should be borne in mind that with increased 
altitude the airspeed must be greater, and this exactly counter- 
balances the decrease in density. 

Power required is the product of drag and velocity. Let H.P.o 
be the horsepower at zero altitude and H.P.^; the horsepower 
required at a;-feet altitude. 

DoX Vo 
550 



H.P.o = 



H.P. 



2).X V, 



550 
But Do = D^ 

and V, = Fov/^ 

^ Px 

Therefore H.P. = ^^°V^ 






H.P. 

Px 

The density becomes less as ascent is made in the atmosphere; 
that is, Px is less than po, so that po/px is greater than 1. With the 
same angle of attack, the horsepower required to move a wing 
forward through the air will always be greater at altitude than 
at sea-level. 

When the angle of attack is changed, keeping the same airspeed, 
a lightly loaded plane is affected differently from a heavily loaded 
plane. At ordinary flying angles, the lift coefficient varies ap- 
proximately as angle of attack. The drag coefficient changes very 
httle with angle of attack at small angles of attack corresponding 
to high speeds. At slow speeds, corresponding to high angles, 
there is a big change of drag coefficient with angle of attack 
changes. 



50 AIRFOILS 

When attempt is made to fly at the same speed at high altitude 
as at the ground, the angle of attack must be increased to obtain 
the greater lift coefficient to offset the decrease in density. The 
larger angle of attack will mean an increased drag coefficient. If 
at sea-level the airplane was being flown at low angle of attack, 
a moderate increase in altitude would mean a very small increase 
in drag coefficient. It is quite possible that the decrease in density 
could be greater in proportion than the increase in drag coefficient. 
In this case, the drag and consequently power required would be 
less at altitude than at sea-level. A plane flying slowly at the 
ground (high angle of attack) and rising to a moderate height, or 
an airplane flying fast at the ground but rising to extreme altitudes, 
will usually result in the drag coefficient increasing disproportion- 
ately more than the decreasing of density so that the drag and 
horsepower will be greater at the higher altitude. 

Example. An airplane has a Clark Y wing 400 sq. ft. in area. It 
flies 100 ft. per sec. at 2° angle of attack. What are the lift, wing 
drag, and horsepower required at sea-level? What are the lift, wing 
drag, and horsepower required at 10,000-ft. altitude? 

At sea-level, L = 0.5 X ^'^^^^"^^ X 400 X 100^ 
= 2,378 lb. 
D = 0.024 X ^-^^1^^^ X 400 X 100^ 

= 114 1b. 
„^ 114X100 
^'^' = 550 

= 20.7 hp. required 

At 10,000-ft. altitude 

L = 0.5 X ^•^^^'^^^ X 400 X 100^ 

= 1,756 lb. 
D = 0.024 X 9^99^1^ X 400 X 100^ 



H.P. = 



84.3 lb. 
84.3 X 100 



550 
= 15.3 hp. 

Example. An airplane weighing 4,000 lb. has a Clark Y wing 350 
sq. ft. in area and is flying at sea-level at 100 miles per hour. What 



FLYING LEVEL AT ALTITUDE 51 

are the wing drag and horsepower required for the wing? If the 
airplane is flying at an altitude of 10,000 ft., at 100 miles per hour, 
what are the wing drag and horsepower required? 
At sea-level 

4.000 



Cl = 



0.002378 



2 
= 0.445 



Therefore /a^42°\ 

and Czr^.0217 



X 350 X (1.47 X 100) 



0.002378 



D = 0.0217 X ^ ^ X 350 X 1.47 X 100' 

= 195 lb. 
jjp ^ 195 X 147 



550 
= 52.2 hp 

At 10,000-ft. altitude 
Cl = 



4,000 



0.001756 
2 
= 0.604 

Therefore a = 3i° 

and Cd = 0.0308 



X 350 X (1.47 X 100)2 



D = 0.0308 X ^-^^i^^^ X 350 X lif 



= 204 lb. 
204 X 147 



H-P-= 550 

= 54.6 hp. 

Problems 

1. An airplane weighing 3,000 lb. has a Clark Y wing 250 sq. ft. in 
area and is flying at sea-level at 150 miles per hour, (a) What are the 
wing drag and horsepower required for the wing? (6) What are the 
wing drag and horsepower if flying at 150 miles per hour at 15,000-ft. 
altitude? (c) What are the wing drag and horsepower if flying at 
150 miles per hour at 30,000-ft. altitude? 

2. An airplane weighing 6,000 lb. has a Clark Y wing 250 sq. ft. 
in area. It flies at 125 miles per hour, (a) What are the wing drag 
and horsepower required for the wing at sea-level; (6) at 15,000-ft. 
altitude; (c) at 25,000-ft. altitude? 



52 AIRFOILS 

3. An airplane weighing 5,000 lb. has a Clark Y wing 275 sq. ft. 
in area. It flies at 4° angle of attack, (a) What are the wing drag 
and horsepower required at sea-level; (b) at 15,000 ft.? 

4. An airplane weighing 4,000 lb. has a Clark Y wing 350 sq. ft. in 
area. It flies at 150 miles per hour, (a) What is the horsepower 
required at sea-level; (b) at 20,000-ft. altitude? 

5. An airplane whose wing loading is 12 lb. per sq. ft. has a Clark 
Y wing and is flying at 200 ft. per sec. at sea-level, (a) What is the 
angle of attack? (6) What is the angle of attack if flying at 20,000- 
ft. altitude at the same speed? 

Lift-Drag Ratio. Air flowing around a wing causes forces to 
come into action, and the resultant of these forces is usually 
expressed in terms of its lift and drag components. In level flight 
the relative wind is horizontal, so that the lift component is vertical, 
the drag component horizontal. The lift component sustains the 
weight of the airplane. The drag component is the resistance to 
forward motion of the wing. In a complete airplane there are 
other parts, such as the fuselage, landing gear, and struts, which 
offer resistance to forward motion through the air. The resistance 
of these other parts of the airplane to forward motion is called the 
parasite drag. The sum of the wing drag and parasite drag 
constitute the total drag of the airplane. This total drag is the 
backward force that must be balanced by the forward thrust of 
the propeller in order to sustain forward movement of the airplane. 

The sole purpose of a wing is to provide a sustaining force for 
the airplane. It is to be expected that a wing will offer resistance 
to movement through the air. The wing that offers the least 
resistance and at the same time furnishes the most lift would be 
the most desirable from this standpoint. 

Wings must be capable of being made structurally strong. A 
very thin wing might have merits from an aerod3rQamic stand- 
point, but it might be so shallow that the spars usable in such a 
wing would be too light to have suflficient strength. The amount 
of movement of the center of pressure on a wing is also important 
in the securing of longitudinal balance. These matters will affect 
the selection of the wing section to be used, so that big lift with 
little drag will not be the sole consideration. 

The term '' efficiency " as used in engineering has a very exact 
meaning, namely the power output divided by the power input. 
In engineering, eflftciency is always less than unity. The term 



LIFT-DRAG RATIO 53 

efficiency cannot be correctly applied to ratio of the lift force to the 
drag force of a wing, since it is a ratio of forces, not of powers. 
The expression '' efficacy of the wing " which has been suggested 
is rather cumbersome, and it is practically universal to employ 
the expression '^ Hft-drag ratio " or '' L over Z)." 

The hft-drag ratio is the same as the ratio of lift coefficient to 
drag coefficient. 

Lift _ ^4^^' Cr. 

and is the tangent of the angle which the resultant force on the 
wing makes with the horizontal plane. 

At small angles of attack, drag coefficient is small but lift co- 
efficient is also small. At large angles of attack, lift coefficient is 
large but drag coefficient is also large. At the angle of attack of 
minimum drag coefficient, it will be found that an increase of a 
couple of degrees in the angle of attack will cause only a slight 
increase in drag coefficient but it will cause a considerable increase 
in the lift coefficient. It will therefore be at an angle of attack 
1J° or 2° greater than the angle of minimum drag coefficient that 
L/D will have its greatest value. 

The values of LID or Cl/Cd for various angles of attack of the 
Clark Y airfoil are shown in Fig. 17. It will be noted that the 
angle of minimum drag coefficient is — 3J°, while the angle of 
maximum L/D is slightly greater, i.e., +|°. 

The angle of maximum L/D is important. For level flight, 
lift is considered as practically equal to the weight. With weight 
constant, lift is constant; the drag will be least when D/L is least, 
that is, when L/D is greatest. 

In a preceding paragraph, it was shown that for level flight there 
is one velocity corresponding to each angle of attack. The velocity 
corresponding to the angle of maximum L/D will be the velocity 
at which the wing will have the least drag. 

Since parasite drag coefficients vary only sHghtly with angle of 
attack, it is approximately correct to state that, if the angle of 
attack of the wing is that of maximum L/D, the whole airplane 
will have less total drag than in any other position. This position 
will require the least thrust force from the propeller. 

It should be noted that power contains the element of speed 



54 AIRFOILS 

or time, so that less power may be required from the engine if the 
airplane is flown at slightly less speed, i.e., sUghtly greater angle 
of attack. This will be treated later in this book. The gasoline 
consumption depends on horsepower-hours, which, in turn, de- 
pends on L/D total. In flying from one point to another, the 
least amount of fuel will be required if the airplane is flown 
entirely with the wing at the angle of attack of maximum L/D. 

For best performance of the engine and propeller and for com- 
fort it is desirable that the longitudinal axis of the airplane be 
horizontal. In order that in that position of the airplane the wing 
shall be at the angle of maximum L/D, the wing is usually set at a 
small positive angle to the axis of the airplane. This angle is 
called the angle of incidence. 

In a race, the nose of an airplane would be depressed, causing 
the angle of attack to be less than that of maximum L/D. The 
drag coefficient will be less at this angle of attack than at the angle 
of maximum L/D, but in order that there shall be sufficient lift 
to sustain the plane in the air, velocity must be greater than at 
the larger angle. Because velocity is greater, the drag is greater 
at the lower angle even though the drag coefficient is less. 

Example. In level flight, what is the least drag of the Clark Y 
wing of an airplane weighing 5,000 lb.? 
Solution. From Fig. 17, (L/Z))max. for Clark Y is 21.5 

L _W _ 5,000 _ ^. - 
D " D ~ D ~ ^^'^ 

P = SI = 2331b. 

Problems 

1. (a) Plot L/D versus angle of attack for the Gottingen 398 airfoil. 
(6) What is the least drag of a Gottingen 398 wing for an airplane 
weighing 4,000 lb.? (c) What is the drag for this airplane when the 
wing is at 6° angle of attack? {d) What is the drag when the wing is 
at —3° angle of attack? 

2. (a) Plot L/D versus angle of attack for the M-6 airfoil. (6) What 
is the least drag of an M-6 wing for an airplane weighing 2,000 lb.? 
(c) What is the angle of attack for least drag? 

3. (a) Plot L/D versus angle of attack for the R.A.F. 15 airfoil. 
(6) What is the least drag of an R.A.F. 15 wing for an airplane weighing 
2,000 lb.? (c) What is the angle of attack for least drag? {d) What 
is^the drag at zero degrees angle of attack? 



POLARS 



55 



Polars. Instead of plotting lift and drag coefficients against 
angle of attack, the information may be given in other ways. 
Quite frequently the lift coefficient is plotted against drag coeffi- 
cient, as in Fig. 25. The curve represents the changes in Co with 



1.5 
1.4 
1.3 
1.2 

1.1 

1.0 
.9 
o'.S 
.7 
.6 
.5 
.4 
.3 
.2 
.1 


















/ 


^^ 




20^^ 


...^ 














/ 


c 




















f< 


y 






















/ 






















/ 


12" 




















/ 


0° 




















/ 


/. 






















/ 






















. 


r 






















/< 















































i 
























/ 


o* 






















h 


10 












■ 










L 















































.02 .04 .06 .08 .10 .12 .14 .16 .18 .20 .22 .24- 
Cd 

Fig. 25. Polar curve for Clark Y airfoil, aspect ratio 6. 



changes in Cl, or vice versa. The angle of attack is designated 
on the curve itself. This form of plotting is called a ipolar curve. 
Since Cd is small in comparison with Cl it is customary to use 
a much larger scale for plotting Cd than for Cl- If the same 
scale were used for plotting both Cl and Cd, a line drawn from the 
origin to any point on the curve would represent the resultant 
coefficient, both in direction and magnitude. Its length could be 



56 



AIRFOILS 



scaled off in the same units as the Cl and Cd scale. Multiplied 
by (p/2)SV^, it would represent the force acting on the wing. 

Using different scales for Cl and Cd, the length and direction 
of a line from the origin to a point on the polar curve are meaning- 
less. However, irrespective of scales, the line which makes the 
largest angle with the horizontal base line, having the greatest 
tangent, will have the highest Cl/Cd ratio. Therefore a line drawn 
from the origin tangent to the polar curve will locate the angle of 
maximum L/D. By reading the coordinates of the point of 
tangency, the maximum value of L/D may be quickly found. 

22 



20 



18 



16 



14 



12 



-J|0 



10 



11 



2 A 6 .8 1.0 1.2 14 16 li 

Fig. 26. L/D versus Cl for Clark Y airfoil. 



Another form of graph is shown in Fig. 26, where L/D is plotted 
against Cl- Because lift coefficient varies almost directly with 
angle of attack up to near the burble point, this graph resembles 
very much the plot of L/D against angle of attack in Fig. 17. 

In comparing one airfoil with another, the actual angles of 
attack are of httle consequence. The important factors are Cl, 
Cd, and L/D. These are all given by one polar curve; the 
information would have to be obtained from three separate curves 
otherwise. 



POLARS 



57 



.Example. An airplane weighing 3,000 lb. has a Clark Y wing 350 
sq. ft. in area. What horsepower is required for the wing when flying 
at 90 ft. per sec? 



Cl = 



W 






3,000 



0.002379 



X 350 X (90)^ 



2 

= 0.891 
From Fig. 25, when d = 0.891, Cd = 0.0561 



H.P. = 



550 



^ 0.0561 X 0.001189 X 350 X (90)^ 

550 
= 31.0 hp. 

















— \^ 




















A 





1 


2 






~^ 










y/s 






















/ 


V 




















/ 


i/ 


^■' 




















/ * 
6/ 




















/ 


'-2 X" 






















r 
























/ 






















J 
























1 
























/ 
/ 
/ 
























/ 
/ 

























.02 .04 .06 .08 .10 .12 .14 .16 .18 .20 .22 .24 

Cd 

Fig. 27. Polar curve for U.S.A.-35A airfoil, aspect ratio 6. 

Example. Compare maximum L/D of a Clark Y airfoil with that 
of a U.S.A.-35 airfoil. 



58 AIRFOILS 

In Fig. 25, the line from origin tangent to curve is tangent at point 

where Cl = 0.43 and Cd]^ 0.020. 

43 
For Clark Y, Maximum L/D = ^-^ = 21.5 

In Fig. 27, the line from origin tangent to curve is tangent at point 
where Cl = 0.55 and Cd = 0.03. 

For U.S.A.-35 max. L/D = ^ = 18.3 
or directly from Fig. 26, 

For Clark Y, Maximum L/D = 21.0 

Example. An airplane weighing 2,000 lb. has a C-80 wing 180 
sq. ft. in area. What horsepower is required for the wing when the 
airspeed is 120 miles per hour? 

Solution. 

W 



Cl = 

Isv^ 

2,000 



0.001189 X 180 X (120 X 1.47)2 
= 0.301 

From Fig. 28, when Cl = 0.301, L/D = 23.2 
For horizontal flight, W = L 

2^0 ^ 23.2 

D = 86.2 lb. 
jy p 86.2 X 120 
^'^' = 375 
= 27.6 hp. 

Problems 

1. An airplane weighing 2,500 lb. has a C-80 wing 200 sq. ft. in 
area. What horsepower is required for the wing when airspeed is 
200 ft. per sec? 

2. An airplane weighing 1,800 lb. has a C-80 wing. What area should 
the wing have in order that only 25 hp. will be required for the wing 
when the airspeed is 150 ft. per sec? 

3. An airplane weighing 4,000 lb. has a C-80 wing. It is flying 
at the angle of attack which has Cl of 0.6. What is the drag? 

4. An airplane has a C-80 wing. When Cl is 0.7 and the drag is 
300 lb., what is the lift? 



POLARS 



59 



6. An airplane weighing 2,400 lb. has a U.S.A.-35A wing 210 sq. 
ft. in area. What is the drag when flying at an airspeed of 90 ft. per 
sec? 

6. An airplane weighing 3,500 lb. has a U.S.A.-35A rectangular 
wing of 54-ft. span and 9-ft. chord. What horsepower is required by 
the wing when the airspeed is 150 ft. per sec? 

7. What horsepower is required by the wing in problem 6 when the 
airplane is flying at 100 ft. per sec? 

24 



_J|Q12 



10 



Pii 

iil 



.2 .4 .6 .8 1.0 

Cl 

Fig. 28. LID versus Cl for C-80 airfoil, aspect ratio 6. 



8. An airplane with a U.S.A.-35A wing has a wing loading of 12 lb. 
per sq. ft. What is the Lj D when airspeed is 100 miles per hour? 

9. An airplane with a U.S.A.-35A wing 600 sq. ft. in area is flying 
at a 6° angle of attack with an airspeed of 90 miles per hour. What 
is the lift and what is the drag? 

10. An airplane with a U.S.A.-35A wing has a wing loading of 14 
lb. per sq. ft. It is flying at a 4° angle of attack. What should be the 
airspeed? 

Absolute Coefficients with Metric Units. The absolute coeflfi- 
cients of lift and drag, Cl and Cz), are dimensionless, that is, they 
are pure numbers. Because of this, Cl and Cb are usable in the 



60 AIRFOILS 

standard equations for lift and drag, provided the other factors 
are consistent. With the Enghsh units: hft and drag forces are 
in pounds, mass density is in slugs per cubic foot, area is in square 
feet, and velocity is in feet per second. With metric units: lift 
and drag forces are in kilograms, density is in metric slugs per 
cubic meter, area is in square meters, and velocity is in meters per 
second. 

Under standard conditions, a cubic meter of air weighs 1.2255 
kg. Since the standard acceleration of gravity is 9.807 meters 
per second per second, the standard mass density of air is 1.2255 -^ 
9.807 or 0.12497 metric slug per cubic meter. 

Example. An airplane having a Gottingen 398 wing 35 square 
meters in area is flying at 4° angle of attack, with a velocity of 40 
meters per second. What is the lift? 

Solution. From Fig. 18, Cl = 0.76. 

Lift = Cl^SV^ 

= 0.76 X 5^ X 35 X 40^ 
= 2,660 kg. 

Example. An airplane with a wing loading of 30 kg. per square 
meter is flying with its Gottingen 398 wing at a 2° angle of attack. 
What should be the airspeed? 

Solution. From Fig. 18, Cl = 0.585. 

V = 




= V30xy^ 



.585 X 0.0625 
= 28.7 meters per second 

Engineering Coefficients. In the United States, it was formerly 
universal custom to employ engineering coefficients, Kx and Kyy 
instead of the absolute coeflScients, Cl and Cd- K^ and Ky in- 
cluded in themselves the standard density of air as well as a factor 
for changing miles per hour into feet per second. With engineering 
coefficients, the standard formulas for lift and drag become 

L = KySV^ S in square feet 
D = KxSV^ V in miles per hour 



ENGINEERING COEFFICIENTS 61 

Ky is the lift in pounds of a wing, 1 sq. ft. in area, traveling at a 
velocity of 1 mile per hour; K^ is the drag of that wing, the move- 
ment taking place through standard air. 

The Ky and K^ coefficients given by characteristic curves for 
various airfoils are for use only when the airfoils are moving 
through air of standard density. They should be more properly 
designated Ky^ and Kx^, since their application is only at zero 
altitude. At altitudes where the air density differs from that at 
sea-level, the Ky and K^ shown on the graphs must be corrected 
for the new air density. The symbol a (sigma) is used to express 
the ratio of the density at any altitude to standard density. The 
corrected Ky and K^ for any altitude is found by multiplying the 
Kyo and Kxo (taken from the graphs) by the a for that altitude. 

Ky = aKy^ = —Kyg 

Po 

Kx = (tKxo = — Kxo 
Po 

By comparing the equations for lift and drag containing absolute 
coefficients with the similar equations with engineering coefficients, 
a relation is found between the two sets of coefficients, as follows. 



K, = C,x(^-^^)x{^''^' 



7 ^ \3;6ooy 

= 0.00256 Cl 
and K:, = 0.00256 Cd 

or Cl = 390.7 Ky 

and Cd = 390.7 K^ 

Cl and Cd, being each multiplied by the same constant of trans- 
formation, 0.00256, to obtain Ky and K^ respectively, L/D is equal 
to the ratio of engineering coefficients, since 

L^Cl^ 0.00256 Cl ^ Ky 
D Cd 0.00256 Cd K^ 

Engineering coefficients are much more convenient to use since 
in practice airspeed is usually measured in miles per hour rather 
than in feet per second. There is incongruity, however, in not 
using the same unit of length in calculating area as in computing 
velocity. Engineering coefficients cannot be used for metric units 
without transformation, 



62 



AIRFOILS 



The characteristics of the Clark Y airfoil are shown in Fig. 29. 
It will be noted that the curves are identical with those in Fig. 17, 
but the scales are different. 



























/ 


"^n 
























/ 




























/ 


/ 






















K 






/ 


























>y 










/ 


nnnR 
















/ 










i 


















/ 


/ 










/ 


















/ 












/ 
















/ 












/ 
















y 










*^^ 


/ 








.0003 










/ 










/ 


















/ 










/ 












.0002 






/ 










/ 




















/ 








y 




















/ 






^ 


y' 




















7 





























.0040 

.0030 

.0020^ 
.0015 
.0010 
.0005 



-4-2 2 4 6 8 10 12 14 16 18 20 
Angle of Attack 

Fig. 29. Characteristics for Clark Y airfoil, aspect ratio 6, 
with engineering coefficients. 

Example. A Clark Y wing having an area of 400 sq. ft. at 4° angle 
of attack is moving through the air at a speed of 125 miles per hour. 
What are the lift and drag at sea-level? At 10,000-ft. altitude? 

Solution. From Fig. 29: 
At 4° angle of attack: Ky = 0.00169 
Kx = 0.000089 

At zero altitude: L = 0.00169 X 400 X 125^ 

= 10,6001b. 
D = 0.000089 X 400 X 12? 

= 556 lb. 
From Table I, a at 10,000 ft. = 0.738. 
At 10,000-ft. altitude: 

L = 0.738 X 0.00169 X 400 X 12? 
= 0.738 X 10,600 
= 7,820 lb. 

D = 0.738 X 0.000089 X 400 X 12? 
= 0.738 X 556 
= 410 lb. 



ENGINEERING COEFFICIENTS 63 

Example. An airplane weighing 5,000 lb. has a Clark Y wing 400 
sq. ft. in area; what should be the airspeed at 4° angle of attack at 
sea-level? At 10,000-ft. altitude? What is the drag under each 
condition? 



Solution. 

At zero altitude: 



'Sf: 



5,000 



400 X 0.00169 
= 86.0 miles per hour 



D = 0.000089 X 400 X 86.0 
= 263 lb. 



At 10,000-ft. altitude: 



-s/ 



5,000 



400 X 0.738 X 0.00169 
X 86.0 



V 0.738 
= 100 miles per hour 
D = 0.738 X 0.000089 X 400 X lOO" 



= 0.738 X 0.000089 X 400 X (v/Q-ygg X 86.0J 

= 0.000089 X 400 X SO' 
= 263 lb. 

Note: With weight fixed, at a constant angle of attack, drag is 
independent of altitude. 

Problems (Use Fig. 29). 

1. A wing 150 sq. ft. in area has a Clark Y section. What is the 
maximum load that can be carried at 50 miles per hour? 

2. For the wing in problem 1, what is the minimum speed to carry 
a load of 1,600 lb.? 

3. For the wing in problem 1, what must be the angle of attack 
for the wing to lift 1,600 lb. when the airspeed is 80 miles per hour? 

4. For a wing loading of 14 lb. per sq. ft., (a) what should be the 
angle of attack for a Clark Y wing at an airspeed of 90 miles per hour 
at sea-level; (b) at 10,000-ft. altitude? 

5. An airplane weighing 2,000 lb. has a Clark Y wing 240 sq. ft. 
in area, (a) What is the drag at 120 miles per hour at sea-level; (6) 
at 10,000-ft. altitude? 

6. For an airplane with a Clark Y wing, what should be the wing 
loading to fly 150 miles per hour at 0° angle of attack: (a) at sea- 
level; (6) at 10,000-ft. altitude? 



64 AIRFOILS 

7. An airplane, with a wing loading of 18.2 lb. per sq. ft., has a 
Clark Y wing whose angle of attack is 7°. For level flight what should 
be the velocity (a) at sea-level; (b) at 10,000-ft. altitude? 

8. An airplane has a minimum speed of 40 miles per hour at sea- 
level, what is its minimum speed at 10,000-ft. altitude? 

9. An airplane has a Clark Y wing 270 sq. ft. in area. With wing 
at 3° angle of attack, what should the airspeed be in order that the 
drag does not exceed 300 lb. (a) at sea-level; (6) at 10,000-ft. altitude? 

10. An airplane weighing 1,800 lb. has a Clark Y wing. What 
should be the wing area, if the airspeed is 130 miles per hour when the 
wing is at 1° angle of attack? 

Power with Engineering Coefficients. With engineering coeffi- 
cients, velocity is in miles per hour units. Power required by the 
wing, being the product of drag and velocity, will be in mile- 
pounds per hour units, if velocity in miles per hour is multiplied 
by drag force in pounds. One horsepower is 550 ft-lb. per sec. or 
375 mile-lb. per hour. Using this factor, the formulas for horse- 
power required by the wing become 

DV 
H.P.req. = ^7^ D in pouuds 

S in square feet 

V in miles per hour 

Example. What is the horsepower required by a Clark Y wing 
225 sq. ft. in area at 2° angle of attack and 125 miles per hour (a) at 
sea-level; (6) at 15,000-ft. altitude? 

Solution. From Fig. 29, at 2° angle of attack, Kx is 0.0000614. 

At sea-level : 

„ ^ _ 0.0000 614 X 225 X 125^ 

M.l'.req. ^ 

= 72.0 hp. 
At 15,000-ft. altitude: 

(0.629 X 0.0000614) X 225 X 125^ 
375 
= 0.629 X 72.0 
= 45.2 hp. 




H.P.req. = 



Problems 

1. What horsepower is required by a Clark Y wing 325 sq. ft. in 
area at 7° angle of attack and airspeed of 85 miles per hour? 



POWER IN TERMS OF Cd/Cl^^'' 65 

2. What should be the airspeed in order that a Clark Y wing 
200 sq. ft. in area should require only 45 hp. at 4° angle of attack? 

3. What should be the area of a Clark Y wing so that 50 hp. is 
required to fly at 90 miles per hour at 5° angle of attack? 

4. What horsepower is required by a Clark Y wing 260 sq. ft. in 
area, on an airplane weighing 1,750 lb., when flying at 3° angle of 
attack (a) at sea-level; (6) at 15,000-ft. altitude? 

5. What horsepower is required by a Clark Y wing 260 sq. ft. in 
area, on an airplane weighing 1,750 lb., when flying at 100 miles per 
hour {a) at sea level; (6) at 15,000-ft. altitude? 

Power in Terms of CbICj}''^. The power required to move a 
wing forward through the air may be expressed in a form which 
does not contain velocity (F). 

but 72 = ^^ 



Cl^S 



and F = 



W 



Cl^S 




I 



Then H.P. =-^Cz)^.S 

55U 2, Q P_g 

550 /p V ,S Cl3/2 

An airplane, with a given weight and wing area, will require the 
least power to move the wing forward at the particular angle of 
attack at which Cd/Cl^^^ is the least. In other words, the angle of 
attack at which Cl^^'^/Cd is the maximum is the angle of attack for 
which the least power is required. 

The above expression also shows that at any one angle of attack, 
the power required varies directly as W^^'^, inversely as the square 
root of the wing area S, and inversely as the square root of the air 
density p. 

Example. An airplane weighs 4,000 lb., the wing area is 300 sq. ft. 
The wing is flying at an angle of attack for which Cl = 1.2 and 
Cd = 0.1. What horsepower is required for the wing? 



66 AIRFOILS 

Solution. 



1 ^ A nnn. A.OOO . . 0.1 



sf' 



' V 300 ^ 



55Q . /0.002378 V 300 ^ 1.23/2 



= 58.6 

Example. An airplane weighing 5,000 lb. requires 100 hp. to 
move the wing at a certain angle of attack. If 500 lb. is added to 
the weight of the airplane, what horsepower is required for the wing 
if flown at the same angle of attack as before? 

Solution. 

H.P.2 - [W^J 

100 ^ / 5,000 \V^ 
H.P.2 ~ V5,500J 
H.P.2 = 115.4 

Problems 

1. An airplane wing requires 90 hp. to be flown at a certain angle 
of attack. If the wing tips are clipped so as to reduce the wing area 
from 300 sq. ft. to 250 sq. ft., and the total weight is the same as 
before, what horsepower is needed to fly at the same angle of attack 
as before? 

2. An airplane requires 40 hp. to fly at an angle of attack at which 
Cl = 0.9 and Cd = 0.06. What power is required for the wing 
when flying at an angle of attack at which Cl = 0.6 and Cd = 0.03? 

3. An airplane weighs 6,000 lb., the wing area is 450 sq. ft., the Cd 
of the wing is 0.02, and Cl is 0.45. What power is required by the 
wing at sea-level? 

4. An airplane weighing 5,000 lb. requires 30 hp. for the wing to 
fly at one angle of attack. What horsepower is needed for the wing 
at that same angle of attack but with 800 lb. less load? 

5. Plot a curve of Cl^^^/Cd versus angle of attack for the Clark Y 
airfoil, aspect ratio of 6. 

Moment Coefficient and Center of Pressure. In studying the 
problem of longitudinal balance and stability, it is necessary to 
know not only the magnitude and direction of the resultant of the 
forces on the wing, but also the position of this resultant. Lift 
and drag forces are perpendicular to each other and are com- 



MOMENT COEFFICIENT AND CENTER OF PRESSURE 67 

ponents of the resultant force. The lift squared plus the drag 
squared equals the resultant squared. Like its two components, 
the resultant varies as the air density, the wing area, and the 
square of the velocity; therefore a resultant coefficient may be 
used in a similar manner to the lift and drag coefficients. 

Resultant = R = Vl^ -\- D^ S is area in square feet 

p F is velocity in feet per second 

The direction in which the resultant force acts is the angle 
cot-i L/D, backward from the direction in which the lift com- 
ponent acts. 

The point on the chord through which the line of action of the 
resultant force passes is termed the center of pressure. This is 
abbreviated as C.P. Its location in percentage of chord length 
is given by a curve which is customarily included among the 
characteristic curves of an airfoil. 

For all unsymmetrical airfoils, that is, those having greater 
camber on the upper than the lower surface, the curves of center- 
of-pressure location bear a close resemblance. At angles close to 
the angle of zero lift, the center of pressure is near the trailing 
edge of the wing. As the angle becomes more positive, the center 
of pressure moves forward. At some angle, usually a few degrees 
below the angle of maximum lift, the center of pressure is at its 
most forward position. The maximum forward position of the 
center of pressure is about 28 per cent of the chord length back 
from the leading edge. 

As the angle of attack is decreased below that for maximum 
forward position, the center of pressure moves backward. The 
resultant is an upward and backward force. At the exact angle 
of zero lift, the resultant instead of being a single force becomes a 
couple tending to depress the leading edge and raise the trailing 
edge, plus the drag component. If the angle of attack becomes 
more negative than the angle of zero lift, the resultant reappears 
as a single force again but acting downward and rearward, with 
the center of pressure near the trailing edge. Depressing the 
leading edge still more, the resultant, as a larger downward and 
backward force, moves forward. 

The above-described movement is termed " unstable " center-of- 
pressure movement. When the wing is balanced at one angle of 



68 AIRFOILS 

attack, if this angle of attack is momentarily increased by a gust 
of wind or otherwise, the forward movement of the upward 
resultant force tends to tip the leading edge of the wing upward 
still more. This increase of attack angle moves the center of 
pressure still further forward, tending to increase the nosing-up 
still more, so that a stall would eventually result. Conversely, 
a decrease in the angle of attack from a previously balanced 
condition would cause a backward movement of the center of 
pressure which, by lifting on the trailing edge, tends to decrease 
the angle of attack more. 

With symmetrical airfoils, that is, those having both surfaces 
convex and of the same camber, there is practically no center-of- 
pressure movement. These airfoils are termed '' stable " airfoils. 
On the more common non-symmetrical airfoils, a reverse curvature, 

^ ^ i.e., having the trailing edge 

>/ i curve upward, gives an airfoil 

rpw^^.^c^^ nW \ with little or no center-of- 

\*~~^ /c p X c X sin a pressure movement. 
^^^-^^^JDrag The moment of the resultant 

Dir ection of relative wm7^C >/^^^ ^ ^rCC On the wing is the product 

^ ^ of the force and the distance 

Fig. 30. Diagram of moments about ^^^^ ^j^g i^^e of action of 
mg e ge. _^^^ force to the point about 

which the moment is taken. Moments which act in a manner 
tending to increase the angle of attack are called stalling moments 
and are designated by a positive sign. Moments which tend to 
decrease the angle of attack are called diving moments and are 
negative in sign. 

When the moment is taken about the leading edge, the sum of 
the moments of the two components may be used; see Fig. 30. 
Moment about leading edge = Mq = C.P. XcXcosaXL 
+ C.P. X c X sin a; X D 

moment is in foot-pounds. 

C.P. is in percentage of chord length, c is chord length in feet, 
L and D are in pounds. 

This may also be written as follows: 

Moment about leading edge = Mo = CmqC -^^^^ 
Cmo = (C.P.) (Cl cos a + Cd sin a), V m feet per second 



MOMENT COEFFICIENT AND CENTER OF PRESSURE 69 

In engineering units : 

Moment about leading edge = Mo = KmoC^^^V^ 
Kmo = (C.P.)(i^y cos a -\- Kx sin a), V in miles per hour 

Since a is always a small angle, cosine a is very close to unity, 
and only a slight error is introduced if the term Cl is used instead 
of Cl cos a. Also since Cd is usually small compared with Cl, 
Cd sin a will be still smaller compared with Cl and only a slight 
additional error will be incurred if the term Cd sin a is dropped. 
Then a sufficiently close approximation for most work is to let 

Cmo = — (CP.) (Cl) V in feet per second 

and Kmo = — {S^^.){Ky) V in miles per hour 

minus signs being used to designate diving moment. 

Equations (Mq = CmoC^SvA and (Mo = KmocSV^) may be 
transformed to read 

Cmo = — V in feet per second 

and Kmo = 0x^2 ^ ^^ miles per hour 

Example. Find moment about leading edge of a R.A.F. 15 airfoil 
of 42-ft. span and 7-ft. chord at a 2° angle of attack with an airspeed 
of 120 ft. per sec. 
Solution. 
From Fig. 21, when a = 2° 

Cl = 0.295 
Cd = 0.0156 
CP. = 36 per cent 
Then: 

Moment about leading edge = 

002*^78 
0.36 X (0.295 cos 2° + 0.0156 sin 2°) X 7 X / X 42 X 7 X T20^ 

= -3740ft-lb. 

Problems 

1. Plot Mc versus angle of attack for an R.A.F. 15 airfoil. 

2. What is the moment about the leading edge of an R.A.F. 15 wing 
of 8 ft. chord and 48-ft. span at a 7° angle of attack with an airspeed 
of 70 miles per hour? 

3. What is the moment about the leading edge of an R.A.F. 15 



70 



AIRFOILS 



wing of 45-ft. span and 7f-ft. chord at a 4° angle of attack with an 
airspeed of 120 miles per hour? 

4. What is the moment about the leading edge of an R.A.F. 15 
wing of 36-ft. span and 6-ft. chord at a 3° angle of attack with an 
airspeed of 90 miles per hour? 

5. What vertical force must be applied 5 ft. back from the leading 
edge to prevent rotation of a R.A.F. 15 wing of 54-ft. span and 9-ft. 
chord at an angle of attack of 0° at an airspeed of 115 miles per hour? 

Center of Pressure. Instead 
of finding the moments about 
the leading edge of the wing, 
the moment may be found 
about a point located back of the 
leading edge. Let pc feet be the 
distance from the leading edge 
to this point, where p is in per- 
centage of chord length; see 
Fig. 31. Then the moment 
about this point will be 




Fig. 31. Moments about point not 
at leading edge. 



-M = {Cl cos a + Cn sin a)^SV^ [(C.P. X c) - pc] 

the minus sign being used because the force shown in Fig. 31 will 
cause a diving or negative moment. An approximate form of the 
above is 

-M = Cl^SV'{cT. Xc-pc) 

dividing by J pSV^c gives 
M 



^^SV^c 



= C.P. XCl- pCl 



letting 



M 



ISV^c 



Cm, moment coefficient about point p, 



-Cm = C.P. XCl- pCl 
That is, for any angle of attack, the moment coefficient about 
any point along the chord is the lift coefficient for that angle 
multiplied by the difference between the position of the center of 
pressure and the position of the point measured in percentage of 
chord. 



AERODYNAMIC CENTER OF PRESSURE 



71 



For most airfoils, if, from experimental data, 1/Cl is plotted 
against C.P., a curve resembling that in Fig. 32 results. It 
closely approximates a straight line up to values of Cl close to 
the burble point. Any deviations from the straight line may be 
accounted as due to experimental error. Assuming that it is a 















/ 










4.0 












/ 


















/ 


1 










3.0 

-J 










/ 




















/ 












2.0 








/ 




















/ 














1.0 








/ 


















/ 






















/ 

















A .5 .6 
C. P from L E. 



8 



.9 1.0 



.1 .2 3 
Fig. 32. 1/Ci, versus C.P. for Clark Y airfoil. 



straight line, it represents an equation of the form x = ay -\- h. 
The intersection of the straight line with the horizontal axis is 
approximately 0.25 for practically every airfoil investigated. 
That is, when 1/Cl equals zero, C.P. equals 0.25, and since the 
straight line has a constant slope, from the experimental data, 
it may be stated that 

— 0.25+_^(i-J 

Cl(C.P. - 0.25) 



C.P. 



or 



h 



Fromthepreviouslyderivedequationthat — Cm = Cl(C.P. — p), 
where Cm is the moment coefficient about a point that is p per cent 
of the chord back from the leading edge, the moment coefficient 
about a point that is a quarter-chord length back from the leading 
edge is 

-Cmo.25 = Ci(C.P. - 0.25) 



72 AIRFOILS 

The constant slope k of the graph of C.P. versus 1/Cl is there- 
fore equal to the moment coefficient about the quarter-chord 
point, Cmo.25. Since the moment coefficient, and consequently 
the moment, about the quarter-chord point are constant, this 
point is called the aerodynamic center of the airfoil. The moment 
coefficient about the quarter-chord point for various airfoils is 
tabulated in Table V. 

Since -Cj/0.25 = Cl(C.P. - 0.25) 



C.P. = 0.25 - 



^^^0.25 



Cl 

substituting this value for C.P. in the equation for moment 
coefficient about any point p per cent of the chord back from the 
leading edge gives 

-Cm- -pCL + CL{o.25-^y 

= -pCl + 0.25 Cl - Cmo.25 
An inspection of this relation shows that, when Cl equals zero, 
the moment coefficient about any point is equal to the moment 
coefficient about the quarter point. Therefore at the angle of zero 
lift the moment coefficient about the leading edge is equal to the 
moment coefficient about the quarter point, which is approxi- 
mately constant for all angles. 

CuiO.^b) = Cmo{Cl=q) 

Example. An airplane whose wing loading is 8 lb. per sq. ft. is 
flying level at an airspeed of 100 miles per hour. If at angle of zero 
lift, Cmo = —0.067, what is the center of pressure? 

Solution. V = 100 miles per hour = 146.7 ft. per sec. 

W 



Cl = 






0.002378 . , ^^^2 



2 X 
= 0.313 



C.P. = 0.25 - 
= 0.25 - 



Cl 
-0.067 



0.313 
= 0.25 + 0.214 
= 0.46 
or 46 per cent chord length back of leading edge. 



AERODYNAMIC CENTER OF PRESSURE 73 

Example. At 6° angle of attack, an airfoil has a Cz, = 0.84, 
Cd = 0.06, and Cmo = —0.27. Find position of center of pressure by 
both approximate and exact methods. 

Solution. 

— Cmo 



Approximate C.P. 



Cl 



Exact C.P. 



/ -0.27 \ 
~ \ 0.84 ) 

= 0.322 

— Cmo 



Cl cos a -\- Cd sin a 

-(-0.27) 

~ 0.84 X 0.995 + 0.06 X 0.105 

_ +0.27 
~ 0.836 + 0.006 
= 0.321 

Problems 

1. An airplane with wing loading of 15 lb. per sq. ft. is flying at an 
airspeed of 125 miles per hour. At angle of zero lift Cmq = —0.06. 
By approximate method, where is the center of pressure? 

2. A monoplane weighing 2,592 lb. with wing span of 36 ft. and 
wing chord of 6 ft. is flying at a speed of 120 miles per hour. If the 
center of pressure is 35 per cent of chord length back of the leading 
edge: (a) what is the moment about the leading edge; (b) what is the 
moment coefficient about the leading edge; (c) what is the moment 
about the leading edge at angle of zero lift? 

3. An airfoil at 3° angle of attack has & Cl = 0.65, Cd = 0.036, 
and Cmo — —0.24. Find location of the center of pressure {a) by 
approximate method; (6) by exact method, (c) Find Cmq at angle 
of zero lift. 

4. At 0° angle of attack a certain airfoil has Cl = 0.16 and the 
center of pressure is 30 per cent of the chord length back from the 
leading edge. What is the moment coefficient about the leading 
edge when lift is zero? 

5. For a certain airfoil, when lift is zero, the moment coefficient 
about the leading edge is 0.06; what is Cl when the center of pressure 
is 40 per cent of chord length back of leading edge? 

6. jWhat value of moment coefficient about the leading edge at 
angle of zero lift will give a center of pressure of 0.52 at a speed of 
150 miles per hour for an airplane weighing 8,000 lb. and having a 
wing area of 750 sq. ft.? 



74 AIRFOILS 

BIBLIOGRAPHY 

U. S. Army Air Corps, Handbook of Information for Airplane Designers. 
N.A.C.A. Technical Reports. 
DiEHL, Engineering Aerodynamics. 



CHAPTER V 
INDUCED DRAG OF MONOPLANES 

In the previous chapter, the action of air in flowing over wings 
was described. All the data given, or used in problems, were, 
however, for wings having an aspect ratio of 6; i.e., the span was 6 
times the chord length. 

By chance, early measurements of the lift and drag of airfoils 
were made on small model wings having an aspect ratio of 6. 
Later when it was desired to know whether a change in the camber 
would improve the characteristics, the models of the newer airfoils 
were made geometrically similar to the older models, i.e., with the 
same aspect ratio, so that any difference in the characteristics 
could be attributed solely to the difference in contour of the airfoil. 
Practically all tests on airfoils are therefore made on models having 
an aspect ratio of 6, and unless a definite statement is made to the 
contrary, it may be safely assumed that data published about 
airfoils refer to airfoils having this standard aspect ratio. 

The changes in characteristics due to changes in aspect ratio are 
now quite definitely known. For convenience, it is becoming more 
customary to furnish data for airfoils having an infinite aspect 
ratio. These data can then be corrected to be applicable to wings 
of any aspect ratio. 

Variation of Lift across Span. The motion of the air in flowing 
around a wing is very complex. Air flows backward over the top 
side and under the lower side. This motion is complicated 
because of other incidental conditions. 

The air pressure on the upper side of a wing in motion is slightly 
less than atmospheric pressure; that on the under side is slightly 
greater than atmospheric pressure. Fluids will always go from 
a region of high to one of low pressure. Therefore, in flight, air is 
spilling out from below the wing tip and up into the region of low 
pressure on top of the wing. Therefore, on the upper side of the 
wing near the wing tips, the pressure is not quite as low as over the 
rest of the wing because of the excess coming up from below partly 
filling this low-pressure area. On the under side of the wing near 

75 



'6 



INDUCED DRAG OF MONOPLANES 



the tip, because of the air passing out and up, the positive pressure 
is not as great as under the inner portions of the wing. 

Lift, being due to the difference in air pressure between the lower 
and upper sides of the wing, is not uniform over the span of even 
a rectangular wing. Tests have been repeatedly run which check 
with these deductions. Pressure measurements, made at various 
points across the span, show that difference in pressure is maximum 
at the center of the span, decreasing by small amounts towards the 
wing tips. 

The distribution of lift along the span of a rectangular wing is 
shown in Fig. 33a. Close to the center of the span the decrease 



i 



a' 


^i 


I" 


J 




\i 






I y 






. y 


,, 






., 


J 


^ J 


ttttf 



Fig. 33a. Lift distribution across span of a rectangular wing. 

from maximum is slight. At a greater distance from the center 
the rate of decrease is more pronounced. 

In calculating the necessary strength of the spars the distribu- 
tion of the supporting forces of lift should be known. The 
Department of Commerce arbitrarily prescribes that, for internally 
braced wings, the lift shall be considered uniform from the center 
of the span to a point one chord length in from the tip, and from 
there it shall decrease to 80 per cent of the uniform lift (see Fig. 




Fig. 336. Department of Commerce lift distribution for internally 
braced wing. 

336). For non-rectangular or externally braced wings, shghtly 
different assumptions are made. Although these lift distributions 
are not exactly correct, they do make some allowances for the de- 
crease in lift at the wing tips and are sufficiently close for strength 
computations. 

It is to be noted that, with a wing of large aspect ratio, the 
decrease in lift at the wing tip is less in proportion to the total 
than with a wing of small aspect ratio. An imaginary wing of 



ACTION AT TRAILING EDGE 



77 



infinite aspect ratio, having no wing-tip losses, would have uniform 
lift across the span. 

Inward and Outward Flow. Because, on the under side of the 
wing, air near the tip is flowing out and upward, air nearer the 
center flows out to replace this air. The air on the under side of 
the wing has therefore not only a backward motion relative to the 
wing but also an outward motion as well. Near the center of the 
span this outward component is weaker than near the tip. The 
resultant motion is illustrated in Fig. 34a. 

On the upper side of the wing, owing to air coming up and 
inward over the tips, there is an inward component, which is 



Leading Edge 

'^11 III \\\\\' 


Trailing Edg 

(a) 

Leading Edg 


e 


:^\\\\ 


//////. 



Trailing Edge 

% 

Fig. 34. Air motion (a) under side, (6) upper side of wing. 

strongest at the tips. The resultant of this inward movement of 
air and the rearward velocity is shown in Fig. 346. 

The theoretical wing of infinite aspect ratio would have no 
inward and outward flow. The flow would be directly backward 
and downward. 

Action at Trailing Edge. At any point on the trailing edge, 
coming over the upper side of the wing, is a streamline of air 
which has a direction backward, downward, and inward towards 
the center of the span, and coming under the lower side of the wing 
is a streamline of air which has a direction backward and outward. 
The juxtaposition of these two streamlines at the trailing edge 
initiates a tiny vortex. These vortices are formed at an infinite 
number of points along the traihng edge. 

On the left wing trailing edge, these vortices are counter- 
clockwise, viewed from the front; on the right wing they are 



78 INDUCED DRAG OF MONOPLANES 

clockwise. These small vortices, immediately after their forma- 
tion, merge into two main vortices, one at each wing tip. These 
vortices are powerful air movements, and in fljdng in close forma- 
tion great care must be taken that the wing of a following airplane 
shall not protrude in a wing-tip vortex of a leading plane. 

The effect of the vortex at each wing tip is to give an upward 
motion to air outside the wing tip and an inward and downward 
motion to air inside it. Since at each wing tip air is coming in- 
ward, the momentums neutrahze each other and can be neglected. 
The downward components of motion are, however, of considerable 
importance. 

On a wing of infinite aspect ratio, there are no wing-tip vortices. 
Air flowing over and under the wing is given a downward deflection 




Fig. 35. Wing-tip vortices. 

as explained in the last chapter. A streamUne, under these 
conditions, is always in a vertical plane parallel to the vertical 
plane of symmetry of the airplane. This type of flow is two 
dimensional and is called profile flow. 

With a finite aspect ratio, there are wing-tip vortices. The 
down flow due to these vortices is in addition to the down flow of 
the profile flow. With smaller aspect ratio the importance of this 
down flow due to the wing-tip vortices becomes greater. The 
downward velocity caused by the wing tips is called the induced 
velocity and is denoted by the symbol w. 

Induced Angle of Attack. The geometric angle of attack of a 
wing, heretofore referred to simply as the angle of attack, has been 
defined as the angle between the relative wind and the wing 
chord. The relative wind is the direction from which the air 




INDUCED ANGLE OF ATTACK 79 

comes in meeting the wing, this direction being the direction of the 
air stream before it has been disturbed by the approaching wing. 

With a wing of finite aspect ratio, the air through which the 
wing is passing has a downward velocity due to the wing-tip 
vortices. Then the wing is not traveUng through air which has a 
relative direction exactly opposite to Relative wind 

the wing's forward direction, but, in- ^^ 
stead, through air which has a motion ^^eanT^/^T — 

both backward and downward past ^'"^• 

the wing. The true velocity of the air ^^«- 36«- Relative wind and 

, ,. , ,1 . • ii ^. , mean relative v.dnd. 

relative to the wmg is the resultant 

of the two velocities, the backward velocity V, equal in magni- 
tude to the airspeed, and the induced downward velocity w; see 
Fig. 36a. This resultant may be considered a true relative wind 
as it is the actual direction of the air passing backward and down- 
ward past the wing. 

The downward velocity w is always small with respect to the 
airspeed V, so that the resultant true velocity is very little different 
in magnitude from the airspeed V. This slight difference is 
negligible, and the magnitude of the true velocity is considered 
to be F. 

The angle between relative wind V and the true relative wind is 
called the induced angle of attack and, though small, is very 
important. It is represented by the symbol o;;. 

The geometric angle of 'attack a may be considered as being 
made up of two parts, the induced angle of attack a,- and the 
effective angle of attack ao, where 

a = ao -{- ai 

With an infinite aspect ratio, there being no wing-tip vortices, 
there is no downward motion of the air in which the wing is moving 
so that the direction of the relative wind measured forward of the 
wing in undisturbed air is the same as the direction of the relative 
wind right at the wing. The induced angle of attack is therefore 
zero, and the effective angle of attack is the same as the geometric 
angle of attack. For this reason the effective angle of attack is 
sometimes called the angle of attack for infinite aspect ratio. 

A wing of finite aspect ratio in traveling forward horizontally is 
always moving in air that has a downward motion. The resultant 
relative wind is therefore the direction of the air which must be 



80 



INDUCED DRAG OF MONOPLANES 



considered, and it is the effective angle of attack which determines 
the air flow and the forces acting on the wing. The Hft and drag 
forces depend on the effective angle of attack rather than the 
geometric angle of attack. 

Two wings, having the same airfoil section, the same area, and 
moving at the same forward speed but having different aspect 
ratios, will have the same forces acting over the surface of the wing 
and the same resultant force provided the effective angle of attack 
is the same for both wings. The wings having different aspect 
ratios, the strength of the wing-tip vortices will be different so 
that the induced angles of attack will be different. The wing 
with the greater aspect ratio will have the smaller induced angle 
of attack. 

Profile and Induced Drag. In Fig. 366, let Cr be the coefficient 
of the resultant of all the small forces acting over the upper and 

lower surfaces of the wing. 
Resolve Cr into two com- 
ponents, Clq perpendicular to 
the true or resultant relative 
wind and Cdq parallel to the 
true relative wind. These 
would be the lift and drag 
coefficients for a wing of 
.infinite aspect ratio. For 
practical wings, the true 
relative wind does not coin- 
cide with the free air direc- 
to a 




Fig. 366. Induced drag coefRcient. 



tion, and these coefficients must therefore be transformed 
different set of axes. 

Resolving these forces into components perpendicular and 
parallel to the undisturbed relative wind, it is found that 



and 



where 



Cl = CiQ cos ai — Cdo sin ai 
Cd = Clo sin ai + Cdo cos ai 

ai = tan ^^ 



The induced angle of attack, a,-, is always small, so that cos a,- 
may be taken as unity. Cdq is small in comparison with C^o, and 
sin a, is a small fraction; therefore Cm sin oi is very small in 



PROFILE AND INDUCED DRAG 81 

comparison with Clo and may be neglected. If angle a, is in 
radians, for small angles the sine is equal to the angle itself; i.e., 
ai = sin ai. The equations thus become: 

Cl = Clo 
and Cd = Cioii + Cm 

or Cd = Cm + Cdo where Cni = Cloli 

The drag coefficient is composed of two parts. One part, 
represented by the coefficient Cm, is caused by the induced down- 
ward velocity of the wing-tip vortices and is called the induced 
drag coefficient. The other part, Cdq, is called the profile drag 
coefficient. In the same way, the drag itself is divided into two 

corresponding parts: the induced drag equal to Cm ly^y^ and the 

profile drag equal to Cm ^SV^. 

When the wing has an infinite aspect ratio, there is no induced 
drag and all the drag is that due to the profile or two-dimensional 
flow. The drag is then due only to the skin friction of the air with 
the surface of the airfoil. The amount of this drag depends 
chiefly on the shape of the airfoil at angles below where excessive 
burbling takes place. It is because of the profile of the airfoil 
determining its magnitude that Do is called profile drag. Profile 
drag is independent of angle of attack, up to the burble point, and 
is independent of aspect ratio. 

With finite aspect ratio, in addition to the profile drag, there is 
the induced drag. The coefficient of induced drag depends on the 
lift — which depends on the angle of attack — and on the induced 
angle of attack, which depends on the effect of the wing-tip 
vortices, which in turn depends on the aspect ratio. 

The induced drag, being a component of Hft, depends not only 
on the lift but also on the slant of the lift backward from the per- 
pendicular to the relative wind. Dr. M. Munk has proved that 
this angle, a,-, is minimum when the downwash is constant along 
the span, and this constant downwash is obtained when the lift 
distribution along the span varies in such a manner that, when 
plotted with the lift as ordinates and distance along span as ab- 
scissas, the resulting curve is a semi-ellipse with the span as the axis. 

When a wing moves forward the air which comes in immediate 
contact with the surface is affected and has motion imparted to it. 



82 INDUCED DRAG OF MONOPLANES 

Farther away from the wing the air is affected less. At great 
distances above or below the wing, there is still an effect on the air 
but so slight as to be negligible. The actual situation is that air 
is affected by a varying amount from close to the wing to infinite 
distances away from the wing where the effect is zero, but for the 
purpose of calculation it is assumed that the effect is the same as 
if a definite area is affected uniformly. This fictitious area t)f 
uniform effect is termed the swept area and is denoted by S\ It 
is a cross-section of the air stream taken perpendicular to the 
direction of motion of the wing. 

Then the mass of air affected by the wing in unit time is pS' V. 
The downward momentum imparted to this mass in unit time is 
pS'V X Vai where Vai is considered equal to w, the downward 
velocity; it is measured in radians. It has been proved mathe- 
matically that the downward momentum in unit time is equal to 
one-half the lift, i.e., 

Then a; = 27sT^2 

Prandtl has shown that, with semi-elliptic lift distribution, the 
swept area is a circle whose diameter is the span. Then 

^ -T 





L 


Cl^SV' 


ClS 


2 


,^^y. 


2p"fF^ 


Trb^ 


pe 


ct ratio 


is equal to ¥/S. Therefore 




ai = 


Cl 

TT A.R. 






Cm = 


CLai 








Cl'S 








7r62 








Cl' 





also 



TT A.R. 

It must be emphasized that these two expressions are true only 
when the lift plotted against span is a semi-elliptic curve. This 



PROFILE AND INDUCED DRAG 



83 



makes at the same all across the span. This particular lift distri- 
bution is obtained from wings that are elliptic in plan view. 
For wings of other shape, such as rectangular or tapered wings, 
these expressions are not exactly correct. The error in using them 
is very slight, however, and they are commonly employed without 
any corrections.* 

The expression given above for ai gives the induced angle of 
attack in radians. Since 1 radian is 57.3° 



ai (degrees) 



Cl 



ttA.R. 

18.24 Cl 
A.R. 



X57.3 



The induced drag coefficient being known, the induced drag may 
be found, as follows 



A = Cm-SV^ 






7r^62 72 



* Glauert has shown that, for rectangular wings, more nearly correct for- 
mulas are 

Cl 



ai 



A.R. 



{l + T) 



and 



CDi = -^ (1 + S) 



ttA.R. 

where the correction factors T and S vary with aspect ratio as follows: 



A.R. 


T 


S 


A.R. 


T 


S 


3 


0.11 


0.022 


7 


0.20 


0.064 


4 


.14 


.033 


8 


.22 


.074 


5 


.16 


.044 


9 


.23 


.083 


6 


.18 


.054 









The neglect of these correction factors never results in an error exceeding 
6 per cent. 



84 



INDUCED DRAG OF MONOPLANES 



7r|62 72 
0.148 L^ 



V in feet per second 

V in miles per hour 



For level flight, the lift is always substantially equal to the 
weight. Therefore for level flight 



«=^(?J «=!'" 



The weight divided by the span is called the span loading. For 
level fhght the induced drag varies directly with the square of the 
span loading and inversely as the dynamic pressure. 

At sea-level (i.e., standard density) 



A= 



1 



TT X 0.00119 VKh 

2 



268 



(!)■ 



125 



©■ 



V in feet per second 

V in miles per hour 



The horsepower required to overcome induced drag is 
DiV 



H.P.z^- = 



550 



V in feet per second 




1,730 1 6^7 
0.00116 W 



PROFILE AND INDUCED DRAG 85 

At standard density, this becomes 

0.488 ^y^' 
H.F.Di = tT — ^ i^ ^^^^ P^^ second 

1251^ 



or H.P. 



Di 



(fS 



375 V 

2 



(il 

3 V 



V in miles per hour 



That is, the horsepower required at sea-level to overcome the 
induced drag is one-third the span loading squared divided by the 
velocity in miles per hour. At any altitude the horsepower 
required to overcome induced drag is the horsepower required at 
sea-level, at that speed, divided by the relative density at that 
altitude: 



H.P.Dj = — X Vt7 V in miles per hour 

p 6 V 

Example. A rectangular monoplane wing has a span of 39 ft. and 
a chord of 6 ft. What are the induced angle of attack and the induced 
drag coefficient, when the lift coefficient is 0.8? 

Solution. 





A.R. 


39 
6 
= 6.5 


CLi 


(degrees) 


18.24 X 0.8 
6.5 
= 2.22° 




Cm 


0.82 
TT X 6.5 

= 0.031 



Example. A monoplane weighing 2,000 lb. has a span of 38 ft. 
What is the induced drag at 10,000-ft. altitude if the airspeed is 80 
miles per hour? What horsepower is required to overcome the in- 
duced drag"* 



86 INDUCED DRAG OF MONOPLANES 

Solution. 

80 miles per hour = 117 ft. per sec. 

Span loading = -r- 



_ 2,000 

38 
= 52.61b. per ft. 

p.- ^^ 

' TT X i X 0.00176 X 1172 

= 73 lb. 



^ ^ 0.00116 X 52.6' 



0.00176 X 117 
= 15.5 hp. 



Problems 

1. A rectangular monoplane wing has 42-ft. span and 5-ft. chord. 
"When Cl = 0.65, what are the induced angle of attack and the 
induced drag coefficient? 

2. A rectangular monoplane wing has 38-ft. span and 7-ft. chord. 
When Cl = 0.72, what are the induced angle of attack and the in- 
duced drag coefficient? 

3. At an airspeed of 90 miles per hour at sea-level what is the in- 
duced drag of a monoplane weighing 4,700 lb. and having a wing span 
of 52 ft.? 

4. The Northrop Delta weighs 7,000 lb., and its span is 48 ft. 
What horsepower is required to overcome induced drag of wing when 
flying at sea-level at 200 miles per hour? 

5. The Taylor Cub weighs 925 lb., and its wing span is 35 ft. 2| in. 
What horsepower is required at sea-level to overcome induced wing 
drag at an airspeed of 60 miles per hour? 

6. The Lockheed Vega weighs 4,750 lb.; its wing span is 41 ft. 
At sea-level, what is the induced drag at 210 ft. per sec? 

7. A Stinson Reliant weighs 3,125 lb.; its wing span is 43 ft. 3| in. 
At 5,000 ft., what is induced drag at 150 ft. per sec? 

8. A Fairchild monoplane weighs 1,600 lb.; its wing span is 32 ft. 
10 in. At 10,000-ft. altitude what horsepower is required to over- 
come induced drag at 110 miles per hour? 

a. A monoplane weighs 3,000 lb. At sea-level, flying at 150 ft. 
per sec, what is the induced drag (a) if span is 35 ft.; (6) if span is 
30 ft.; (c) if span is 25 ft.? 

10. A monoplane weighs 3,000 lb. ; its span is 30 ft. What is the 
induced drag, at sea-level, (a) at airspeed of 100 ft. per sec; (6) at 
200 ft. per sec? 



CORRECTIONS FOR ASPECT RATIO OF MONOPLANE 87 



If engineering coefficients are used, since Ky = 0.00256 Cl, the 
expression for a,- becomes 

,, , 18.24 Ky 

ai (degrees) = ^.00256 A.R. 

^ 7130 Ky 
A.R. 

Also in engineering coefficients, the coefficient for induced drag 
is K^i, where K^i = 0.00256 Cm. 
Since 

Cl' 



C 



Di 



ttA.R. 



0.00256 Coi = 0.00256 



7_^Y 

\0.00256/ 



A.R. 



K. 



K/ 



0.00256 7rA.Il. 



125 K/ 
A.R. 



Corrections for Aspect Ratio of Monoplane. As shown in a 
previous paragraph, the geometric angle of attack is made up of 
two parts, the effective angle of attack and the induced angle of 




i.oo 



-5 5 ^ 10 15 20 

Degrees 

Fig. 37. Effect of aspect ratio on Clark Y characteristics. 



attack. Two wings of the same area, same airfoil section, and 
same airspeed will have the same lift, provided the effective angle 
of attack is the same in both cases. If these two wings have a 
different aspect ratio, the one having the smaller aspect ratio will 



88 INDUCED DRAG OF MONOPLANES 

have a larger induced angle of attack and will consequently need 
to have a greater geometric or total angle of attack. 

These two wings having the same effective angle of attack will 
have the same profile drag, but since the induced drag is greater 
on the wing of smaller aspect ratio, the total drag will be greater 
on that wing. 

If the total angle of attack is known that gives a certain lift 
coefficient with a wing of one aspect ratio, the total or geometric 
angle of attack that will be needed for a wing of a different aspect 
ratio to give the same lift coefficient can be found as follows. 
The difference between the geometric angles will be the difference 
between the induced angles. Let A and B be the two aspect 
ratios, a^ being the total angle of attack for wing of aspect ratio A 
which gives the same hft as the total angle of attack as gives for 
the wing of aspect ratio B. 

Then 

18.24 Cl 18.24 Cl 
aA-aB= —^ ^r— 

= 18.24 C.[^-i; 

The difference in total drags will be the same as the difference in 
induced drags. Then for the two wings of the preceding para- 
graph, A and B being the two aspect ratios and the wings being 
the same airfoil section and same area, the total drags for each 
will be 

Cd(A) — Cdo{A) + CuiiA) 

Cd{B) = Cdo(,b) + Cdub) 

The profile drags, being independent of aspect ratio, are the 
same for both wings. Then 

Cdu) — Cd{B) = CmiA) — Cdub) 
ttA ttB 



^71 -1^ 



Example. An airfoil, with aspect ratio of 6, at an angle of attack 
of 3°, has Si Cl = 0.381 and Cd = 0.0170; find, for the same airfoil 
section, the angle of attack and the Co that will correspond with the 
Cl of 0.381 if the aspect ratio is 4. 



COEFFICIENTS FOR INFINITE ASPECT RATIO 89 

Solution. 

Approximate: 

ae-a,= 18.24 X 0.381 (i - i) 
= -0.579 
3° - (-0.579) = 3.579° = angle of attack for A.R. 4 

CDie) - Cm,) = ^^^ Xii-i) 

TT 

= -0.00385 
0.0170 - (-0.00385) = 0.0208 Cd for A.R. 4 

Exact (using Glauert corrections) : 

ae-ai = 18.24 X 0.381 (J^ - ^^ 

= -0.611 



^ ^ 0.381Y1.054 1.033\ 



= -0.0038 

Coefficients for Infinite Aspect Ratio. With infinite aspect 
ratio, there is neither induced angle of attack nor induced drag. 
Lately it has become customary to give the characteristics of an 
airfoil as if the airfoil were of infinite aspect ratio. When they 
are so given, the angle of attack is identical with the effective 
angle of attack and the drag coefficient is the profile drag coeffi- 
cient. To find the characteristics for a finite aspect ratio, to the 
effective angle of attack is added the induced angle of attack for 
that aspect ratio to give the geometric angle of attack, and to the 
profile drag coefficient is added the induced drag coefficient for 
that aspect ratio to give the total drag coefficient. 

Example. For a certain airfoil of infinite aspect ratio at 9° angle of 
attack, the Cl is 1.03 and Cd is 0.067. Find corresponding charac- 
teristics for an aspect ratio of 8. 
Solution. 

Approximate : 

a = ao -\- ai 

_ Qo , 18.24 X 1.03 
"^ "^ 8 

= 9° + 2.3° 
= 11.3° 
Cd = Cdo + Coi 

= 0.067 + ifi3 

= 0.067 + 0.042 
= 0.109 



90 



INDUCED DRAG OF MONOPLANES 



Wing of aspect ratio 8 at angle of attack of 11.3° will have lift coeffi- 
cient of 1.03 and drag coefficient of 0.109. 



1.6 
















_^^ 








Cdq Profile Drag Coefficient -- 


>^ 


'*/' 


/- 










12 
1.0 
.8 

V 

1-2 

-.2 
-.4 
-.6 
-.8 










^ 




/ 


















/ 




/ 


















/ 




> 


/ 


















/ 




/ 


-—Angle of Attack Of, 















/ 


/ 


/ 




















1 


/ 






















/ 


/ 






















Y 






















/ 


\ 






















/ 


\ 


V 


















/ 






^ 


-^ 
















[/ 
















' ■ 











'16 



-12 


-8 
.01 


-4 4 
Degrees 
.02 


8 
.03 


12 


16 
.04 



.05 .06 

Fig. 38. Characteristics of Clark Y airfoil, infinite aspect ratio. 



Example. From Fig. 18, the lift and drag coefficients for the 
Gottingen 398 airfoil for an aspect ratio of 6, at an angle of attack of 
2°, is Cl = 0.293 and Cd = 0.031. Find corresponding characteristics 
for infinite aspect ratio. 



Solution. 



Q;(A.R.6) = ao(A.R.6) + a»(A.R.6) 



2° = ao + 



18.24 X 0.293 
6 



Qjo = 2° - 0.89 

= 1.11°, angle of attack for infinite aspect ratio 
to give Cl of 0.293. 



COEFFICIENTS FOR INFINITE ASPECT RATIO 91 

Cz)(A.R.6) = Cdo + C'l>,-(A.R.6) 



0.031 = Cdo + ^'^^^ 



TT X 6 
Cz)o = 0.031 - 0.0045 

= 0.026 drag coefficient for infinite aspect ratio 
when Cl = 0.293 (i.e., angle of attack 1.11°). 

Example. A certain airfoil with aspect ratio of 6, Ky = 0.00245 and 
Kx = 0.00017, at 6° angle of attack; with aspect ratio of 9, what will 
be the angle of attack for Ky of 0.00245, and what will be i^«? 

a(A.R.6) - q:(A.r.9) = 7130 X 0.00245 {\ - 4) 
= 0.97° 

a(A.R.9) = 5.03° for Ky = 0.00245 
K:c(A.R.6) - ii::c(A.R.9) = 125 X 00024? (i - i) 
= 0.0000417 
Kx{K.R.^) = 0.00017 -^0.00004 
= 0.00013 

By using these formulas, if the characteristic curves for one as- 
pect ratio are known, the characteristic curves for any other aspect 
ratio can be calculated. Figure 37 shows the lift and drag coefl&- 
cient curves for an aspect of 8, as calculated from data for an aspect 
ratio of 6. 

Note that variation in the aspect ratio has no effect on the angle 
of zero lift. 

Problems 

1. An airfoil of infinite aspect ratio has a Cd of 0.01 when Cl is 
0.7. (a) What is the Cd for a similar airfoil with aspect ratio of 9, 
when Cl is 0.7? (6) What is the L/ D ratio? 

2. For the same airfoil section as in problem 1, but with an aspect 
ratio of 8: (a) what is Cd when Cl is 0.7; (6) what is the L/D ratio? 

3. For the same airfoil section as in problem 1, but with an aspect 
ratio of 6: (a) what is Cd when Cl is 0.7; (6) what is the L/ D ratio? 

4. An airfoil with aspect ratio of 6, at 2° angle of attack has Cl = 
0.78 and Cd = 0.047. (a) At what angle of attack will this airfoil 
have the same Cl if the aspect ratio is infinity? (6) What will be the 
Cd under these conditions? 

5. An airfoil with aspect ratio of 6, at 10° angle of attack has Cl = 
1.32 and Cd = 0.110. (o) With an aspect ratio of 9, at what angle of 
attack will Cl = 1.32? (6) What will be Cd'^ (c) What will be 
L/D? 



92 INDUCED DRAG OF MONOPLANES 

6. An airfoil with aspect ratio of 6, at 8° angle of attack has Cl = 
0.85 and Cd = 0.046. (a) With aspect ratio of 8, at what angle of 
attack will Cl = 0.85? (6) What will be Cd'^ 

7. An airfoil with aspect ratio of 6, at 0° angle of attack has Cl = 
0.64 and Cd = 0.035. (a) With aspect ratio of 8.5, at what angle of 
attack will Cl = 0.64? (b) What will be Cz?? 

8. An airfoil with aspect ratio of 6, at 1° angle of attack has Cl — 
0.58 and Cd = 0.031. (a) With aspect ratio of 4, at what angle of 
attack will Cl = 0.58? (b) What will be Cd? 

9. An airfoil of aspect ratio of 6, at angle of attack of 8°, has Ky 
= 0.0025 and Kx = 0.000175. (a) With aspect ratio of 8, at what 
angle of attack will Ky = 0.0025? (6) What will be Kx? 

10. From Fig. 28 obtain the data and plot the lift and drag coeffi- 
cient curves for an aspect ratio of 8.5. 

Total Drag of Monoplane Wings from Model Data. In practical 
calculations, it is often desired to find the drag of a wing from the 
characteristics of the model, without knowing the corresponding 
velocities. The L/D of the monoplane can be found, and then 
use is made of the fact that, in level flight, lift equals weight. 

The formula for induced drag 



may be rewritten as 



A 


L2 

7rq¥ 




A 


l{Cl^SV^) 


ir^V'W 






LxCl 






LxCl 





TT (A.R.) 

The drag being composed of two parts, profile drag independent 
of aspect ratio and induced drag dependent on aspect ratio, the 
drag of any monoplane may be written as follows 

Cd (monoplane) = Cd (model) — Cdi (model) + Czji (monoplane) 

= Cd (model) ^ r + ^^ 



TT (A.R.model) TT (A.R. monoplane) 



TOTAL DRAG OF MONOPLANE WINGS 93 

then 



(^) monoplane = j^ -^ 



1 



TT VA.JLV.model A. xv. monoplane/ 

Knowledge of L/D of the monoplane wing enables the drag to 
be found quickly, since, in level flight, Uft substantially equals 
weight. 

Example. An airplane weighs 2,000 lb.; the wing area is 180 sq. 
ft.; the wing span is 39 ft. What is the wing drag at 4° angle of 
attack? Wing section is U.S.A.-35A. 

Solution. From Fig. 27 at 4° angle of attack Cl = 0.84, Cd = 
0.058. 



(x) 



model = ^^ = 0.069 
0.84 



39 
A.R. monoplane = — - = 8.45 
loO 

L 1 



D ^^^^ 0.84/1 1 \ 

0.069-— (g-g;^j 

1 



0.069 - 0.0129 

1 



0.0561 
Since lift is equal to weight 



2,000 1 



D 0.0561 
D = 2,000 X 0.0561 
= 112.2 lb. 

It is to be noted that span is always the distance from wing tip to 
wing tip. The term chord used in connection with the aspect ratio 
of a tapered wing refers to the average or mean arithmetic chord and 
should not be confused with the mean geometric or mean aerody- 
namic chords. The wing area is total area including ailerons and is 
assumed to extend through the fuselage. 

Problems 

1. A monoplane weighing 3,000 lb. has a U.S.A.-35A wing, of 
35-ft. span, 4-ft. chord. What is the wing drag at 2° angle of attack? 
2. A monoplane weighing 2,600 lb. has a U.S.A.-35A wing, of 27-ft. 
span, 3j-ft. chord. What is the wing drag at 6° angle of attack? 



94 INDUCED DRAG OF MONOPLANES 

3. A monoplane weighing 700 lb. has a U.S.A.-35A wing of 36-ft. 
span and 4-ft. chord. What is the wing drag at 8° angle of attack? 

4. A monoplane weighing 2,700 lb. has a rectangular wing of 42- 
ft. 5-in. span and 6-ft. 0-in. chord. If Gottingen 398 airfoil is used, 
what is the wing drag at 5° angle of attack? 

5. A monoplane weighing 9,300 lb. has a Gottingen 398 wing 
574 sq. ft. in area. The span is 66 ft. What is the wing drag at 8° 
angle of attack? 

6. A monoplane whose weight is 5,400 lb. has a C-80 wing 285 sq. 
ft. in area. The span is 44 ft. What is the wing drag at the same 
angle of attack for which model airfoil (A.R. 6) has a Cl of 0.7? 

7. A monoplane whose weight is 4,360 lb. has a Clark Y wing 292 
sq. ft. in area. The span is 42 ft. 9 in. What is the wing drag at 8° 
angle of attack? 

8. A monoplane weighing 7,000 lb. has a U.S.A.-35A wing 363 
sq. ft. in area. Span is 48 ft. What is the wing drag at 10° angle of 
attack? 

9. A racing plane weighing 6,500 lb. has a C-80 wing of 211 sq. ft. in 
area. Span is 34 ft. 3 in. What is the wing drag at 2° angle of attack? 

10. What is the wing drag of the airplane in problem 9 at 10° angle 
of attack? 

Effect of Taper on Induced Drag. Whenever the term area has 
been used in connection with aspect ratio, it refers to the total 
area of a wing including ailerons. Nothing is subtracted for 
fuselage or nacelle; the leading edge is assumed to continue 
across in a straight line from where the leading edges of each wing 
intersect the fuselage, and the same assumption is made for the 
trailing edge. 

With the airfoil section decided upon, the area is fixed by the 
landing speed. The largest possible span will, of course, give the 
least possible induced drag. The long wing spars necessary for a 
large span will have to be very deep to prevent sagging or hogging, 
so that there are very practical limits to the length of span. Rec- 
tangular wings are cheapest to manufacture since each rib is 
identical; however, there are many advantages to tapering a 
wing. 

There are three ways of tapering wings. The first is tapering 
in plan form only. This means using the same airfoil section 
throughout the wing. The chord is lessened from the root to the 
tip; this entails decreasing the thickness from root to tip. The 
second method is to taper in thickness only. This means a rec- 



EFFECT OF TAPER ON INDUCED DRAG 95 

tangular wing, the chord remaining the same from root to tip. 
The airfoil section is changed from a thick wing at the root to a 
thin wing at the tip. The third method is a combination of the 
first two. 

The first method of tapering in plan form is the only one com- 
monly used. Besides constructional advantages and improvement 
in performance, if the upper surface is flat, the under surface 
slopes outward and upward, which gives an effective dihedral. 
This feature is explained in the chapter on stability. 

By decreasing the chord near the tip, the lift and drag loads are 
made small at the outer extremities of the wing spars. This 
makes it possible to design a spar that is properly effective, being 
deep where the greatest stresses occur. 

Example. When Cl = 0.9, find the difference in induced drag for 
a rectangular wing of 36-ft. span and 6-ft. chord and a tapered wing 
of the same area, the tapered wing having a 6-ft. chord at the root, the 
leading and trailing edges being tangent to a circle of 2-ft. radius at 
the tip, and the fuselage being 3 ft. wide. The airspeed is 100 miles 
per hour. 

Solution. 

Area = ^ = 36 X 6 = 216 sq. ft. 

Rectangular wing: 

D- = 
' TT X A.R. 

^ 0^9^ X 0.00256 X 216 X 100^ 
TT X 6 

= 236 lb. 

By laying-out tapered wing conforming to above description, span is 
found to be 44 ft. 

^, ^ 09^ X 0.00256 X 216 X 100^ 



44 
^><2l6 



= 158 lb. 



Problems 

1. When Cl is 1.2, what is the induced drag coefficient of a wing 
which has a span of 40 ft.? The chord at the root is 5 ft. Both 



96 INDUCED DRAG OF MONOPLANES 

leading and trailing edges are tangent to the arc of a circle of 1^-ft. 
radius at the tip. The fuselage is 3 ft. wide. 

2. When Cl is 0.9, what is induced drag coefficient of a wing of 
74-ft. in. span? The chord at root is 15 ft. in. Both leading 
and trailing edges are tangent to circular tips, 3-ft. 6 in. radius. 
Fuselage is 4 ft. wide. 

3. When Cl is 0.8, what is induced drag coefficient of a Douglas 
Transport whose span is 85 ft. in.? The center section of the wing 
is 28 ft. in. wide and has a 17-ft. in. chord. The trailing edge is 
straight. The leading edge sweeps back. Both leading and trailing 
edges are tangent to circular tips of 3-ft. in. radius. 

4. When Cl is 0.85, what is induced drag coefficient of a General 
Aviation Transport wing of 53-ft. in. span? Fuselage is 4 ft. 6 in. 
wide. Root chord is 11 ft. in. Wing is tapered from fuselage out, 
both leading and trailing edges being tangent to circular tip of 2-ft. 
in. radius. 

5. When Cl is 0.9, what is induced drag coefficient of Northrop 
Delta wing? Span is 48 ft. in. Rectangular center section is 11 ft. 
in. wide. Root chord is 9 ft. in. Wing tapers from center section 
to tip, leading and trailing edges being tangent to circular tip of 
2-ft. in. radius. 

Correction Factor for Aspect Ratio. By making certain ap- 
proximations, a correction factor may be obtained by which a 
correction may be applied to the angle of attack of a model airfoil 
to obtain the angle of attack with the same lift coefficient of an 
airfoil with a differing aspect ratio. 

Earlier in this chapter it was pointed out that the change in 
angle of attack for a change in aspect ratio is due to the change 
in the induced angle of attack. 

,_ 57.3 Cl 57.3 Cl 

"" ""^ ttA.R. "^ ttA.R.' 

a = original angle of attack (in degrees) 
a' = angle of attack for new aspect ratio 
A.R. = original aspect ratio 
A.R.' = new aspect ratio 

It will be noticed that, when Cl is plotted against angle of attack, 
the curve is a straight line from the angle of zero lift up to near 
the burble point. For this portion of the curve the following is 
true 



CORRECTION FACTOR FOR ASPECT RATIO 97 

^ -T — = slope of lift curve 

C, = ^^ X azL ^" 

Aa ■ ' az.L. = angle of attack measured from 

angle of zero lift 
Substituting this value of Cl in the previous equation: 



"K!?)... , 5-(t) 



az.L. 



a'z.L. = «z.L. ^J^ + ^ ^^, 

Examining a large number of hft curves for airfoils with aspect 
ratio of 6, it will be found that the slope is practically the same 
and ACl/Aq: = 0.0718. Substituting this value of slope and the 
value of 6 for original aspect ratio the equation becomes 

57.3 X 0.0718 X azx. , 57.3 X 0.0718 X azL. 
az.L. - O.Z.L. —^ + ^^j^, 



57.3 X 0.0718 X o;z.l. 
= azx. 



\6 A.R.7 



\6 A.R.7 
= «zx.(0.782+gl,) 



= «Z.L. ~~ 1.311 q;z.l 
= az.L. 1 



q;z.l. 



0.782 + 1-311 



A.R/ 

Q^Z.L. 

A.R.' 



0.782 A.R/ + 1.311 



or 



«'z.L. = ^^ where Fa.r. = 



^A.R. 0.782 A.R.' + 1.311 

This factor Fa.r. is much used in stability computations, 
especially in connection with finding the reactions at the tail sur- 
faces. It is rarely necessary to consider the condition of large 
angle of attack of the tail surfaces, and the small errors incidental 



98 



INDUCED DRAG OF MONOPLANES 



to the use of this approxunation are unimportant. 
-^A.R. is plotted against aspect ratio in Fig. 39. 



The factor 



«1.10 
115 
II 

pi.oo 



o 90 
^ .80 



70 



2 3 4 5 6 7 
E. M.A.R. 



9 10 

Fig. 39. Fa.r. plotted against aspect ratio. 



This factor, i^A.R., is used by the Army Air Corps and is based 
on the average slope of a large number of airfoils being 0.0718, 
where a is measured in degrees. The Department of Commerce 
advocates a sHghtly different value, based on theoretical results 
deduced for thin airfoils. 

The Department of Commerce uses the symbol m to represent 
the slope of the lift curve : m = ACl/Aq;, where a is measured in 
radians. From the theory of thin airfoils, the slope of the lift 
curve for a wing of infinite aspect ratio is 2 tt. 

- — = m^ = 2t 

A«oo 

ACl = 2 TT Aa^ 

Cl 



Then 
but 



+ 



or 



A(x = Aa^ + 



TT X A.R. 
ACl 



TT X A.R. 

2 7rAa. 



Therefore 


^" ^""+.XA.R. 




= ^""(l+A.R.) 


Then, for aspect ratio of 6, 




Aae = Aa„ ^1 + gj 




4^ 


or 


Aa» =|Aa6 



CORRECTION FACTOR FOR ASPECT RATIO 99 



3 3 
Then m& = jm^ = j (2 tt) if a is in radians, or me = 


37r 


2 X 57.3 


= 0.0822 if a is in degrees. 
Returning to the earlier equation 




A« = A«„(i+^y 




Substituting from above 




A„=|Aae(l+^y 




- Aae ^ 




Then 

ACl ACl 4 
Aa Aae ^ , 6 




using the Department of Commerce notation 




4 
m = m&X r 





3+i^ 



Example. A certain airfoil has a lift coefficient of 0.76 at 8° angle 
of attack when the aspect ratio is 6. The angle of zero lift is —2°. 
What is the slope of the lift curve for an aspect ratio of 9? 

Solution. 



^76 
= 0.076 



^6 - 8° _ (_2)° 



m 



= 0.0829 

Problems 

1. An airfoil has a lift-curve slope of 0.0712 for aspect ratio of 6. 
What is slope of lift curve for aspect ratio of 4j? 

2. An airfoil has a lift-curve slope of 0.073 for aspect ratio of 6. 
What is slope of lift curve for infinite aspect ratio? 



100 INDUCED DRAG OF MONOPLANES 

3. With aspect ratio of 6, an airfoil has a Cl of 0.7 at 5° angle of 
attack. Angle of zero lift is —3°. What is Cl at 3° angle of attack 
when aspect ratio is 8? 

4. With aspect ratio of 6, slope of lift curve is 0.076; what is Cl 
for 4° when aspect ratio is 8j, if angle of zero lift is — 2°? 

5. For a symmetrical airfoil (angle of zero lift = 0°), slope of lift 
curve for aspect ratio 6 is 0.074; what is Cl at 5° angle of attack when 
aspect ratio is 9^? 

Chord and Beam Components. In analyzing the stresses in 
spars, it is necessary to know the force produced by the air flowing 
around the wing. Whereas, for performance calculations, com- 
ponents are used which are perpendicular and parallel to the 
relative wind (lift and drag components), for stress calculations, 
components are desired which are perpendicular and parallel to 
the chord. 

If the lift and drag components are known, the components 
about different axes may be found by resolution. Since the result- 
ant at high angles of attack is only a small angle back of the 
vertical, the chord component at high angles of attack actually 
acts in the direction of the leading edge. 

Calling forces upward and rearward positive, the chord com- 
ponent coefficient Cc and the beam or normal coefficient On may 
be expressed in terms of lift and drag coefficients, as follows, 

Cc = Cd cos a — Cl sin a 
Cn = Cd sin a -{- Cl cos a 

Department of Commerce Method of Aspect-Ratio Correction. 

In correcting angle of attack and drag coefficients for the effect of 
aspect ratio, instead of using the formulas 

and CoA-Cn, = ^{j-l) 

since it is customary for data to be given for airfoils with aspect 
ratio of 6, the Department of Commerce uses formulas in the form 

^ lo OA zrry ^ = angle of attack 

a = as + 18.24 KCl n a txi - ^ c 

, ^ ^ I n 010 7^/> 2 Cd ^ drag coefficient for as- 

and Cd = C^e + 0.318 KCj} ^ .. . ,, 

pect ratio of the wmg 



METHOD OF ASPECT-RATIO CORRECTION 



101 



K=~ 



1 


1 
6 


1 

R~ 


- 0.1667 



and 



The Department of Commerce uses a standard form for tabulat- 
ing the correctional data. Below appears a data sheet worked 
out for a Clark Y monoplane, aspect ratio of 9, the data for the 
second, fifth, and sixteenth lines being taken from Fig. 17. K is 
equal to 1/9 - 0.1667 or -0.0556; a is 24.1 per cent chord. 



TABLE IV 

Computation of Airfoil Characteristics 
Department of Commerce Method 



1 


Cl 


0.2 


0.4 


0.6 


0.8 


1.0 


1.2 


1.4 


2 


ae 


-2.2 


0.6 


3.4 


6.2 


9.0 


12.2 


15.4 


3 


Aa 


-0.2 


-0.4 


-0.6 


-0.8 


-1.0 


-1.2 


-1.4 


4 


oc 


-2.4 


0.2 


2.8 


5.4 


8.0 


11.0 


14.0 


5 


Cdg 


0.012 


0.019 


0.030 


0.046 


0.068 


0.098 


0.132 


6 


ACoi 


-0.0007 


-0.0028 


-0.0064 


-0.0113 


-0.0177 


-0.0256 


-0.0346 


7 


Cd 


0.0113 


0.0162 


0.0236 


0.0347 


0.0503 


0.0724 


0.0974 


8 


COS a 


0.9991 


1.0000 


0.9988 


0.9955 


0.9903 


0.9816 


0.9703 


9 


sin a 


-0.0419 


0.0035 


0.0489 


0.0941 


0.1392 


0.1908 


0.2419 


10 


Cl cos a 


0.200 


0.400 


0.599 


0.796 


0.990 


1.178 


1.358 


11 


Cd sin a 


-0.0005 


0.0001 


0.0012 


0.0033 


0.0070 


0.0138 


0.0235 


12 


Cn 


0.2005 


0.4001 


0.6002 


0.7993 


0.9970 


1.1918 


1.3815 


13 


Cl sin a 


-0.0084 


0.0014 


0.0293 


0.0752 


0.1392 


0.2290 


0.3387 


14 


Cd cos a 


0.0113 


0.0162 


0.0235 


0.0346 


0.0497 


0.0711 


0.0945 


15 


Cc 


0.0193 


0.0148 


-0.0055 


-0.0406 


-0.0895 


-0.1579 


-0.2445 


16 


C.R 


0.60 


0.41 


0.35 


0.32 


0.31 


0.30 


0.29 


17 


Cuc/i 


-0.070- 


-0.064 


-0.0599 


-0.0559 


-0.0598 


-0.0596 


-0.0553 


18 


Cua 


-0.0718 


-0.0676 


-0.0654 


-0.0631 


-0.0688 


-0.0702 


-0.0677 


19 


CDi 


0.0014 


0.0056 


0.0128 


0.0226 


0.0354 


0.0512 


0.0692 


20 


Cdo 


0.0099 


0.0106 


0.0108 


0.0121 


0.0149 


0.0212 


0.0282 



Explanation 

Item 2, from characteristic curve, Fig. 17. 

Item 3, Aa = 18.24 KCl- 

Item 4, a = (2) + (3). 

Item 5, from characteristic curve, Fig. 17. 

Item 6, ACdz = 0.318 KCl^. 

Item 7, Cd = (5) + (6). 

Item 8, cos a = cos (4). 



102 INDUCED DRAG OF MONOPLANES 

Item 9, sin a. = sin (4). 

Item 10, Cl cos a = (1) X (8). 

Item 11, Cb sin a = (7) X (9). 

Item 12, C^ = (10) + (11). 

Item 13, Cl sin a = (1) X (9). 

Item 14, Cd cos a = (7) X (8). 

Item 15, Cc = (14) - (13). 

Item 16, C.P. is for aspect ratio 6, from Fig. 17. 

Item 17, Cmc/4. = (0.25 - (16)) X (12). 

Item 18, Cua = (17) + {a - 0.25) X (12). 

Item 19, Cdi = {Q)/KR = ^:5i|^' = %'. 

Item 20, Cdo = (7) - (19). 

Note: Item 12, the beam or normal component, and item 15, the chord 
component, are used in stress analysis but are not needed in performance 
calculations. 

Problems 

1. Prepare a table similar to above for a Clark Y wing with an 
aspect ratio of 8. 

2. Prepare a table similar to above for a wing with an aspect ratio 
of 7.5. 



METHOD OF ASPECT-RATIO CORRECTION 

TABLE V 
Aerodynamic Characteristics of Airfoils 



103 





C'imax. 


CMc/4 


a 


M 


Airfoil 


Maximum 

Lift 
Coefficient 


Moment 
Coefficient 

about i 
Chord Point 
where Cl = 


Aerodynamic 

Center in 

Percentage of 

Chord 


dCL 

da 

for 

A.R.6 


Qark 










Y 


1.56 


-0.068 


24.2 


0.0716 


YM-15 


1.58 


-0.068 


24.1 


0.0722 


YM-18 


1.49 


-0.065 


23.6 


0.0716 


Curtiss 










C-72 


1.62 


-0.084 


23.8 


0.0739 


Gottingen 










387 


1.56 


-0.95 


23.9 


0.0745 


398 


1.57 


-0.083 


24.4 


0.0734 


N-22 


1.60 


-0.074 


25.0 


0.0743 


N.A.C.A. 










0006 


0.88 





24.3 


0.0748 


0012 


1.53 





24.1 


0.0743 


2212 


1.60 


-0.029 


24.6 


0.0753 


2409 


1.51 


-0.044 


24.7 


0.0753 


2412 


1.62 


-0.044 


24.6 


0.0743 


2415 


1.55 


-0.040 


24.3 


0.0743 


2418 


1.43 


-0.037 


24.0 


0.0726 


4412 


1.65 


-0.089 


24.5 


0.0736 


CYH 


1.47 


-0.027 


24.5 


0.0740 


M-6 


1.40 


0.002 


25.0 


0.0744 


M-12 


1.25 


-0.022 


25.0 


0.0705 


R.A.F. 










15 


1.21 


-0.052 


23.2 


0.0719 


U.S.A. 










27 


1.59 


-0.077 


23.7 


0.0718 


35-A 


1.48 


-0.111 


23.4 


0.0730 


35-B 


1.69 


-0.076 


24.5 


0.0749 



From tests in the variable density tunnel of the National Advisory 
Committee for Aeronautics. 



CHAPTER VI 
INDUCED DRAG OF BIPLANES 

Biplane Mutual Interference. The air flowing over and under 
a wing causes the pressure to be less than atmospheric on the upper 
side of the wing and slightly more than atmospheric on the under 
side of the wing. If another wing is placed over the first wing, 
the gap being relatively small, the low-pressure area on the upper 
side of the lower wing will be affected by the high-pressure area 
on the under side of the upper wing, and vice versa. Owing to the 
proximity of the upper wing, the pressure on the upper side of the 
lower wing will not be as low as if there were no upper wing. The 
pressure on the under side of the upper wing will not be as high as 
if there were no lower wing. 

In addition the vortices on each wing have interaction on each 
other, so that added drag is produced on each wing by the presence 
of the other wing. This has the same effect as reducing the aspect 
ratio. 

Equivalent Monoplane Aspect Ratio. Prandtl has proved that 
in a biplane the added induced drag on one of the wings caused 
by the other wing is 

7rqbih2 

where Li and 61 are the lift and span respectively of one wing, 
L2 and 62 are the lift and span of the other wing, q is (p/2) F^, and a 
is a dimensionless factor dependent on the ratio of gap to average 
span and on the ratio of the shorter to longer span, a is called 
the Prandtl interference factor. 

The added drag on the upper wing produced by the lower wing 
is the same as the added drag on the lower wing produced by the 
upper wing. The total added drag has twice the value of that 
for a single wing. Then the total induced drag of a biplane is 

* Tvqbi^ 7rqbib2 irqh'^ 
104 



EQUIVALENT MONOPLANE ASPECT RATIO 105 

If Ci and C2 are the two chords, the subscripts corresponding 
with the subscripts for hft and span, then, assuming there is no 
decalage, so that the angles of attack and consequently the lift 
coefficients are the same for both wings, 

* Trqbi^ 7rqbih2 



irqb 



= + + 

TT TT TT 

but since Di = Cdh^SV^ where S = wing area 

(J 2 

then Cdi = —^ (ci^ + 2 o-CiC2 H- Ci^) 

this bears a strong resemblance to the expression for induced drag 
coefficient for a monoplane, viz.. 

Cm = 



A.R. 



TT 



The expression ^ , ^ ■ — - is therefore called the equiva- 

Ci^ + 2 aCiC2 + C2^ 

lent monoplane aspect ratio of a biplane, abbreviated as E.M.A.R. 
E.M.A.R. = ,_^^ ^ , , 

Cr + 2 (7CiC2 + C2^ 

Where the chords of both wings are the same length, ci = C2, 
and the expression becomes 

S 



E.M.A.R. = 



2c2(l+<7) 



Figure 40 gives values of a- plotted against ratio of gap to mean 
span for various values of fx where ^ is the ratio of the shorter to 
longer span. Using values of Prandtl's interference factor from 
this figure, the E.M.A.R. may be found. 

It is to be noted that the area, S, in expressions for E.M.A.R. is 
the total area of wings including ailerons. Where there is a fuse- 



106 INDUCED DRAG OF BIPLANES 

.24 

.23 

.22 



^1 



.20 



.19 



.18 



^.17 

c 

3 

2.16 

a 

<3 

o 

2.15 

(X 



14 



13 



.12 



.11 



.10 



.09 



.08 





w 












\ 


\v 


y 










\ 


\ \ 


1^ 


// _ Shorter Span 
' Longer Span 




\ 


\ 


\\ 










\ 


\ 


\v 












\ \ 


y 












\ 


\ \ 










\ 


\ \ 








\ 


\ 












\ 


\ 




\ 








\ 


\ 




\\ 










i 




\\ 










\ 




\ \ 


\ 








\ 




\ 


\ V 










\\\ 


•cp\ 


A\ 








\ 




\ 


\^ 


\\ 






^ 


\ 


\ 


\ 


W 



.35 40 .45 .50 .55 .60 

Values of <r= Prandtl's Interference Factor 

Fig. 40. Prandtrs interference factor. 



65 



.70 



EQUIVALENT MONOPLANE ASPECT RATIO 107 

lage or nacelles, the wing is assumed to be continuous through the 
fuselage or nacelle, the leading and trailing edges being considered 
as continuing across in straight lines from where they intersect 
the fuselage. 

Another expression for E.M.A.R. is derived as follows. The 
ratio of the lift of the wing with the shorter span (L2) to the lift 
of the wing with the longer span (Li) is called r. Then L2 = rLi 
and the total lift L = Li + rLi = Li(l + r) = L2 + L2/r. The 
ratio of the shorter span 62 to the longer span 61 is called ijl: /jl = 
h^/hi] jjL is never greater than unity. 

Making these substitutions in the original expression for induced 
drag of a biplane 

irqbi^ 7rg6i62 Trgh^"^ 



1^ 

Tvq 



(r+-r} , ^"(l^rjir+'r) , (l 



rL 






6i(m6i) ' (m6i) 



%^#^,+ 



Di 



m(1 + rf ' ix\\ + r)2j 
^ U I II"- + 2 o-zxr + r\ 
-KqhA mH1 + 0' / 
The coefficient of induced drag for a biplane may then be found. 

n .o^ JCLgSy (m^ + 2 cr^r + r^) 
^""'^^ ~ Trqhi^ mHI + ry 

Cl'S / m' + 2 (T/xr + rn 

7r6i2 V mHI + ry ) 

The expression -^ ( ^ ^ IJr~2) ^^ another form of writing 

the E.M.A.R. and is more widely used than the one previously 
given. 

Example. Find the equivalent monoplane aspect ratio of a biplane 
with rectangular wings; upper span 40 ft., upper chord 4 ft. 10 in., 
lower span 32 ft., lower chord 3 ft. 9 in., gap 4 ft. 6 in. 

Solution. 

Gap ^ 4^ ^ Q ^25 
Mean span 36 



108 INDUCED DRAG OF BIPLANES 

From Fig. 40, c = 0.56 

Area, upper wing = 40 X 4.83 = 193.2 sq. ft. 

Area, lower wing = 32 X 3.75 = 120.2 sq. ft. 



E.M.A.R. = 



Total area = 313.4 sq. ft. = S 
313.4 



4.83' + 2 X 0.56 X 4.83 X 3.75 + 3.75^ 

313.4 

~ 23.4 + 20.3 + 14.0 
= 5.4 

Problems 

1. Find E.M.A.R. of a biplane with rectangular wings; upper span 
31 ft. 6i in., upper chord 4 ft. 8 in., lower span 28 ft. 4j in., lower 
chord 4 ft. in., and gap 4 ft. 6 in. 

2. Find E.M.A.R. of a biplane with rectangular wings; upper span 
28 ft. in., upper chord 4 ft. in., lower span 25 ft. 3 in., lower chord 

3 ft. 6 in., and gap 50 in. 

3. Find E.M.A.R. of a sesquiplane with rectangular wings; upper 
span 25 ft. in., upper chord 4 ft. 6 in., lower span 15 ft. in., lower 
chord 3 ft. in., gap 45 in. 

4. Find E.M.A.R of a biplane with rectangular wings; upper span 
27 ft. in., upper chord 4 ft. 6 in., lower span 22 ft. in., lower chord 

4 ft. 6 in., gap 4 ft. 6 in. 

5. Find E.M.A.R. of a biplane with rectangular wings; upper span 
38 ft. in,, upper chord 9 ft. in., lower span 38 ft. in., lower chord 
9 ft. in., gap 9 ft. in. 

Best Lift Distribution in a Biplane. The two wings of a biplane 
cause mutual induced drag, and the amount of this drag depends on 
several factors. The best performance will be secured when these 
factors are so selected that the induced drag is the least. 

The reasons for not having an airplane designed as a monoplane 
are several. If the weight is too great to be carried on a single 
wing, the designer resorts to a biplane. A biplane has greater 
maneuverability than a monoplane. Probably some airplanes are 
built as biplanes merely because the designer has a predilection 
for biplanes. 

The simplest form of biplane would be one having equal spans 
and equal areas. It is sometimes thought advisable to have wing 
panels interchangeable, so that an upper right wing, for example, 
may be used as a lower right wing, etc. It is also desirable in 



BEST LIFT DISTRIBUTION IN A BIPLANE 109 

some designs to make the lower wing smaller to improve the 
visibility in landing. Sometimes the reason for making the wings 
of unequal size is to aid stability. 

The greater the gap between the wings the smaller will be the 
interference between the wings. The biplane having an infinite 
gap between the wings will have no mutual wing interference and 
the induced drag will be solely the induced drag of each wing. 
Practically, of course, the gap must be finite. Each wing might be 
built as a cantilever monoplane wing, but the internal bracing 
would be excessive. Very simple struts between wings will 
permit very moderate wing spans to be used. Struts act as 
columns, and, if the gap is unduly big, the struts are excessively 
long and the structure is weak. It is usual to make the gap ap- 
proximately equal to the mean chord. If the gap/chord ratio is 
greater than 1 the structure becomes weak; if the gap/chord ratio 
is less than 1 the mutual drag interference becomes excessive. 
Any departure from gap/chord ratio of unity is usually for reasons 
of visibility. 

The expression for E.M.A.R. = -^ ( „ , ^ — -A contains 

>S V/x^ + 2 (Jiir + r I 

three factors m? ^j and a. To find the relation of these variables 

which will give the minimum induced drag, the expression for drag 

may be differentiated. If it is differentiated with respect to /x, 

the ratio of spans, the solution for r, ratio of lift, is inapplicable. 

Differentiating with respect to r, the following results. 

dr Trqhi^ dr\ /^^(l + r)^ ) 



= J^f- (1 + r)((Tfj, + r) - (m' + 2 a fir + r^) ] 

TrqbAu^' (l + ry J 

_ L^ r 2 (afj, + r + c/xr + ^^ — M^ ~ 2 o-fxr — r^)' 



Trqbi^ Ifj,'^ 



(1 + ry 

r(l - o-m) - (m^ - o-m) " 

(1 + ry 



u 


?.«- 


-")('-f:;;) 


irqW 




(1 + ry 



110 INDUCED DRAG OF BIPLANES 

This differential of the induced drag —f^ is equal to zero when 
T = infinity or when 

^ — (TfJi 

r =-. 

Mathematically when r has this value the induced drag may be 
either a maximum or a minimum. It should be noted that m is the 
ratio of the smaller to the larger span, so that ju is always either 
unity or less than unity. Prandtl's interference factor o- is always 
less than unity. Then the product c/x is always less than unity 
and the quantity (1 — aij) is always positive. If r is greater in 

value than -zr-^ , the expression ( r — -— 1 is positive and 

-—■ is positive; that is, the slope of A plotted against r is positive. 

If T is less than -^ , the expression [r — ) is negative, 

\ — uix \ I — 0-/X/ 

and the slope of A plotted against r is negative. Therefore, 

2 

when r = -z , the induced drag of a biplane is a minimum. 

1 — <ffJL 

In the expressions for induced drag and induced drag coefficient, 
r is the ratio of Hfts of the two wings. If the lift coefficients of the 
two wings are the same, then the ratio of lifts is the same as the 
ratio of the areas. Actually, even if the same airfoil section is 
used for both wings, there will be interaction between the two 
wings, so that neither Clu, the lift coefficient for the upper wing, 
nor Cll, the lift coefficient for the lower wing, are the same as Cl, 
the lift coefficient of the biplane. 

Clu = Cl ^ ^Clu 
Cll = Cl =F ACll 

= Cl^ AClu-^- 

When both wings are the same area, the increments AClu and 
ACll are equal and of opposite sign. 

AClu = K, + K^Cl 

where Ki and K2 are functions of stagger, overhang, gap/chord, 
decalage, and wing thickness. The reader is referred to N.A.C.A. 
Report 458 for details of relative wing loading. 



BEST LIFT DISTRIBUTION IN A BIPLANE 111 

Common practice ignores this correction to lift coefficients and 
assumes that r, the ratio of lifts, is also the ratio of wing areas. 
Then if unequal spans have been decided upon for a biplane, the 
areas should be divided so that there is least induced drag, that is, 

r is made equal to -^ = ti • If unequal areas have been 

1 — (T^X 1/ {X — <T 

chosen, the spans should be so selected that, when the ratio of the 
smaller to larger span, less the interference factor, o-^ is divided 
by the ratio of the larger to smaller span, less the interference 
factor 0-, the quotient is equal to the ratio of smaller to larger wing 
area. This ratio being known, the proper chords can be found. 

Example. A biplane is to have a 30-ft. span on the upper wing and 
a 27-ft. span on the lower wing; the gap is to be 4 ft. 6 in. What 
should be the ratio of lower wing to upper wing? If rectangular 
wings are used, what should be the chords? 

^ = 0.9 |i=l.ll 

<^ap 4.5 _ ^ . -^ 

Mean span ~ 28.5 ~ ' ^^ 
From Fig. 40, a = 0.538 

0.9 - 0.538 
"^ ~ 1.11 - 0.538 
= 0.632 

Therefore the area of the lower wing should be 0.632 times the area of 
the upper wing. The area of both the wings will be 1.632 times the 
area of the upper wing, or the area of the upper wing will be 1/1.632 
or 0.613 of the total area. 

^ = 0.632 
^= 0.632 |l 
- = 0.632 X 1.11 

Cl 

= 0.702 
Therefore lower chord should be 0.702 times the upper chord. 

Problems 

1. A biplane with rectangular wing is to have upper span 30 ft., 
lower span 24 ft., and gap 4 ft. What should be the ratio of areas and 
chords? 

2. A biplane is to have 45-ft. upper span, 38-ft. lower span, gap 
55 in. What should be the ratio of areas? 



112 



INDUCED DRAG OF BIPLANES 



3. A biplane is to have 40-ft. upper span, 24-ft. lower span, gap 
6 ft. (a) What should be the ratio of areas? (b) For rectangular 
wings, what should be the ratio of chords? 

4. A biplane is to have 28-ft. upper span, 20-ft. lower span, and 
gap 45 in. What should be the ratio of areas? 

5. A biplane is to have 28-ft. upper span, 20-ft. lower span, and 
gap 5 ft. What should be the ratio of areas? 

E.M.A.R. for Best Lift Distribution. Of the two expressions for 
equivalent monoplane aspect ratio, namely, 

S 



E.M.A.R. = 

and 
E.M.A.R. = 



Ci^ + 2 (TC1C2 + C2^ 
^2(1 + r)2 






M = 7- = ratio of spans 

-, L2 02 ,. /. 

r = y- = -^ = ratio of areas 
2 M oi 



+ 2 arfjL + r 

The second may be modified by introducing the condition for best 
lift distribution, namely, that 

/JL — (T 



' 1/m - (T 

^ M^ — g-M 

1 — ajJL 

Making this substitution the expression becomes 

E.M.A.R. (Min.Z?i) 



hi 

s 



s 






\ 1 - 0-/X / 



)u2(l - o-m)' + 2 o-mHm - cr)(l - (Tfi) + mHm' - 2 (t^ + cr^) 



(1 - ctm)^ 
(1 - 2 (7/. + M^)2 



^&i^r (1 - 2 (7/. + M^)^ 1 

,S L(l - atxY + 2 (7(m - (7)(1 - afx) + (iu2 - 2 C7M + o-^)] 

_6i^r (1 - 2 (TM + M^)^ 1 

S Li-2(7m+o-V+2(7m-2o-2-2o-V+2o-3m+^2_2 0-^+^2] 

^vr (1-2^M + My 1 

,S [1 - 2 0-/X + m' - <^' + 2 o-V - c^Vj 
^bi^r i - 2(7M + M^ l 



E.M.A.R. FOR BEST LIFT DISTRIBUTION 113 

Since o- < /x 

then < fjL — a 

< fjL^ - 2 (Tfj, + a^ 
and 1 - 0-2 < 1 - 2 o-M + m' 

Therefore the fraction -— — ^ is always greater than 1. 

Comparing a monoplane with a biplane whose greater span is 
the same as the monoplane, and both monoplane and biplane 
having the same lift (area), the equivalent monoplane aspect ratio 
of the biplane will be greater than the aspect ratio of the mono- 
plane, since the E.M.A.R. of the biplane is the A.R. of the mono- 
plane multiplied by a fraction which is greater than 1. Since the 
induced drag depends on the aspect ratio, the biplane will have 
less induced drag than the monoplane. 

■(1 -2(7M + M^) 

1-0-2 

applicable only for biplanes which have the best lift distributions 
between the wings. 



T 2 P/ 

The above expression for E.M.A.R. = -^ - 



IS 



I 



If T- = fJihe substituted in the equation, it becomes 

Oi 

L 2<7&2 , h^^\ 



E.M.A.R.(Min. A) ^ "o"! rn — 2 

_ W -2 ahM + hi" 

If spans are equal bi = 62 and 

E.M.A.R.(Min.Z)i) = 



S{1 


-<r^) 


2 62(1 - 


-") 


8(1- 


<^) 


2 62 





/S(l + 0-) 
6, 



Also if 61 = 62, r^ = /x = 1, and the ratio of wing areas for best 



114 INDUCED DRAG OF BIPLANES 

lift distribution becomes 





m' 


- GIX 




1 


- afj, 


= 


1 - 
1 - 


(T 
(T 




= 1 





The spans being equal and the areas (lift) being equal, the chords 
of each wing are equal and the aspect ratios of each individual 
wing are the same. The area of one wing is one-half the total 
area. The actual aspect ratio of each wing is 2 b^/S. The equiva- 
lent monoplane aspect ratio of the biplane is the aspect ratio of 
one wing divided by (1 + cr), where a is a factor depending on 
the gap. 

If the spans are not equal, // is less than 1. Then for any ratio 
of areas, the aspect ratio will be less and the induced drag more 
than for the same ratio of areas with equal spans. 

Although it is always desirable to have equal spans, it is some- 
times necessary for various design reasons to have one wing, 
usually the lower, shorter span than the other. With unequal 
spans, the best efficiency is obtained not with equal areas but with 
the wing of greater span having the greater area. 

Example. A biplane has a span of 40 ft. for upper wing and 36 ft. 
for lower wing and a gap of 5 ft. The total wing area is 400 sq. ft. 
Assuming that the area of each of the wings has been selected for best 
lift distribution, what is the E.M.A.R.? 

Solution. 





b2 

bi 


36 

~40 ~ 


0.9 




Gap 


5 

38 


0.132 


f 


Mean span 


'rom Fig. 40, cr = 


0.580 






E.M.A.R 


40' (1 - 


2 X 0.580 X 0.9 + 0.92) 


~ 400 


1 


- 0.580' 




40' 
-4,,X1.16 





= 4.64 



EQUIVALENT MONOPLANE SPAN 115 

Problems 

1. Assuming that wing areas are for best distribution, what is 
E.M.A.R. of a biplane with total area of 380 sq. ft., upper wing 3o-ft. 
span, lower wing 28-ft. span, gap 5 ft.? 

2. Assuming wing areas for best distribution, what is E.M.A.R. 
of a biplane with total wing area of 350 sq. ft., upper span 30 ft., 
lower span 25 ft., and gap 4 ft.? 

3. What is the E.M.A.R. of a biplane with equal spans of 24 ft., 
each wing having an area of 90 sq. ft., gap being 3 ft.? 

4. What is the E.M.A.R. of a biplane whose upper span is 31 ft. 
7 in., lower span 29 ft. 5 in., total area 298 sq. ft., gap 75 in.? Assume 
best lift distribution. 

5. What is the E.M.A.R. of a biplane the span of both upper and 
lower wings of which is 30 ft., the chord of each wing 5 ft. 3 in., and 
the gap 4 ft. 10 in.? 

6. What is the E.M.A.R. of a biplane whose total area is 245 sq. ft., 
upper span 30 ft. in., lower span 26 ft. 4 in., and gap 4 ft. 10| in.? 

7. What is the E.M.A.R. of a biplane whose total area is 379 sq. ft., 
upper span 38 ft., lower span 35 ft., and gap 64 in.? 

8. What is the E.M.A.R. of a biplane whose total area is 376 sq. ft., 
upper span 40 ft., lower span 38 ft. 5 in., and gap 61 in.? 

9. What is the E.M.A.R. of a biplane whose total area is 1,182 
sq. ft., both spans being 79 ft. 8 in. and gap 110 in.? 

10. What is the E.M.A.R. of a biplane whose total area is 1,596 
sq. ft., both spans being 90 ft. in. and gap 126 in.? 

Equivalent Monoplane Span. The aspect ratio for a monoplane 
being ly^/S and the equivalent monoplane aspect ratio of a biplane 



6i2 [;,2(i _|_ ^)2] ___,_ , I M^l + r) 



V M^ + 



being -^ — r-— 7^ -. — -„ , the expression \/ , , ^ -. — -^is called 

the '^ apparent span factor " or '' Munk's span factor '' and is 
usually represented by the symbol K. 



=vVl 



^ _ . , -W + r) 



+ 2 aril + f- 
m(1 + r) 
Vju' + 2 (7rM + 7-2 



Using this factor K^ the expression for E.M.A.R. of a biplane 
becomes 

E.M.A.ll. = (^ 



116 INDUCED DRAG OF BIPLANES 

The expression for induced drag coefficient becomes 

The expression for induced drag becomes 
D- = ^' 

Being used in these expressions exactly in the same manner as h, 
the span in the case of a monoplane, the term {Kh-^ is called the 
'' apparent span " or '' equivalent monoplane span." 

Example. Find equivalent monoplane span of a biplane whose 
upper span is 32 ft., lower span 29 ft., gap 4.63 ft., area upper wing 
152 sq. ft., and area lower wing 120 sq. ft. 

Solution. 

on 

M = H = 0.879 

Gap _4.63_Q^^^ 
Mean span 30.5 

From Fig. 40, o" = 0.54 
. = 1-1 = 0.9 

0.879 X 1.79 
K. = 



Vo.879' + 2 X 0.54 X 0.79 X 0.879 + 0.79^ 
= 1.07 
Kbi = 1.07 X 32 
= 34.2 ft. 

Problems 

1. Find equivalent monoplane span of an MB-3 airplane; area 
upper wing 124 sq. ft., area lower wing 122.9 sq. ft., upper span 26 ft., 
lower span 24 ft. 6 in., gap 4 ft. 5 in. 

2. Find equivalent monoplane span of a PW-8 airplane; area 
upper wing 172 sq. ft., area lower wing 125 sq. ft., upper span 32 ft., 
lower span 32 ft., gap 4 ft. 7j in. 

3. Find equivalent monoplane span of a Vought training plane, 
area upper wing 143 sq. ft., area lower wing 140.7 sq. ft., upper and 
lower spans each 34 ft. 5j in., gap 4 ft. 8 in. 

4. Find equivalent monoplane span of a Curtiss observation air- 
plane; area upper wing 214 sq. ft., area lower wing 150 sq. ft., upper 
span 38 ft., lower span 35 ft., gap 5.35 ft. 



INDUCED DRAG OF A BIPLANE 117 

5. Find equivalent monoplane span of a Curtiss bomber; area of 
each wing 795 sq. ft., upper and lower spans each 90 ft., gap 9 ft. 3 in. 

6. Find equivalent monoplane span of a Sperry Messenger; area of 
each wing 77 sq. ft., length of each span 20 ft., gap 3 ft. 10 in. 

7. Find equivalent monoplane span of a Douglas observation plane; 
area upper wing 192 sq. ft., area lower wing 186 sq. ft., upper span 
40 ft., lower span 38 ft. 5 in., gap 6 ft. 

8. Find equivalent monoplane span of a Stearman biplane; area 
upper wing 150 sq. ft., area lower wing 135 sq. ft., upper span 32 ft., 
lower span 27 ft. 10 in., gap 5 ft. 

9. Find equivalent monoplane span of a Stearman sportster; 
area upper wing 175 sq. ft., area lower wing 150 sq. ft., upper span 
33 ft., lower span 26 ft., gap 4 ft. 10 in. 

10. Find equivalent monoplane span of a Consolidated Courier; 
total wing area 295 sq. ft., upper and lower spans 34 ft. 5f in., upper 
and lower chords equal, gap 4 ft. 11§ in. 

Effect on Aspect Ratio of Gap Variation. In previous para- 
graphs, the gap has been specified and changes in aspect ratio 
with different areas studied. It was shown that with a given 
gap, the least induced drag (i.e., maximum aspect ratio) would be 
obtained with equal spans and areas. It is evident, however, 
from a study of Fig. 40 that o- is a function of gap as well as both 
spans, and with both spans fixed, a decrease in gap will mean an 
increase in u. As the gap is decreased the mutual interference 
increases. When the gap has decreased to zero, a becomes 1, the 
two wings of the biplane merge and coincide, becoming a mono- 
plane; the spans will then be equal, the expression for E.M.A.R. = 

2 62 , 2 62 62 

becomes 



S{1 + c) S{1 + 1) s 

Induced Drag of a Biplane. The equivalent monoplane aspect 
ratio of a biplane being known, the induced drag may be found 
from the formula 



7r(E.M.A.R.) 



L 



P 



where q = ^V^,Vm feet per second 



TrqiKhiY — vf 2 

For level flight, lift must equal the total weight of the airplane. 

Example. A biplane is to weigh 4,000 lb. It has been decided 
that a wing area of 250 sq. ft. will be sufficient to give proper landing 



118 INDUCED DRAG OF BIPLANES 

speed and that the lower span should be 0.8 of the upper span, which 
is to be 40 ft. What are best chords, and what is the E.M.A.R., if 
the gap is chosen as one-sixth of the longer span? What is the in- 
duced drag at sea-level, at airspeed of 100 miles per hour? 

Solution. 

r = 0.167 

then 

0.167 

From Fig. 40, a = 0.477 
For best lift distribution 

_ (0.8)2 _ 477 y^ Q g 
^ ~ 1 - 0.477 X 0.8 
0.64 - 0.382 
~ 1 - 0.382 
= 0.417 

0.417 ='-^M 

Cl 

'-^ = 0.522 

Cl 

Total area (S) = 250 sq. ft. 
Siil +r) = 250 

^1 



250 



1.417 
= 176.5 sq. ft. 
S2 = 0.417 X 176.5 
= 73.5 sq. ft. 

61 = 40 ft. 

176.5 

= 4.41 ft. 

62 = 0.8 X 40 
= 32 ft. 

_ 73.5 
^~ W 

= 2.3 ft. ^ 

^ ^ 1 - 2 X 0.477 X 0.8 + 0.8' 

1 - OAff 
= 1.138 



INDUCED DRAG OF A BIPLANE 119 



E.M.A.R. = ^ X 1.138 

= 7.28 



4000 

A = 



TT X 0.00119 X 147 X 1.138 X 40" 
= 109 lb. 

It is to be noted that the formula K^ = 7—^ — 2 is applicable 

only when the ratio of areas is such that the induced drag is minimum; 
if the areas are arbitrarily chosen without regard to this, the general 

formula K^ = o , o i — 2 must be used. 

Example. A biplane weighs 4,000 lb.; the upper span is 40 ft., 
lower span 32 ft., upper chord 3.45 ft., lower chord 3.5 ft., and gap 
6.67 ft. What is the induced drag at 100 miles per hour at sea-level? 

Solution. 

32 
^^=40 
= 0.8 
G 2 X 6.67 



i (61 + 62) 40 + 32 
= 0.185 

From Fig. 40, a = 0.477 

32 X 3.5 
40 X 3.45 
_ 112 
~ 138 
= 0.81 



r = 



^ (0.8)^ X (1 + 0.81)2 

^ ~ (0.8)2 -f- 2 X 0.477 X 0.81 X 0.8 + (0.81)^ 



= 1.096 
Di 



4,000' 



IT X 0.00119 X Tif X 1.096 X 40^ 
= 113 1b. 

Problems 

1. A biplane weighing 4,000 lb. has a total wing area of 250 sq. ft., 
upper span 40 ft., lower span 32 ft., upper chord 4.41 ft., lower chord 
2.3 ft., gap one-seventh of longer span; what is induced drag at air- 
speed of 100 miles per hour ao sea-level? 

2. What is the induced drag for the airplane in problem 1, if gap 
is one-ninth of longer span? 



120 INDUCED DRAG OF BIPLANES 

3. What is the induced drag for the airplane in problem 1, if gap 
is one-eighth of longer span? 

4. A biplane weighs 4,000 lb. and has a total wing area of 250 sq. 
ft., upper span 40 ft., lower span 32 ft., gap one-seventh of longer span, 
and chords such as to give minimum drag. What is the induced drag 
at 100 miles per hour at sea-level? 

5. What would be the induced drag for the airplane in problem 4 
if the gap were one-eighth of the longer span and the chords such as 
to give minimum drag? 

6. What is the induced drag for the airplane in problem 4 if gap is 
one-fifth of longer span, and the chords are chosen to give minimum 
drag? 

7. A biplane weighing 4,000 lb. has an upper span of 40 ft., lower 
span 32 ft., total wing area 250 sq. ft., gap one-seventh of larger span. 
What should be chords and E.M.A.R. for minimum drag, and what is 
drag at airspeed of 100 miles per hour at sea-level? 

8. A biplane weighing 2,932 lb. has a total wing area of 257 sq. ft., 
the upper span 31.58 ft., lower span 26 ft.; chords are chosen for 
minimum drag, and gap is 4.19 ft. What is drag at 150 miles per 
hour at an altitude of 10,000 ft.? 

9. A biplane weighing 10,350 lb. has a total wing area of 1,154 
sq. ft., upper and lower spans are each 66.5 ft., the chords are equal, 
and the gap is 9.17 ft. What is the drag at 120 miles per hour at sea- 
level? 

10. The Curtiss Condor weighs 17,370 lb., its total wing area is 
1,510 sq. ft., each span is 91 ft. 8 in., the gap is 10 ft. 6 in., the chords 
are equal. What is the induced drag at 110 miles per hour at sea- 
level? 

Biplane Characteristics with Engineering Coefficients. The 
expressions for induced drag coefficient 



and 






may be transformed by substituting Cm = 391 Kxi and Cl = 391 Ky 



_ 391 X/(ci^ + 2 0-C1C2 + C2') 
^ 125 X/(ci^ + 2 (7CiC2 + C2^) 



TOTAL DRAG OF BIPLANE WINGS FROM MODEL DATA 121 



K^i - IZb Ky ^^^y MHl + r)2 J 



125 X/ ^ 



The expressions for E.M.A.R., equivalent monoplane span, and 
span factor are functions of areas and lengths and are the same 
whether absolute or engineering coefficients are used. 

The expression for induced drag is as follows. 

Di = KxiSV^ V in miles per hour 



[l25^/(^]^^^ 



Z^ 2 C2 y 4 



125 L 

"( 



DiX V 



V in miles per hour 



375 



SV\KhiJ 



It will be noted that the expressions for induced drag and for 
horsepower to overcome induced drag are independent of lift and 
drag coefficients and are therefore the same whether absolute or 
engineering coefficients are used in the airfoil data. 

Total Drag of Biplane Wings from Model Data. With the veloc- 
ity not known, the total drag may be found from model data. 
The simplest method, devised by F. W. Herman, is first to find 
the L/D of the combination of wings and then make use of the 
fact that, in level flight, hft equals weight. 

D - L^ 
LXCl 

IT 



s 

LXCl 



7r(E.M.A.R.) 



122 INDUCED DRAG OF BIPLANES 

Let Dib be the induced drag of the biplane wings and Db the 
total drag of the biplane wings. 

I^b = -Dmodel " -^t model + A"6 

LXCl , LxCl 



■^model / A n \ T 



(^) =- 

\£^/biplane /D 



7r(A.R.)model 7r(E.M.A.Il.)bipla 

1 



\L/model TT \E.M. A.R.biplane A.R.model/ 



V^/model TT \{Khiy A.R.model/ 

mgineer 



With engineering coefficients, this becomes 
^ 1 

'biplane 



(f ) + 125 Ky(j^, j^ ) 

\-tv/model \ (.A Oi; biplane A.It.model/ 

In level flight, lift is practically equal to weight; so that, if the 
weight and the L/D at that angle of attack are known, the drag 
may be found. 

Example. A biplane weighs 3,116 lb.; the upper wing area is 
154.8 sq. ft., lower wing area 103.67 sq. ft., upper span 30.1 ft., lower 
span 24.33 ft., gap 4.7 ft., wing is Clark Y. What is the total wing 
drag at 6° angle of attack? 

Solution. 

2 G^ 2X4.7 



h + 62 24.33 + 30.1 
= 0.173 

From Fig. 40, a = 0.493 

m'(1 + r)2 



K' = 



ju^ + 2o-r/i + ^^ 



0.808^ (1 + 0.67)^ 



E.M.A.R. = 



0.808' + 2 X 0.493 X 0.67 X 0.808 + 0.67 
= 1.03 

1.03 X (30.1)2 



103.7 H- 154.8 
= 3.61 



TOTAL DRAG OF BIPLANE WINGS FROM MODEL DATA 123 



Clark Y, A.R.6, at 6° angle of attack; Cl = 0.791; L/D = 17.5. 
1 

'biplane 



G). 



1 0.791 / 1 1\ 

17.5"^ w V3.61 qJ 



1 



0.0849 
D = 3,116 X 0.0849 
= 268 lb. 

If the spans are equal, the E.M.A.R. is ^t^ — ; — r or the aspect 

/b(l + (T) 

ratio of one wing divided by (1 + a). This simplifies the expression 
for total drag to 

-t^biplane — -'>'model "T ^ VA"R, .x~A"R ,,/ 

TT \-f^--Cv. (one wing) -fi--tv. model/ 

Example. Find wing drag of a D.H. at 6° angle of attack. Weight, 
5,000 lb.; total wing area, 450 sq. ft.; equal spans of 42 ft.; gap 6 ft., 
Clark Y section. 

SoliUion. 

/x = l 

Gap 6_ 

Span ~ 42 
= 0.143 
From Fig. 40, a = 0.574 

42^ 
A.R. (one wing) = ^ 

= 7.84 
L 1 



fiA7_l\ 
V7.84 6/ 



i) ~ J_ 0.791 

17.5"^ TT 

~ 0.0655 
D = 5,000 X 0.0655 
= 328 lb. 

Problems 

1. Biplane weighs 3,000 lb. Wing section, Clark Y. Area, upper 
wing 120 sq. ft., lower wing 100 sq. ft., upper span 30 ft., lower span 
24 ft., gap 3.5 ft. What is total wing drag at 8° angle of attack? 

2. A Thomas Morse observation airplane weighs 4,104 lb. It uses 
Clark Y airfoil section. Area upper wing 212 sq. ft., area lower wing 
146.7 sq. ft., upper span 39.75 ft., lower span 37.6 ft., gap 5.64 ft. 
What is wing drag at 6° angle of attack? 



124 INDUCED DRAG OF BIPLANES 

3. A Consolidated training plane weighs 2,445 lb. and uses a Clark 
Y airfoil. Total wing area is 300.6 sq. ft., equal spans are 34.46 ft., 
gap is 4.72 ft. What is total wing drag at 8° angle of attack? 

4. Plot LlT> versus Cl for the D.H. biplane. Wings from data 
in sample problem. 



CHAPTER VII 
PARASITE DRAG 

When an object moves through air, a resistance is encountered. 
The air flowing around the object exerts a force on the object. 
This force always has a component acting parallel to the direction 
of movement of the body with respect to the air. This component 
is drag. 

Struts, wires, landing gear and other parts of the airplane offer 
resistance to forward motion of the airplane. The force of the air 
on these parts is all drag. There is no upward or lift component. 
This is called parasite drag. 

Formerly the total drag of an airplane was considered as being 
composed of two parts, the wing drag and the parasite drag. 
Classified in this way the parasite drag was the drag of all parts 
of the airplane except the wings. 

The wing drag, however, is composed of two parts, the induced 
drag and the profile drag. It is now the custom to consider the 
profile drag of the wing as part of the parasite drag. Parasite 
drag being drag that is not associated with the production of 
useful lift, it is quite correct to consider profile drag as parasite. 

Parasite drag is caused not only by skin friction but is also 
due to turbulence. Every care must therefore be exercised in the 
design to eliminate protuberances or sharp corners. To this end, 
'' fairing " of balsa wood or sheet metal is frequently used. Wher- 
ever possible the airplane parts that would cause parasite drag 
are made streamline in shape. 

Sometimes, but rarely, parasite drag is referred to as structural 
drag. 

Units of Measurement of Parasite Drag. It was formerly the 
custom to give data on parasite drag in terms of equivalent flat plate 
area. That is, the actual drag in pounds at a given airspeed being 
known, the area of a flat plate normal to the wind, which at the 
same airspeed would have the same drag, would be calculated. 

This unnecessarily complicated the computation, and the con- 
vention of referring to a fictitious flat plate has been discarded. 

125 



126 PARASITE DRAG 

Some data of old tests are still recorded as being in '' equivalent 
flat plate area." The resistance in pounds of a flat plate at sea- 
level is 0.0032 aV^, where a is the area in square feet and V is in 

miles per hour; or the resistance is 1.28 ^ aV^, where p is the air 

density in slugs per cubic foot, a is the area in square feet, and V 
is in feet per second. 

Another method of stating the drag of an airplane part is to 
give the drag of that part at some defined airspeed. If the drag 
is given at an airspeed of 1 mile per hour, this drag multiplied by 
the square of the airspeed is the drag at that airspeed. Since the 
drag of minor parts of an airplane is usually quite small, the drag 
is often recorded for an airspeed of 100 miles per hour. The drag 
at any other airspeed will be the drag at 100 miles per hour 
multipHed by the square of the airspeed and divided by 
10,000. 

Another method of designating drag is to use a parasite drag 
coefficient. 

The total drag of the wing, Co ^ ^SF^, is made up of two parts: 

the profile drag CiJo^/SF^ and the induced drag Coi^xSV^. 

The total drag of the airplane is the wing drag plus the parasite 
drag. The calculation of drag is therefore much simplified if the 
parasite drag coefficient is put in such a form that when multiplied 

by ^SV^, the product is parasite drag. The parasite drag coeffi- 
cient in this form can then be added directly to the profile drag 
coefficient and the induced drag coefficient. 

If a is the equivalent flat plate area of all the surfaces except 
wings, which produce drag, then the coefficient of parasite drag is 

1 28 X a 
Cdp = — — o ^ i^ wing area in square feet 

Then the total drag, exclusive of wing drag, is 

Dp = Cdp ^SV^ V in feet per second 

and 

Dp = 0.00256 CDp&y Y in miles per hour 



STREAMLINING 127 

If the drag of an airplane part is given in pounds at an airspeed 
of 100 miles per hour, the coefficient of parasite drag is 



Cn* = 



_ -PlOO m.p.h. 

'^^ " 0.00256 S X 10,000 

^ 0.0392 X Aoom.p.h. 
S 

Streamlining. The purpose of streamlining is to reduce as 
much as possible the turbulence in the rear of objects moving 
through the air. The air flowing along the sides of a streamlined 
object experiences no sudden change in direction. If the object 
always met the air in a direction parallel to the axis of the object, 
there might be merit in having the nose of the object pointed or 
knife-shaped. A very sHght change in direction of an object 
whose entering surface is so shaped would increase the resistance 
or turbulence enormously. It is therefore best to have a blunt 
nose. Since an object whose rear end is blunt leaves a wake when 
passing through air, and the air rushes into this region of low 
pressure in a turbulent manner, the contour of a streamlined body 
must taper back to a '' tail." 

Any non-streamlined body can have its resistance or drag 
greatly reduced merely by the addition of a blunt nose and a 
tapered tail. If the contour is also a continuous curve the shape 
approaches the ideal streamline. 

Fineness. Fineness is the ratio of the length of a body, i.e., 
parallel to the usual direction of motion, to its maximum thick- 
ness. Best results can be obtained with a fineness ratio of about 
4. A fineness less than 4 shows a big increase in drag coefficient. 
A fineness ratio greater than 4 gives a slight increase probably due 
entirely to the added skin friction. 

The preceding remarks about streamlining and fineness apply 
both to solids of revolution Hke the gas-bag of dirigibles and to 
the cross-section of objects like struts about which the flow is 
two-dimensional. 

Graphical Method for Streamline Shape. The following pro- 
cedure describes a method of obtaining a Joukowski symmetrical 
airfoil section, which serves as a very good contour for a streamline 
section. If the section which is obtained appears to be too thin at 
the rear for strength, it may be arbitrarily thickened there and, 



128 



PARASITE DRAG 



provided this change is not too great, there will be little increase in 
the low air resistance of the section. 




Fig. 41. Graphical method for Joukowski symmetrical airfoil. 

If D is the depth from front to back of the desired streamline 
section and F is the fineness ratio, the maximum thickness is D/F. 
Before starting the graphical construction it is first necessary to 
calculate three dimensions : a, h, and r. 



a = 



0.6495 F - Vo.4219 F' - 1 
D/F 

ar 



2a + r 



For example, it is desired to have a streamline section 4 in. deep 
with a fineness ratio of 5; then the maximum thickness will be 
I in. or 0.8 in., and 



- = 0.6495 X 5 - Vo.4219 X 25 - 1 
r 

= 0.1578 

0.8 



a = 



5.196 
0.1540 in. 



CYLINDERS AND ROUND WIRES 129 

0.1540 



r = 



6 = 



0.1578 
= 0.976 in. 
a 



2 a/r + 1 
0.154 
2 X 0.158 + 1 
= 0.1170 in. 

On a baseline AB, with as center and r as radius, a circle is 
drawn. This circle is divided into an even number of equal divi- 
sions. The distance OOi equal to a is laid off on baseline AB io 
the right of 0. With Oi as center and with r + a as radius, a 
second circle is drawn. The distance OO2 equal to h is laid off on 
baseline ABio the left of 0. With O2 as center and r — 6 as radius, 
a third circle is drawn. All three circles will be mutually tangent. 

With as center, radial lines are drawn through the division 
marks on the first-drawn circle. These radial lines are prolonged 
to intersect the outer circle. 

These radial lines are treated in pairs. The radial line above the 
baseline, and making a given angle with it, is paired with the radial 
line below the baseline, and making the same angle with it. In 
Fig. 41, ORi and OR2 are two such radial lines. Where ORi 
intersects the outer circle, a line is drawn parallel to OR2. Where 
OR2 intersects the inner circle, a line is drawn parallel to ORi, 
The intersection of these two lines is a point on the streamline 
contour. By continuing this construction, sufficient points are 
obtained through which a smooth curve is drawn. The trailing 
edge will be a distance of 2 r to the left of while the leading edge 
will be a distance of 2 r(l + a'^/r^) to the right of 0. 

Cylinders and Round Wires. Circular cylinders and wires are 
rarely used in aircraft construction. If round tubes are used, 
they are faired with falsework of balsa or sheet metal. 

The drag of a cylinder in an air stream whose direction is 
perpendicular to the axis of the cylinder is 

D = drag per foot of length 
D = 0.00026 XdXV^ d = diameter in inches 

V = airspeed in miles per hour 
or 

D = 0.000121 XdXV^ V = airspeed in feet per second 



130 PARASITE DRAG 

The drag of round wires may be found by the above equations. 
If the wire is less than 0.2-in. diameter, the result will be slightly 
high. 

Stranded cable will have greater drag than smooth wires, and 
the following equations should be used : 

D = 0.00031 XdXV^ 7 in miles per hour 

D = 0.000144 XdXV 7 in feet per second 

If two parallel wires are in the air stream and one wire shields 
the other by being directly behind it, the total drag will be less 
than for two single wires, and will be approximately IJ times the 
drag of a single wire. If the distance center to center of the wires 
is greater than 3 times the diameter of the wire, the shielding 
effect is negligible and the drag of each individual wire should be 
counted separately. Quite often thin strips are fastened on the 
double wires, in which case the pair of wires has approximately 
the same resistance as a single wire. 

" Streamline " Wire. A solid wire or rod is manufactured 
commercially which is termed '^ Streamline " wire. It is not 
truly streamlined, but it is much better from the standpoint of 
decreased air resistance than round wire. The drag of this wire 
is found by the equation : 

D is drag in pounds per foot of length 
D = 0.000089 XT XV^ T is thickness in inches 

V is airspeed in miles per hour 

or 

D = 0.0000414 XT XV^ 7 is airspeed in feet per second 

There is usually present, in flight, some vibration of the wires. 
Owing to sudden rapid reversal of stresses, hard wire is liable to 
fatigue and failure with little or no warning. Stranded wire cable 
will usually stretch, and one or two wires will fray so that the need 
of repair will be noted. 

Struts. Round tubing is the most efficient structural shape, 
but its drag is high. Nowadays all struts in the air stream have a 
streamline-shaped cross-section. The drag of a streamline strut 
depends on the fineness ratio. In air of standard density, the drag 
in pounds per foot of length of a strut whose leading edge is per- 



STRUTS 131 

pendicular to the direction of the air stream is as follows: 

w is width in inches 
D = KwV^ V in miles per hour 

Fineness K 

2.5 0.0000194 

3.0 0.0000180 

3.5 0.0000175 

4.0 0.0000171 

4.5 0.0000175 

or 



D = K'wV^ 


V 


in feet 


per second 


Fineness 






K' 


2.5 






0.00000902 


3.0 






0.00000836 


3.5 






0.00000812 


4.0 






0.00000795 


4.5 






0.00000812 



By comparing coefficients for round tubes and for streamlined 
struts, it will be seen that, for the same width, the round tube has 
approximately 15 times the drag of a streamlined strut. 

If a strut is in tension only, its strength depends on its cross- 
sectional area. The area of a solid circular rod is approximately 
one-half the area of a soHd streamline-section strut of the same 
width. A streamline-section strut to have the same cross- 
sectional area as a round rod would have less thickness and more 
depth; so that the drag of the streamline strut will be about -j^ 
of the soUd circular rod of the same area. 

Hollow streamline struts are made of tubes rolled to the proper 
shape. The cross-sectional area of metal will be the same for 
both shapes and the tensile strength will be the same, but the 
streamline form will have jV the resistance. 

In direct compression, the strength of a piece depends on area 
only. Struts as used in aircraft act as columns and are subject 
to compression and bending. Failure will occur about the axis 
which has the least moment of inertia. A streamline strut will 
have a greater moment of inertia about any axis than will a round 
rod of the same diameter (thickness). If comparison is made 
between round and streamline rods having the same area and 
consequently the same weight, the streamline strut will be found 
to have a greater moment of inertia about its short axis than the 
round strut; but about its long axis, the streamline strut will have 
less moment of inertia than the round rod. 



132 PARASITE DRAG 

Example. Find difference in drag at 100 miles per hour, at sea- 
level, of four main struts each 6 ft. long if they are round tubes f in. 
in outside diameter or streamline f in. in width and have a fineness of 
3.5 

Solution. 
Round: 

D = (0.00026 X i X Too') X 4 X 6 
= 46.8 lb. 
Streamline : 

D = (0.0000175 X f X Too') X 4 X 6 
= 2.6 lb. 

Problems 

1. What is resistance in pounds of eight standard streamline wires 
each J in. thick and 21 ft. long, at an airspeed of 90 miles per hour at 
sea-level? 

2. What is the resistance of the wires in problem 1 at 12,000-ft. 
altitude? 

3. What is the drag of the four interplane struts of a biplane if 
they are streamline in shape with fineness of 3, thickness is f in., and 
length of each is 5 ft. 10 in.? Airspeed 100 miles per hour. 

4. What is the drag of 40 ft. of standard ^' streamline " wire, f in. 
thick, at 125 miles per hour airspeed? 

5. A radio antenna mast is a round tube 6 in. in diameter. It is 
40 ft. high. What is the force on the mast in a 50-mile wind? 

Fittings. At the end of a strut, where it meets the wing or the 
fuselage, the air flow is distributed by the conjunction of the two 
surfaces. In a carefully designed plane the end of the strut is 
buried inside the wing and the interference effect is practically 
negligible. In the ordinary planes, the socket on the end of the 
strut and the fitting holding it to the wing surface cause extra drag. 

A very good way to allow for the extra resistance is to follow the 
practice of the U. S. Navy, as follows: For resistance of cables 
and wires, '^ Add one foot to length of cable for turnbuckle, and 
one foot for the eye and fitting. . . . For total resistance of struts, 
use total length including space occupied by sockets and fittings, 
and add three feet per strut for the additional resistance of the two 
end fittings. '^ 

Another rule for fittings is to find the projected area on a plane 
normal to the air stream. This projected area is multiplied by 2 
and figured as flat plate area. The reason for figuring on double 
the area is the interference effect of the nearby surfaces. 



FUSELAGES 133 

Fuselages. The drag of a fuselage is very difficult to predict. 
There are usually many projections, such as windshields, control 
wires, and pinned joints, to break up the smooth flow of air. The 
most accurate way of finding the drag is to build a scale model 
and actually make wind-tunnel tests. This model to be of value 
should be an exact reproduction and should have all excrescences 
such as windshields and filler-cap covers duplicated in miniature. 

Building a scale model to exact scale requires meticulous care 
and is quite costly. If such a proceeding is not feasible, the 
designer endeavors to ascertain the drag of the fuselage of an al- 
ready built airplane whose fuselage most nearly resembles his own 
design. Knowing the cross-sectional area of the fuselage and the 
airspeed at which the drag was measured, he finds a coefficient 
he may use to determine the drag of his own-designed fuselage. 

Example. An airplane fuselage has been designed which resembles 
the fuselage of a Vought (VE-9) training plane. The Vought fuselage 
has a cross-sectional area of 8.9 sq. ft.; at 110 miles per hour its drag 
is 102 lb. The unknown fuselage has a cross-sectional area of 12 
sq. ft.; what is its drag at 145 miles per hour? 

Solution. 

From data on Vought: 

Coefficient = ^ 

8.9 X 110^ 

= 0.00095 
For newly designed airplane fuselage: 

Drag = coefficient X A X T^ 
= 0.00095 X 12 X T45^ 
= 239 lb. 

With wheels, struts, tie wires, etc., there is usually very little 
difference whether the relative wind is exactly parallel or at a 
small angle to the longitudinal axis of the airplane; with fuselages, 
there may be a difference in the drag coefficients with different 
angles of attack. More exactly, the fuselage drag coefficient de- 
pends on the lift coefficient, since a major portion of the fuselage 
and the tail surfaces is always in the downwash from the wings 
and the downwash angle relative to the longitudinal axis of the 
airplane depends on the lift coefficient. For a very rough approxi- 
mation, the drag coefficient for the fuselage may be assumed 
constant; but for any degree of accuracy, use should be made of 
Fig. 42, where a factor F is plotted against the ratio of airspeed 



134 



PARASITE DRAG 



(V) to stalling or minimum speed (F5). This factor F multiplied 
by the fuselage drag coefficient at high speed (i.e., low angle of 
attack) gives the drag coefficient at any airspeed V. It may be 
considered as the ratio of the flat plate area of the fuselage at any 
speed to the flat plate area at high speed. 






1.0 11 1.2 1.3 1.4 1.5 16 
-^ , Speed Ratio 



1.8 1.9 2.0 



Vs 



Fig. 42. 



Parasite drag factor for fuselages, etc. 

Example. The top speed of a DH-4 is 124 miles per hour at sea- 
level; its stalling speed is 61 miles per hour. The wing area is 440 
sq. ft.; at a speed of 100 miles per hour, the fuselage has a drag equiva- 
lent to a flat plate 3.06 sq. ft. in area. What are the equivalent flat 
plate area and drag coefficient at staUing speed and at maximum 
speed? 

Solution, 

From Fig. 42, F, at V/Vs = 1.64, is 1.02 
F at stalling speed is 4.00 
F at maximum speed is 1.00 

At stalling speed: 

3.06 X 4 



Equivalent flat plate area = 
At maximum speed: 

Equivalent flat plate area = 
At 100 miles per hour: 

C/) fuselage = 



1.02 

3.06 X 1.00 
1.02 

3.06 X 1.28 



= 12.0 sq. ft. 



3.0 sq. ft. 



At stalling speed : 



At maximum speed : 



440 
= 0.0089 



Cz? fuselage — 



0.0089 X 4 



1.02 
= 0.0349 



^ _ 0.0089 X 1.00 

'^£> fuselage p^^ 

= 0.00874 



LANDING GEAR 



135 



Landing Gear. In the fixed type of landing gear, the drag is 
composed of the resistance of the struts, the wheels, the mutual 
interference of the struts and wheels, and the interference caused 
by the fuselage and wings. Merely estimating the resistance of 
the individual struts and the wheels will not take into account the 
interference which may cause a major part of the drag. 

If feasible, the landing gear should be incorporated with the 
wind-tunnel model of the fuselage, and the total drag of fuselage 
with landing gear measured. If wind-tunnel tests are not obtain- 
able, an approximate idea of the drag may be gained by totaling 
the resistance of the struts and the wheels, using Table VI. 





TABLE 


. VI 




High-Pressure Tires and Wheels 




Tire 


Drag in pounds per wheel at 100 miles per hour 


size 










Bare wheel 


Usual fairing 


Full fairing 


26X3 


10.8 


7.2 


3.6 


28X4 


14.1 


9.4 


4.7 


30X5 


18.1 


12.0 


6.0 


32X6 


22.8 


15.2 


7.6 


36X8 


33.8 


22.5 


11.2 


44X10 


51.2 


34.1 


17.0 


54X12 


75.9 


50.6 


25.3 


64X14 


105.6 


70.4 


35.2 



Within the past few years, it has become common to use low- 
pressure tires. With a smaller wheel the same landing shock can 
be taken up as with a larger wheel using high-pressure tires. The 
low-pressure tires decrease the chances of ground looping since 
they grip the ground better, they permit a smoother take-off and 
landing on a rough field, and they afford a better traction on a 
wet, muddy field. Brakes operate well, and owing to their pro- 
portions the drag is less than for a wheel with high-pressure tire 
to take the same landing load. The SJ in. by 10 in. low-pressure 
tire and wheel have become the most widely used. The drag is 
6.1 lb. at 80 miles per hour. An extra-low-pressure tire and wheel, 
sometimes referred to as the ''doughnut " wheel, are also used; 
the tire dimensions are 11^ in. by 25 in.; the drag is 7.1 lb. at 80 
miles per hour. 



136 PARASITE DRAG 

Retractable landing gear is now being employed with remarkable 
success on transport planes. It was first used in the United States 
on the Verville-Sperry racing plane which won the Pulitzer 
Trophy Race in 1921. In flight, the wheels are retracted into the 
wings or into the fuselage. Lowering the landing gear not only 
increases the drag but it also lowers the line of action of the total 
drag force, so that, if the airplane is balanced before lowering, 
there is a tendency to nose over when the gear is caused to descend. 
This presents no problem since the landing gear is lowered when 
the pilot wishes to descend for a landing. During the introduction 
of retractable landing gear, great difficulty was encountered by 
pilots' forgetting to lower the gear when landing. Warning 
devices and the presentation of " dumb-bell " trophies to pilots 
who land on the belly of the fuselage appear to be of value in 
aiding the pilot to remember to lower his gear. 

Engine Cowling. It is possible to improve the airflow aroimd 
the engine and thus reduce the parasite resistance, by proper 
cowling. The Townend ring was developed in England for this 
purpose. Research at McCook Field by Lieutenants Breene and 
Greene and later work at the N.A.C.A. produced other forms of 
cowling. Radial air-cooled engines have considerable drag owing 
to the eddies around the fins and valves, so that cowling is chiefly 
beneficial for this type of engine. 

It is difficult to predict the decrease in drag which may be 
attained by cowling the engine, and it is almost impossible in 
small wind-tunnel models to reproduce in miniature all the little 
details of the engines which initiate the air disturbances. Full- 
size flight tests appear to be the only method of finding out exactly 
the precise value of any particular cowl. Tests would need to be 
run at several different speeds since a cowling might be helpful 
at one speed and detrimental at another. 

Polar Diagram. The plotting of Cl against Co has special merit 
when used in combination with the parasite coefficient, the latter 
being expressed as a function of the wing area. It was pointed 
out in Chapter IV that, with d plotted against Cd, if a straight 
line is drawn through the origin tangent to the curve, the point of 
tangency will locate the maximum L/D for the airfoil alone. Lay- 
ing off the parasite drag coefficient in the form of 1.28 a/S, to the 
left of the origin, gives a point such as P in Fig. 43. A line through 
point P tangent to the curve locates the maximum L/D of the 



POLAR DIAGRAM 



137 



entire airplane. In Fig. 43, it is assumed that the airplane has a 
Clark Y wing 36 ft. by 6 ft., and that the parasite resistance has an 
equivalent flat plate area of 8 sq. ft. Then the distance OP is 
1.28 X 8/216 or 0.047 unit. The point of tangency shows that the 
angle of attack for this combination of wing and parasite that has 
maximum L/D total is 8°. The ordinate {Cl) for the point of tan- 



1.6 




















1 




1.5 
1.4 
















y 


^8° 


H^ 


— ^ 














/ 


C 






















/ 












1.3 












y 


^° 






















/ 














1.2 










J 


/ 






















/r 














1.1 










/ 






















// 


10° 
















1.0 
.9 








// 






















-i 


i 


















C,.8 




y 


f 




















.7 
.6 




/ 
























// 


> 




















/ 
























.5 


/ 
























A 


























/ 


r 






















/.a 

/i 1 


/ 
























i 


2° 












































A_4o 























.02 .04 .06 .08 .10 .12 .14 .16 .18 .20 .22 .24 
Co 

Fig. 43. Polar curve for Clark Y airfoil with parasite drag. 

gency is 0.93 and the abscissa, measured from P, (Cz? total), is 0.108, 
so that the L/D for the entire airplane is 0.93/0.108 or 8.51. This 
should be compared with Fig. 25, where the tangent drawn 
from the origin would locate the angle of maximum L/D for the 
wing alone as 1° and the maximum L/D of the wing alone as 
21.5. 



CHAPTER VIII 
ENGINES 

The universally employed aircraft engine is the internal-com- 
bustion type. It is either air cooled or liquid cooled. The 
cylinders may be arranged either radially or vertically in line or 
in two rows in the form of a letter V. 

The essential requirements of an airplane engine are Hght weight 
per horsepower, extreme reliabihty, low fuel consumption, and 
low frontal area. In principle, aircraft engines do not differ 
materially from automobile engines. At full throttle, the auto- 
mobile engine has usually a higher number of revolutions and 
greater piston speed than the airplane engine, but whereas the 
average automobile engine rarely is run wide open and the bulk 
of its operation is at less than half speed, the airplane engine is 
frequently at top speed and most of its operation is at three- 
quarter speed or better. 

The Department of Commerce before approving a new type of 
airplane engine requires a 100-hour endurance test. For 50 hours 
of this test the engine must be run at wide-open throttle, and for 
the other 50 hours at not less than 75 per cent of rated horsepower. 
This test is more severe than those required by other countries, 
and it is not believed that any automobile engine has sufficient 
reliability to pass it. 

The requirement of lightness in weight is a severe handicap in 
obtaining reliability. It would not be so difficult to obtain de- 
pendability if all the parts could be made sturdy and rugged. 
Reducing weight to a minimum means that the stresses of every 
part must be carefully analyzed so that there shall be no weakness. 

In automobile operation, after the engine is started, the car is 
slipped into low gear. After the automobile is moving, the speed 
is slowly increased, and it is usually not till some time has elapsed 
that the engine develops full power. With airplane operation, 
the engine is wide open for the take-off and wide open for the climb. 
It is not till the altitude selected for ffight has been reached that 
the engine is throttled back. 

138 



AIR-COOLED VERSUS LIQUID-COOLED 139 

For reliability all engines of more than 100 hp. are required by 
the Department of Commerce to have dual ignition systems and 
two spark plugs per cylinder. 

Mounting engines at the rear of the wing as in the pusher type 
airplane brings the center of gravity of the airplane far back and 
makes the problem of balance quite difficult. It is only in fljdng 
boats where the hull extends a good distance forward of the wing 
to offset to some extent the weight of the engine that the pusher 
type is permissible. Having the engine in the rear of the pilot 
presents a hazard in a crash. 

To some degree in the pusher type but chiefly in the tractor 
type is the question of frontal area important. In the early 
days of aviation, the width of the engine largely determined the 
width of the fuselage. With improvements in the V-type engine, 
it became the breadth of the pilot's shoulders that restricted the 
narrowness of the fuselage. With radial engines, it is still the 
frontal area of the engine which sets the limit on the decrease in 
cross- sectional area of the fuselage, as the body is streamlined 
backwards from the size at the motor. 

Air-Cooled versus Liquid-Cooled. The term cooling is some- 
what a misnomer since it is not desired to make the engine cold 
or even moderately cool. In the cylinders, gasoline is burned; 
and unless the heat is removed, the engine parts would soon reach 
a temperature at which they would have no strength. The pur- 
pose of cooling is to keep the temperature sufficiently low that the 
engine will function properly. 

All engines are air cooled. Some types are cooled directly by 
air; other types are cooled indirectly by having liquid remove 
heat from the cylinders and then having the liquid cooled by air. 
If an air-cooled and a water-cooled engine are of the same power, 
it is to be expected that the same amount of heat would need to be 
taken from each. The rate at which heat passes from a surface 
to air circulating past the surface depends on the difference in 
temperature of the air and the surface, so that, if the two engines 
are to run at the same temperature, the same volume of air per 
minute should go past the cooling fins in one engine and through 
the radiator in the other. Then, if there were no burbling or 
eddy currents, it might be expected that the drag of each would be 
the same. 

In order that each cylinder could be properly cooled, the first 



140 ENGINES 

air-cooled engines had radial cylinders. This arrangement has an 
advantage of very short length which is desirable in not having 
the center of gravity of the plane too far forward. In recent 
years, air-cooled engines have been built both in-hne and V-type, 
and no difficulty has been experienced in keeping the rear cyhnders 
properly cool. Water-cooled engines have the disadvantage of 
extra weight of the water, the radiator, pump and plumbing — 
which on a 400-hp. engine would amount to upward of 180 lb. 
Air-cooled engines save this weight, but in large horsepowers the 
parasite drag, even with the benefit of cowling, is more than that 
of water-cooled engines of the same horsepower. 

Because of the great weight of water, in the past few years, 
ethylene glycol has been used as a coolant. The weight of ethy- 
lene glycol to cool an engine properly need be only one-quarter 
the weight of water that would be necessary. The size of radiator 
needed for ethylene glycol would be about one-quarter that of a 
water radiator, thus greatly reducing the parasite drag. Because 
the boiling point of this chemical is considerably above that of 
water, the engine can be run at a temperature of 300° F. At this 
temperature the mixture in the intake manifold is absolutely dry, 
whereas, if water cooled, " wet gas " often is drawn into the 
cylinders. More horsepower is actually obtained for this reason 
from an engine when ethylene glycol cooled than when water 
cooled. Greater care is needed with the plumbing, as this chemical 
will seep through tiny cracks that would be impervious to water. 

There appears to be a definite limit to the horsepower per 
cylinder of an internal-combustion engine. Good practice dictates 
that the stroke should be approximately the same length as the 
diameter of the cylinder. The diameter of the piston is limited 
by the problem of cooling ; with a large piston it is difficult to keep 
the center of the piston-face from getting too hot. The displace- 
ment of each cylinder being restricted and the mean effective 
pressures on standard engines not exceeding 150 lb. per sq. in., 
the power obtainable is rarely above 90 hp. per cylinder on present 
standard engines. To obtain large power means a large number 
of cylinders, and nine cylinders are the most that can be crowded 
around the crankcase of a radial engine. Resort has been made, 
with notable success, to a double bank on one American-made 
engine. 

Engine failures may be classified under four heads : fuel trouble, 



STEAM POWER 141 

lubrication trouble, electrical trouble, and cooling trouble. With 
liquid cooling there is possibiHty of leaky radiator, water boiUng 
away, hose bursting, pump failing, etc. With air cooling, one of 
the sources of possible trouble is eliminated. 

Steam Power. Langley used a steam engine in his small suc- 
cessful flying model airplane. 

In considering the practicabihty of a steam engine for aircraft 
use, one must consider the entire unit of boiler, engine, and con- 
denser, since obviously a non-condensing engine is not feasible. 
Some German torpedo boats have been powered with steam 
engines, and on the basis of the entire unit, the weights were 13.2 
lb. per hp., with a fuel consumption of 0.573 lb. per hp-hr. These 
were large power plants, and always efficiencies in power and sav- 
ings in proportionate weight can be effected in large units that 
cannot be realized in smaller units. Authorities on steam-engine 
design have, however, announced their belief that the weights 
of units adapted to aircraft use can be reduced to 9 lb. per hp. 

For ordinary flying, this weight would bar out all consideration 
of steam, since gasoline units have been built weighing less than 
1 lb. per hp. These were special jobs, and the average internal- 
combustion aircraft engine can be taken to weigh around 2 lb. 
per hp. 

Lately considerable thought has been given to stratosphere 
flying. The steam engine does not lose horsepower with altitude. 
If there is any difference, the low temperature at high altitudes 
will give better condensing and there may be a slight increase in 
power. With unsupercharged gasoline engines, there is a decrease 
in power with altitude. At 36,000-ft. altitude the horsepower of 
an unsupercharged gasoline engine is approximately one-fifth of 
its horsepower at the ground. It has been estimated that at an 
altitude of 66,000 ft. all the power of an ordinary gasoline engine 
will be used up in overcoming mechanical losses, so that the 
effective power will be zero. Supercharging will enable the 
gasoline engine to deliver full power at moderate altitudes, but 
there is a definite limit to supercharging. The weight remaining 
the same with decreasing power, the weight per horsepower 
increases so that it may safely be said that, at altitudes higher 
than 40,000 ft., steam engines are lighter than gasoline engines. 

With a steam power plant, energy is obtained from the fuel and 
stored up as pressure energy, to be drawn upon later. This is 



142 ENGINES 

considered extremely hazardous as high-pressure superheated 
steam is very dangerous. Against the objectionable features of 
greater weight at ordinary altitude and peril from boiler explosion 
may be weighed the following advantages. A much cheaper fuel 
may be used, and it may contain more energy per unit weight. A 
steam engine has no carburetion difficulties, nor has it spark 
plugs, distributors, or other electrical devices. Eliminating these 
removes sources of trouble which have caused many forced 
landings. 

Steam turbines have received little consideration since for effi- 
cient operation they must be in such large units or must be run at 
such high speed as to be unsuitable for aircraft use. 

Rotary Gasoline Engines. In the early days of gasoHne engines, 
since the ordinary four-stroke cycle means one power stroke per 
four strokes, it was thought that a flywheel was absolutely neces- 
sary. To eliminate extra weight and still have a rotating mass, 
an engine was built with the cylinders rotating about a stationary 
crankshaft. The propeller was affixed to the cylinders. The 
cyhnders were arranged radially about the hub and were air 
cooled. 

In order to furnish the gaseous mixture to the cyhnders the 
crankshaft was made hollow. Through the hollow shaft, the 
mixture passed into the cylinders. The mixture in the crankcase 
was very dangerous. A backfire might result in the crankcase 
exploding, tearing the engine apart. 

The gyroscopic action of the rotary engine is quite pronounced. 
On certain wartime airplanes, the pilots found that pushing the 
rudder for a right turn started a precession so powerful that the 
plane would be nosed down. Pushing the left rudder caused the 
plane to nose up to a dangerous stall. As the enemy pilots were 
not unaware of this action, they would approach these planes 
from the left side. The pilots of the rotary-engined plane were 
thus faced with the alternatives of attempting a left turn with its 
possibility of loss of control or of making a right turn of 270° or 
more while completely exposed to enemy fire. 

Another objection to the rotary engines was that, on stopping, 
the lower cylinders became filled with oil. Because of the above- 
listed objections, and because on modern engines it is found that 
no flywheel action in addition to that of the propeller is needed, 
rotary engines are no longer manufactured. 



DIESEL ENGINES 143 

Diesel Engines. Engines operating on the Diesel principle 
have been used successfully as stationary engines and as power 
plants for large marine vessels. They have always been very 
heavy in pounds per horsepower compared with engines operating 
on the Otto-cycle principle. 

For reliable performance, the ordinary aircraft engine requires 
a high-test gasoline. With high compression, special compounds 
are desirable to prevent pre-ignition. An especially volatile 
gasoline will cause trouble by vapor-lock in the lines unless the 
plumbing is specially designed to guard against this. 

Because of the high cost of special gasolines for the standard 
aircraft engines, thought has been given recently to the adaptation 
of the Diesel cycle to a light-weight aircraft engine. Not only 
will the Diesel engine use a very cheap oil, but also, since it 
depends on compression for ignition, electrical troubles are 
banished. The fuel consumption of a Diesel is a little more than 
half the pounds-per-horsepower-hour of a gasoline engine. 

Two manufacturers have to date produced aircraft Diesel 
engines, having no greater size than a radial gasoline-fueled engine 
of the same horsepower. Because of the low weight of fuel con- 
sumed per horsepower-hour, an airplane equipped with a Packard- 
Diesel engine now holds the world's endurance record of 84 hours 
and 32 minutes. 

In the Diesel engine, clean air is drawn into the cylinder and, 
as the piston moves up, the air is compressed to a pressure of 
1,100 lb. per sq. in. While the piston is at the end of its stroke, 
oil fuel is sprayed into the cylinder under great pressure. Since, 
owing to compression, the air is at a high temperature, the oil 
immediately ignites. 

As the engine is running at high speed, the time at the end of 
the compression stroke, during which the oil is injected, is infini- 
tesimally short. During a minute fraction of a second, the drops 
of oil must be broken up into droplets and mixed with the air. 
Unless the oil is finely atomized in the brief working stroke, a drop 
of oil will not burn completely inward to its core and there will be 
incomplete combustion. 

For various reasons, airplane designers have been chary of 
adopting the Diesel engine as standard equipment. With further 
development and improvement, however, it undoubtedly will be 
viewed with more favor. 



144 ENGINES 

Brake Horsepower. The power of an engine depends on the 
force exerted on a piston being transmitted through a connecting 
rod to a crank, to produce a turning or twisting moment on the 
shaft. If a pressure-indicating instrument were connected to 
each cyhnder and the pressure measured throughout the cycle it 
would be possible to calculate the horsepower of the engine. This 
is the indicated horsepower. 

Owing to various friction losses in the engine itself, not all the 
indicated horsepower can be used for outside work. The indicated 
horsepower less the power required to overcome the friction in the 
engine leaves as remainder the brake horsepower, which is the 
power of the engine to do outside work. The ratio of brake horse- 
power to indicated horsepower is the mechanical efficiency of the 
engine, usually about 90 per cent. The indicated horsepower 
multiplied by the mechanical efficiency gives the brake horsepower. 

The mean effective pressure, in pounds per square inch, multi- 
plied by the area of the piston, in square inches, gives the average 
force acting on the piston during a cycle. If this force is multiplied 
by the stroke, in feet, the product is the work, in foot-pounds, 
applied to the piston during a cycle. If the work per cycle is 
multiplied by the number of cycles per minute, which in the four- 
cycle engine is one-half the revolutions per minute, the product 
is the work per minute. This may be expressed as follows 

P = mean effective pressure 
L = stroke in feet 
Horsepower (indicated) A = piston area of one cylinder in 
_ PLANn square inches 

33,000 N = working strokes per minute 

= J X revolutions per minute 
n = number of cylinders 

Horsepower (brake) = Mechanical efficiency X Horsepower 

indicated 

Increasing the compression ratio will have the effect of increasing 
the mean effective pressure and thus increasing the horsepower. 

From the foregoing formula, it would appear that horsepower 
varies directly with revolutions per minute. Within a range close 
to the rated revolutions per minute, this is true within a reasonable 
degree of accuracy, below rated power and speed. Above rated 
speed, the frictional losses get to be very large and the mechanical 



EFFECT OF ALTITUDE ON HORSEPOWER 



145 



I 



efficiency is very poor, so that above rated speed the horsepower 
no longer increases directly as speed. 

The variation of horsepower with engine speed has been tested 
for a large number of engines, both air and water cooled, and the 
results conform closely with the relation described above. For 
any particular engine, the manufacturer can furnish information 
of the relation of horsepower to speed for its type. 

The load on the engine is furnished by the propeller. The pro- 
peller is chosen for the engine so that at a selected airspeed and 
rated engine speed the propeller will offer enough torsional resist- 
ance to absorb all the horsepower of the engine and the engine will 
not speed up. At an airspeed less than that for which the pro- 
peller is assigned, the propeller will offer more resistance to turning, 
so that even if the throttle is at the same setting as for full rated 
power, the revolutions per minute will be less. Slowing up the 
engine will have the effect of decreasing the brake horsepower. 
At less than design airspeed, the brake horsepower will be less. 



no 




60 70 80 90 100 
Percent Design Airspeed 



Fig. 44a. Variation of revolutions per minute with airspeed. 



Effect of Altitude on Horsepower. An engine is an apparatus 
for burning fuel. If insufficient oxygen is supplied, the fuel cannot 
burn properly. Each drop of fuel needs a certain quantity of air 
for complete combustion; that quantity not being supplied, 
combustion will not be complete. By quantity is meant mass or 
weight, not volume. 

In a carburetor, air picks up gasoline vapor to form a mixture 
which is drawn into the engine cylinders. Under normal condi- 
tions, the mixture drawn into the engine is at sea-level density; 
and provided the carburetor is adjusted properly, there will be an 
adequate supply of air to support the combustion of the gasoline 
vapor. 

At altitudes, the air is of less density, so that although the same 
volume of air is drawn into each cylinder, the mass of air will be 



146 



ENGINES 



less. The mixture will be richer than at ground level, and a good 
pilot leans his mixture (i.e., changes his carburetor adjustment 
from the cockpit) and opens his throttle. 

The brake horsepower decreases with altitude. Tests of engines 
have been made in an altitude chamber, which is a room from 
which air can be pumped so that the air densities in the chamber 
can simulate various altitudes. As a result of these tests, it 
appeared that at constant revolutions per minute, i.e., maintaining 
the same r.p.m. as at the ground: 



B.H.P. 
B.H.P.o 



1.15 



at constant temperature 



and 



Then 



1.00 

.90 

1.80 

« 

t-70 

<u 
nj 

1.50 
o 

|.40 
o 

^.30 

(0 

a .20 
.10 



B.H.P. ./To . \ ^ 

T3 TT p = V/ -^ at constant pressure 



.H.P.o Vpo/ ^ \T,j 



B.H.P. 
B 



^ 










































N 


s. 










































\ 


^ 








































V 












































'^ 


^ 










































""^ 
















































^ 



































































































































10,000 20,000 30.000 40,000 

Altitude in Feet 

Fig. 446. Variation of brake horsepower with altitude at constant r.p.m. 
In standard atmosphere 

,4.256 



and 



Po \T,) 

P. = (P.\ 
PO \Po/ 



EFFECT OF ALTITUDE ON HORSEPOWER 147 

Substituting these values in the expression for ratio of brake 
horsepower at altitude to brake horsepower at sea-level 

B.H.R ^ /^\ 1.15/0.81 /p,\-o^/4^ 

B.H.P.o W Vpo/j" 

Some slight modifications have been made to this relation at 
higher altitude, and the variation of horsepower and altitude that 
is believed to be most nearly correct is given graphically in 
Fig. 446. 



CHAPTER IX 
PROPELLERS 

Function. The air screw propeller is the device which changes 
the torque power of the engine into forward thrust, thus impelUng 
the airplane forward. In the United States it is termed simply 
the propeller; in Europe, the air screw. 

A wood screw forces its way into wood by the back face of the 
screw thread pressing against the wood. A bolt enters a fixed 
nut by the back face of the bolt thread sliding against the front 
face of the nut thread. With a marine propeller, it is the rear 
face of the blades which pushes against the water. The very 
first aircraft propellers were built with this viewpoint ; very shortly 
thereafter, however, it was conceived that the blades might be 
considered as wings, or rather as a number of wings of infinitely 
short span set end to end. Treating the blades as airfoils makes 
of primary importance the front face of the blades, i.e., the upper 
surface of the airfoils. 

A propeller should be efficient in transforming the rotary power 
imparted to it into forward tractive power. Since the propeller 
is attached to the engine, at whatever speed the engine is run, the 
propeller must be able to absorb the power furnished by the 
engine. The propeller cannot be most efficient at all engine speeds 
and all airspeeds. It is therefore designed to give maximum 
efficiency at the rated revolutions per minute of the engine and 
some one airspeed, either maximum speed or whatever is decided 
upon as cruising speed. At other rates of revolution and air- 
speeds, the propeller will be less efficient than for the design 
conditions. 

Momentum Theory. The first fundamental propeller theory 
was the momentum theory developed by R. E. Froude. It deals 
with the changes in energy of the mass of air affected by the pro- 
peller. 

In this theory, the propeller is assumed to be a disk which 
exerts a uniform pressure or thrust over the cross-section of the air 
column passing through the disk. It is further assumed that the 

148 



MOMENTUM THEORY 149 

disk imparts no twisting or rotation to the air column, and that 
flow remains streamline while coming to the disk, passing through 
it, and beyond it. 




t 



Fig. 45. Momentum theory of propeller action. 

Referring to Fig. 45, X is a point in the air stream far enough 
in front of the propeller so that the pressure is atmospheric; 
F is a point in the air stream far enough to the rear of the pro- 
peller so that the pressure is atmospheric. 

'TA Let A = propeller disk area. 

Ax = cross-sectional area of air column at X. 
Ay = cross-sectional area of air column at Y, 
V = velocity at X. 
F(l -f- a) = velocity at propeller disk. 
F(l + h) = velocity at F. 

p = atmospheric pressure = static pressure at X 

and F. 
pi = static pressure just ahead of propeller disk. 
P2 = static pressure just behind disk. 

Since the volume of air flowing through each area in unit time 
is the same, 

AxV = AV{l-\-a) = AyV(1 + h) 

~^t^ No energy is added or subtracted to the air from X to a point 
immediately in front of the propeller disk, so Bernoulli's equation 
may be applied, and the sum of static and velocity heads at these 
two points placed equal. 

P+^V' = Pi+^VKl + ay 



150 PROPELLERS 

No energy is added or subtracted to the air from a point im- 
mediately in the rear of the disk to point Y, so the total heads 
at these points may likewise be placed equal. 

P2 +1 F2(l + ay = p +1 7^(1 + hy 

P2 = p + ^vn{i + hy - a + ay] 

It will be noted that the velocity immediately in front is as- 
sumed to be the same as the velocity immediately in the rear of 
the disk, otherwise there would be an instantaneous increase in 
velocity and an infinite acceleration, which cannot be. The 
propeller disk does exert thrust on the air column, and the thrust 
(jT) is equal to the difference in pressure created by the propeller 
times the area of the propeller disk. 

T= ip2- pi)A 
Substituting values of pi and p2 already found 

T = A p+^VK[l + hr-[l + aY) - p - I P[l - (1 + ayij 



^AV'i¥ + 2h) 



^ 



Thrust is also equal to the rate of change of momentum which 
is the mass of air affected per second times the change in velocity. 
The mass of air handled per second is the density multiphed by 
the volume, and the volume is the cross-sectional area at any point 
times the velocity at the same point. At the propeller disk the 
area is A and the velocity F(l -\- a), so the mass of air passing 
through the propeller disk in 1 sec. is pAV{l + a). At point X, 
the velocity is V; at point Y, the velocity is F(l -\- h) or V + hV ; 
therefore the change in velocity imparted to the air column by the 
propeller is hV. 

Then 

T = pAV{l + a)hV 
= pA72(l + a)h 



t 



MOMENTUM THEORY 151 

Equating this value for T to the one previously found: 
pAV\l + a)h=^A V\h^ + 2 6) 

6 + a6 = W + 2 6) 
26 + 2a6 = 62 + 26 
2a = 6 
6 
^ = 2 

This means, that, of the total increase in velocity imparted to 
the air column, one-half is added before the air passes through 
the propeller disk. 
"^r^ To find the ideal efficiency, the power exerted by the propeller 
is divided by the power absorbed. 

The power exerted by the propeller is the thrust times velocity 
or 

Power output = TV 

= pA72(i + a)67 
= pAFHl + a)6 

The power absorbed by the propeller is the work done on the 
air column in unit time, which is the kinetic energy added in unit 
time to the kinetic energy at X, to give the kinetic energy at Y. 
Kinetic energy is one-half the mass times the square of the ve- 
locity, and the mass can be measured at any point. 

The kinetic energy at Y is 

K.E.K = Mp4F(JLrho)] [T^(l + 6)? 
= |^y»(l + a)(l+6)^ 

The kinetic energy at Z is yv,»^ 

KE.X = MpAT^dT^)]!^ 

The difference in kinetic energies at Y and X which represents 
the work absorbed by the propeller in unit time is the power input. 

Power input = | A Y\\ + a)(l + 6)^ - | A Y\\ + a) 
= |A7Kl + a)(6^ + 2 6) 



152 



PROPELLERS 



Substituting h = 2 a: 



Power input = | A 7^(1 



Therefore 



Efficiency 



a){2ah + 2h) 
pAV%(l + aY 

Power output 



Power input 

pAV%(l +a) 
pAV'b{l +ay 



1 + a 



This is the ideal efficiency of a perfect propeller having no losses 
due to rotation of the sHp stream, profile drag of the blades, inter- 
ference of blades, or radial flow. The only loss considered in the 
momentum theory is the kinetic energy loss in pure translation. 
The momentum theory would apply equally well to any device 
which applies thrust or additional pressure to an air column. 

Blade-Element Theory. As the blade-element theory was first 
put in practical form by Drzewiecki, this conception of the action 
of the propeller is usually referred to as the Drzewiecki theory. 
This treatment involves considering the blades as being composed 
of an infinite number of small airfoils. The span of each airfoil 
is an infinitely small distance along the length of the blade. All 
these small airfoils are joined together, wing tip to wing tip, to 
form one twisted or warped wing which is the propeller blade. 



^-^ 








^^t^--^^ 






dT 


^^.^.---''^ 






^ 


r 




> 



27rRN 



Fig. 46. Blade element theory of propeller action. 

Each small section of the blade may be studied separately and 
the results summed up to get the total action of the blade. Figure 
46 shows one of these elements of the blade, which is located a 
distance of U feet from the axis of the propeller. In one rotation 
of the propeller, this blade element travels a distance of 2 -kU 
feet around a circle in the plane of the propeller disk. If the 



BLADE-ELEMENT THEORY 153 

propeller is turning over N revolutions per second, the linear 
velocity of the element in the plane of rotation is 2 wR N feet per 
second. If the airplane has an airspeed of V feet per second, the 
propeller is moving forward with that speed. Since the motion of 
the airplane is practically in the direction of the thrust line of the 
propeller, here it is sufficiently accurate to state that the motion of 
the blade element is the resultant of a velocity oi 2 tRN feet per 
second in the plane of the propeller disk combined with a velocity 
of V feet per second perpendicular to the plane of the propeller 
disk. This resultant velocity will be in a direction making an 
angle <^ with the propeller disk such that 



tan<^ = 



2 ttRN 

The magnitude of this resultant velocity (W) is 

V 2 7rRN 
W = - — - = — 

sm <^ cos 

The blade element has a cross-section which has the contour 
of an airfoil section, usually modifications of the Clark Y or 
R.A.F.6 sections. The angle which the reference chord of this 
section makes with the plane of rotation of the propeller is the 
blade angle /3. The angle of attack a of the blade element is 
chosen as the angle of maximum L/D of the airfoil section used. 
It is to be noted that every blade element moves forward with the 
same velocity V, but the linear speed of rotation is greater for blade 
elements nearer the tip of the propeller, so the relative velocity W 
will be greater at the tips. The angle 4> is less for blade elements 
near the tip than for those nearer the center. Since 

a = ^ - <t> 

if the same angle of attack is maintained along the blade, the blade 
angle /3 becomes less as the tip is neared. This is why a propeller 
blade has its warped or twisted appearance. 

The air flows around each element and causes a resultant as 
with any airfoil. The area of the element is the span, dR, times 
the chord, which is 6, the blade width at that point. The velocity 
is W, The lift force on the element is 

dL^Ci^hdRW^ 



154 PROPELLERS 

The resultant force is acting in a different direction from the 
lift component; let the small angle between these two directions 
be called angle B. Then 

cote =-j^ 

If the propeller blade is so constructed that under design con- 
ditions (i.e., normal revolutions per minute and design airspeed) 
every airfoil section is meeting the air at the same angle of attack 
(a) ; then, if the same airfoil section is used throughout, 6 will be 
the same all along the blade. In practice, for strength, sections 
near the hub are thicker than those near the tip. 

The resultant force on the blade element will be 

^ cos 9 

CL^hdRW^ 
cos 6 

This resultant force, instead of being divided into lift and drag 
components, is divided in components parallel to the propeller 
axis and parallel to the plane of rotation. The component parallel 
to the propeller axis is called the thrust component; that parallel 
to the plane of rotation is called the tangential or torque component. 

The lift component is perpendicular to the direction of the rela- 
tive wind (TF), and the relative wind makes an angle of </> with the 
plane of rotation; therefore the lift component makes the same 
angle <f) with the direction of the propeller axis. The resultant 
force is at angle (</> + 6) with the direction of the propeller axis. 
The thrust component is then 

dT = dFR cos {(!> + 6) 

= C4hdRW^'-^^^^ 
2 cos^ 

The torque or tangential component is 
dF = dFR sin (0 + 6) 

2 cos ^ 

The torque itself, which is the moment which must be applied, is 



BLADE-ELEMENT THEORY 155 

the tangential force times its moment arm R. Letting Q repre- 
sent torque 

dQ = R dFR sin (0 + d) 

The power absorbed will be the torque component force times 
the distance traveled per second: 

dPa = 2 ttNR dFR sin (0 + d) 

The efficiency of each element is the power output divided by 
the power input or 

VdT 



Efficiency 



2 7rNdQ 

V dFR cos (0 + 6) 
2 7rNRdFRSm((t>-{-d) 



V 

Substituting ^ — ^-^ = tan cf) 

L 'Kti IS 

■y^rn ' tan <f) 

^^''''''^ = tan (0 + e) 



Tangent is the ratio of the airspeed V to the tangential ve- 
locity of the blade element. Every element moves forward with 
the same airspeed, but the tangential velocity is greater nearer the 
blade tip. For elements near the tip tangent (f> is smaller than for 
elements nearer the hub. Cotangent 6 is L/D for the blade ele- 
ment. If the same airfoil section is used at the same angle of 
attack throughout the blade, will be constant. By differen- 
tiating the efficiency with respect to and placing result equal to 
zero 

For the Clark Y section, used frequently for propellers, the 
maximum L/D is 22.5, making have a value of 2J°. Then the 
maximum efficiency will be when = 45° — li°, and the efficiency 
will be 

^ tan 43f ° 
tan 46i° 
^ 0.957 
1.045 
= 91.5 per cent 



156 



PROPELLERS 



This will be for the most efficient element. Nearer to or farther 
from the tip the blade elements will be less efficient, so the total 
efficiency of the entire blade will be less than the maximum 
efficiency of the optimum blade element. 

Returning to the thrust of a single blade element, 



dT = 



CL^hdRW^co^{<l> + e) 
cos 6 



The blade width (6) instead of being expressed in feet might be 
given as a fractional part (h/R) of the radial distance out to that 
particular blade element. The ratio h/R is a pure number, but 
this ratio is different for each element. The blade width is then 
R{h/R). 

Also W, the relative velocity, is equal to 2 tt NR/ cos <^. The 
angle d is always small so that little error is introduced if cos 6 is 
assumed to be unity. The distance R to each blade element is 
expressed as R/Ri X Ri or R/Ri X D/2, where Ri is the radius 
of the tip and D is the diameter. Making these substitutions, the 
expression for thrust becomes 



dT 



-i[ 



^ h cos (0 + d) 

COS^ 



R 



dJ^I] 



NW^ 




.2 4 ... 6 .8 10 

Fig. 47. Typical thrust and torque force curves. 



The quantity in the brackets contains terms which vary for 
each blade element. Values of the bracketed quantity may be 
plotted against (R/Ri) as in Fig. 47, and a curve of thrust varia- 



BLADE-ELEMENT THEORY 157 

tion along the radius may be drawn. All the terms in the bracket 
depend on the airfoil section used and its angle of attack, on the 
variation of blade width with respect to radius, and on the angle 0. 
For geometrically similar propellers, the bracketed quantity de- 
pends on the angle 4>, which is the angle whose tangent is the air- 
speed divided by the peripheral speed of the tip, since the angle 
of any blade element on one propeller will be the same as the angle 
(J) of the blade element of a similar propeller located the same frac- 
tional distance out from the center. The thrust curve depends 
therefore on V/iirND). It is common to drop the constant I/tt 
and state merely that the thrust curve depends on the V/{ND) 
of the propeller. V being airspeed in feet per second, N being 
the revolutions per second, and D being the diameter in feet, the 
expression V/(ND) is dimensionless. 

The area under the thrust curve of Fig. 47 is integrated and 
multipHed by the number of blades to give the thrust coefficient 
(Tc) for the entire propeller. The equation for the thrust of the 
whole propeller is 

The expression for the power absorbed by an individual blade 
element may be modified in a similar manner. 

dPa = 2 ttNR dFR sin (0 + 6) 

= 2TNRC4hdRW^ sin^0+^) 
2 cos^ 

2 L 4 R cos2 \RiJ Rij 

The bracketed quantity is plotted against R/Ri in Fig. 47. 
Integrating the area under the curve and multiplying by the num- 
ber of blades gives a power coefficient (P^). The equation for 
power absorbed by a propeller becomes 

Pa=^PcN'D' 

It will be noted, in Fig. 47, that the blade elements inside of 
20 per cent R contribute no thrust. To withstand centrifugal 
force the blades are made extremely thick near the hub and are 



158 PROPELLERS 

consequently very poor airfoil sections. This, together with the 
low relative wind speed and large angle 0, makes for small 
thrust. 

The efficiency (??) of a propeller may be found in terms of these 
coefficients, as follows 



TV 
P 


{^iTcmD^y 


^PcNW' 


Tc^ V 
Pc ND 



The simple blade-element theory neglects the inflow velocity. 
It also neglects the fact that air passing around a wing is given a 
downward velocity, or if circulating around a propeller blade is 
given a similar flow. The blade of a propeller, instead of meeting 
a relative wind from the direction W, as shown in Fig. 46, is 
meeting air disturbed by the previous blade passage. The theory 
also neglects tip losses. 

A combination of the two basic theories, the momentum and 
blade-element theories, with certain corrections has been found to 
give excellent results in the design or selection of propellers. 

Thrust and Torque Coefficients. The coefficients for thrust 
{Tc) and power {Pc) described in the simple blade-element theory 
are used with certain modifications in the complete propeller 
theory. These coefficients depend not only on the shape and 
proportions of the blades, but also on the airspeed and speed of 
rotation; i.e., on the V/{ND) of the conditions under which the 
propeller is run. They are useful in comparing the thrust de- 
veloped and power required by different propellers of the same 
diameter. For use on an airplane, a propeller is selected for a cer- 
tain V and a certain N, and these coefficients, though suitable for 
comparing propellers at the same V/(ND), are not applicable for 
propellers for the same V and N but different D. 

The coefficients Tc and Pc are non-dimensional. This can be 
proved by putting the other terms of the equations for thrust and 



BLADE-ELEMENT THEORY 



159 



power into the basic dimensions of mass (ikf), length (L), and 
time {t). 



^80 

c 
w 
O 

I 60 
I 40 

u 

20 













1 
5" Bladi" 








_/<. 


=-<--^' 






/', 


^ \^ 


angle 






v/ 


^ 


\ 


ri 




A^ 




I. 






'// 






I-. 


/' 


7 






-15° 


i~ 


// 












/ 












r 




_ 









0.2 0.4 06 



0.8 
NO 



1.0 \2 1.4 



Fig. 48. Typical eflSiciency curves. 



For a given propeller (or family of geometrically similar ones), 
the thrust and power coefficients as well as the efficiency may be 
plotted against F/(iVi)). The 
efficiency is a maximum at one 
particular value of V/{ND), de- 
creasing for either larger or small- 
er values. The value of 7/ ( ND) 
for a given propeller which 
gives the maximum efficiency 
is called the design V/{ND). 
Curves showing how a propeller 
efficiency changes with V/(ND) 
are presented in Fig. 48. 

The thrust and power coeffi- 
cients of a given propeller will vary with the airspeed and revolu- 
tions per minute, i.e., with V/{ND). For a typical propeller they 
are plotted in Fig. 49. This plot will apply to all propellers which 
are geometrically similar to the one for which the data are plotted. 

It will be noted that the thrust coefficient decreases as the 
V/{ND) increases. At the V/{ND) of 1.3 for this propeller the 
thrust coefficient becomes zero. This would correspond to a 
steep dive with power on. At this V/(ND) of zero thrust, the 
torque coefficient still has a positive value. The thrust coefficient 
being zero means that the relative wind is meeting the blades at 
such an angle that the resultant force is parallel to the plane of 
the propeller disk. Any further increase in airspeed will mean 
that the relative wind will meet the blades at an angle which will 
produce negative or backward thrust. The condition of zero 
thrust corresponds with angle of zero lift for an airfoil and means 



160 



PROPELLERS 



that the blades are moving forward the greatest possible distance 
per revolution without developing negative thrust. Pitch being 
the forward distance per revolution, the V/{ND) for zero thrust, 
multiplied by I/tt, is called the experimental mean pitch, since it is 
the ideal or maximum value of pitch for the particular propeller. 



.20 



.18 



16 



.14 



.12 



.10 



.08 



06 



.04 



.02 



0.2 0.4 0.6 0.8 1.0 1.2 1.4 

ND 

Fig. 49. Typical thrust and power coefficient curves. 







^ 






















V 










^ 


"TT^ 


^ 


v \ 
















N^ 


V 
















V 
















\ 


\ 
















\\ 
















V 


I 














\ 


\\ 
















w 





When the airplane is stationary, V is zero, making V/{ND) 
zero. Under this condition, both thrust and power coefficients 
have large positive values. The value of the thrust coefficient for 
V/{ND) = is a measure of the ability of an airplane to acceler- 
ate from a standstill position. 

Power-Speed Coefficients. The expression V/{ND) is a non- 
dimensional quantity. For a given propeller (or geometrically 
similar ones) there are only one Tc, one Pc, and one efficiency for 
each V/(ND) condition. The coefficient P^ may therefore be 
properly divided by the corresponding V/{ND) or a power thereof 



PITCH AND PITCH RATIO 



161 



to obtain another coefficient of different magnitude. Using 
V/{ND) as a divisor, the diameter is ehminated from the equation 
for power absorbed. 




n. 



Substituting 

C -^ 

N'Cs' 

The coefficient Cs is called the power-speed coefficient, since, if 
this coefficient is known together with the forward speed and 
speed of rotation, the power may be found. Conversely, if the 
available power (P), the rated revolutions per minute of the 
engine (N), and the design airspeed of the airplane (F) are known, 
the Cs can be found. A propeller is then chosen of the proper 
diameter and pitch to give the maximum efficiency with this 
power-speed coefficient (Cs). 

Pitch and Pitch Ratio. The pitch of a single-thread machine 
screw is the distance the screw is advanced in one revolution or 
the distance between threads. Using the convention that a 
screw is an inclined plane wrapped around a cylinder, the pitch 
would be the slope of this inclined plane. In appl3dng the ter- 
minology of mechanics to an air screw, the difficulty is met that 
air is not a solid medium and the blade angle differs along the blade. 

" Geometric pitch " is the term applied to the calculated dis- 
tance an air propeller would move forward in one revolution if 
there were no slip, i.e., if air were solid. This definition cannot 
be applied without amplification. Each blade element is set at 
a different blade angle. For uniform geometric pitch the blade 
angle would decrease as the radius increased. On account of 
engine interference near the hub the air flow is slower and the 
pitch is usually made less there. Owing to the non-uniformity of 
the pitch, the blade element two-thirds the radius was formerly 



162 PROPELLERS 

used to measure geometric pitch. For correct performance, it is 
necessary to have the proper diameter as well as proper pitch, 
and since it is physically impossible for a propeller manufacturer 
to stock an infinite number of different diameters as well as an 
infinite number of pitches, if the user cannot obtain the exact 
diameter propeller he desires, he saws off and trims the tips of a 
larger-diameter propeller. Reducing the diameter woujd change 
the location of the two-thirds point on the radius, and the meas- 
ured geometric pitch would be different. It has now become the 
custom generally to measure the geometric pitch at the blade 
element at a point 42 in. from the axis and call this the standard 
or nominal pitch. The geometric pitch is the tangent of the blade 
angle multipHed by 2 tt times the radial distance of the blade 
element from the axis. 

" Effective pitch " is the actual distance the propeller travels 
forward in one revolution and is equal to V/ N feet. 

" Slip " is the difference, in feet, between the geometric pitch 
and effective pitch. 

" Pitch ratio '' is the ratio of the geometric pitch to the propeller 
diameter. 

With a fixed-blade propeller the geometric pitch is permanently 
fixed. Adjustable-pitch propellers have blades with circular 
cross-sections near the hub which are gripped in sleeves. By 
loosening clamping rings, the blades may be turned in their sockets, 
so that they have a new blade angle and consequently new geo- 
metric pitch. An adjustable-pitch propeller can have its blade 
angles changed only when the engine is at rest, i.e., when the 
plane is on the ground. A controllable-pitch propeller is one whose 
blade angles may be changed in flight. This is accomplished by 
electrical or oil-pressure-operated devices. The term variable- 
pitch propeller may be applied to either adjustable- or controllable- 
pitch propellers. 

With a fixed-blade propeller, the effective pitch varies: it is 
zero when the airplane is stationary, and it has a finite value as 
the airplane moves forward. In level flight, the effective pitch 
has a value approximately 75 per cent of the geometric pitch; 
in a dive, it increases still further and may even exceed the geo- 
metric pitch. 

Propellers are selected for certain design requirements, i.e., 
for the V/{ND) of the desired flight conditions. While the pro- 



DETERMINING PROPER POWER-SPEED COEFFICIENT 163 

peller is operating under these design conditions, the effective 
pitch is D times the design V/{ND), and the angle of attack (a) 
of the blade elements is the angle of maximum L/D. The blade 
angle of a fixed-blade propeller being constant, when the effective 
pitch is decreased by flight conditions, with the decreased V/{ND) 
there is an increase in the angle of attack and consequently a de- 
crease in L/D. When the airplane is stationary, the effective 
pitch and V/{ND) being zero, the angle of attack of each element 
is equal to the blade angle of the element. For the portion of the 
blade near the center, the angle of attack is greater than the burble 
point, and the efficiency is therefore low. 

In climbing, the airspeed is low while the engine is running at 
full r.p.m. so that the V/{ND) is much less than the design 
V/{ND). The effective pitch during climb is such that the angle 
of attack a of the blades is greater than the angle of maximum 
L/D, so the efficiency is low. 

Improvement in efficiency which would mean obtaining greater 
thrust with the same horsepower expenditure can therefore be 
gained in take-off and in climb by using a controllable-pitch 
propeller. With a small blade angle in take-off or climb, the angle 
of attack can be nearly if not exactly the angle of maximum L/D. 
After the altitude is reached where level flight is to be maintained, 
the blades can be reset to a new geometric pitch, which will give 
the correct blade angle for level flight conditions. 

Determining Proper Power-Speed Coefficient. The equation 
for power using the speed-power coefficient Cs is 

P in foot-pounds per second 
p _ pV^ p in slugs per cubic foot 

N^Cs^ V in feet per second 

N is revolutions per second 

In order to employ the more usual units, this formula is modi- 
fied, as follows. 



PX 
550 X H.P. = 



/ 88 M.P.H. y 

\ 60 ; 



i^h- 



jjp _ 44.5 X p X M.P.H.- 
R.P.M.' X C/ 



164 PROPELLERS 

For sea-level conditions, where p is 0.002378 slug per cubic foot 
„ ^ 44.5 X 0.002378 X M.P.H.' 

ti.r. = — 



R.P.M.' X C. 



^ 0.106 X M.P.H 
Solving for Cs gives 



R.P.M.' X C/ 



V R.P.M.^ 



a = M.P '''' 



XH.P. 
0.638 X M.P.H. 



R.P.M.i X H.P.^ 

Therefore, the available horsepower, engine revolutions, and 
airspeed being known, Cs may be found. 

This formula does not lend itself readily to slide-rule operation. 
It may be solved by logarithms or by use of the chart of Fig. 50. 

In using the chart, a line is drawn from the origin through the 
intersection of lines representing velocity and revolutions per 
minute. The intersection of this line with the horsepower line 
gives the value of Cs. 

Example. The engine gives 400 hp. at 1,900 r.p.m. What should 
be the Cs of the propeller for an airspeed of 150 miles per hour? 

Solution. See dotted lines on Fig. 50. From draw diagonal line 
through intersection of lines representing 150 miles per hour and 1,900 
r.p.m. Where diagonal line intersects line representing power of 
400 hp., read on Cs scale 1.40. 

Problems 
(Use logarithms and check by nomogram.) 

1. A Stearman plane uses a Kinner engine which is rated 210 hp. 
at 1,900 r.p.m. What should be the Cs of a propeller for an air- 
speed of 100 miles per hour? 

2. A Monocoupe uses a Warner engine rated 125 hp. at 2,050 
r.p.m. What should be the Cs of the propeller for an airspeed of 112 
miles per hour? 

3. A Taylor Cub cruises at 65 miles per hour. It is powered with 
a Continental engine, giving 37 hp. at 2,550 r.p.m. What should be 
the Cs of the propeller? 

4. An Aeronca Collegian cruises at 65 miles per hour, with an engine 
rated 36 hp. at 2,400 r.p.m. What should be the Cs of the propeller? 



DETERMINING PROPER POWER-SPEED COEFFICIENT 165 



Brake Horsepower 
o p in o 




ir> •>* 00 rsi -H -. 
Brake Horsepower 



Fig. 50. Chart for finding Q. 



166 PROPELLERS 

5. A Laird Speedwing is equipped with a Wasp Junior engine of 
300 hp. at 2,000 r.p.m. What is the Cs of a propeller suitable for 
cruising at 150 miles per hour? 

6. For the airplane described in problem 5, what is the Cs of a pro- 
peller for racing at 190 miles per hour? 

7. A Douglas Airliner cruises at 185 miles per hour, equipped with 
two Wright engines each giving 700 hp. at 1,900 r.p.m. What should 
be the Cs of suitable propellers? 

8. A Lockheed Vega cruises at 150 miles per hour, equipped with 
a Wasp engine of 420 hp. at 2,000 r.p.m. What is the Cs of a suitable 
propeller? 

9. A Bellanca Pacemaker cruises at 125 miles per hour with a 
Wasp Junior of 300 hp. at 2,000 r.p.m. What is the Cs of a suitable 
propeller? 

10. A Northrop Delta has a maximum speed of 221 miles per hour 
with a Wright Cyclone engine of 710 hp. at 1,900 r.p.m. What is the 
Cs of a suitable propeller? 

Determining Blade Angle and Diameter. The design power, 
airspeed, and revolutions per minute being known, the power-speed 
coefficient can be found as shown in the preceding section. V and 
N being known, the effective pitch {V / N) can be found. It is 
desired to have an efficient propeller for the design conditions, but 
there are two variables, the diameter and the geometric pitch. 

The diameter is limited in size, in that too large a diameter will 
not allow sufficient ground clearance in landing and taking-off. 
Also with given number of revolutions per minute, a larger diame- 
ter will mean greater tip speeds. With tip speed above 900 ft. 
per sec. the tip losses are very great and cut down on the efficiency. 
If design V and N are fixed, increasing the diameter decreases the 
design 7/(ArD). 

Increasing the V/(ND), by decreasing the diameter, means that 
a large geometric pitch (large blade angle) will be necessary to 
secure efficiency under design conditions. A large blade angle will 
give a low static thrust resulting in poor take-off and poor climb. 

Examining Fig. 48, it will be seen that propellers with larger 
pitch ratios (blade angles) have greater maximum efficiencies than 
propellers with smaller pitch ratios; also that these greater maxi- 
mum efficiencies for the large pitch ratio propeller are for greater 
V/{ND) values. This would seem to indicate that a very small 
diameter would be desirable accompanied by large pitch ratio. 
This might be true if the propeller always operated at design 



1 



DETERMINING BLADE ANGLE AND DIAMETER 167 

V/{ND). In level flight, throttling the engine to give slightly 
fewer revolutions per minute will decrease the airspeed approxi- 
mately the same ratio, and since D is constant, V/{ND) will be 
approximately the same as design V/(ND). In climb, however, 
with full throttle, the airspeed will decrease greatly so that 
V/(ND) will be much less than design V/{ND). Referring to 
Fig. 48 , if a propeller has been picked which has a higher efficiency 




Fig. 51. Efl&ciency versus Cg for various blade angles. 

than any of the other propellers plotted for some given V/(ND), 
at a smaller V/{ND) a propeller with less pitch will show a higher 
efficiency. 

The efficiency versus V/{ND) curves are practically horizontal 
near the maximum point, then drop rather sharply. If a propeller 
is selected which has peak efficiency for the V/{ND) of level 
flight maximum speed, at the V/{ND) of cruising conditions the 
propeller will have almost the same efficiency. 



168 



PROPELLERS 



In Fig. 51 is shown efficiency rj plotted against Cs for a metal 
propeller with a type J-5 engine mounted in a high-wing cabin 
monoplane. With a different engine or mounted in the nose of 
a different type of airplane, owing to the different interference 




Fig. 52. Cs versus blade angle for various values of V/{ND). 



immediately behind the central part of the propeller, the efficiency 
curves will be altered slightly. The curves of Fig. 51 are quite 
typically representative, however. 

Selecting the peak efficiency of any one blade setting, it will be 
noted that, at the same value of Cs, a higher efficiency is obtain- 



DETERMINING BLADE ANGLE AND DIAMETER 169 

able with a larger blade angle. If the utmost efficiency is desired 
for a given Cs then a propeller with the blade angle giving the 
highest efficiency for that Cs should be chosen. At slightly lower 
values of Cs, the efficiency of the bigger-blade-angled propeller 
will have dropped sharply whereas the efficiency of the smaller- 
blade-angled propeller will have decreased only slightly. 

If high speed is the only objective, the Cs being known, the blade 
angle is chosen which gives the greatest efficiency for that Cs. 
This will result in an extremely poor take-off. 

The blade angle which has peak efficiency for this Cs will permit 
of a reasonably good take-off but owing to the smaller efficiency 
will reduce the topspeed. 

A propeller with a blade angle which is the mean between those 
described in the last two paragraphs will result in a take-off 
approximating that for the peak efficiency while the maximum 
speed obtainable will be practically that of the propeller selected 
for high speed. This is the propeller usually selected. 

With the blade angle decided on, resort is made to a chart similar 
to Fig. 52, where Cs is plotted against V/(ND) for various blade 
angles. Two reference lines are on this diagram, one designating 
the maximum efficiency for racing planes, the other suitable for 
general purposes. The use of Figs. 51 and 52 can best be explained 
by an illustrative example. 

Example. A cabin monoplane has a top speed of 150 miles per hour. 
Its engine gives 400 hp. at 1,900 r.p.m. Find diameter, blade angle, 
and efficiency of the propeller for general utility. What is its effi- 
ciency if airplane ffies 100 miles per hour at 1,600 r.p.m.? 

Solution. From previous work: 

Cs = 1.40 

From Fig. 51, for Cs = 1.40; blade angle of 19° is at peak efficiency, 
7] = 0.79; blade angle of 20° gives efficiency of 0.80, blade angle of 
22° gives 7] = 0.81. 

From Fig. 52, for Cs 1.40 and blade angle 19° 

-^ = 0.725 

but 

F = 150 miles per hour 

= 220 ft. per sec. 
N = 31.7 r.p.s. 



170 PROPELLERS 

T. 220 

^^^^ ^ = 31.7 X 0.725 

= 9 ft. 7 in. 
for V = 100 miles per hour 

= 146.7 ft. per sec. 
and N = 26.7 r.p.s. 

_F 146.7 

ND ~ 26.7 X 9.57 
= 0.574 
From Fig. 52 for 19° blade angle and V/{ND) of 0.574 

Cs = 1.08 
From Fig. 51 for 19° angle and Cs = 1.08 

V = 0.75 

Problems 

1. Find diameter, blade angle, and efficiency for a general-utility 
propeller for V = 120 miles per hour, 200 hp., and 1,800 r.p.m. 
• 2. Find diameter, blade angle, and efficiency for a general utility- 
propeller for V = 140 miles per hour, 200 hp., and 1,800 r.p.m. 

3. Find diameter, blade angle, and efficiency for a general-utility 
propeller for V = 180 miles per hour, 200 hp., and 1,800 r.p.m. 

4. Find diameter, blade angle, and efficiency for a general-utility 
propeller for V = 120 miles per hour, 400 hp., and 1,900 r.p.m. 

5. Find diameter, blade angle, and efficiency for a general-utility 
propeller for 70 miles per hour, 90 hp., and 1,400 r.p.m. 

Finding Blade Angle with Diameter Fixed. It is quite often 
that the size of the propeller is influenced by the need of having 
proper ground clearance. Also it is becoming more common to 
use a variable-pitch propeller, in which case the problem resolves 
itself into selecting the proper blade setting for a given diameter. 

Example. What should be blade setting and efficiency for a vari- 
able-pitch propeller 8 ft. 6 in. in diameter for F = 150 miles per hour 
and 400 hp. at 1,900 r.p.m.? 

Solution. From previous work Cs = 1.40 and 

_V_ 220 

ND ~ 31.7 X 8.5 - 

= 0.816 
From Fig. 52 for Cs = 1.40 and V/{ND) = 0.816; 

Blade angle = 25° 
From Fig. 51 for Cs = 1.40 and blade angle 25°: 

17= 0.81 



i 



PROPELLER PERFORMANCE AT ALTITUDES 171 

Problems 

1. Find blade angle for a 7-ft. propeller for 110 miles per hour, 210 
hp., at 2,000 r.p.m. 

2. Find blade angle for an 8-ft. propeller for 150 miles per hour, 
200 hp., at 1,800 r.p.m. 

3. Find blade angle for an 8-ft. propeller for 150 miles per hour, 
300 hp., at 2,000 r.p.m. 

4. Find blade angle for a 9-ft. propeller for 150 miles per hour, 300 
hp., at 2,000 r.p.m. 

5. Find blade angle for an 8-ft. propeller for 70 miles per hour, 
90 hp., at 1,400 r.p.m. 

Propeller Performance at Altitudes. The expressions for thrust, 
torque, and power all contain the term p, the air density. They 
therefore vary directly with density just as do the lift and drag of 
a simple wing. Efficiency of a propeller does not depend on den- 
sity, merely on the V/{ND) condition. As long as the propeller 
advances the same distance per revolution, the efficiency will 
remain the same, regardless of altitude. As pointed out in the 
chapter on engines, the brake horsepower decreases with altitude 
but at a greater rate than the density decreases. 

At the ground, the engine turns over at a certain speed. That 
speed is the speed at which the torque power or resistance of the 
propeller just balances the engine power. At that particular 
speed of rotation, the propeller exerts a definite thrust which pulls 
the airplane forward at an airspeed which is the velocity at which 
the total drag, or resistance of the airplane to forward motion, just 
balances the thrust. 

At an altitude, if the throttle is left at the same opening, the 
engine would tend to turn over faster, because the thinner air 
would offer less resistance to the propeller; i.e., the power absorbed 
by the propeller would be less. Speeding up an engine gives 
greater power, but, since at an altitude the actual power per revo- 
lution drops off as the result of lessened volumetric efficiency and 
this decrease is at a greater rate than the increase in power due to 
putative increased revolutions per minute, the engine actually 
runs slower at altitude. 

If, at the ground, the airplane is cruising at less than full power, 
at an altitude the throttle may be opened wider so that the engine 
speed is the same as it previously was at the ground. In doing 
this the horsepower deUvered by the engine is made equal to the 



172 PROPELLERS 

torque horsepower of the propeller at that engine speed at that 
altitude. 

The drag of an airplane in level flight at any one airspeed varies 
directly with the air density. Drag equals thrust in level flight. 
At any altitude the thrust required to maintain a given airspeed 
compared with the thrust required at the ground for that same air- 
speed will be directly as the ratio of the air densities. The power 
required to turn the propeller at this airspeed at the altitude would 
be related to the power required at the ground as the relative densi- 
ties only if the revolutions per minute were the same. The engine 
cannot at altitude furnish power proportionate to the relative 
density compared with the power at the ground at the same 
revolutions per minute. Therefore, for a given airspeed, the 
revolutions per minute will be less at altitude than at the ground. 

The advance per revolution, or V/(ND), not being the same for 
a given airspeed at altitude as at the ground, the efficiency of the 
propeller will be different. The thrust horsepower at altitude 
compared with the thrust horsepower at the ground for the same 
airspeed is affected in two ways : first, the brake horsepower of the 
engine is less; and second, since the engine speed is less the effi- 
ciency of the propeller is different. 

Propeller Selection for Altitude Flying. Previously in this 
chapter the method of selecting the proper diameter and blade 
angle was shown on the basis of flying at or near sea-level. If the 
engine delivers approximately the same horsepower at altitudes, 
it is conceivable that the propeller might not offer enough resist- 
ance to prevent the engine running over speed. 

The formula used for determining the power-speed coefficient 

^ _ 0.638 X M.P.H. 



H.P.'/' X R.P.M.'/' 
was derived from the more basic formula 

^5 _ 44.5 X p X M.P.H.^ 



Cs = 



WY. X R.P.M.' 
2.136 X p'^' X M.P.H. 
H:p:'/' X R.P.M.'/' 



The power-speed coefficient varies with the fifth root of the air 
density, provided horsepower, airspeed, and revolutions per min- 



PROPELLER SELECTION FOR ALTITUDE FLYING 173 

ute are unchanged. The nomogram in Fig. 50 being based on 
sea-level density or 

0.638 X M.P.H. 



Cs = 



-1/5 



-2/5 



H.P.'^' X R.P.M. 

2.136 X (0.002378)1/^ X M.P.H. 



H.P.'/' X R.P.M 



-2/5 



the power-speed coefficient at altitude (Csa) may be obtained by 
multiplying the power-speed coefficient for sea-level (Cso) by the 
fifth root of the relative densities (pa/po). 



= <^'" 



TABLE VII 





4 A* 




4 Z^'' 


Altitude in 


Altitude in 


feet 


v^ 


feet 


V.7 





1.0 


10 000 


0.941 


1000 


0.994 


11000 


.935 


2 000 


.988 


12 000 


.929 


3 000 


.982 


13 000 


.923 


4 000 


.977 


14 000 


.917 


5 000 


.971 


15 000 


.911 


6 000 


.965 


16 000 


.906 


7 000 


.959 


17 000 


.900 


8 000 


.953 


18 000 


.894 


9 000 


.947 


19 000 


.888 


10 000 


.941 


20 000 


.882 



Example. What should be the blade setting and diameter of the 
propeller which makes an airspeed of 150 miles an hour at an altitude 
of 10,000 ft.? The engine is supercharged to give 400 hp. at 1,900 
r.p.m. at the altitude of 10,000 ft. 
Solution. 

Csio,m = 0.941 X C^o 
= 0.941 X 1.40 
= 1.33 

From Fig. 52 for Cs = 1.33, blade angle = 19|°, V/ ND = 0.70, 

220 



D = 



31.7 X 0.7 



= 9.9 ft. 



174 PROPELLERS 

The propeller, 9.9 ft. in diameter, with a blade angle of 19j°, will 
absorb the 400 hp. at 10,000-ft. altitude. 

If the propeller that was selected previously for 150 miles per hour, 
400 hp. at 1,900 r.p.m. at sea-level, with 9.57-ft. diameter and blade 
angle of 19°, is a fixed-blade propeller, then at 10,000-ft. altitude, 
since V and N are the same, the V/{ND) will be the same as at the 
ground. The Cs which is a function of the V/(ND) and blade angle 
will be the same. The power absorbed by this propeller at 10,000-ft. 
altitude is 

H p = ^^-^ ^ ^10,000 X M.P.H.' 



R.P.M.' X C/ 
44.5 X 0.001756 X 150^ 



1900^ X 1.40^ 



= 295 



This horsepower is not sufficient to hold back the 400-hp. engine so 
that the engine would tend to race if not throttled. 

A propeller suitable for flying at altitude is not efficient near the 
ground. A good propeller for ground conditions is not suitable for 
altitude work. For this reason a variable-pitch propeller is highly 
desirable. 

Problems 

1. Find the blade angle for a 9-ft. propeller for flying at 15,000-ft. 
altitude, with an airspeed of 110 miles per hour, 210 hp. at 2,000 
r.p.m. at that altitude. 

2. Find the blade angle for a 9-ft. propeller for flying at 15,000-ft. 
altitude with an airspeed of 150 miles per hour, 300 hp. at 2,000 r.p.m. 
at that altitude. 

3. Find the blade angle for a 9-ft. propeller for flying at 20,000-ft. 
altitude with an airspeed of 150 miles per hour, 300 hp. at 2,000 
r.p.m. at that altitude. 

4. Find the blade angle for a 9-ft. propeller for flying at 15,000-ft. 
altitude with an airspeed of 150 miles per hour, 250 hp. at 1,800 r.p.m. 
at that altitude. 

5. Find the blade angle for a 9-ft. propeller for flying at 20,000-ft. 
altitude, with an airspeed of 150 miles per hour, 250 hp. at 1,800 
r.p.m. at that altitude. 

Geared Propellers. Propellers with high geometric pitch, i.e., 
large blade angle, have high efficiencies. A high geometric pitch 
postulates a large V/{ND). The horsepower of airplane engines 
is proportional to the speed of revolution, so that, to keep down the 



GEARED PROPELLERS 175 

size and weight per horsepower, the revolutions per second should 
be high. With a fixed-diameter propeller, for a given airspeed, 
increasing N has the effect of decreasing V/{ND). The increase 
in horsepower with increased revolutions per minute increases 

the denominator of tt p 1/5 \/ p p iv/r 2/5 ^^^ therefore decreases 

the value of Cg. In addition, high revolution speed with large- 
diameter propeller means high tip speeds with consequent loss of 
efficiency. High engine speed is desirable, but high propeller 
speed is undesirable. 

Geared propellers have the following disadvantages. There is 
added weight and loss of power in gear friction. Owing to high 
engine speed and low propeller speed, cooling of the engine becomes 
a problem. The necessary greater diameter involves greater 
landing-gear height. There is an added cost. 

The advantages of gearing-down propellers are as follows. 
There is increase in performance of the airplane due to greater 
efficiency of the propeller consequent on larger pitch ratio, on the 
absence of tip losses, and on the larger diameter involving a pro- 
portionately less interference effect from the fuselage and engine. 
A geared propeller is practically noiseless owing to the low tip 
speed. 

With gearing, the propeller must be mounted outboard of the 
bearings, and since the propeller shaft is very short, it must be 
lined up perfectly or the wear on the gears will be excessive. The 
gear teeth must be cut accurately, and the gears must be well 
lubricated at all times. 

The effect of gearing is best shown by an illustrative problem. 
In a previous illustration, it was worked out for a plane making 
150 miles per hour, with an engine of 400 hp. at 1,900 r.p.m., that 
the efficiency was 80.5 per cent. Considering that gearing will 
give greater efficiency and therefore greater horsepower let it be 
assumed that the airspeed will be increased to 160 miles per hour. 

Example. A cabin monoplane has a speed of 150 miles with a 400- 
hp. engine at 1,900 r.p.m. If the propeller is geared 2 to 1, what will 
be Cs, D, and efficiency? 

Solution. 

_ 0.638 X 150 

= 1.86 



176 PROPELLERS 

From Figs. 51 and 52, for C. = 1.86, V/{NB) = 1.08, blade angle = 
27°, 7? = 0.86: 



15.8 X 1.08 
= 12.9 ft. 



Summarizing for the geared and direct-driven propellers gives this 
comparison: 

Gear ratio Direct 2:1 

R.p.m. of propeller 1,900 950 

Cs 1.41 1.86 

V/{ND) 0.76 1.08 

Diameter, feet 9.1 12.9 

Blade angle 21° 27° 

Efficiency 0.805 0.86 

Tip speed, feet per second 920 650 

Body Interference. The presence of the engine and fuselage 
immediately in the rear of the propeller modifies the flow of air 
through the propeller disk. The air coming through the central 
part of the disk is diverted outward, causing a change in the di- 
rection of stream flow back of the propeller. 

The net effect of this interference is to increase the Cs slightly 
for all values of V/{ND). It also causes the peak value of 
efficiency to occur at a slightly higher V/{ND) than it would with- 
out the interference. 

Propeller Performance. The brake horsepower of an airplane 
engine varies almost directly with the speed of rotation. The 
amount of power which a propeller is capable of absorbing depends 
on its speed-power coefficient and its V/(ND). The full-throttle 
engine speed is dependent on the torque resistance offered by the 
propeller. 

The thrust horsepower of an engine-propeller unit is the brake 
horsepower of the engine multiplied by the efficiency of the pro- 
peller. It is the thrust horsepower that produces and maintains 
the velocity of the airplane. 

The speed of the engine may be reduced either directly by throt- 
tling or by reducing the airspeed by pulling back on the control 
stick. At a reduced airspeed, the numerator of the expression 
V/{ND) becomes less. With smaller V/(ND), the power coeffi- 
cient of the propeller becomes greater (see Fig. 49). For a two- 
bladed propeller Cs = V/{ND V^c), if the power coefficient 



PROPELLER PERFORMANCE 177 

Pc is increased, the fifth root of the reciprocal of the power coeffi- 
cient will be less. With a decreased airspeed, the power coefficient 
will decrease at a faster rate than the airspeed. The power re- 
quired to rotate the propeller being 

0.106 



H.P. = . , 

R.P.M.' 



/ M.P.H. V 
V Cs ) 



and Cs decreasing faster than the miles per hour, the horsepower 
absorbed would increase if the revolutions per minute remained 
the same. Actually putting more load on the engine will slow 
it down. 

Propeller performance is the effectiveness with which the pro- 
peller changes engine power into thrust power at all airspeeds. 
Propeller performance is either stated in the form of a table or as 
a curve of thrust horsepower at various airspeeds.. 

It usually is assumed that brake horsepower varies directly as 
revolutions per minute, or, letting B.H.P.o be rated horsepower 
and R.P.M.o be rated speed, for unsupercharged engines, 

B.H.F. ^ R.P.M. 
B.H.P.o KP.M.o 

Airspeeds are chosen for 10-mile-per-hour intervals throughout 
the range of airspeed it is expected that the airplane will fly. 
Since at airspeeds less than design airspeed, the engine will run 
slower than design revolutions per minute, to find the engine speed 
at each airspeed it is necessary to interpolate. 

At each airspeed at least three engine speeds are assumed in the 
neighborhood of which the engine speed might be expected to be. 
For each of these engine speeds a value of Cs is found. These 
calculations are for a fixed-pitch propeller so that the blade angle 
does not change. If the Cs corresponding to each engine speed is 
known, the V/{ND) may be found from Fig. 52 for each engine 
speed. With V and D known, N may be found giving a second 
value of engine speed for each engine speed originally assumed. 
By plotting the originally assumed engine speeds against Cs and 
the calculated engine speeds against Cs and locating the inter- 
section of these two plots, the actual revolutions per minute and 
Cs are found for this airspeed. An illustrative example follows. 

Example. A cabin monoplane has an engine rated 400 hp. at 1,900 
r.p.m. The propeller has a blade angle of 21° and the diameter is 9.1 



178 PROPELLERS 

ft. The design airspeed is 150 miles per hour. What are the revo 
lutions per minute and thrust horsepower at 120 miles per hour? 
Solution. For 1,850 r.p.m.: 

HP = 400 X ^-^ 

= 389.5 

0.638 X 120 



Cs = 



1,8502/5 X 389.5^/5 
= 1.146 

From Fig. 52, V/{ND) = 0.63 for 21° angle and Cs = 1.146. 

120 X 88 



N 



9.1 X 0.63 
1,842 r.p.m. 



For 1,800 r.p.m. 



H.P. = 400 X ''''" 



Cs = 



1,900 
= 379 

0.638 X 120 



1,8002/5 X 3791/5 
= 1.165 

From Fig. 52, V/{ND) = 0.64 for 21° angle and C, = 1.165. 

120 X 88 
^ ~ 9.1 X 0.64 
= 1813 

For 1,700 r.p.m.: 

400 X 1,700 
^' ' ~ 1,900 
= 358 

_ 0.638 X 120 
^' ~ 1,7002/5 X 358^/5 
= 1.205 

From Fig. 52, V/{ND) = 0.66 for 21° angle and Cs = 1.205. 

120 X 88 
^ 9.1 X 0.66 
= 1,758 r.p.m. 

In Fig. 53 are plotted the assumed revolutions per minute against the 
corresponding Cs values and a line is drawn through these points. 
The calculated revolutions per minute are also plotted against Cs and 
a line is drawn. The intersection gives 1,830 r.p.m. for an airspeed of 
120 miles per hour, and corresponding to these conditions Cs is 1.153. 



PROPELLER PERFORMANCE 



179 



From Fig. 51, the efficiency for this Cs and a blade angle of 21 
76 per cent. 

B.H.P. = 400 X 



IS 



1,830 



1,900 

= 385 
Thrust H.P. = B.H.P. X v 

= 385 X 0.76 

= 293 hp. 
The foregoing method when worked out for each 10-mile-per- 
hour interval is a series of calculations of some magnitude. A less 
exact method, often employed, 
makes use of the assumption 
that for most propellers the 
percentage of maximum effi- 
ciency is the same at the same 
percentage of design V/ (ND). 
The curve in Fig. 55a shows 
this relation for metal propel- 
lers. If wooden propellers are 
used the curve has a some- 
what different slope. The 



1850 



1800 



1750 



1700 




1.14 1.15 1 16 1.17 1 18 1.19 1.20 
Cs 

Fig. 53. Interpolation for illustrative 
example, 
ordinates of this curve are not 
efficiencies but percentages of the maximum efficiency; likewise 



yu 




































^ 
















80 


















/• 


^ 






















/ 






















/ 


/ 






















60 


/ 
























/ 
























en 



























1.0 1.1 1.2 1.3 1.4 



Fig. 54. 



.5 .6 .7 .8 .9 

Design ^ 

Maximum efficiency versus design V /{ND). 



the abscissas are not values of V/(ND) but percentages of design 
V/{ND). The use of this graph will be illustrated in performance 
calculations later in this book. 



180 



PROPELLERS 



In preliminary calculations of performance where meticulous 
accuracy is not required, an approximate idea of the maximum 
efficiency of a propeller may be obtained by use of the graph in 
Fig. 54. This gives the maximum efficiency for various values of 
design V/{ND). 



1 80 

LU 

1 70 

-So 










^ 


^^^ 


^^ 


\ 










/ 








\ 


\ 






/' 












\ 




/ 














\ 


/ 





































40 50 60 70 80 90 100 110 120 130 
Percent Design ^ 

Fig. 55a. Variation of efficiency with V/(ND). 



§ 1.00 

0} 

J^-95 
1°" .90 



.85 









— ' 


--— 


^^ 


















R.P 


M. at 


consta 


it airsp 


eed — 


>-^ 






























\ 



8000 12000 16000 

Altitude in Feet 



4000 

Fig. 556. Variation of R.P.M. with altitude 



20000 24000 



Construction. The first airplane propellers were made of wood. 
This material offers the simplest form of construction. Many 
different kinds of wood have been tried. Walnut, oak, and birch 
are considered the best, birch usually being thought superior to 
the others. The propeller is not made from a single block of wood, 
but for greater strength is made of several layers firmly glued 
together under pressure. These laminations extend across from 
one blade to the other, being continuous through the hub. The 
tips are covered with thin sheet brass, this protective material 
extending inward along the leading edge for some distance. 

Wood and the glue uniting the laminations are both affected 
by heat and dampness. Also the blades have to be thicker if made 
of wood than if made of stronger material; being thick they can- 



CONSTRUCTION 181 

not be made an efficient shape in cross-section. Wooden propellers 
find their chief use in small, inexpensive airplanes. 

The first advance in propeller material was the use of Micarta. 
Micarta is stronger than wood; it is made of layers of canvas 
strongly impregnated with a Bakelite composition, united in one 
piece and formed under heavy pressure in a mold. Each blade is 
made separately, the inner end being cylindrical with a shoulder 
or projecting rim on the extreme butt. The inner blade ends are 
firmly held in metal sockets attached to the metal hub. Micarta 
is not affected by moderate heat or by dampness and is some- 
what lighter in weight than a wooden propeller. 

One of the first forms of metal propellers was made by shaping 
a flat piece of aluminum sheet and twisting it near the hub to the 
desired pitches while cold, then heat-treating to relieve the 
stresses. The hub section was reinforced by clamping a round 
block of wood on each side. This type of propeller was a vast 
improvement over the wooden one. If a plane equipped with a 
wooden propeller nosed over in landing, the propeller invariably 
shattered. With a thin metal propeller, the tip might be bent 
at right angles in nosing-over without breaking. An airplane has 
even been flown with a blade in this shape, but the blade can be 
roughly hammered flat to be later sent to the manufacturer for 
straightening. 

Later one-piece duralumin propellers were made by forging. 
Magnesium alloy propellers are now being employed to some ex- 
tent. Hollow steel propellers that are seamless and have no welds 
have been made. They are fabricated by drawing a seamless 
steel tube through rollers to taper it. The small end is then spun 
closed, and the other end is peined out to form a shoulder which 
can be gripped in the hub. After machining to obtain proper 
wall thickness, the tapered tube is squeezed into final shape in iron 
dies. Since the forming is done while cold, the blades are heat- 
treated to relieve stresses. 

When a propeller blade whirling at high speed strikes a drop of 
rain, the rain drop acts as if it were a solid bullet. A wooden 
propeller flown through a rainstorm will be badly pitted. Metal 
propellers withstand the action of rain much better. Aluminum 
parts are adversely affected by salt air or salt spray unless they 
are protected by a special treatment. 

The Department of Commerce requires that any new type of 



182 



PROPELLERS 



propeller must pass a whirl test. This consists of 10 hours of 
whirling at an overload speed as specified below: 

Nt = revolutions per minute during test 
Nt = CaNa Na = r.p.m. in level flight at which propeller 
will absorb rated horsepower 

Ng = r.p.m. on ground at which propeller will 
]^j, = CgNg absorb rated horsepower 

Ca and Cq = constants from following table 





Tested on 
Electric Meter 


Tested on In- 
ternal Combustion 
Engine 




Ca 


Cg 


Ca 


Cg 


Non-metallic (fixed pitch) 

Metal (fixed pitch) 


1.05 
1.15 
1.15 


1.20 
1.30 
1.30 


1.00 
1.10 
1.10 


1.20 
1.25 


Variable pitch (any material) . 


1.25 



A further requirement of the Department of Commerce is that 
propellers should be so designed that they will limit the speed of 
the engine at full throttle to 105 per cent of the oflacial rated 
engine speed. 

BIBLIOGRAPHY 

N.A.C.A. Reports. 

Weick, Aircraft Propeller Design. 



CHAPTER X 
AIRPLANE PERFORMANCE 

The performance of an airplane is its ability to do certain things. 
Its functioning in respect to some characteristics is difficult to 
measure quantitatively, and the stability, maneuverability, and 
ease of control can be described in only a general or qualitative 
manner. Other features of its behavior in the air which can be 
specified quite exactly are termed the performance. The items 
usually listed as performance characteristics are maximum and 
minimum speed at sea-level and at various altitudes, absolute 
ceiling, service ceiling, time to climb to various altitudes, best 
climbing speeds, angle of glide, radius of glide, range, and en- 
durance. 

The present state of our knowledge permits the prediction, with 
a high degree of accuracy, of the performance of an airplane while 
in its design stage. This is especially true if wind-tunnel measure- 
ments have been made of the drag of the fuselage and landing- 
gear assembly. 

The stalhng speed of an airplane can be found if only the wing 
loading and the maximum lift coefficient of the wing section are 
known. All other performance characteristics of an airplane 
require that data on the engine and further aerodynamic data on 
the airplane be known. In studying performance, there are two 
primary conditions: first, the power required to move the air- 
plane through the air; secondly, the power that the engine can 
furnish while the airplane is executing that movement. 

Horsepower Required at Sea-Level. The total horsepower 
required to move the airplane forward through the air is the sum 
of the horsepowers required to move the various parts of the air- 
plane through the air. The parts may be grouped in various 
ways. 

If a graph is available giving the characteristics of the airfoil 
section data for the aspect ratio of the actual wing, it is probably 
simpler to find the horsepower needed for the wing and that for 

183 



184 AIRPLANE PERFORMANCE 

the parasite. The sum of these two is the total horsepower re- 
quired. 

It is not common, however, to have at hand characteristic 
curves for the proper aspect ratio. Curves for airfoils of infinite 
aspect ratio, which usually are obtainable, serve in calculating the 
profile drag of the wing. This may be joined with the parasite 
drag of the struts, landing gear, and wheels, etc., which like the 
profile wing drag does not vary with the angle of attack. The 
horsepower required to overcome this combined drag varies only 
as the cube of the airspeed. As the parasite drag of the fuselage 
and tail surfaces is presumed to change with angle of attack as 
well as with the cube of the velocity, the horsepower needed for 
this is computed separately. The horsepower to overcome the 
induced wing drag is calculated by itself. 

Since the increased drag of the fuselage and landing gear with 
angle of attack is manifest only at high angles, i.e., low airspeeds, 
where ample engine horsepower is available, little error is pro- 
duced in the final result if these drags are assumed to change not 
with angle of attack but only with velocity. Making this assump- 
tion permits all the parasite resistance to be grouped and simplifies 
the computation. This approximation causes very small error 
at high speeds but introduces inaccuracy at low speed. Each 
method is exemplified in an example. 

Example 1. (First method.) Find the horsepower required for a 
monoplane weighing 2,000 lb. and having a Clark Y rectangular wing 
36 ft. by 6 ft. The parasite drag has an equivalent flat plate area 
of 3.8 sq. ft. Result to be given as plot of horsepower versus velocity. 

Solution. From Fig. 17, read off and, as shown in table below, 
tabulate Cl and Cd for various angles of attack from near the zero 
lift angle to beyond the burble point. 

The velocity is found by using the formula 



v.. 



W 



, ^.00256 >S 

VCl 



/ w/s 

For any particular airplane, l/ n 0025fi ^^ constant throughout the 



series of calculations. 



HORSEPOWER REQUIRED AT SEA-LEVEL 185 

For this airplane: 

^^^. . ,. 2,000 

Wing loading = -oTa" 



= 9.26 lb. per sq. ft. 
V 0.00256 V 0.( 



9.26 



.00256 
= 60.2 

Horsepower required for the wing is determined for each angle of 
attack by the formula 

HP . ^X^ 

^ Cp X 0.00256 X ^ X 7^ 
375 

For any given problem, the term in parentheses is constant; for this 
problem 

0.00256 S 0.00256 X 216 
375 ~ 375 

= 0.00147 
and • H.P.wing = 0.00147 CdV^ 

For flat plates 

„^ 0.00327 X « X T^ 
^•^- = 375 

For this airplane 

0.00327 X 3.8 X V^ 
H.P.par. - 375 

= 0.0000331 F3 

A variation of the above method is to make use of the L/ D curve of 
Fig. 17. Since in level flight lift is equal to weight: 

^ Weight 



J 



186 



AIRPLANE PERFORMANCE 
TABLE VIII 



1 


2 


3 


4 


5 


6 


7 


a. 


Cl 


Cd 


y 


H.P.req.wing 


H-Prreq.par. 


H.P.req.total 


-4 
-3 


0.07 
0.14 


0.010 
.010 


228 
161 


174 
61 


392 
138 


566 
199 


_2 

-1 


0.215 
0.285 


.012 
.014 


130 
112 


38 
29 


73 

46.7 


111 

76 



4 


0.36 
0.645 


.017 
.033 


100 
74.9 


26.5 
20.3 


33.6 
13.9 


60 
34 


8 
12 


0.93 
1.19 


.060 
.095 


62.3 
55.2 


21.4 
23.6 


8.0 
5.5 


29 
29 


16 
18 


1.435 
1.545 


.139 
.164 


50.2 
48.1 


25.6 
29.5 


4.2 
3.7 


30 
31 


19 
20 


1.560 
1.540 


.180 
.206 


48.1 
48.5 


29.5 
34.5 


3.7 
3.7 


33 

38 



Columns 2 and 3 obtained from Fig. 17. 

Columns 5, 6, and 7 plotted against column 4 in Fig. 56. 

Velocity is found as above. Horsepower for the wing is 



H.P. 



wing - 375 



375 LjD 



For this airplane 



HP . _2£00 J^ 



= 5.33 X 



LID 



Horsepower required for the parasite is found as above. 



HORSEPOWER REQUIRED AT SEA-LEVEL 
TABLE IX 



187 



1 


2 


3 


4 


5 


a 


Cl 


L/D 


V 


H- "'req.wing 


-4 
-3 


0.07 
0.14 


7.0 
14.0 


228 
161 


174 
61 


-2 

-1 


0.215 
0.285 


17.9 
20.4 


130 
112 


38 
29 




4 


0.36 
0.645 


21.2 
19.5 


100 
74.9 


26.5 
20.3 


8 
12 


0.93 
1.19 


15.5 
12.5 


62.3 
55.2 


21.4 
23.6 


16 

18 


1.435 
1.545 


10.3 

8.7 


50.2 

48.1 


25.6 
29.5 


19 
20 


1.560 
1.540 


8.6 
7.5 


48.1 
48.5 


29.5 
34.5 



Example 2. (Second method.) Find the horsepower required for 
a monoplane weighing 2,000 lb. and having a rectangular wing 43.2 
ft. by 5 ft. The parasite assumed to vary with angle of attack, con- 
sisting of fuselage and tail surface, has an equivalent flat plate area 
of 1.6 sq. ft. The parasite not varying with angle of attack, namely, 
struts, landing gear, and wheels, etc., has an equivalent flat plate area 
of 2.2 sq. ft. 

Solution. 



Aspect ratio 


43.2 
~ 5 




= 8.64 


CDi 


~T X 8.64 




= 0.0368 Ci? 



From Fig. 17 or Fig. 25, Czmax. = 1.56 



ain. - y Q 



2.000 



,00256 X 1.56 X 216 
= 48.1 miles per hour 



188 



AIRPLANE PERFORMANCE 



Equivalent flat plate area of parasite drag varying with angle of 
attack, Ae = 1.6 sq. ft. 

., , . , , „ 1.6 X 1.28 
At high speed, Cdpi = 2I6 

= 0.00948 



240 
220 
200 
180 
160 

|,40 

a. 
1 120 

X 

100 
80 
60 
40 
20 



40 60 80 100 120 140 160 180 200 220 240 
Miles per Hour 

Fig. 56. Horsepower required for various airspeeds. (Example 1.) 

Equivalent flat plate area of parasite drag constant with angle of 
attack, Ae = 2.2 sq. ft. 

„ 2.2 X 1.28 

^^^^ = 216 
= 0.0130 

The velocity being assumed, the corresponding value of Cl can be 
found by 

W 
^^ ""0.00256 X SV^ 
2,000 



_l_l 


tt i- 


tt J^ 


4-t 2 


tt t 


Jj^ t 


44^ €- 


tt J- 




ii 7 


z/^7 




_,^ 



For this problem, Cl = ^^^^^ ^ 2I6 X V^ 

3,640 

- 72 



HORSEPOWER REQUIRED AT SEA-LEVEL 



189 



Since the coefficient of parasite drag is expressed as a function of wing 
area S, it may be added to the coefficients of induced and profile drag 
to give an all-inclusive drag coefficient, Cd- 

Cp X 0.00256 XSV^ 
H.P.totai = 375 

For this problem 

0.00256 X S XCdV^ 



H.P.totai = 



375 
= 0.00147 Cz? 7^ 











TABLE X 








1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


V 


V/Vr, 


Cl 


Cdi 


Cdo 


F 


Cdpi 


Cdp 


Cd 


H.P.req. 


48 
50 


1.00 
1.04 


1.58 
1.45 


0.0920 
.0773 


0.0446 
.0310 


4.0 
2.5 


0.0380 
.0237 


0.0410 
.0367 


0.1876 
.1450 


30.5 
26.7 


60 
70 


1.25 
1.46 


1.01 
0.74 


.0376 
.0202 


.0140 
.0120 


1.3 
1.1 


.0123 
.0104 


.0253 
.0234 


.0769 
.0556 


24.4 
28.0 


80 
90 


1.67 

1.87 


0.57 
0.45 


.0118 
.0074 


.0105 
.0100 


1.0 
1.0 


.0095 
.0095 


.0225 
.0225 


.0448 
.0399 


33.8 

42.8 


100 
110 


2.08 
2.29 


0.36 
0.30 


.0048 
.0033 


.0100 
.0099 


1.0 
1.0 


.0095 
.0095 


.0225 
.0225 


.0373 
.0357 


54.8 
69.9 


120 
130 


2.50 
2.71 


0.25 
0.215 


.0023 
.0017 


.0098 
.0099 


1.0 
1.0 


.0095 
.0095 


.0225 
.0225 


.0346 
.0341 


87.4 
110.5 


140 


2.92 


0.186 


.0013 


.0099 


1.0 


.0095 


.0225 


.0337 


135.8 



Explanation of Table 

Column 2 obtained by dividing items in column 1 by Tnim. (= 48). 
Column 3 obtained by dividing 3,640 by items in column 1 squared. 
Column 4 obtained by multiplying items in column 3 squared by 0.0368. 
Column 5 obtained from Fig. 38. 
Column 6 obtained from Fig. 42. 

Column 7 obtained by multiplying CdpiH.S. (= 0.0095) by items in column 6. 
Column 8 obtained by adding Cdp2 (= 0.0130) to items in column 7. 
Column 9 obtained by adding items in columns 4, 5, and 8. 
Column 10 obtained by multiplying 0.00147 by items in column 9 by 
items in column 1 cubed. 



190 



AIRPLANE PERFORMANCE 



In Fig. 57 velocity (column 1) is plotted against horsepower re- 
quired (column 10). 




70 80 90 100 no 

Velocity (Miles per hour) 



140 



Fig. 57. Horsepower required for various airspeeds. (Example 2.) 



Example 3. Find the horsepower required by a biplane weighing 
4,225 lb. The upper span is 38 ft., the lower span is 35 ft., the gap is 
5.35 ft. The area of upper wing is 214 sq. ft., the area of lower wing is 
150 sq. ft. Both wings are Clark Y airfoils. At high speed, the 
parasite has an equivalent flat plate area of 9.4 sq. ft. of which 3.2 
sq. ft. varies with angle of attack. 

Solution, 









M 


62 

35 
38 

= 0.92 








Gap 


5.35 




Mean span 


~36.5 










= 0.146 


From Fig. 


40, <T = 0.56 






r 


150 
~214 
= 0.737 








K" 






(0.92)2 


(1 + 0.737)2 


~ (0.92)2 
= 1.185 


+ 2 X 0.56 X 0.737 X 0.92 + (0.737)^ 


E.M.A.R. 


~ S 










1.185 X 38^ 








~ 150 + 214 








= 4.7 









Cm 



Fmir 



HORSEPOWER REQUIRED AT SEA-LEVEL 
Cl' 



191 



TT X 4.7 
= 0.0677 (7z,2 






TF 



00256 Czmax.AS 



4225 



0.00256 X 1.56 X 364 
= 54 miles per hour 
Equivalent flat plate area of parasite drag varying with angle of at- 
tack, Ae = 3.2 sq. ft. 



380 
360 
340 








































/ 




















/ 




















/ 


300 

280 

^ 260 


















1 








































1 


















1 






^220 

?00 
















/ 


















1 


1 






180 














/ 








160 












/ 


^ 








140 
120 












/ 


















/ 












100 
80 


V 






/ 














V 




^ 

















50 60 70 80 90 100 110 120 130 140 150 

Velocity (Miles per hour) 

Fig. 58. Horsepower required for various airspeeds. (Example 3.) 

3.2 X 1.28 



At high speed, Ci>^i(h.s.) = 



364 
= 0.0113 



192 



AIRPLANE PERFORMANCE 



Equivalent flat plate area of parasite drag constant with angle of at- 
tack, Ae = 6.2 sq. ft. 

6.2 X 1.28 

= 0.0218 
If velocity is assumed, corresponding value of Cl can be found by 

W 



For this problem 



Cl = 



Cl 



0.00256 X S X V^ 

4,225 

0.00256 X 364 X V^ 

4,550 
72 



For this problem 



xi.r .req total — 



Cd X 0.00256 X ^ X T^ 



•req lulju QV^ 

_ Cp X 0.00256 X 364 X y^ 

375 
= 0.00248 Cd 73 











TABLE 


XI 








1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


V 


V/Vrr^. 


Cl 


Cm 


Cdo 


F 


Cdp^ 


Cdp 


Cd 


H.P.req. 


54 


1.00 


1.56 


0.1640 


0.0446 


4.00 


0.0452 


0.0670 


0.2756 


104 


60 


1.10 


1.26 


.1070 


.0200 


1.72 


.0194 


.0412 


.1682 


90 


70 


1.30 


0.93 


.0588 


.0135 


1.20 


.0136 


.0354 


.1077 


91 


80 


1.47 


0.71 


.0342 


.0120 


1.08 


.0122 


.0340 


.0802 


102 


90 


1.67 


0.56 


.0212 


.0105 


1.00 


.0113 


.0331 


.0648 


117 


100 


1.85 


0.45 


.0140 


.0100 


1.00 


.0113 


.0331 


.0571 


142 


110 


2.04 


0.37 


.0095 


.0100 


1.00 


.0113 


.0331 


.0526 


174 


120 


2.22 


0.32 


.0070 


.0099 


1.00 


.0113 


.0331 


.0500 


214 


130 


2.40 


0.27 


.0049 


.0098 


1.00 


.0113 


.0331 


.0478 


260 


140 


2.59 


0.232 


.0036 


.0099 


1.00 


.0113 


.0331 


.0466 


317 


150 


2.78 


0.202 


.0028 


.0099 


1.00 


.0113 


.0331 


.0458 


384 



In Fig. 58, velocity (column 1) is plotted against horsepower required 
(column 10). 



HORSEPOWER AVAILABLE 193 

Horsepower Available. The horsepower available for moving 
the airplane is the brake horsepower of the engine multiplied by the 
efficiency of the propeller. The propeller is selected for one air- 
speed, either maximum, cruising, or sometimes best climbing 
speed. At any other speed, the propeller will be less efficient; 
and because proportionately more torque load is put on the 
engine, so that the engine runs slower, less brake horsepower is 
developed. This means that, in an airplane flying at less than 
design airspeed, the engine will develop less than its rated horse- 
power even though the throttle is full open. 

The first requirement is to select the most suitable propeller. 
Propeller choice depends on rated horsepower, speed of the engine, 
and design airspeed. Engine speed and power will have to be 
furnished in order to make the performance curves, but the maxi- 
mum airspeed is not known exactly until the horsepower-available 
curve is drawn. 

Two procedures may be followed. An arbitrary top speed may 
be assumed based solely on similarity to existing airplanes of the 
airplane for which performance curves are being drawn. On the 
basis of this tentative maximum speed, a propeller is chosen, 
enabling the horsepower available to be drawn. The completion 
of this curve will give a maximum airspeed, which is the top speed 
which can be obtained using the chosen propeller. If the top 
speed obtained agrees closely with the assumed airspeed, the pro- 
peller is the correct one to use. If the assumed airspeed is different 
from that indicated by the curves, the propeller selected on the 
basis of the assumed airspeed is not suited for the engine-airplane 
combination, and higher airspeeds will be obtained by substituting 
another propeller. The second method of procedure differs only 
in that arbitrarily a design efficiency is assumed for the propeller. 
Usually 82 per cent efficiency is selected as this is about the highest 
efficiency obtainable from propellers. Applying this factor to the 
rated brake horsepower gives a tentative thrust horsepower. 
The total horsepower required curve is then consulted to find the 
velocity at which the horsepower required corresponds to this 
thrust horsepower. 

The velocity obtained from the horsepower required curve is 
assumed to be the design airspeed. With this trial airspeed the 
propeller is selected as described above. 

With design airspeed and propeller tentatively chosen, the effect 



194 



AIRPLANE PERFORMANCE 



of changes in airspeed on engine speed is next found. An exact 
method was described in the chapter on propellers, but it is 
usually considered sufficiently accurate to make use of the curve 
of Fig. 44a. 

If the engine manufacturer furnishes curves or other information 
as to the drop in horsepower with decrease in engine speed, they 

"D p T\/r 

should be utilized. Otherwise use should be made of -p' '^ = 

ix.r.Ni.2 

■D XT p 

■pVr'p' , which is representative of most engines. This gives the 
ij.xl.ir.2 

brake horsepower. 

At each airspeed there will be a corresponding engine speed, and 
these determine the V/{ND) of the propeller. At each value of 
V/{ND), the propeller efficiency is found, either from Fig. 48 or 
more quickly but less exactly by Fig. 55a. 

The thrust horsepower or horsepower available is the product of 
the propeller efficiency and the brake horsepower. The method is 
illustrated by the following example. 

Example. For the airplane described in example 1 in this chapter, 
an engine rated at 150 hp. at 1,800 r.p.m. is installed. Find the data 
for horsepower available versus velocity curve. 

140 




70 80 90 100 no 

Velocity (Miles per hour) 



Fig. 59 

Solution. 



Horsepower required and available curves. 

Assume t] for propeller as 82 per cent. 
First approximation of H.P.avaii. = 150 X 0.82 = 123. 
From Fig. 56, 123 hp. is required at 135 miles per hour. 

0.638 X M.P.H. 



(Example 1.) 



Cs = 



(H.P.)i/5 X (R.P.M.)2/5 
0.638 X 135 



150^/^ X 1,8002/5 
= 1.57 



HORSEPOWER AVAILABLE 



195 



From Fig. 52, a propeller is chosen with a blade angle of 23° with 
V/iND) of 0.87 (r; = 82.5 per cent). 



D = 



N X 0.87 

_ 135 X 88 
~ 1,800 X 0.87 

= 7.59 ft. 
TABLE XII 



1 


2 


3 


. 4 


5 


6 


7 


8 


9 


10 


V 
m.p.h. 


% 

design 

V 


% 
design 

A^ 


N 
r.p.m. 


Brake 
H.P. 


V/ND 


% 
design 

V/ND 


% 
max. 

V 


V 


H.P. 

avail. 


50 


37.1 


87.0 


1565 


131 


0.371 


42.6 


52.0 


0.430 


56.3 


60 


44.5 


87.0 


1565 


131 


.445 


51.1 


61.5 


.507 


66.5 


70 


51.9 


87.2 


1570 


131 


.517 


59.5 


70.5 


.580 


76.0 


80 


59.3 


88.0 


1583 


132 


.587 


67.5 


79.0 


.651 


86.0 


90 


66.7 


89.0 


1600 


134 


.653 


75.0 


86.0 


.710 


95.2 


100 


74.0 


90.8 


1633 


136 


.710 


81.6 


91.8 


.757 


103.0 


no 


81.5 


93.0 


1673 


139 


.759 


87.2 


95.8 


.790 


109.9 


120 


88.9 


97.5 


1751 


146 


.795 


91.5 


98.0 


.808 


118.0 


130 


96.2 


99.0 


1781 


149 


.848 


97.5 


99.5 


.820 


122.1 


135 


100.0 


100.0 


1800 


150 


.870 


100.0 


100.0 


.825 


123.8 



Explanation of Table 

Column 2 is the items of column 1 divided by design speed (135 miles 
per hour). 

Column 3 is obtained from Fig. 44a. (Note: A more accurate method is 
described in the chapter on propellers.) 

Column 4 is the items in column 3 multiplied by design r.p.m. (1,800). 

Column 5 is the items in column 3 multiplied by design horsepower (150) 
(assuming power varies as engine speed). 

Column 6 is the items in column 1 multipHed by 88 and divided by the 
corresponding items in column 4 and the diameter (7.59). 

Column 7 is the items in column 6 divided by the design V/iND) 
(0.870). 

Column 8 is obtained from Fig. 55a. 

Column 9 is the items in column 8 multiplied by design efficiency (0.825). 

Column 10 is the items in column 5 multiplied by corresponding items 
in column 9. 



196 AIRPLANE PERFORMANCE 

Items in column 10 are plotted against corresponding items in column 1 
on the same graph as the horsepower-required curve; see Fig. 59. It will 
be noted that the horsepower-required curve and horsepower-available 
curve intersect at 135 miles per hour. This is, then, the correct maximum 
airspeed. If the assumed maximum airspeed, on which the propeller was 
picked, differs materially from the maximum airspeed found by the inter- 
section of the two curves a second set of calculations must be gone through. 
It must be borne in mind that if an incorrect assumption was made of maxi- 
mum airspeed, whether too high or too low, the propeller selected was un- 
suited to this airplane engine combination. The maximum airspeed ob- 
tained by the intersection of the curves is low, since the horsepower-avail- 
able curve is based on using an improper, poor-performing propeller. With 
a more nearly suitable propeller the horsepower-available curve will have 
larger ordinates and the intersection with the horsepower available will 
show a higher top speed. Allowance should be made for this in a second 
assumption of maximum airspeed to select the more suitable propeller. 

Maximum Speed at Sea-Level. The maximum speed is found 
by the intersection of the curves of horsepower available and 
horsepower required. At velocities less than the maximum, the 
ordinate of the horsepower-required curve is less than the ordinate 
of the horsepower-available curve, which means that more horse- 
power is available than is needed for level flight so that the engine 
may be partially throttled. At velocities greater than the maxi- 
mum, the ordinate of the horsepower-required curve is larger than 
the ordinate of the horsepower-available curve, meaning that more 
horsepower is needed at this speed than is available. The air- 
plane cannot fly at this speed, and if it is placed at the angle of 
attack corresponding to this velocity, the thrust horsepower will 
not be sufficient to equal the drag at this speed. Since this 
velocity cannot be attained in level flight the lift will not equal the 
weight, and the airplane will lose altitude. 

Velocities for Minimum Fuel Consumption at Sea-Level. It 

can be proven quite easily that the minimum horsepower is 

required for a wing when flying at the angle of attack where 

Cd/Cl^^^ is the minimum. For the entire airplane, minimum 

1.28 a 
^D n o — 

horsepower is required for level flight when rrjji ^^ 

minimum. 

Examining the horsepower-required curve, it will be seen that, 
as speed decreases from the maximum, the horsepower decreases 
to a minimum value, but if the speed decreases further the horse- 



VELOCITIES FOR MINIMUM FUEL CONSUMPTION 197 

power increases. The minimum horsepower is found by drawing 
a horizontal Hne tangent to the horsepower-required curve (see 
hne AB, Fig. 59). The velocity at the point of tangency, which 
is the velocity of minimum power, corresponds to the angle of 

c. + ^« 

attack where 77-^7^ is the minimum. 

For the Clark Y wing with aspect ratio of 6, it was shown that 

the maximum value of Cl^^^/Cd (or minimum value of CdICj}^'^') 

will occur at 3° angle of attack, corresponding to 79.8 miles per 

(73/2 
hour. If a curve is plotted having as ordinates, ; 

Cd + —^ 

abscissas, angle of attack; the curve will resemble the curve for 

the wing alone, but the maximum ordinate will be at a higher 

angle of attack. For the airplane whose performance is shown in 

(73/2 
Fig. 59, the maximum value of ., ^^ is at 11° angle of 

Cd + ^ 

attack, corresponding to an airspeed of 56.6 miles per hour. 

This velocity is the airspeed for minimum horsepower required, 
which means the least fuel consumption in gallons per hour. If a 
plane is to make an endurance record for time in the air, the fuel 
supply should be used as economically as possible. The airplane 
should then be flown so that the wing is at the angle of attack 

for maximum ^-^r^ — . In the example of the monoplane 

weighing 2,000 lb. with Clark Y wing, the pilot should regulate his 
throttle and stick so that the airplane maintains level flight at 
56.6 miles per hour airspeed as long as the airplane's total weight 
is 2,000 lb. As time elapses, fuel consumed will make the airplane 
lighter. Always, during the endurance flight, the pilot will want 

1.28 a 

(11° for this Clark Y wing and ratio of parasite to wing area); 
but as the airplane's weight decreases, the velocity necessary for 
level flight at this angle of attack will decrease. With less weight, 
the horsepower required will be less, so that the pilot can throttle 



to keep his wing at angle of attack of maximum 



198 AIRPLANE PERFORMANCE 

his engine more. With less throttle-opening, the fuel consump- 
tion (in gallons per hour) will be diminished, so the hourly decrease 
in weight will be less. The pilot should have a precalculated 
schedule of best airspeeds to fly at various hours during his flight. 

Flying at the angle of maximum ., „„ of the wing will 

mean that the least gasoline is consumed per hour, so that an 
airplane can stay the longest time in the air on a given supply of 
fuel. Flying either faster or slower will involve a greater fuel 
consumption per hour. The foregoing is true whether in still 
air or in wind. Airspeed, not ground speed, is important. 

If the object of the flight is not greatest duration, but to cover 
the greatest distance on a given gasoline supply or, what is more 
usual, to fly to a destination burning the least possible amount of 
fuel, the airplane should be flown faster. To cover a given dis- 
tance, the gasoline consumption depends not only on the con- 
sumption per hour but on the total hours required for the flight. 
It is therefore advantageous to fly at greater speed, more spe- 
cifically to fly so the wing is at the angle of attack of maximum 

L/Dtotal- 

Drawing a line tangent to the horsepower-required curve from 
the origin (0 miles per hour, hp. ; see line CD, Fig. 59) determines 
the speed and horsepower for minimum fuel consumption for a 
given distance. The point of tangency will be the point where the 
ratio of horsepower to velocity is least. Neglecting any variation 
in fuselage drag, the horsepower required is 

H.P.req. = 3^5 (^^ >< 0.00256 SV + ^^ X 0.00256 SV'\ 

To cover a given distance at a velocity V requires ^^ — hours. 

The gasoline consumption depends on horsepower-hours ; therefore 

XT p 

Total consumption = Constant X Distance X -^ — ^^^ 



V 



1^^- = ^(CdX 0.00256 SV'-{- ^-^ X 0.00256 SV'\ 



W 

But V = 



Cl X 0.00256 S 



VELOCITIES FOR MINIMUM FUEL CONSUMPTION 199 
Then 



This has a minimum value when 



375 V Cl J 



Cl 



IS mmimum. 



With -^ — constant, for any airfoil section, there is only one 
o 



^'Hm 



angle of attack where p^ has a minimum value. This 

angle of attack will be slightly greater than the angle of attack for 
minimum Cd/Cl, i.e., maximum L/Dj but as the parasite area is 
made smaller with respect to wing area, the difference between 

^«+(^) .. c 

angle of minimum t; and angle of mmimum 7p 

becomes less. 

With the Clark Y wing alone; the angle of maximum L/D is 1°; 
with a Clark Y monoplane and the parasite of the illustrative 
example, the angle of maximum L/Aotai is 5°, the corresponding 
velocity being 71.2 miles per hour. This checks with the point 
of tangency of line CD in Fig. 59. 

The pilot maintains this speed as long as the weight remains the 
2,000 lb. The angle for best L/ Aotai is constant for the airplane, 
since the wing area and parasite area are fixed. The speed for 
any given angle of attack varies as the square root of the wing 
loading. As fuel is consumed, and the airplane is lightened, the 
airspeed for least fuel consumption per distance decreases as the 
square root of the weights. 

The foregoing discussion of most economical airspeed apphes to 
still air. If a head wind holds back a plane, it will take a longer 
time to reach its destination; a tail wind means a quicker trip. 
The formula for total fuel consumption must therefore be modified 
so that the horsepower-hours contains the expression Fg, the 
ground speed, for divisor instead of F, the airspeed. 



200 AIRPLANE PERFORMANCE 

Horsepower-hours = Distance X \/^^^' 



375 Vo 



i'^'-mU 



W 



_ Distance \ "^ ' S /Cl\ ClX 0.00256 >Sf 
375 Vg 

The horsepower-hours or total fuel consumption will be a minimum 

, CzJtotal ... . Cl'^'Vg . 

when ^ ,/„Tr IS minimum or when ., ^^ is a maximum. 

Cl^^Wg ri I 1-28 a 

A chart or table may be made out for a given airplane for vari- 
ous strength winds and for various angles of the wind to the air- 
plane's heading. The ground speed, Vg, is found by trigonometry. 

W is windspeed in miles 

^ ^ ^ IS angle of wind from 

dead ahead 

The desirable changes in airspeed on account of wind are sur- 
prisingly small. For the illustrative problem with a 10-mile head 
wind the optimum angle of attack is 5°, for a 10-mile tail wind the 
best angle is still 5°, so that the airspeed should be maintained as 
in still air at 71.2 miles per hour. For a 20-mile head wind the 
best angle becomes 4°, meaning an airspeed of 75 miles per hour; 
for a 20-mile tail wind it is best to fly at 6° with an airspeed of 68.1 
miles per hour. With a 30-mile head wind an airspeed of 85.4 
miles per hour is best; with a 30-mile tail wind an airspeed of 68 
miles per hour is best. 

Problems 

1. For the monoplane of Clark Y wing, 36 ft. by 6 ft., with 3.8 sq. 
ft. equivalent flat plate area, what should be the airspeed for mini- 
mum horsepower when fuel has been burned so that total weight is 
1,800 lb.? 

2. For the monoplane of problem 1, weight 1,800 lb., what should 
be airspeed in still air to cover a given distance with least possible 
expenditure of fuel? 



RANGE 201 

3. For the monoplane of problem 1, weight 1,800 lb., what should 
be the airspeed for minimum gas consumption, if there is a 35-mile- 
per-hour head wind? 

4. What should be airspeed for problem 3 if it is a 35-mile-per-hour 
tail wind? 

5. What should be airspeed for problem 3 if the weight were only 
1,6001b.? 

Range. The calculation of the range or distance that can be 
flown non-stop is not simple. The airplane leaves the ground with 
a given weight, but, as fuel is consumed, the weight decreases. 
To achieve the greatest possible distance, the airplane should be 
flown constantly at the angle of maximum L/Dtotai but, as the 
weight decreases, both velocity and horsepower required for this 
angle decrease, for constant angle of attack V varies as W^^^ and 
H.P. varies as W^^^. With less power being used, the rate of fuel 
consumption is less, that is, the rate of change of weight is less. 
Also, with less power required, the throttle is closed more and more, 
which decreases the revolutions per minute of the engine. Since 
the revolutions per minute decrease at a different rate from the 
airspeed, the V/iND), and consequently the efficiency, of the 
propeller changes. A close approximation to the range may be 
obtained by assuming an average fuel consumption and an average 
propeller efficiency. 
Let W = total weight of airplane at take-off. 

Wt = total weight of airplane at the end of t seconds after 
take-off. 
Q — total weight of fuel at take-off. 
Qt = weight of fuel consumed at the end of t seconds. 
C = fuel consumption (lb. per hp. per hr.). 

(B.H.P.)« = brake horsepower available at time t 
In a short time interval dtj the weight of fuel consumed is 

ex (B.H.P.), 

'"^' poo — '^ 

The horsepower available (>; X B.H.P.) must equal the horsepower 
required at any time t. 

,(B.H.P.).=g 

W,V 



(S 



550 



202 AIRPLANE PERFORMANCE 

Then 

(B.H.P.). = ^^^ 



ns 



550 



Substituting in the initial equation gives 

dQt = 7-TT at 

3,600 X 550 77 f^j 



1,980,000 77 ^^^ 

7Ji = 1,980,000^ X^x|^' 

Let 22 be the total number of miles traveled (i.e., the range). 
Then 

dR = Vdt 

= 1,980,000 ^ X ^ X ^^ 

= 1,980,000^ X^X^^ 

Integrating and bearing in mind that at take-off the quantity- 
consumed is zero and at the end of the flight Qt = Q 



B = 1,980,000 ^X^/^^ 

v L W 

= 1,980,000 ^ X -^ log, jY _Q ^^^* 



= 1,980,000 X 2.303 ^ X ^ X logio r^ _ n ^eet 

1,980,000 X 2.303 v ^, L ^, W ., 

= 5;280 C^D^ ^^S- TT^Q ^^^^^ 

n L W 

= 863.5 7^ X -^ X logio jY _ n ^i^^^ 

The above equation is known as Breguet's formula for maximum 
range. 



RANGE 203 

The range as calculated by Breguet's method is the ultimate that 
can be achieved. The speed is much less than the maximum. 

Amount of fuel X high speed 



At high speed: R^^ = 



Rate of fuel consumption at high speed 



Since it is rarely that an airplane will be flown at the low speed 
corresponding to the absolutely maximum range, the range is cus- 
tomarily considered to be between the maximum and the range at 
high speed. Common practice is to regard the maximum range as 

RmsiX. = 0.7 5 (Rb — Rvm) + Rvm 

where Rb = range by Breguet formula. 

While specific fuel consumption is slightly higher when engine is 
run at rated power than when the engine is throttled, only a 
shght error is involved by using average fuel consumption. 

Example. An airplane weighs 4,000 lb. and takes off with 80 gallons 
of fuel. It has a Clark Y wing of 216 sq. ft. area and has 3.8 sq. ft. 
equivalent flat plate area of parasite. It has a 180-hp. engine. The 
maximum efficiency of the propeller is 78'per cent. Assume fuel con- 
sumption to be 0.55 lb. per B.H.P. per hr. and maximum velocity to 
be 135 miles per hour. Find the range. 

Solution. 

Weight of fuel (Q) = 80 X 6 
= 480 lb. 

«^ = 863.5 I X 5;;;^ X log ^^r^Q 

= SC3.5gxn.9 1o.g-J 
= 808 miles 



■tivm — 



480 X 135 



180 X 0.55 
= 655 miles 

R = 0.7 5 {Rb - RvJ + Rvn. 
= 0.75(808 - 655) + 655 
= 770 miles 

Problems 

1. A Martin Clipper weighs 51,000 lb. and takes off with 4,000 
gallons of fuel. Its top speed is 180 miles per hour. It has four twin- 



204 AIRPLANE PERFORMANCE 

row Wasp engines of 800 hp. each. Assume propellers to have a 
maximum efficiency of 82 per cent, the maximum L/D to be 14.2. 
The fuel consumption is to be 0.48 lb. per hp. per hour. What is the 
range? 

2. An Aeronca with a total weight of 1,000 lb. takes off with 8 
gallons of fuel. Its top speed is 93 miles per hour. The engine is 
rated at 36 hp. Assume propeller to have a maximum efficiency of 75 
per cent, the maximum L/D to be 9.5, and the fuel consumption to be 
0.56 lb. per hp. per hour. What is the range? 

3. A Curtiss Robin with total weight of 4,200 lb. takes off with 
320 gallons of fuel. Its top speed is 118 miles per hour and its Wright 
Whirlwind engine is rated at 165 hp. Assume propeller to have a 
maximum efficiency of 75 per cent, the maximum L/D of the airplane 
to be 9.0, and fuel consumption to be 12.5 gallons per hour. What is 
the range? 

4. A Bellanca Pacemaker with total weight of 5,600 lb. takes off 
with 200 gallons of fuel. Its maximum speed is 160 miles per hour, 
and it has a 420-hp. engine. Assume propeller to have 80 per cent 
maximum efficiency and maximum L/D to be 11.5. The fuel con- 
sumption is 0.55 lb. per hp. per hour. What is the range? 

6. A Lockheed Electra, weighing 10,000 lb., takes off with 194 
gallons of fuel. Its maximum speed is 210 miles per hour and it has 
two Wasp Junior engines rated at 400 hp. with fuel consumption 0.48 
lb. per hp. per hour. Assume maximum propeller efficiency to be 
82 per cent and maximum LID of airplane to be 14.5. What is the 
range? 

Rate of Climb at Sea-Level. Examining the curves of horse- 
power available and required versus velocity as shown in Fig. 59, 
it will be seen that for any velocity less than maximum there is 
greater horsepower available than is required for level flight. If 
level flight is desired, the engine may be throttled so that it is 
furnishing only the horsepower required for level flight. If it is 
desired to increase the altitude, the engine may be opened to full 
throttle when it will give the thrust horsepower shown on the 
horsepower-available curve. The extra power, which is the differ- 
ence in ordinates for horsepower available and horsepower re- 
quired at any velocity, represents the power available to do the 
work of raising the airplane. 

Power is ability to do work in unit time. By definition, a 
horsepower is the power to do 33,000 ft. -lb. of work in 1 min. 
Then the rate of climb, in feet per minute, can be found by the 



RATE OF CLIMB AT SEA-LEVEL 205 

formula 

R.C. (ft.permin.) = (H.P-a.an. - HJ'.^J X 33,000 

^ Excess H.P. X 33,000 
W 

Another conception of the conditions in climb is shown in Fig. 
60. The airplane in the sketch is ascending along a path which 
makes an angle 6 with the horizontal. The direction opposite to 
the flight path is the direction of the relative wind. The wing is 
at angle of attack a to the relative wind. Lift is perpendicular 
and total drag is parallel to 
the relative wind. For con- 
venience, there is assumed to 
be zero angle of incidence so 
that the wing chord is parallel 
to the propeller axis. For 
equilibrium, the forces paral- 
lel to the flight path must 
balance as must the forces per- Fig. 60. Forces in climb, 

pendicular to the flight path. 

Neglecting the small force on the tail which is ordinarily considered 
negligible, 

T = thrust 
T cosa = D + W sine D = total drag 

Tsina+L = W Gosd ' W = weight 

L = lift 

Since a is small, cos a may be considered unity and T sin a may 
be neglected; the two equations become 

T = D -{-W sind 
L = W cosd 
T - D 




W 



sm 6 = 

and, since d is small, 

L = W approximately 
But F, = F sin ^ 

Vc = vertical velocity (miles per hour) 

V = velocity along flight path (miles per hour) 



206 



AIRPLANE PERFORMANCE 



Then 



R.C. (ft. per min.) 



W 



^xvx5f» 



„. n re . Useful power output 

Since propeller einciency rj = ^ , , ; 7- 

^ ^ *^ Total power input 

Thrust X Velocity 



Brake horsepower X 375 
TV = 'nX B.H.P. X 375 

TV 

^ = .XB.H.P.. 

== ii-t^-avail. 

Referring to the equation for rate'of climb 



R.C. (ft. per min.) = 



D 



X FX88 



W 
(TV - DV)X 



X375 



375 X W 

(H.P.avaU. — H.P. 



req. 



X 



33,000 
W 



1000 



E 800 



600 



400 



At every velocity less than maximum there will be a different 
rate of climb. For the illustrative problem, this rate of climb is 
shown in Fig. 61. It will be seen that, for this particular com- 
bination of propeller, engine, 
and airplane, the maximum 
rate of climb is achieved, at 
sea-level, of 790 ft. per min. 
when the airspeed is 83 miles 
per hour. 

CUmb may be made at any 
speed below maximum, but if 
the pilot desires the greatest 
rate of climb at sea-level, he 
should open the throttle wide 
and pull back on his stick till 
the airspeed indicator reads 83 miles per hour, the best climbing 
speed. 

This rate of climb is for the propeller which was chosen for 
maximum efficiency at airspeed of 135 miles per hour. At other 
airspeeds the efficiency is much less so that the horsepower avail- 
able is low. 



J 



J 

o 

"S 200 



- 






/ 


/* 






- 






/- 




\ 




- 




// 






\ 


k 


- 


/ 










\ 


/ 












\ 



20 



40 60 80 100 
Velocity, Miles per hour 



120 140 



Fig. 61. Rate of climb at sea-level. 



ANGLE OF CLIMB AT SEA-LEVEL 



207 



The propeller instead of being chosen for 135-mile-per-hour 
airspeed might have been chosen for best performance at a lesser 
airspeed. In Fig. 62 are shown the available-horsepower curves 
for the 150-hp. engine equipped with propeller chosen for design 
airspeed of 110 miles per hour and 83 miles per hour. It will be 
noted that while the rate of climb is improved, as shown by the 
greater excess horsepower, the maximum airspeed is much de- 
creased. 



140 




40 50 60 70 80 90 100 110 120 130 140 

Velocity (Miles per hour) 

Fig. 62. Horsepower available with propellers of differing pitch. 



Angle of Climb at Sea-Level. The rate of climb is another name 
for vertical velocity. In Fig. 61 it is shown for the illustrative 
example that the greatest rate of climb, 790 ft. per min. or 8.95 
miles per hour, is obtained when the airspeed is 83 miles per hour. 
The sine of the angle of climb is then 8.95 -^ 83, and the angle of 
climb is 6.2°. By flying slower, the actual rate of climb will be 
less, but the ratio of climb to forward speed may be greater. By 
drawing a line from the origin tangent to the rate of climb curve, 
line OA in Fig. 61, the point of tangency will give the maximum 
angle of climb. 

For the airplane in the illustrative example, the point of tan- 
gency is 66-miles-per-hour airspeed, and 700-ft.-per-min. rate 
of climb. This gives an angle of chmb of 6.9°. 

In military maneuvers, rate of climb is very important. While 
" dog-fighting," the airplane that can gain altitude on its opponent 
will have an advantage even if a wide circle is necessary to attain 
this altitude. 

In taking-off, an airplane goes straight, and it is the angle of 
cHmb, not the rate of climb, that is the decisive factor in clearing 



208 AIRPLANE PERFORMANCE 

obstacles. In taking-off from a small field, it is better to fly at the 
best angle of climb rather than at the best rate of climb. For the 
illustrative example, with wide-open throttle, the stick should be 
pulled back until the airspeed indicator reads not 83 but 66 miles 
per hour. 

On a calm day the angle of climb is as deduced from the 
graph. A pilot always takes-off into the wind, so if there is any 
wind the actual angle of climb will be greater than that given. 

By diving the airplane, greater speed can be attained than in 
level flight. If, from a dive, the stick is suddenly pulled back, the 
airplane, owing to momentum, retains momentarily some of this 
excess speed, so that a cHmb can be made for a few seconds at a 
greater rate and a greater angle than from level flight. This 
maneuver is called a zoom. 

Take-off Distance. There are three phases to the take-off of 
an airplane. First, there is a very short period during which the 
tail is being raised from the ground. Second, there is a compara- 
tively long period during which the airplane is gaining speed with 
the tail up so that the wing is at a low angle of attack. In the final 
phase the stick is pulled back to put the wing at a big angle of 
attack so that the plane is lifted into the air. If the field is 
sufficiently long, the airplane may be kept in the second position 
until it has attained enough speed to lift itself while flying at the 
low angle of attack, thus ehminating the third phase described 
above. 

In the following description, it is assumed that the take-off is at 
sea-level with no wind. If the airplane has retractable landing- 
gear, it is naturally not folded up till the plane is in the air so the 
drag on the ground is with the gear extended. 

At the end of the run, when the wing is suddenly put at a high 
angle of attack, the wing drag will become very big, tending to 
slow down the airplane. Because of this and also because excess 
horsepower will be needed for climbing, the stick is not pulled 
back until a speed has been attained which is somewhat greater 
than the minimum or stalling speed. It is customary in calcula- 
tions to use a speed which corresponds to 90 per cent of maxi- 
mum Cl- 

In starting, the throttle is opened wide, the brakes released, and 
the tail raised. The plane starts to roll along the ground with the 



TAKE-OFF DISTANCE 209 

wing at the angle of attack of maximum L/Dtotzi- At first, the 
entire weight of the airplane rests on the ground, the lifting force 
on the wing being negligible. As the speed increases, more and 
more of the airplane weight is carried by the wing, so that the 
plane rests less heavily on the ground. Since the weight on the 
ground is less, the friction of the wheels with the ground becomes 
less and less engine power is needed to overcome this friction. 
When the airplane is just about ready to leave the ground, practi- 
cally the entire weight is being borne by the wing. 

The coefficient of friction varies with the surface of the runway, 
depending on whether it is the smooth deck of a ship or a soft 
gravel field. The accepted values of the coefficient of friction (/x) 
are as follows : 

Concrete runway or wooden deck fi = 0.02 

Hard turf, level field /x = 0.04 

Average field, short grass /x = 0.05 

Average field, long grass fx = 0.10 

Soft ground M = 0.10 - 0.30 

With a fixed pitch propeller, the thrust is quite different when 
the airplane is stationary from the thrust when the airplane is 
moving at design speed. Use is made of a graph such as Fig. 63a, 
which is for the average two-bladed metal propeller, to find the 
static thrust coefficient (Kto) and the static thrust is found by 
the formula, for static thrust (To) 

^ Kto X B.H.P . 
' R.P.M. X D 

The initial accelerating force (Fo) is the static thrust of the 
propeller minus the friction of the wheels with the ground. 

Fo = To- Wn 

As the speed increases, the ground friction becomes less but the 
air resistance becomes greater. Just prior to take-off, ground 
friction has become negligible, the sole retarding force being the 
air drag. The propeller thrust has decreased from its static value 
to a new value which depends on the airspeed. The take-off is at 
the angle of attack where Cl is 90 per cent of maximum Cl- The 



210 AIRPLANE PERFORMANCE 

velocity at take-off (Vt) will be 



.20 



w 



0.90Ci:n.ax.^^ 



Vo.91 



.90 

1.054 Fxnin. 



IIU.UUU 


S^ 
























100,000 


\ 


s 
























\ 






















90,000 






\ 
























\ 




















80,000 






) 


v 
























\ 


















70,000 








\ 


[ 
























\ 
















60,000 










\ 


























\ 














50,000 












\ 


























\ 












40,000 














\ 


\ 
























\ 


\, 








30,000 




















\ 


s^ 
























■^ 




20.000 



























.40 



.60 



.80 



.00 



1.20 



1.40 



— for Maximum Efficiency 



Fig. 63a. Variation of static thrust coefficient with design V/ND. 

The thrust at take-off (Tt) is found by first obtaining the thrust 
horsepower at take-off speed by means of Fig. 636 and then mak- 



TAKE-OFF DISTANCE 



211 



ing use of the relation 

550 X T.H.P. 



or 



Tt = 



Tt = 



375 X T.H.P. 



V in feet per second 

V in miles per hour 



1.00 
















--^^ 


.90 
.80 

,.70 

a. 

" .60 
.50 










/ 


^ 












/ 














/ 












/ 


/ 














.40 


/ 
















.30 



















.20 .30 



.40 .50 60 



.70 



.90 1.00 



Fig. 63&. Variation of thrust horsepower with velocity. 

The accelerating force at take-off {Ft) is the propeller thrust 
minus the air drag of the entire airplane. 



\ J^ /max. 



It is assumed that the accelerating force varies linearly with the 
airspeed. Although this assumption is not exactly correct, the 
error involved is so slight as to be negUgible. Then, if the force is 
i^o when V is zero, and the force is Ft when the velocity is Vt, the 
force F at airspeed V will be 



F ^F,- 



(Fo - FtW 

Vt 



M-[^']^.) 



212 



AIRPLANE PERFORMANCE 



Letting 



K = 



Ft 



f^f.{i-kI) 



The acceleration produced by this varying force {¥) is equal at 
any instant to the force divided by the mass {W /g) of the airplane. 



Acceleration = -rr = —r- X -77 = ?77 
dt ds dt W 



but 



dt 
VdV 
ds 
VdV 

vr 

This is in the form 

xdx r xdx 



V 



= w^X~^y) 



' J a 



= ^Fods 



= r2 [« + ^^ 



a loge (a + hx)]. 



a -\- bx ' J a -{- bx b 
Therefore integrating the above gives 

and between the limits of V and zero 

-i-.[-¥-(t'-('-^)] 

When V = Vt 



Vt in feet per second 
s in feet 



Example, Find the take-off run for a monoplane, weighing 2,000 
lb., having a Clark Y wing 216 sq. ft. in area, and 3.8 sq. ft. parasite, 
powered with an engine rated at 150 hp. at 1800 r.p.m. and a propeller 
7.59 ft. diameter and blade angle of 23°. Runway is smooth concrete. 
Neglect preliminary distance before tail is up. 



TAKE-OFF DISTANCE 213 



Solution. 

From Fig. 63a Kto = 46,000 

Kto X B.H.P. 



Static thrust, To = 



r.p.m. X D 
46,000 X 150 



~ 1,800 X 7.59 
= 505.0 lb. 
Initial accelerating force, 

Fo = To - ixW 

= 505 - 2,000 X 0.02 
= 465 lb. 
Thrust horsepower at design conditions, 
T.H.P. = B.H.P. X Tj 

= 150 X 0.825 (see p. 195) 

= 123.8 
Take-off speed, Vt = 1.054 X Fmin. (see p. 186) 

= 1.054 X 48.1 
= 50.7 miles per hour 
Ratio, take-off speed to maximum speed, 

Vt _ 507 
Fmax. " 135 
= 0.376 
T.H.P.f/T.H.P.design = 0.580 (from Fig. 57) 

T.H.P.i = 123.8 X 0.580 
= 71.8 H.P. 
375 X T.H.P.; 



Thrust at take-off, Tt 



Vt 
375 X 71.8 



50.7 
= 531 lb. 
Accelerating force at take-off, 



\ ^ /max. 

- 531-^£^ 
~ ^'^^ 11.9 

= 363 



214 AIRPLANE PERFORMANCE 

Fo — Ft 



K = 



465 - 363 



465 
= 0.219 
(50.7X1.47)2 2,000r 1 / . 1 , .. ^«..xM 

= 436 ft. 

Problems 

1. Find the run in still air, after tail is up, to take-off an airplane 
weighing 4,000 lb., the maximum L/Dtotai being 12.5, the stalling speed 
being 55 miles per hour, the design top speed being 150 miles per hour; 
the engine giving 200 hp. at 1,850 r.p.m. and the 9-ft. diameter pro- 
peller having an efficiency of 80 per cent under design conditions. 
M = 0.03. 

2. Find the distance in still air, after tail is up, to take-off an air- 
plane weighing 7,000 lb., the maximum L/Dtotai being 10.8, Fmin. being 
50 miles per hour, Fmax. being 140 miles per hour; the engine giving 
300 hp. at 1,800 r.p.m., the 8-ft. propeller having 81 per cent efficiency 
under design conditions. Field is hard turf, ju estimated as 0.04. 

3. Find the distance in still air, after tail is up, to take-off a DH 
weighing 4,300 lb., the maximum L/Dtotai being 6.4, Fmin. being 61 miles 
per hour, Ymax. being 124 miles per hour; the engine giving 425 hp. at 
1,750 r.p.m., the 9. 8-ft. propeller having 75 per cent efficiency under 
design conditions. Field is hard turf, fx estimated as 0.04. 

4. Find the distance in still air, after tail is up, to take-off a Sperry 
Messenger, weighing 1,076 lb., the maximum L/Dtotai being 5.44, Fmin. 
being 27 miles per hour, "Fmax. being 90 miles per hour; the engine 
being rated at 63 hp. at 1,830 r.p.m., the 6.5-ft. propeller' having 
75 per cent efficiency under design conditions. It is estimated /x 
is 0.04. 

5. Find the distance in still air, after the tail is up, for a Grumman 
Fighter weighing 4,800 lb., to take-off if the maximum L/Dtotaiis 9.3, 
T^min. is 65 miles per hour, Fmax. is 216 miles per hour; the engine giving 
770 hp. at 2,100 r.p.m.; the 9.5-ft. propeller having 82 per cent 
efficiency under design conditions. Take-off is from the deck of a 
naval vessel, which is stationary. 

Gliding Angle at Sea-Level. There is no marked difference in 
the meaning of glide and dive. A very steep glide is called a dive. 
Ordinarily a glide is considered to be with power partly or com- 
pletely off. A dive may be either with power on or off, although 



GLIDING ANGLE AT SEA-LEVEL 



215 




Fig. 64. Forces in glide. 



usually the word dive by itself means the maneuver with the power 
off, and with power on the term power-dive is used. 

Gliding angle is the angle below the horizontal of the flight path 
when the airplane descends with the engine either completely 
throttled or ** turning-over " so 
slowly that there is no appre- 
ciable thrust . The forces acting 
on the airplane, neglecting a 
small force on the tail, are 
weight, lift, and total drag. 
These forces are shown in Fig. 
64. The component of weight 
parallel to the flight path is the 
force that pulls the airplane 
along the flight path. When 
the airplane attains a steady speed along the flight path, the 
weight multiplied by the sine of the glide angle just equals the 
total drag. 

The lift is equal and opposite to the component of the weight per- 
pendicular to the flight path. If the lift is less than W cos 6j 
the airplane will " squash " or settle, and the flight path will be 
steeper. If the lift is greater than W cos 6, the airplane will not 
descend on that angle of glide but will tend to level out, so that the 
glide angle will be flatter. For any airplane there will be an angle 
of glide associated with each angle of attack determined by the 
following equations. 

Wcosd = L 

= ClX 0.00256 SV 
W cos d 



and 



Then 



V 


Cl X 0.00256 S 


Wsind 


= Aotal 






= CdX 0.00256 SV + 1.28 a X 0.00256 V^ 




= (Cd-\- 


■ ^'^^ "") X 0.00256 SV' 


tan^ 


Aotal 

L 






CD + 


1.28 a 

S 



Cl 



216 



AIRPLANE PERFORMANCE 



A table may be calculated as follows, which is for the illustrative 
example of a monoplane with Clark Y wing, 36 ft. by 6 ft., and 
parasite of 3.8 sq. ft. E.F.P.A. weighing 2,000 lb. 



TABLE XIII 





Cl 


Cd 


Cd+ ^ 


tan 6 


6 


cos 6 


Tf COS0 


72 




a 


0.00256 S 


V 


-4 


0.07 


0.010 


0.032 


0.456 


24.5 


0.910 


3 300 


47 100 


217 


-3 


0.14 


.010 


.032 


.229 


12.9 


.975 


3 540 


25 300 


159 


-2 


0.215 


.012 


.034 


.158 


9.0 


.988 


3 580 


16 650 


129 


-1 


0.285 


.014 


.036 


.126 


7.2 


.992 


3 600 


12 620 


112 





0.36 


.017 


.039 


.108 


6.2 


.994 


3 610 


10 000 


100 


1 


0.43 


.020 


.042 


.098 


5.6 


.995 


3 615 


8 400 


92 


2 


0.50 


.024 


.046 


.092 


5.3 


.996 


3 620 


7 240 


85 


3 


0.57 


.028 


.050 


.088 


5.0 


.996 


3 620 


6 350 


80 


4 


0.645 


.033 


.055 


.085 


4.9 


.996 


3 620 


5 610 


75 


6 


0.715 


.038 


.060 


.084 


4.8 


.996 


3 620 


5 060 


71 


6 


0.785 


.045 


.067 


.085 


4.9 


.996 


3 620 


4 610 


68 


7 


0.857 


.052 


.074 


.086 


4.9 


.996 


3 620 


4 230 


65 


8 


0.93 


.060 


.082 


.087 


5.0 


.996 


3 620 


3 890 


62 


12 


1.19 


.095 


.117 


.099 


5.6 


.995 


3 615 


3 040 


55 


16 


1.435 


.139 


.161 


.105 


6.0 


.994 


3 610 


2 520 


50 



In Fig. 65a are plotted angle of glide versus angle of attack and 
velocity versus angle of attack, for this example. It is to be 
noted that the angle of glide is big for negative angles of attack, 
that as angle of attack is increased the angle of glide decreases 
until at one angle of attack, in this case 5°, the angle of glide is a 
minimum of 4.8°. Further increase of angle of attack increases 
the angle of glide. 

In the illustrative example, with a dead engine, the pilot pushes 
forward on the stick. The airspeed will increase. When it 
reaches 71 miles an hour, the pilot will manipulate the stick to 
maintain that speed. The angle of attack will then be 5° and the 
angle of glide will be the flattest possible, namely, 4.8°. 

By maintaining the flattest possible glide, the greatest hori- 
zontal distance can be traveled in descending. The pilot can thus 
reach an emergency field and land, even though the field is at a 
considerable distance from the spot where his engine quit. The 
actual horizontal distance that may be achieved is a function of 



GLIDING ANGLE AT SEA-LEVEL 



217 



the altitude of the airplane when the engine quits. If h is the 
original altitude, 

Horizontal gliding distance = 7 a 

tan u 

The minimum ghding angle, as found from a graph similar to 
that of Fig. 65a, will give the flattest possible glide. It will be 
noted that this angle of glide is a function of the lift and drag 




2 4 6 8 10 12 14 

Angle of Attack 

Fig. 65a. Angles and velocities in glide. 

coefficients of the airfoil section and of the ratio of parasite area to 
wing area. It is independent of weight. That is, an airplane 
will have the same optimum ghding angle whether empty, partly 
loaded, or fully loaded. 

The velocity plotted in Fig. 65a is velocity along the flight path. 
This velocity varies as the square root of the weight. A fully 
loaded airplane will therefore glide faster than a partly loaded 
plane even though it glides at the same angle. The slowest descent 
along the flight path would be at a large angle of attack. The 
danger of getting out of control would mean that some angle less 
than that of maximum lift should be used. 



218 



AIRPLANE PERFORMANCE 



Vertical speed of descent is the sine of the angle of ghde multi- 
plied by the velocity along the flight path. 

F, = 7 sin 
For the illustrative example at the flattest angle of glide, 4.8°, 
the airspeed is 71 miles per hour. The rate of vertical descent is, 
then, 

71 X sin 4.8° = 71 X 0.084 

= 5.9 miles per hour 

At a greater angle of attack, the 
vertical descent may be slightly 
slower; for example, at a 16° angle 
of attack, while the angle of glide 
is 6°, the airspeed is only 50 milss 
per hour and the rate of vertical 
descent is 




50 X sin 6° 



50 X 0.104 

5.2 miles per hour 



Decreasing the angle of attack 
below that for minimum gliding 
angle will make the angle of glide 
increase and the airspeed increase. 
At the angle of zero lift, tangent 6 
will be infinite, that is, 6 will be 
90° and the airplane will be in a 
vertical dive. For this case, the 
weight pulls the airplane down- 
ward faster and faster, until the 
total drag equals the weight, when 
the airplane will not go down any 
faster. This speed is called the terminal velocity. 

= (Cd + i^) X 0.00256 ;S (Fterminal)^ 



Fig. 656. Polar diagram of 
velocities in dives. 



w 



w 



terminal 



Cd + 



1.28 a 



S 



)xo. 



00256 S 



For the illustrative example 



'terminal — \/ {(\ 



2.000 



(0.032) X 0.00256 X 216 
= 338 miles per hour 



GLIDE TESTS 219 

The data on glides and dives are frequently presented in the form 
of a polar diagram, as shown in Fig. 656. The radius is the velocity 
to scale of the airplane when gliding or diving in the direction from 
the horizontal of the radius. 

Problems 

1. Plot angle of attack versus angle of glide for an airplane with a 
Clark Y wing 36 ft. by 6 ft., having 2 sq. ft. equivalent flat plate 
area of parasite. 

2. If the airplane in problem 1 above weighs 2,500 lb., what is the 
airspeed for flattest gliding angle? 

3. What horizontal distance can be traveled if the airplane in 
problem 1 glides from 5,000 ft.? 

4. If the airplane in problem 1 weighs 3,000 lb., what is the airspeed 
for flattest ghding angle? 

5. What is the airspeed of the airplane in problem 4 when gliding 
down on a 45° path? 

6. What is the terminal velocity of the airplane in problem 4? 

7. Plot a polar diagram of angle of glide and airspeed for an air- 
plane weighing 2,500 lb. and having a Clark Y wing 36 ft. by 6 ft. 
and 5 sq. ft. equivalent flat plate area of parasite. 

Glide Tests. After an airplane has actually been constructed, 
quite often tests are made to determine the full-scale lift and total 
drag coefficients. These determinations are useful for reference 
in designing future airplanes. They are also of primary impor- 
tance if minor changes are being made in streamlining, engine- 
cowling, etc. 

The airplane is tested in flight and the airspeed in various glides 
noted. The angle of ghde cannot be measured directly; the rate 
of descent is found from a calibrated barograph, or by timing the 
readings on an altimeter. The airspeed is held constant in any 
one glide. The difficulty encountered in obtaining useful results 
from this sort of test is that of eliminating or making proper 
allowance for thrust. With the engine throttled, thrust may be 
positive, zero, or negative. With the engine stopped, the propel- 
ler produces additional drag ordinarily absent in flight. 

Example. An airplane weighs 4,650 lb. Its wing area is 460 sq. 
ft. The wing section is Clark Y, aspect ratio 6. With airspeed in- 
dicator constant at 118 miles per hour, the airplane glides from 
1,000-ft. to 500-ft. altitude in 26 sec. Neglect propeller thrust and 
drag. Find equivalent flat plate area of parasite. 



220 AIRPLAiqE PERFORMANCE 

Solution. 

Vv= 500 ft. in 26 sec. 
= 19.2 ft. per sec. 
= 13.1 miles per hour 

Angle of glide, 



d 


. 1 13.1 

- ^^^ 118 




= sin-i 0.1115 




= 6.4° 


Cl 


TFcos^ 
~ 0.00256^72 




4,650 X cos 6.4° 


1.28 a 

S 


0.00256 X 460 X (118)2 
= 0.282 

= Cl tan d 




= 0.282 X 0.112 




= 0.0316 



Cd + 



From Fig. 17, for Clark Y airfoil, aspect ratio 6, when Cl is 0.282, 
Cowing is 0.014 

Cd + i^ = 0.0316 

0.014 + —^ = 0.0316 

a = 6.25 sq. ft. 

Problems 

1. A monoplane weighs 5,000 lb. ; its wing area is 400 sq. ft., aspect 
ratio of 6, airfoil section U.S.A. 35A. The parasite has an equivalent 
flat plate area of 7 sq. ft. Plot a polar diagram of velocity at various 
gliding angles. 

2. An observation airplane weighs 4,225 lb., airfoil section Clark Y, 
total wing area 365 sq. ft. Equivalent flat plate area of parasite 8.4 
sq. ft. Find terminal velocity in a vertical dive. (Since dive is at 
zero lift, there is no induced drag correction.) 

3. A pursuit airplane weighs 2,932 lb.; airfoil section Clark Y; 
total wing area 264 sq. ft.; equivalent flat plate area of parasite 7.1 
sq. ft. Find terminal velocity in a vertical dive. 

4. A pursuit airplane weighs 2,548 lb.; airfoil section Clark Y, 
total wing area 247 sq. ft.; equivalent flat plate area of parasite 7.7 
sq. ft. Find terminal velocity in a vertical dive. 

5. A monoplane with Clark Y airfoil has an aspect ratio of 5.23; 
it weighs 3,500 lb.; its total wing area is 300 sq. ft. At an airspeed of 
125 miles per hour, it glides at an angle of 8°. What is equivalent 
flat plate area of parasite? 



EFFECT OF CHANGING WEIGHT 221 

Effect of Changing Weight. By changing weight, wing loading 
and power loading are changed. 



v^. 



W 



Cl X 0.00256 S 
375H.P.req. = (Cd X 0.00256 S + 1.28 X 0.00256 a)V^ 



= (0.00256 CdS + 1.28 X 0.00256 a) ( 



■pf \3/^ 



ClX 0.00256 iSy 

At any one angle of attack, V varies as the square root of the 
weight and required horsepower as the cube of the square root of 
the weight. 

V varies as W^^^ 
H.P.req. varics as W^^^ 

If the horsepower-required curve has been found for an airplane 
as in Figs. 56, 57, and 58, any point on that curve gives the velocity 
for level flight at some one angle of attack (the abscissa) and the 
horsepower needed for that same angle of attack (the ordinate). 
If weight is added to the airplane, all other dimensions remaining 
the same, for each point on the original curve there will be a point 
on a new curve with the relations that the abscissas are to each 
other as the square root of the relative weights, and the ordinates 
are to each other as the three-halves power of the relative weights. 

Putting the matter in another way, at the same velocity, for 
the same airplane, the Hft coefficients must vary directly as the 
weights. Therefore, if load is added to an airplane, at any given 
speed the airplane must fly at a higher angle of attack. A higher 
angle of attack always means a greater drag coefficient throughout 
the flying range. An airplane will require more horsepower when 
more heavily loaded, since horsepower required at constant speed 
depends on drag coefficient, wing and parasite area. This means 
that if horsepower curves are drawn for a lightly loaded plane and 
the same plane heavily loaded, at any value of velocity, the horse- 
power needed for the heavier plane will always be greater than for 
the lighter airplane, therefore the two curves wili never cross. 

In detail, if an airplane is loaded more heavily, the landing speed 
will be increased. As the maximum lift coefficient and wing area 
are the same in both instances, the landing speeds will vary as the 
square root of the weights. With same engine and propeller, the 
maximum speed will be decreased slightly. The maximum speed 



222 



AIRPLANE PERFORMANCE 



is the intersection of the horsepower-required curve and thrust- 
horsepower curve, and no exact relation can be expressed for 
change in top speed for change in weight. 

The excess horsepower at any speed will be decreased with in- 
creased weight. The rate of climb, being the excess power divided 
by the weight, will be decreased with increased weight. The 
speed of best climb will be increased as will the speed of maximum 
angle of climb. 

The flattest gliding angle will be always the same for any air- 
plane regardless of weight. The velocity at any angle of glide, 
including terminal velocity, will vary as the square root of the 
weight. 

Figure 66 shows the effect of adding weight in changing the 
horsepower required. With a total weight of 4,500 lb., the horse- 
power-required curve is tangent to the horsepower-available 



180 

160 

140 

|l20 

|100 

80 

60 




















/ 


/ 


t 










H.F 


. Required: 
W 45001b.- 
W 4000 1b.- 
W 3000 1b.- 
/^ 200OIb.- 




/ 


/ 


/ 


















/ 


















y 




^ 


/ 
















^ 


^ 


V 
















^ 


'^ 


-< 


// 


/ 












^ 


^. p. Av 


ailabU 


y 


> 


/ 














_ 




^^ 


y 
















40 
20 




1 — 




_ 























60 60 70 80 90 100 110 120 130 140 150 160 170 
Miles per Hour 

Fig. 66. Effect of weight on horsepower required. 



curve; this is the limiting weight; for this weight, the ceiling is 
at sea-level. It is to be noted that, as weight is increased, mini- 
mum speed is no longer determined by C^max. but by lower inter- 
section of horsepower curves. 

Example. An airplane weighs 3,000 lb. and has a landing speed 
of 50 miles per hour. What is landing speed with 500 lb. additional 
load? 



EFFECT OF CHANGE IN WING AREA 223 

Solution. 

W 



= 50 



v^ 



3,500 



3,000 
= 53 miles per hour 

Example. An airplane, weighing 3,000 lb., requires the least horse- 
power to fly level, 40 hp. when flying at 80 miles per hour. If 500 lb. 
load are added, what are velocity and power for minimum horsepower 
in level flight? 

Solution. At same. angle of attack 



7i ~ V ^1 

V = 8oy/; 



3,500 
3,000 



= 86.5 miles per hour 
H.P . /TFV/2 
H.P. 



= 50.4 hp. 

Problems 

1. An airplane weighs 2,485 lb. Its landing speed is 45 miles per 
hour. What is landing speed when 300 lb. extra load are added? 

2. An airplane weighs 4,500 lb. Its landing speed is 48 miles per 
hour. What is landing speed when 250 lb. of fuel have been burned? 

3. An airplane weighing 3,900 lb. uses minimum horsepower for 
level flight at an airspeed of 90 miles per hour. What is velocity for 
minimum horsepower with 600 lb. of load removed? 

4. The Barling Bomber, fully loaded, had a gross weight of 42,500 
lb. and a landing speed of 52 miles per hour. What is landing speed 
with 3,000 lb. less load? 

5. A basic training plane weighing 4,060 lb. has a landing speed of 
55 miles per hour. What is landing speed after 250 lb. of fuel have 
been burned? 

Effect of Change in Wing Area. With constant weight, de- 
creasing the wing area has the effect of increasing the wing loading. 
Airspeed at any angle of attack varies inversely as the square root 



224 



AIRPLANE PERFORMANCE 



of the wing area. Horsepower required at any angle of attack 
varies inversely as the square root of the area. 

1 



V varies as 



Vs 



H.P.req. varies as 



1 

Vs 



For a given velocity, an airplane with a smaller wing area will 
have to fly at a larger angle of attack. A larger angle of attack 
means a greater drag coefficient, but drag coefficient does not 
vary lineally with angle of attack. At high speeds (small angles 
of attack) the drag coefficient varies very little with angle of 
attack. At slow speeds, the slope of the drag coefl&cient curve 
versus angle of attack is very steep. At constant airspeed, wing 
drag varies as drag coefficient multiplied by wing area. At 
high speed, since lift coefficient varies Hneally with angle of attack, 
angle of attack must vary inversely with wing area. At angles 



200 
180 




















/ 








Horsepower Required; 






/ 


/ 


/ 


160 

140 

% 120 




216 sq.ft. Wing A 
150 sq.ft. Wing A 






v 










ea — 
ea- — 


IT? 


^ 


/ 




















/^ 


^ 




■/ 


|ioo 

80 




Hor 


>epowe 


r Avail 


able-. 


>^ 




/ 


/ 


^ 


7 










.^ 




^-'^ 


/ 


r 


/ 


/ 






An 




^ 


^ 






/ 






y 








40 




i 




...^ 


^ 


^ 
















^ c 




^^ 


^ 


















20 







40 50 60 70 80 90 100 110 120 130 140 150 160 
Velocity 

Fig. 67. Effect of wing area on horsepower required. 

near the angle of minimum drag, an airplane having the same 
weight as another airplane but less wing area will at the same air- 
speed have to fly at a larger angle of attack. This increased angle 
will not mean a proportionate increase in drag coefficient. The 
airplane with smaller wing area will have less drag and less horse- 
power required. 



EFFECT OF CHANGE IN WING AREA 225 

At lower speeds, the increase in drag coefficient with increased 
angle of attack is much greater. Near stalling speeds, drag 
coefficient varies approximately as the square of the angle of at- 
tack. An airplane having the same weight as another plane but 
a smaller wing will have a greater drag and greater horsepower 
required at high angles of attack. 

The horsepower curves will therefore cross as shown in Fig. 
67. This means that an airplane's top speed can be increased by 
clipping the wings, provided that there is ample engine power. 

Decreasing wing area increases landing speed. 

Decreasing wing area increases the minimum horsepower and 
the velocity for minimum horsepower. 

Decreasing wing area increases the minimum gliding angle, 
decreases the horizontal gliding distance, and increases the ter- 
minal dive velocity. 

All the above effects are assuming that the parasite resistance 
remains the same. 

Example. An airplane with 300 sq. ft. wing area has a landing 
speed of 40 miles per hour. If wing area is reduced to 250 sq. ft., 
what is the landing speed? 

Solution. 



7min. = 40y/25o 

= 43.8 ftiiles per hour 

Problems 

1. An airplane with 340 sq. ft. of wing area lands at 40 miles per 
hour. It is desired to reduce landing speed to 35 miles per hour; 
how much area should be added to the wing? 

2. A certain airplane with wing area of 400 sq. ft. is flying at angle 
of best L/D, when airspeed is 70 miles per hour. What airspeed 
corresponds to angle of best L/D when wings have been clipped to 
350 sq. ft.? 

3. A certain airplane with 450 sq. ft. of wing area flies with least 
horsepower of 40 hp. at an airspeed of 90 miles per hour. After wings 
have been clipped to 410 sq. ft., what is least horsepower and what 
is corresponding velocity? 



226 AIRPLANE PERFORMANCE 

4. An airplane with 450 sq. ft. of wing area lands at 50 miles per 
hour; what will be landing speed if wings are clipped to 400 sq. ft.? 

5. An airplane with 630 sq. ft. of wing area lands at 45 miles per 
hour; what will be landing speed if 50 sq. ft. are added to the wing 
area? 

Effect of Change in Engine. A more powerful engine means a 
larger and heavier engine. To consider the effect of increased 
power alone, it must be assumed that weight and parasite are 
unchanged, that is, any increase in engine weight is offset by a 
decrease in payload. 

A change in power-available curve makes no change in the 
power-required curve. Provided there is no added weight or 
added drag, a more powerful engine will not affect the landing 
speed, nor will it affect the speed of minimum horsepower required, 
minimum gliding angle, or terminal velocity. It will increase 
maximum speed, rate of climb at any speed, and angle of climb 
at any speed. 

When an airplane is at its top speed, it is flying at a low angle 
of attack. At small angles of attack the change in Cd with change 
of angle is quite small, and the change in parasite drag coefficient 
is negligible. When an airplane is flying level, the rate of climb 
is zero and thrust is equal to total drag. 

-n X B.H.P. X 550 = TV 
= DV 

V X B.H.P. X 550 

Ki X V X B.H.P. , ^ 550 
= („. + Iff). """''■ = ^2 

Since most propellers have approximately the same design 
efficiency, and assuming the drag coefficients as constant, the 
above expression may be reduced to 



73 = 



V'- 



„.;b,h.p. 



s 

The maximum airspeed may therefore be said to vary approxi- 



POWER LOADING AND WING LOADING 227 

mately as the cube root of the engine horsepower, provided that 
the change in horsepower is not excessive. 

Example. An airplane with a 200-hp. engine has a maximum speed 
of 120 miles per hour. If total weight is unchanged, what is maximum 
speed with a 250-hp. engine? 

Solution. 



V ~\ B.H.P.' 



= 129 miles per hour 

Problems 

i. An airplane with a 200-hp. engine has a maximum speed of 130 
miles per hour. If a 165-hp. engine is substituted without change in 
total weight, what is maximum speed? 

2. An airplane with a 400-hp. engine has a top speed of 150 miles 
per hour. Substituting a 500-hp. engine, what is the maximum 
speed? 

3. An airplane with a 220-hp. engine has a top speed of 140 miles 
per hour. With total weight unchanged, what horsepower engine is 
needed to attain 150 miles per hour top speed? 

4. An airplane with a 300-hp. engine has a top speed of 180 miles 
per hour. With total weight unchanged, what will be maximum speed 
with a 350-hp. engine? 

5. An airplane with a 300-hp. engine has a top speed of 180 miles 
per hour. With total weight unchanged, a 250-hp. engine is sub- 
stituted. What is maximum speed? 

Power Loading and Wing Loading. Power loading is total 

weight per brake horsepower. Wing loading is the total weight 

per square foot of wing area. In the preceding section, the maxi- 
mum speed was found to be 



^^ ,B.H.P. 



This may be rewritten as 



v-K ir^KL 

A high wing loading accompanied by a low power loading will 
give a high maximum speed. A high wing loading, however, 



228 AIRPLAXE PERFORMANCE 

means a high landing speed. If landing speed is fixed, high speed 
depends on ^ „ , the reciprocal of power loading. 



The rate of climb is given by 
R.C. (ft. per min.) = 



33,000 (H.P.g - H.P.req.) 



w 

33,000 r? (B.H.P.) 60 Z)7 
W W 

33,000 rj 60 F 



TT; B.H.P. L/D 



33,000. 60\AV^ 



^/ B.H.P. Ci 



^^^1.28 a 



S 

1.28 a 
33,000 77 60 ^^ ^ ^ fw 



TF/B.H.P. f-p^ Cl'^' Vs 



.... ^„ 



The first term contains the reciprocal of the power loading; 
the second term, the square root of the wing loading. The rate of 
climb depends on the difference between the first and second terms. 
For good rate of chmb, an airplane should have a small power 
loading in order that the first term should be large and a small 
wing loading in order that the second term be small. 

Therefore a high wing loading helps top speed; a low \\ing load- 
ing helps climb. A low power loading helps both top speed and 
climb. 

Span Loading and Aspect Ratio. The total drag is the force 
which is overcome by the propeller thrust. Assuming that fuse- 
lage, landing gear have been " cleaned up " so that parasite drag 
is reduced to a minimum, since the profile drag of most wings is 
about the same, the only other way in which drag may be de- 
creased is by reducing induced drag. 

It \^ill be recalled that both parasite and profile drag vary as 
the square of the velocity, and the horsepower required to over- 
come these drags varies as the cube of the velocity. The induced 
drag varies inversely as the square of the velocity, and the horse- 
power inversely as the velocity. The induced drag therefore 



I 



I 



SPAN LOADING AND ASPECT RATIO 229 

becomes less and less important as airspeed gets higher. The 
aspect ratio or span loading is related to the induced drag only. 

The profile drag depends on the thickness and camber of the 
wing. The wing should be as thin as possible consistent with 
structural considerations. 

At high speeds parasite drag is responsible for as much as 70 
per cent of the total drag. Good streamlining is essential for high 
speeds, and it is of paramount importance for racing planes. 

At speed less than the maximum, the induced drag becomes more 
important, and a small span loading or high aspect ratio becomes 
essential. This is true at climbing speeds and, as will be shown in 
the next chapter, is partly true for high speeds at high altitude. 

Because a high-aspect ratio and a low-wing loading are both 
desirable for performance at low speeds, in England it is customary 
to find the ratio of these two characteristics and refer to this ratio 
as the span loading. 

Wing loading ^ W/S ^ W 
Aspect ratio b^/S h^ 

It is to be noted that in English textbooks span loading refers to 
weight divided by the span squared whereas in American textbooks 
span loading refers to weight divided by the span. 



CHAPTER XI 
PERFORMANCE AT ALTITUDE 

Effect of Altitude on Horsepower Required. The performance 
of an airplane is affected by the density of the air. Lift, drag, and 
horsepower required are all functions of air density. The thrust 
and power absorbed by the propeller are functions of air density. 
The brake horsepower of an internal-combustion engine depends 
on density. 

The fundamental equation for lift is 

L = Cl^SV^ 

For level flight lift equals weight. The wing area, S, is fixed. 
Then if air density (p) decreases, either Cl or V or both must 
increase. A given airplane flying at a higher altitude must either 
fly at a bigger angle of attack or it must fly faster or both. 

Flying at the same angle of attack, Cl constant: 

Po = sea-level mass density 

., , , TT7 n Pon^T^ Po = air density at altitude a 
At sea-level: W ^ Cltt^^q ^7 i -^ i . ] 

2 Fo = velocity, feet per second, 

Ataltitudea: W = CL^SVa' ,, f sea-level 

2 Va = velocity, feet per second, 



at altitude a 



Then 



Pa 



V Pa 



Since the density is always greatest at sea-level, v po/pa is 
always greater than unity, so that, at the same angle of attack, the 
velocity at altitude must always be greater than at sea-level. 

n = (rr. I l-28a \poo^^^ 

i^o y^D^ S ) 2 "^ Do = drag in pounds at sea-level 

-^ /^ , 1.2S a\ Pa ciTT •> Da = drag in pounds at altitude a 

230 



EFFECT OF ALTITUDE ON HORSEPOWER REQUIRED 231 
But 



Va'=^V, 



Pa 



Therefore 






= Do 

That is, whatever the altitude, at the same angle of attack, the 
drag or thrust required for level flight is the same. While the 
density is less, the airspeed must be greater and the product 
remains constant. 

_DoVo 

n.r.req.O - 37^ 

H.P.req.a = ~of^ H.P.req.o = horsepower required at 

t — sea-level 

DoVoy— H.P.req.a = horsepower required at 

» Pa 



375 

= H.P.rea.O X \/^ 

» Pn 



•req.o 

altitude a 



•req. 



That is, at the same angle of attack, the horsepower varies in- 
versely as the square root of the density. 

The minimum speed, Vs, increases with altitude, since 



W 



^imax. n ^ 



W ^so = stalling speed at sea-level 



r Pa Q Vsa = stalUng speed at altitude a 

'^Lmax. TT *J 



'Lmax. n 



> Pa 



The square root of the reciprocal of the densities is tabulated 
in Table I. If the power requirements are known for sea-level 
conditions, the requirements for any altitude are found by making 
use of the factor from Table I. For any point on the total 



232 



PERFORMANCE AT ALTITUDE 



horsepower versus velocity curve for sea-level, a point may be 
found, for the same angle of at tack, for any altitude bymulti- 
plying the abscissa by Vpo/pa and the ordinate by Vpo/po to 
give coordinates of the horsepower required versus velocity curve 
at the altitude a. 

Example. For the airplane described in the first example in the 
preceding chapter, find the horsepower-required curves for 10,000-ft. 
altitude and 15,000-ft. altitude. 

Solution. From Table I: 

V^ po/Pa = 1.16 
Vpo/pa =_1.26 



for a = 10,000 ft.; 
for a = 15,000 ft.; 



TABLE XIV 



From preceding chapter: 


For 10,000-ft. 


For 15,000-ft. 


For sea-level 


altitude 


altitude 


a 


V 


H.P.req, 


V 


H.P.req. 


V 


H.P.req. 


-4 


228 


566 


264 


656 


288 


715 


-3 


161 


199 


187 


231' 


203 


251 


-2 


130 


111 


151 


129 


164 


140 


-1 


112 


76 


130 


88 


141 


96 





100 


60 


116 


70 


126 


76 


4 


74.9 


34 


87.0 


39 


94.4 


43 


8 


62.3 


29 


72.3 


34 


78.5 


37 


12 


55.2 


29 


64.0 


34 


69.5 


37 


16 


50.2 


30 


58.2 


35 


63.4 


38 


18 


48.1 


31 


55.8 


36 


60.6 


39 


19 


48.1 


33 


55.8 


38 


eo.6 


42 



These curves are plotted in Fig. 68. It is to be noted that, since 
V and H.P. are always in the same proportion for one angle of attack, 
a line drawn from the point velocity, hp., through a point on the 
horsepower-required curve for sea-level, if prolonged, will pass through 
the point corresponding to that same angle of attack on the horse- 
power-required curve for all altitudes. 

Horsepower Available at Altitude. At altitudes, the brake 
horsepower of the engine drops off. The variation in brake 
horsepower is approximately as the 1.3 power of the density. 



HORSEPOWER AVAILABLE AT ALTITUDE 



233 



Owing to the decrease in density, the propeller tends to turn 
over faster; but since the engine power drops faster than the 
density, the propeller will actually run slower. 



200 




80 90 100 110 120 
Velocity, Miles per Hour 

Fig. 68. Performance at altitude. 



150 160 



A quick way to find the horsepower available at altitude is as 
follows. In Fig. 556 is shown the decrease in revolutions per min- 
ute with altitude at the same airspeed as at sea-level. Since for 
most engines the brake horsepower varies directly with revolutions 
per minute, the same factor which is used in reducing the revolu- 
tions per minute can be used in reducing the brake horsepower. 
That is, on account of the fewer revolutions per minute less power 
will be developed. In addition, owing to the reduced density of 
air entering the carburetor, etc., the engine will develop less power 
even if running at the same revolutions per minute. This cor- 
rection factor is found in Fig. 446. 

Because the propeller is turning over at a different speed, its 
V/(ND) for various airspeeds will be different from the V/{ND) 
for those airspeeds at sea-level. The propeller efficiencies will 
consequently differ at altitude from the efficiencies at sea-level 
for the same airspeeds. 

Example. Find the horsepower available at 10,000-ft. and 15,000-ft. 
altitude for the 150-hp. engine used on the monoplane of the first 
illustrative example in the preceding chapter. 



234 



PERFORMANCE AT ALTITUDE 



TABLE XV 

Calculations for 10,000-ft. Altttude 



Sea-level 


10,000 
ft. 


Sea- 
level 


10,000 ft. 


(1) 


(2) 


(3) 


(4) 


(5) 


(6) 


(7) 


(8) 


(9) 


(10) 








B.H.P 






% 


% 


% 
eff. 




7 


A^ 


N 


at re- 
duced 

N 


B.H.P 


V/ND 


design 

V/ND 


design 
eff. 


HP.avaU. 


50 


1565 


1522 


128 


87 


0.381 


43.8 


53.3 


44.0 


38.2 


60 


1565 


1522 


128 


87 


.456 


52.5 


62.8 


51.8 


45.1 


70' 


1570 


1530 


128 


87 


.530 


61.0 


72.0 


59.5 


51.8 


80 


'1583 


1540 


129 


87 


.602 


69.2 


80.5 


66.5 


58.0 


90' 


1600 


1550 


130 


88 


.670 


77.0 


88.0 


72.5 


63.8 


100 


1633 


1590 


132 


89 


.728 


83.8 


94.0 


77.5 


69.0 


no 


1673 


1630 


135 


91 


.778 


89.5 


97.0 


80.0 


72.8 


120, 


1751 


1705 


142 


96 


.815 


93.7 


99.0 


81.6 


78.5 


130 


1781 


1735 


145 


98 


.870 


100.0 


100.0 


82.5 


81.0 


135 


1800 


1755 


146 


99 


.890 


102.2 


100.0 


82.5 


81.7 



Calculations for 15,000-ft. Altitude 



Sea-level 


15,000 

ft. 


Sea- 
level 


15,000 ft. 


(1) 


(2) 


(3) 


(4) 


(5) 


(6) 


(7) 


(8) 


(9) 


(10) 


V 


N 


N 


B.H.P 

at re- 
duced 

N 


B.H.P 


V/ND 


% 
design 

V/ND 


design 
eff. 


% 
eff. 


HP.avaU. 


50 

60 

70 

80 

90 

100 

110 

120 

130 

135 


1565 
1565 
1570 
1583 
1600 
1633 
1673 
1751 
1781 
1800 


1490 
1490 
1495 
1510 
1525 
1555 
1590 
1 670 
1700 
1715 


124 
124 
125 
126 
127 
130 
133 
139 
142 
143 


67 
67 
67 
68 
68 
70 
71 
75 
76 
77 


0.390 
.468 
.542 
.616 
.685 
.745 
.797 
.835 
.890 
.914 


45.0 
53.8 
62.3 
71.0 
78.7 
85.5 
91.5 
96.0 
102.2 
105.0 


55.0 
64.0 
73.5 
82.3 
89.3 
94.5 
98.0 
99.5 
100.0 
99.0 


45.4 
52.8 
60.6 
68.0 
73.7 
78.0 
80.9 
82.1 
82.5 
81.7 


30.4 
35.4 
40.6 
46.2 
50.1 
54.6 
57.5 
61.6 
62.8 
62.8 



MINIMUM HORSEPOWER AT ALTITUDE 235 

Explanation of Table 

Columns 1 and 2, V and N from sea-level calculations, last chapter. 

Column 3, items in column 2 multiplied by altitude factor from Fig. 556: 
0.975 for 10,000-ft. altitude; 0.952 for 15,000-ft. altitude. 

Column 4, 150 multiplied by ratio of items in column 3 to 1,800 r.p.m. 

Column 5, items in column 4 multiplied by altitude factor from Fig. 446: 
0.676 for 10,000-ft. altitude; 0.537 for 15,000-ft. altitude. 

Column 7, items in column 6 divided by design V/(ND) (= 0.870). 

Column 8, from Fig. 55a, using items in column 7 as entrants. 

Column 9, items in column 8 multiplied by design efficiency (82.5 per 
cent). 

Column 10, items in column 5 multiplied by items in column 9. 

Maximum Speed at Altitude. The maximum speed at any 
altitude is found by the intersection of the total horsepower re- 
quired at that altitude with the total horsepower available at that 
altitude. It is quite possible with propeller designed for altitude 
work, or other special conditions, that the maximum speed may 
be slightly higher at moderate altitudes than at sea-level. Ordi- 
narily the maximum speed is less at altitude than at the 
ground. 

Minimum Speed at Altitude. The stalling speed increases with 
altitude, that is, an airplane must fly faster to keep from stalling. 
The stalhng speed or minimum speed at any altitude is to the 
minimum speed at sea-level as the reciprocal of the square root 
of the relative density. 

The ordinary pitot static airspeed indicator measures impact 
pressure. The scale of the airspeed indicator gives the airspeed 
corresponding to the impact pressure at the ground of that air- 
speed. At altitude, the impact pressure will be less for any air- 
speed than the pressure for that airspeed at sea-level so that the 
airspeed indicator will read low. The indicated airspeed at 
altitude will be the true airspeed multipHed by the square root of 
the relative density. 

This " error " of the airspeed indicator is rather fortunate from 
a pilot's viewpoint. The minimum speed at sea-level being known, 
whenever the airspeed indicator reads that airspeed, the airplane 
is flying at minimum velocity. At altitude, the airplane is flying 
faster than the airspeed indicator shows but it is necessary for the 
airplane to fly faster to keep from stalling. 

Minimum Horsepower at Altitude. For any given airplane, 
there is one angle of attack at which CL^^^/CDtotai is maximum, and it 



236 



PERFORMANCE AT ALTITUDE 



is at this angle of attack that the least horsepower will be required 
to maintain level flight. At any altitude, this angle of attack for 
minimum horsepower will be the same, but the velocity correspond- 
ing to this angle of attack will be greater at higher altitudes. 
The velocity will be to the velocity at sea-level for minimum horse- 
power as the reciprocal of the square root of the relative density. 

It is also true that the angle of attack for maximum LID will be 
the same at all altitudes, but at higher altitudes greater velocity 
must be maintained for this same angle of attack. 

As described for stalling speeds, the readings of the airspeed 
indicator will be the same at all altitudes when the airplane is 
flying at the angle of attack of minimum horsepower required. 
The airspeed indicator has the same reading at all altitudes when 
the airplane is flying at the angle of attack of maximum LjD. 

Rate of Climb at Altitude. The ability to climb is determined 
by the excess of the horsepower available at any speed to the 
horsepower required at that speed. At altitudes above sea-level, 
the ordinates of the total horsepower-required curve become 
greater and the ordinates of the horsepower-available curve 
become less. The difference between these ordinates at the same 
speed represents the power for climbing. 



800, 
700 



600 



5- 500 

8 

u. 400 



300 



200 



100 



^^^ 


















^^s^ 








^^-^^ 










^^^ 








^^^^^ ; 


bsolute Ceiling 






"^k 








Service Ceiling.,^^ j^"*^^.^^^^ 





5000 



10000 
Altitude 



15000 



Fig. 



Variation of rate of climb with altitude. 



The greatest difference in ordinates gives the maximum climb- 
ing power at that altitude. As the altitude is increased, the 
power decreases. The airspeed for maximum climb is greater 
with increased altitude and is very close to the airspeed for mini- 
mum horsepower required, that is, the angle of attack for best 



ABSOLUTE AND SERVICE CEILINGS 237 

climb is almost the same as the angle of maximum C l^^^ / C DtotsX- 
Therefore with increasing altitude the speed for maximum climb 
increases. 

If the maximum rate of climb is plotted against altitude as in 
Fig. 69, it will be seen that the resulting curve is practically a 
straight line. Actually the slope of the rate of climb curve varies 
lineally with altitude. 

It is customary to assume that the rate of climb varies lineally 
with altitude. 

Absolute and Service Ceilings. If the rate-of-climb versus 
altitude curve is continued to intersect the base line, this inter- 
section marks the absolute ceiling or altitude where the rate of 
climb is zero. This is the highest altitude that it is possible for 
the airplane to reach. 

At this altitude, the horsepower-available curve is tangent to the 
horsepower-required curve. The airplane can be flown at only 
one speed. At either greater or less velocity, there will not be 
sufficient available horsepower to maintain level flight. This one 
velocity of flight will be very close to the best climbing speeds at 
altitude near the ceiling. 

At the absolute ceiling control is very sluggish. The rate of 
climb has been steadily decreasing with altitude, so that as one 
climbs very near to the absolute ceiling the rate of climb becomes 
infinitesimally small. It will therefore require an infinite time to 
reach the absolute ceiling, and no airplane ever reaches there unless 
one conceives of an infinite fuel supply. 

More practical is the service ceiling which is the altitude where 
the rate of climb is 100 ft. per min. 

If the rate of cHmb is assumed to change lineally with altitude 



Hs = 



H = absolute ceiling, feet 
H[(R.C.)o — 100] Hs = service ceiling, feet 

(Il.C.)o (R.C.)o = rate of climb at sea-level, 

feet per minute 



Example. For an airplane weighing 4,000 lb,, the excess horse- 
power at sea-level is 60 hp.; at 10,000-ft. altitude there is 17 excess 
horsepower. What is service ceiling? 

Solution. 

60 X 33,000 

= 495 ft. per min. 



238 



PERFORMANCE AT ALTITUDE 



(R.C.) 10,000 = 



17 X 33,000 



4,000 
= 140 ft. per min. 
H - 10,000 _ 140 
495 



H 
495iy - 4,950,000 = 140 H 

355 H = 4,950,000 
Absolute ceiling = H = 13,970 ft. 

ff. = 13,970 (g|) 

= 11,150 ft. 

Problems 

(Assume that rate of climb varies lineally with altitude.) 

1. An airplane weighs 3,500 lb.; its rate of climb at sea-level is 
1,000 ft. per min.; its absolute ceiling is 16,000 ft. What is its 
service ceiling? 

2. The service ceiling of an airplane is 14,000 ft. Its rate of climb 
at sea-level is 950 ft. per min. What is the absolute ceiling? 

3. An airplane weighs 2,500 lb. What is the excess horsepower at 
its service ceiling? 

4. The absolute ceiling of an airplane is 17,000 ft. The rate of 
climb at sea-level is 1,150 ft. per min. What is the rate of climb at 
10,000-ft. altitude? 

5. The service ceiling of an airplane is 18,000 ft. The rate of climb 
at sea-level is 1,000 ft. per min. What is the rate of climb at 10,000- 
ft. altitude? 

Time to Climb to Altitude. Assuming rate of climb to vary 
lineal with altitude, the slope of the rate-of-climb versus altitude 
curve is constant, and is equal to — (R.C.)o/^. The rate of 
climb at any altitude, /i, is 

/i(R.C.)o 



RC.A = (R.C.)o 



H 



= (R.C.)o[^] 
in different! 
(R.C.)o[^^] 



If rate of climb is expressed in differential form as dh/dt 

dh 
dt 

H dh 



dt = 



{R.C.)o{H - h) 



FINDING CEILING BY TIME-TO-CLIMB FORMULA 239 

Integration gives 
Hdh 



=/: 



(R.C.)o(^ - h) 



H_ r° dh t = time to climb to 



(R.C.)oJk H — h altitude, h, in 

^ minutes 

H = absolute ceiling 



^ log.(^ - h) 



(R.C.) 



^ l\0geH-\0ge(H-h)] ^^^"^^ 



(R.C.)o'^ ^' '' (R.C.)o = rate of climb at 

H , H sea-level, in feet 

(R.C.)o ^^' H -h per minute 



= 2-^^^(RSyoi^^4^J 



Example. At sea-level an airplane's rate of climb is 1,000 ft. per 
min. Its absolute ceiling is 15,000 ft. How long will it take to cHmb 
to 7.000-ft. altitude? 



Solution. 



t = 2.303 X 3^ X 
= 9.5 min. 

Problems 



/ 15,000 \ 
^^^ VI 5,000 - 7,000/ 



1. At sea-level, an airplane weighing 4,000 lb. has 120 excess horse- 
power. Its absolute ceiling is 9,000 ft. {a) How long will it take to 
climb from sea-level to 5,000 ft.? (h) How long will it take to climb 
from sea-level to 6,000-ft. altitude? 

2. A Heath monoplane has an absolute ceiling of 12,000 ft. Its rate 
of climb at sea-level is 450 ft. per min. How longVill it take to climb 
from sea-level to 10,000-ft. altitude? 

3. A Waco airplane, whose ceiling is 16,000 ft., climbs 1,050 ft. per 
min. at sea-level. How long will it require to climb from sea-level 
to 10,000 ft.? 

4. A Bellanca airplane, whose ceiling is 17,000 ft., climbs 900 ft. 
per min. at sea-level. How long will it require to climb from sea-level 
to 12,000 ft.? 

5. A Douglas airliner has a ceiling of 22,000 ft. It climbs 1,100 ft. 
per min. at sea-level. How much time will it take to climb from 
10,000- to 15,000-ft. altitude? 

Finding Ceiling by Time-to-Climb Formula. Since the time to 
climb is a function of ceiling, it is possible to use this relation to 



240 



PERFORMANCE AT ALTITUDE 



find the ceiling of an airplane by noting the altitudes attained at 
two times in a continuous climb. The simplest procedure is to 
use two equal time intervals, so that the time interval counted 
from instant of take-off to the time of reading the second altitude 
is twice the time interval from take-off to the time of taking the 
first altimeter reading. Let h be the time in minutes from take-off 
to the attainment of altitude of hi feet, and ^2 be the time in minutes 
from take-off till altitude of /12 feet is reached, then 



ti = 2.303 



2.303 



H 



{RC)o 

H 
{RC)o 



logi 
logi 



IH -i 





r ^ 1 


iOgio 


[h -h2 




- H ' 


iOgio 


[h - hj 



^2 1 
rlog 



H 



H -hi 

t2/h 



= log 



H 



\H - h) 
If ^ is twice ti, Ulk = 2 

\H - hj 



H -h2 

H 



H -h 



H 



H^ - Hh = 

2 Hhi - Hh2 = 

H = 



H ■ 

h,^ 



h2 

■2Hhi + hi^ 



h'' 



2hi - h2 



Example. An airplane takes 7 min. 30 sec. to reach 8,000 ft. 
altitude. In that same time interval (i.e., 15 min. from sea-level) it 
reaches 13,600 ft. altitude. What is the ceiling? 



H = 



(8,000)^ 



2 X 8,000 - 13,600 
= 26,667 ft. ceiling 



Problems 

1. An airplane climbs in 10 min. from sea-level to 7,780 ft. altitude; 
continuing the climb 10 min. later the altitude is 12,500 ft. What is 
the ceiling? 



STRATOSPHERE FLYING 241 

2. An airplane climbs in a certain time to 9,000 ft. altitude; in 
double that time counted from sea-level the airplane reaches 14,000 ft. 
altitude. What is the ceiling? 

3. An airplane, with a ceiling of 20,000 ft., climbs to a height of 
6,000 ft. in a certain time. What height will it attain in twice the 
time? 

4. An airplane, with a ceiling of 16,000 ft., can climb to 10,000 ft. 
in 20 min. What height will it have reached in 10 minutes? 

5. An airplane climbs in a certain time to 2,000 ft.; in twice that 
time it has climbed to 3,500 ft. What is the ceiHng? 

Stratosphere Flying. An airplane which flies at a certain speed 
at the ground must, at altitude, either fly faster or fly at greater 
angle of attack; since weight and wing area are constant, either 
Cl or V must increase to compensate for decrease in density. 
If the airplane is flown at the same angle of attack at altitude as 
at sea-level, the velocity at altitude must be to the velocity at the 
ground inversely as the square root of the relative densities. The 
horsepower required at altitude will be to the horsepower required 
at the ground in that same ratio, inversely as the square root of 
the relative densities. 

When flying at ground level (except at angles greater than that of 
maximum Cl^''^/Cd tot&i), an increase in velocity means an increase 
in horsepower required. At low angles of attack, there is very 
little change in Cd with angle of attack; therefore the change in 
horsepower is very closely as the cube of the velocity. 

Therefore if it is desired to increase the speed of flying, one may 
either stay at the same altitude and decrease the angle of attack 
or climb to a higher altitude. In each case more horsepower will 
be required, but in the former the horsepower must be increased 
approximately as the cube of the airspeed and in the latter it must 
be increased directly as the airspeed. 

In Fig. 70 are shown the horsepower-required curves for the 
airplane used as example 1 in the preceding chapter, for sea-level; 
for 10,000-ft., 15,000-ft., 30,000-ft., 40,000-ft., and 50,000-ft. 
altitude. It will be noted that to fly this airplane at 160 miles 
per hour at sea-level requires 196 hp.; at 10,000-ft. altitude, 
148 hp.; 15,000-ft. altitude, 133 hp.; 30,000-ft. altitude, 94 hp.; 
40,000-ft. altitude, 73 hp.; and 50,000-ft. altitude, 74 hp. 

Viewed in another aspect, with, say, 100 hp. available, at sea- 
level, 124 miles per hour can be flown; at 30,000-ft. altitude, 164 



242 



PERFORMANCE AT ALTITUDE 



miles per hour; at 50,000-ft. altitude, 210 miles per hour. At 
any altitude the least horsepower required is that corresponding 
to the angle of attack of maximum Cl^''^/Cd total- For the speci- 
men example this means a speed of 59 miles an hour at sea-level, 
but 154 miles per hour at 50,000-ft. altitude. The increase in 
speed, 154 ^ 59 or 2.6 times, means an increase of 2.6 times 

340 
320 
300 



280 
260 
240 

220 

o 

5200 
180 



§160 
140 
.120 
100 
80 
60 
40 



20 

























1 






















\ 
























/ 
























/ 


1 


Horsepower Required at 15,000 ft. Altitude -^ 

1 ' '^ 


f 


1 


S 




1 
Horsepower Required at 10,000 ft. / 


1 i 
Utitude 1 1 






/ 




Horsepower Required at Sea Level 


\ 






1 


/ 












\ 


V 




/ 


/ 




/ 


/ 












/ 


// 




/ 


1 




/ 












1 


' 


/ 




/ 


/ 












1 


k 


/ 


/ 


/ 


/ 


/ 














\/ 


/ 


/ 


/ 


/ 












/ 




/ 


/ 


C< 


/ 

HOfM 


power 


Requir 


edat 
de ~ 






c 


/j 


V- 


n 


.^^ 




~ 50.000 ft. Altitu 






^ 


^ 


^rs 


epovier Required at 40,000 ft. Altitude 

1 1 1 1 1 




^ 


^ 


9^ 




^Horsepower Required at 30,000 ft. Altitude 

1 1 1 1 1 1 





40 60 80 



100 120 140 160 180 200 220 240 260 
Velocity ( Miles per hour) 



Fig. 70. Horsepower required at high altitude. 

(74 -^ 29) the horsepower. If 154 miles an hour were flown at 
sea-level, it would require 176 hp. or 6 times the horsepower. 
Flying at high altitude is conditional on having the sufl^cient 
horsepower available. Unsupercharged engines drop off in 
horsepower with altitude. Propellers which are proper for sea- 
level density are unsuitable for high altitude. Superchargers can 



RAPID ESTIMATION OF PERFORMANCE 243 

provide air at sea-level density at high altitudes; one proposed 
German airplane is to be provided with three superchargers, one 
which is to be put into service at 25,000-ft. altitude, a second which 
is to be added at 35,000 ft., and the third to be added at 45,000 
ft. Variable-pitch propellers will of necessity be used on planes 
designed for stratospheric flying. 

Rapid Estimation of Performance. For a clear understanding 
of the factors affecting airplane performance, the performance 
curves described in Chapters X and XI should be drawn and 
studied. For rapid estimation, however, various methods have 
been evolved. The method of W. Bailey Oswald is described 
below; for a fuller explanation the reader is referred to N.A.C.A. 
Report 408, from which this material was extracted. 

Dr. Oswald assumes that the profile drag coefficient {Cdq) and 
the parasite drag are each independent of the angle of attack. 
While this is not true at high angles of attack, it is substantially 
correct at high speeds (low angles of attack) with which this 
method is chiefly concerned. By combining the profile drag of the 
wing (Cz)op/2 SV^) and the parasite drag of the remainder of the 
airplane (1.28 p/2 aV^), the total drag exclusive of the induced 
drag of the wing is 

Cdo ^ SV^ + 1-28 I aV^ = {Cd,S + 1.28 a) | V^ 

Dr. Oswald divides the airplane weight by the term in parentheses 
above, and terms the quotient, the parasite loading (Lp), 

L ^ 

"" CdoS + 1.28 a 

The weight divided by the square of the effective span is called 
the '^ span loading " (L^) by Dr. Oswald. This does not conform 
with the accepted meaning of the term, see Chapters V and VI, 
that span loading is the weight divided by the effective span. 
For a rectangular monoplane, the " span loading " is 

For a biplane it is (see Chapter VI) 

T ^ 



244 PERFORMANCE AT ALTITUDE 

The weight divided by the design thrust horsepower is termed the 
thrust-power loading {Lt) in the Oswald method. The design 
thrust horsepower is the brake horsepower at rated revolutions 
per minute, multiplied by the propeller efficiency under design 
conditions. 

W W 



u = 



■D.-tl.-T. rated X ^design -L .-tl-i^ 'design 



Oswald also uses a factor i^T^) to represent the ratio of thrust 
horsepower at any speed (F) to the thrust horsepower at design 
(i.e., maximum) speed (7m). 

_ T.H.P. at velocity 7 gg^je^el) 

'- " T.H.P. at velocity F„ ^^^ '^^ '^"^^'^ 

The factor Ta is the ratio of thrust horsepower at some altitude (a) 
to the thrust horsepower at sea-level, both at the same velocity. 

^° = T:H:pr^^ea-le"vtl (at same velocity F) 

A large number of cases have been examined of modern planes 
with unsuper charged engines, and it has been found that 

V 
Ty = R/" where R^ = ^ 

' m 

m = 0.65 for prop with design Cs = 0.9 

m = 0.61 for prop with design Cs = 1.2 

m = 0.55 for prop with design Cs = 1.6 

m = 0.55 for props with peak efficiency 

and 

^ a- - 0.165 
0.835 

= 1.198(0- - 0.165) 

The rate of climb at any speed and at any altitude is the excess 
power at that speed and altitude divided by the weight. 



R.C. (ft. per min.) 



(H.P.avail. - H.P.,eq) X 33,000 

w 



RAPID ESTIMATION OF PERFORMANCE 

Xl. Jr. avail. -H-'-l • req. \ 



245 



II.C. = 33,000 (■ 
= 33,000 



W W I 

B.H.P.rated X ^des. X TaT^ 



W 



1 CdoS + 1.28 a p 



= 33,000 



550 

J- aJ- V 



w 



v 



w 



550Tf|^^P^,^^ 



h 550 L, 550^^^^ 



substituting R 



VJ""- po 



( 



= 33,000 



J- a^^ V 



550 Lp 



Ls 



bbO-ira^RV, 



At sea-level, when flying at maximum velocity, the rate of climb 
is zero and 



a = R = Ta = Ty = 1. 



= 1-2 



POy 3 



Lt 550 Lp po y. 

550 TT 2" l^r 



PoT^Ti 



550 L, 



550x^^7, 



550 



LgLt 



po 



yjL, 



550x^7. 



L/L^ ^ 550 L^^L,^ 



2 
LgLt 



Po 
2 



550 TT?? 7 



L//^ f PO I 7^ 



c 



2 
LsLt 



1/3 



PO 
2 



550 7r?:V^ 



246 



PERFORMANCE AT ALTITUDE 



substituting 






A = 77.3^(1- 0.487 ^^y' 



LsLt 



{ 



A = 52.8-jPll „.™ 

y m \ y m 



0.332 



LshV" 



V in feet per second 

V in miles per hour 



In Fig. 71a, A is plotted against Vm/L Lt, and from this graph 
the maximum velocity at sea-level (Vm) may be found if the other 
parameters are known. 

From one of the above equations, an expression can be found for Vm 
in terms of the three ^4oadings" {Lp, Ls, and Lt), which if substi- 
tuted in the equation for rate of climb will permit a simplification. 

Since 



Po 



V 3 



550 Lp Lt 

550 L 



550 7r|«F^ 



Vr, 



LsLt 



Po J 

2^* 



550 TT ^ 7. 



Substituting this value for VrJ in the equation 



c^RWJ 



R.C. = 33,000 



= 33,000 




550 7r I' o-i^F^y 



33,000 
Lt 



(tj, - 



(7R' + 



(jR^LsLt 
550 7r|F. 



LLt 



550 7r??cr/?F, 



V in miles per hour 



RAPID ESTIMATION OF PERFORMANCE 



247 



At altitudes above sea-level, when flying at maximum velocity, 
the rate of climb is zero. 



12 

11 

10 
9 
8 
7 

5 
4 
3 
2 

1 




















































"i 


'f- 


a function of A 






1 




















\ 




















\ 




















\ 






















V 




















\ 


\, 




















\ 


\ 
























— 





























10 15 20 25 30 35 40 45 50 

A 



Fig. 71a. Performance prediction curves. 
0.332 L,L,(1 - (T^^) 



Vr. 



1 aJ- v^^vrti P ■tivm 



V ~ ^-^^^ 1 - cT^B 4 



where 
but 



^vm 



at altitude 



'?!(•-»■»#')" 



^max. at sea-level 
A = 52.8 

Substituting the value of LsLt/Vm found above gives 
A = 52.8 (3.014 ZIeZ^^^!!^^^^ _ 0.332 ^Y'' 

1.198 ((7 - 0.165)crig,^^+i _ ^2R^j / _^L,L,V/3 



= 159.1 



248 



PERFORMANCE AT ALTITUDE 



In the above equation, A is expressed in terms of R^m, the relative 
density a and certain parameters of the airplane. In Fig. 716, 
R^m is plotted against A for various altitudes. 



1.00 



.95 



-90 



.85 



i.80 



.75 



.70 



.65 



.60 













^.« 


■awpI 








v 


^^ 




N, 














\ 


S. 10,000 ft 


V, 


^ 


00 ft. 










\ 


N 


\ 






\ 










\ 


^ 


\ 








\ 








V 


\ 




\ 






\ 








\ 


Y's.ooo 


A 








\ 




', 


\ 


\ 


) 


\ 






' 






\ 




\ 




\ 












' 




\ 
















































1 










2C 


.000 f 


. 




























RvmJ 


s a fur 


iction c 


fA 













































5 10 15 20 25 30 35 40 45 50 
A 

Fig. 716. Performance prediction curves. 

In a similar manner, equations may be deduced for other items 
of performance, and curves plotted. In Fig. 71c, Lf(R.C.)A is 
plotted against A, so that by the aid of this graph, the rate of climb 
at any altitude h may be found. 

As an illustration of the Oswald method, the performance of an 
airplane is partially worked out below. For comparison, the air- 
plane of the first example in Chapter X is used. 

Example. Find the maximum velocity and rate of climb at sea- 
level, 5,000 feet, and 10,000 feet altitude, for a monoplane weighing 



RAPID ESTIMATION OF PERFORMANCE 



249 



2,000 lb. and having a Clark Y rectangular wing 36 ft. by 6 ft. The 
parasite drag has an equivalent flat plate area of 3.8 sq. ft. The engine 
is rated at 150 hp. at 1,800 r.p.m., and the propeller uiider design con- 
ditions has Cs = 1.57 and efficiency of 82 per cent. 



I 



17.000 
16.000 
15.000 
14.000 




\ 




















\ 




I,Ch as a function of A at various altitudes 




> 


\ 




















\ 
















13.000 






\ 


















V 




\Sea level 












12.000 
11.000 




\ 




\ 
















\ 




\ 


s 




















\ 












10,000 

^ 9.000 
I 

' 8.000 
7.000 
6.000 
5.000 
4.000 
3,000 
2.000 
1.000 






Wooo «. 


\ 














\ 


\ 






\ 












\ 




\ 




\ 


s. 








10,0 


)Oft\ 


i 


\ 






\ 










\ 


\ 




\ 






\ 








\ 


\ 




\ 


V 




\ 


\ 






\ 




\ 




\ 






\ 


\ 




\ 


\ 


\ 




> 


\ 






\ 




\ 


\ 


5,000 
ft. 


\ 




\ 








20'000\ 


\ 


V 


\ 






\ 










\ 


\ 




\ 




\ 


s 




) . 


5 1 


1 


5 2 


2 
/ 


5 3 


3 


5 4 


a 4 


5 50 



Fig. 71c. Performance prediction curves. 

Solution. 

For Clark Y airfoil Cdo = 0.01 (see Fig. 38) 

T ^ 

^ " CdoS + 1.28 a 

2,000 



0.01 X 216 + 1.28 X 3.8 
= 284.7 



250 



PERFORMANCE AT ALTITUDE 







W 










^s = -^ 














2,000 








~ (36)2 








= 1.543 








L ^ 








^' -^ B.H.P. X -n 








2,000 




^ 




" 150 X 0.82 








= 16.26 
















1.543 X (16.26)4/3 








(284.7)1/3 








= 9.66 








LsU = 1.543 X 16.26 








= 25.09 








Level flight 


CHmb 


Altitude 


















Vm 


Rvm 


Vmh 


L,(R.C.) 


R.C. 





133.0 






14,900 


916.3 


5,000 




0.970 


129.0 


11,250 


691.8 


10,000 




0.935 


124.4 


7,550 


464.3 


15,000 




0.884 


117.6 


4,250 


261.3 



Explanation of Table 

Column 2 is obtained from Fig. 71a (for A = 9.66, Vm/LsLt = 5.3). 

Column 3 is obtained from Fig. 71&. 

Column 4 is obtained by multiplying items in column 3 by F^ ( = 133.0). 

Column 5 is obtained from Fig. 71c. 

Column 6 is obtained by dividing items in column 5 by Lf ( = 16.26) . 



CHAPTER XII 
TURNS 

Centrifugal Force. Heretofore the flight of an airplane has 
been considered in one direction. If direction is being changed, 
centrifugal force while the turn is being accomplished must be 
considered. Before the turn starts and immediately after the 
turn ceases, centrifugal force is not acting. 

Acceleration is the change of velocity. Velocity has not only 
magnitude but also direction. Even if magnitude is unchanged 
but there is a change in direction of a velocity, that change is 
acceleration. That acceleration is always radially inward towards 
the center about which the object is circling at that instant. 
The force which causes the body to accelerate inward in a turn 
is measured by the mass times the acceleration. 

In constant circular motion, the acceleration is oi^R where co 
is the angular velocity, or the acceleration is V^/R where F is the 
linear velocity. The centrifugal force of an airplane in a turn is 
equal in magnitude and opposite in direction to the accelerating 
inward (centripetal) force 

g = acceleration of gravity in feet per second^ 

C.F. = ^ V = airspeed in feet per second 

R = radius of turn in feet 

Banking. In straight flight, the force of gravity, called weight, 
is the only exterior force acting on the plane. It acts vertically 
downward, i.e., towards the center of the earth. 

In a turn centrifugal force acts also on the airplane, and the 
exterior force acting on the airplane is the resultant of centrifugal 
force and weight. In turning in a horizontal plane, centrifugal 
force is outward, so the resultant will be outward and downward. 
Lift must balance this resultant force, so lift must act inward and 
upward and must be equal in magnitude to this resultant force. 

In turns it is customary to bank or depress the inner wing. 
If a flat turn is made, that is, without banking, centrifugal force 
will cause the airplane to skid outwards. Except in flying in 

251 



252 



TURNS 



formation, this is not especially objectionable but is considered 
poor flying technique. 

The proper angle of bank depends on the 
airspeed and the sharpness of the turn. If 
the angle of bank is insufficient the airplane 
will skid, that is, move outward. If the 
angle of bank is too much, the airplane will 
slip, that is, move inward and downward. 

Lift must equal in magnitude the resultant 
of weight and centrifugal force. If lift is not 
as great as this resultant, the airplane will 
squash or settle down in the direction of the 
resultant force. If lift is greater than the 
resultant, the airplane will execute a cHmbing 
turn or spiral. 




Fig. 72a. Forces 
a banked turn. 



From examination of Fig. 72a, if j8 is the angle of bank, 



tan jS 



C.F. 
W 

g R 
w 

Yl 

gR 



V in feet per second 
R in feet 



It will be noted that the angle of bank /? is independent of weight. 
A big, heavy bomber and a Hght sport-plane, if they have the 
same airspeed, require the same angle of bank for the same radius 
of turn. Angle of bank is also independent of wing area, airfoil 
section, etc. 

If the angle of bank is correct, lift will act in the opposite di- 
rection to the resultant. The magnitude of the lift force must 
equal the magnitude of the resultant force, or 



L = 



W 

cos 13 

^ g R 

sin 13 

For any angle greater than zero, the cosine is less than unity; 
therefore, if the weight is divided by a number less than 1, the 



BANKING 253 

quotient is greater than the weight. For straight, level flight, 
lift equals weight. For a turn, lift must be greater than the weight. 
The lift may be increased either by increasing the airspeed or by 
increasing the angle of attack or both. 

In low-powered airplanes or in gliders, it is quite often necessary 
to push forward on the stick in order by diving to get the added 
velocity for the turn. 

Whether the added lift is gained by increasing the airspeed or by 
increasing the angle of attack, more power is required in a turn 
than in level flight. 

Example. An airplane is flying straight at 150 miles per hour. 
Its weight is 6,000 lb. Its wing area is 200 sq. ft. (A.R.6), and Clark 
Y airfoil is used. The parasite has an equivalent flat plate area of 
2 sq. ft. (a) What angle of attack is needed? (6) What horsepower 
is required? The airplane turns at 150 miles per hour airspeed with a 
30° angle of bank, (c) What is the centrifugal force? (d) What lift 
is needed? (e) What is the proper radius of turn? (/) What angle 
of attack is needed? (g) What horsepower is required? 

Solution. 

W 
^^^ ^^ ^ 0.00256 iS 72 

6,000 

~ 0.00256 X 200 X 150^ 
= 0.522 

From Fig. 17, a = 2.1° 

(6) From Fig. 17, 

Cd = 0.025 

1.28 a\ 0.00256 ^73 



H.P.=(c.+lf^)' 

= (0.025 +i^) 



375 

0.00256 X 200 X 150^ 



375 

= 175 hp. 

(c) C.F. = T7 tan iS 

= 6,000 X 0.577 

= 3,460 lb. 

W 

{d) Lift = ^ 

^ ' cos p 

6,000 

~ 0.866 

= 6,930 lb. 



254 TURNS 

(e) tan/5=^ 



(/) C£ 



gr tan /8 
(150 X 1.47)2 
32.2 X 0.577 
= 2,600 ft. 

Lift 



0.00256 X 200 X 150^ 
= 0.603 



From Fig. 17, 

a = 3.4° 

(g) From Fig. 17, 
Cd = 0.03 
__^ /^ , 1.28 a\ 0.00256 >S 73 



375 

3 



/'r.r.nr. . 1-28 X 2 \ 0.00256 X 200 X 150 
- [y.OSO + 200 J 375~" 

= 203 hp. 

Problems 

1. A plane of 2,000 lb. gross weight is turning at 100 miles per hour 
with an angle of bank of 45°. (a) What is the centrifugal force? 
(b) What is the lift? (c) What should be the radius of turn? 

2. An airplane is making a 40° banked turn of 800-ft. radius. 
What should be the airspeed? 

3. An airplane weighing 2,500 lb. has a Clark Y wing 250 sq. ft. 
in area (A.R.6) and parasite with equivalent flat plate area of 1.5 
sq. ft. What horsepower is required in straight flight at 120 miles 
an hour and in a banked turn of a quarter-mile radius at 120 miles 
per hour? 

4. A racing plane weighing 1,800 lb. is rounding a pylon at 340 
miles per hour. The radius of the turn is 100 ft. What should be 
angle of bank, and what is centrifugal force? 

5. An airplane weighing 5,000 lb. has a Clark Y wing of 300 sq. ft. 
area (A.R.6) and parasite with equivalent flat plate area of 3.0 sq. ft. 
(a) What horsepower is required to fly straight at 100 miles per hour? 
(6) What horsepower to make a 35° banked turn if the same angle of 
attack is maintained in the turn as in the straight flight? 

Limiting Radius. In turns more horsepower is required than in 
straight flight because more lift is needed, and this greater lift is 



LIMITING RADIUS 255 

attained by increasing either velocity or angle of attack, both of 
which increases drag. Since maximum speed in straight flight 
entails using all the available horsepower, it is impossible at 
top speed to make a perfect turn, i.e., without shpping or 
squashing. 

At speeds less than maximum, the excess horsepower can be 
used to accomplish turns. This extra horsepower is a fixed 
quantity for each airspeed, so that for each airspeed there is a 
limit to the horsepower that can be used on turns. The smaller 
the radius of turn, the greater horsepower is required; therefore, 
if the horsepower available is fixed, there is a definite limit on the 
sharpness of the turn. 

The horsepower available at any velocity being known, the drag 
coefficient may be computed. The angle of attack and lift coeffi- 
cient are then known. With lift coefficient and velocity known, 
the lift may be found. The greatest angle of bank may be then 
computed, and from that, the corresponding minimum radius of 
turn for a series of different velocities; it will be discovered that 
at one airspeed, which is the same airspeed as that for best climb, 
the radius of turn is smallest. With the assigned power plant 
this will be the sharpest possible turn, though with increased 
power a turn of even smaller radius can be made. 

Example. Find the minimum radius of turn for an airplane weigh- 
ing 2,000 lb., having a Clark Y wing 36 ft. by 6 ft. and 3.8 sq. ft. 
equivalent flat plate area of parasite and a 150-hp. engine rated at 
1,800 r.p.m. 

Solution. This is the same airplane as in the first illustrative ex- 
ample in the chapter on airplane performance for which the horse- 
power-available curve is plotted in Fig. 59. 

- Calculations are made for various airspeeds of which the following 
is a sample. 

F = 110 miles per hour 
= 162 ft. per sec. 
H.P.avaii. = 110, from Fig. 59 

HP =rC i 1-^8 a \ 0.00256 X^SXT^ 



110 



375 
1.28 X 3.8\ 0.00256 X 216 X 110^ 



fr I ^-^^ X3.8 \0 
V^"^ 216 J 375 

Cd = 0.0335 



256 



TURNS 



From Fig. 17, 




<x = 


= 4.1° and Cl = 0.65 


Lift = 


= Cl X 0.00256 /SF2 


= 


= 0.65 X 0.00256 X 216 X lio' 


= 


4,330 lb. 




W 


cos /3 = 


Lift 




2,000 




4,330 


= 


0.461 


^ = 


62° 33' 


tanjS = 


1.925 




72 


tan/5 = 


gR 


R = 


162^ 


32.2 X 1.925 


= 


424 ft. 



The results of these computations are shown in Fig. 726. The mini 
mum permissible radius of turn has its smallest value when the air 
speed is 65 miles per hour. 

500 



c 400 



S300 



oc 200 



100 
50 













y 












^ 


y 






\ ^ 




^ 


X 



















At any airspeed, a wider turn 
can be made than shown in 
Fig. 726, which, since it will re- 
quire less horsepower, will per- 
mit throttling the engine. 



60 



70 80 90 100 

Miles per Hour 



10 120 



Fig. 726. 



Minimum radius of properly 
banked turn. 



Spiral Glide. In discuss- 
ing a turn without loss of alti- 
tude, it was shown that more 
power is needed in turns 
than in level flight. In case 
of engine failure, the force that causes motion along the flight path 
is a component of the weight of the airplane itself. If a turn is 
desired while descending in a glide, for example, in order to land 
up-wind, greater lift must be attained or squashing will result. 
In general, lift can be gained either by increasing the angle of 
attack or by increasing the velocity. With a dead engine, velocity 
can be increased only by diving more steeply, i.e., increasing the 
angle of glide. If without the turn the airplane was descending at 
minimum angle of glide, in turning the angle of glide will be 
increased. 



SPIRAL GLIDE 257 

In a spiral glide, the airplane is descending on a helical path, 
this helix being on the surface of an imaginary circular cylinder 
whose radius is r. At any instant, the tangent to the flight path 
is at an angle d to the horizontal tangent to the cyHnder at that 
point. The component of the weight in the direction of the flight 
path which causes the motion along that path is W sin 6, and this 
equals the total drag as in a straight glide. Because the wing is 
banked, the direction of lift is inward and it is the component of lift 
in the vertical plane through the longitudinal axis of the airplane 
(L cos jS) that is equal and opposite to the component of weight 
(TF cos 6) perpendicular to the drag. 

While centrifugal force is actually acting in a direction slightly 
below the horizontal, it is usual to consider only its horizontal 
component {WV^ cos^ ^/S'^)? which is balanced by the horizontal 
component of lift (L sin /3). 

Summing up the above statements, using D to represent total 
drag of the airplane, 

D = TF sin 

L cos jS = TF cos d 

r . ^ TFF2cos2(9 
L sm jS = 

Dividing the first by the second equation gives 

tan Q = 7 

L cos jS 

And from the third equation 

TFF2 cos2 e 



r = 



gL sin j8 



Since 2 irr is the horizontal distance traveled in a complete turn, 
the altitude lost in a complete turn is 2 irr tan d. Then the mini- 
mum altitude will be lost per turn when conditions are such that 
r tan 6 is minimum. 



gL sin /3 Cl cos /3 
gL sin /3 cos ^Cl 



258 TURNS 

2 W cos2 d 



g^s(^^y sin ^ cos ^ 



cos^ 6 



/2 foj^^"'^ 



Since in all cases the airplane would be flown at the flattest 
possible angle of glide, d will be small and cos 6 will be nearly equal 
to one. The slowest rate of descent wifl be when both Cl^/Cd and 
sin 2 j8 have their largest values. An investigation of standard air- 
foils shows that Cj}ICd has a maximum value close to the burble 
and sin 2 ^ will have its greatest value when jS is 45°. The above 
expression shows that with a bigger wing loading (TF/aS), the 
minimum rate of descent is decreased. 

Comparison may be made of the angle of glide ^ in a simple 
glide, where B = cot~^ L/D, with the angle of ghde in turning 
where 6 = cot~^ cos (3 {L/D). Whereas in the straight ghde, the 
angle of attack is chosen for maximum L/D, in the turning ghde 
the angle of attack is chosen to be near the stalling angle and the 
L/D is for that angle. When /? = 45°, cosine ^ = 0.707. For the 
example of straight glide given earlier in this book, it was shown 
that the minimum gliding angle would be 4.8°. For the same 
airplane Cl^/Cd is maximum at 18°, where Cl/Cd is 8.3. The 
cotangent of d is 8.3 X 0.7 1 = 5.9 and d is 9.6° . The velocity 
during the glide wiU be Vw sin d/iCn [p/2] S) (where W/S = 
2000/216), so that for this example the velocity is 83.6 ft. per sec. 
(57.0 miles per hour). The radius of turn corresponding to a 45° 
bank and a ghde of 9.6° is V^ cos d/{g tan ^) or 214 ft. The loss 
of altitude in one complete turn (2 irr tan 6) is 227 ft. 

Loading in Turns. In straight level flight, the lift on an airplane 
is equal to the total weight of the airplane plus or minus a small 
tail load. Practically the lift is considered equal to the weight. 
This is called unit load, or since the force is equal to the mass of 
the airplane multipHed by the normal acceleration due to gravity, 
it is called a force of one g (where g is 32.2 ft. per sec. 2). 

During a turn, the lift must equal the weight divided by the 
cosine of the angle of bank, so that the greater the angle of bank, 
the greater must be the lift force. The ratio of lift or load during 
any maneuver to the unit load is called the load factor. Since 
the mass of the airplane does not change, it may be stated that in 
a maneuver an airplane is stressed to 3 ^, 4 g, etc., meaning that 



LOADING IN VERTICAL BANKS 259 

the force or loading is the mass of the plane multiplied by 3gr, 4gr, 
etc. 

In turns, the load factor is ■:fj7 = 3. 

' W cos p 

Example. In a 45° banked turn, what is the load factor? 



Solution. 



1 



cos 45° 0.7071 
= 1.41 

Load factor is 1.4 or plane is stressed to lAg. 

Problems 

1. Airplane is making a turn of f-mile radius at a speed of 150 
miles per hour. What is the load factor? 

2. Airplane is making a turn of 150-yd. radius at a speed of 180 
miles per hour. What is the load factor? 

3. Airplane is making a turn of 100-yd. radius at a speed of 175 
miles per hour. What is the load factor? 

4o Airplane is making a turn of 200-ft. radius at a speed of 190 
miles per hour. What is the load factor? 

5. Airplane is making a turn of 150-ft. radius at a speed of^200 miles 
an hour. What is the load factor? 

Loading in Vertical Banks. In a turn wherein the wings are 
banked up vertically, lift acts horizontally. At high speeds un- 
doubtedly some small amount of lift is derived from the sides of 
the fuselage, but this cannot be sufficient to sustain the weight 
of the airplane, so that sideslip will take place. In a turn of this 
character, centrifugal force becomes very great. 

In flying into such a turn, no matter if the immediately previous 
flying was at high speed and low angle of attack, the airplane 
would probably '' squash " or settle outward. This movement 
would change the angle of attack to a very high angle and the 
increasing drag would cut down on the airspeed; an accurate 
computation of the loading in a vertical bank becomes exceedingly 
complicated. Some idea of the loading may be obtained by the 
following considerations. 

In a vertical bank, lift acts inward; centrifugal force acts out- 
ward. Neglecting " squashing," lift should equal centrifugal 

force. 

o WV^ 



260 TURNS 

In straight level flight, slowest speed, i.e., stalHng speed, is 
flown when the angle of attack is that of maximum lift coefficient 
and 

Substituting this value in the above equation gives 



Then 



^4^^^ = ,R 



C^LmsiX. y s 
L = 



gR 



and R=^X -^• 

9 Cl 

The radius of turn will be minimum when Cl is equal to Cimax., 
that is, when the angle of attack is that of maximum lift. If, 
owing to excessive centrifugal force, the airplane squashes out- 
ward, the horizontal curve described will be of greater radius 
than if the stick is pushed forward somewhat, so that the angle 
between the flight path and the wing is the angle of maximum 
lift. 

This kind of turn is not to be confused with that described in the 
previous section; as in a vertical bank, the drag will be excessive 
and the speed will decrease during the turn. 

Since the minimum radius of turn will be when Cl = Czmax., the 
turning radius will be 

J? — ^ 

J^min. — 

The load factor in a vertical bank will be 

J CLmax. 2^^^ 



= 11 

It is to be noted that minimum speed varies with altitude, so 
at altitude the minimum radius will be greater. 



DIVES 261 

Example. An airplane, whose stalling speed is 40 miles per hour, 
makes a vertically banked turn of minimum radius. The initial 
speed entering the turn is 125 miles per hour. What is the radius, 
and what is the load factor? 



Solution. 



■timin. — r~ 



Q 

(40 X 1.47)2 
32.2 
= 107 ft. 

Load factor = -77-^ 

= 9.8 

Problems 

1. A Curtiss pursuit plane whose stalling speed is 58 miles per hour 
goes into a vertically banked turn of minimum radius at a speed of 170 
miles per hour. What is the radius, and what is the load factor? 

2. A Gee-Bee Supersportster whose stalling speed is 91 miles per 
hour goes into a vertically banked turn of minimum radius at a speed 
of 270 miles per hour. What are the radius and the load factor? 

3. A Supermarine racing plane whose landing speed is 107 miles 
per hour goes into a vertically banked turn of minimum radius at 409 
miles per hour. What are the radius and load factor? 

4. An airplane whose stalling speed is 38 miles per hour goes into a 
vertically banked turn of minimum radius at 80 miles per hour. 
What are the radius and load factor? 

5. A Curtiss Condor whose landing speed is 55 miles per hour goes 
into a vertically banked turn of minimum radius at 140 miles per hour. 
What are the radius and load factor? 

Dives. When a plane is flying horizontally and the pilot pushes 
the stick forward moving the elevator down, the first action is for 
the tail to move up. If the stick movement is fairly quick so that 
the tail is thrown up suddenly, momentum carries the airplane 
onward in a horizontal path. The relative wind strikes the upper 
side of the wing, that is, momentarily there is a negative angle of 
attack, with a consequent negative lift. During that instant, mo- 
mentum takes the place of weight and acting horizontally forward, 
with respect to the axis of the plane it acts in a direction which in 



262 TURNS 

normal flight would be forward and upward. This loading situa- 
tion is called inverted flight condition since it essentially duphcates 
flying upside down. When actually flying upside down, with the 
conventional unsymmetrical airfoil, flight must be at a high nega- 
tive angle of attack with consequent big drag, so that the airspeed 
will be comparatively low. The load factors obtained in the in- 
verted flight condition at the beginning of a dive are much greater 
than those obtained in actual inverted flight, so the former load 
factors are those used in stress analysis of the airplane structure. 

As the airplane goes into the dive the pilot keeps the nose of the 
airplane depressed, or else with increased velocity the increased Hft 
will pull the airplane out of the dive. 

In coming out of the dive, the pilot pulls back on the stick which 
moves the elevator up tending to throw the tail downward with 
respect to the flight path. Momentum tends to make the plane 
settle or squash so that momentarily, at least, the airplane is at 
high angle of attack while still retaining the high speed attained 
in the dive. If V is the velocity when the airplane is pulled out of 
the dive into the high angle of attack position, then 



but 



and 



L = CLmax.2^^^ 



W = CLm...^SVs' 



Load factor = tt? = 

72 



The expression for load factor is the same as in a vertical bank, 
but an airplane usually goes into a bank from level flight whereas 
in a dive the velocity may be much greater than maximum level 
flight velocity. Coming out of a dive causes one of the most 
severe stresses put on airplanes, and unless care is taken the wings 
may be pulled of£. By coming out of a dive gradually, the stresses 
are much less. 



LOOPS 263 

Problems 

1. A Waco plane whose landing speed is 49 miles per hour is pulled 
out of a dive at 175 miles per hour. What is the load factor? 

2. A Lockheed plane whose landing speed is 64 miles per hour is 
pulled out of a dive at 250 miles per hour. What is the load factor? 

3. An Aeronca whose landing speed is 36 miles per hour is pulled 
out of a 120-mile-per-hour dive. What is the load factor? 

4. A Supermarine racer whose stalling speed is 107 miles per 
hour is pulled out of a dive at 450 miles per hour. What is the load 
factor? 

5. A Northrop airplane whose stalling speed is 58 miles per hour is 
pulled out of a dive at 220 miles per hour. What is the load factor? 

Loops. In looping, a turn is executed in a vertical plane. 
Since, at the start of this maneuver, the pilot pulls back on the 
stick putting the airplane at a high angle of attack, with big drag 
which decreases the airspeed, it is important that the velocity be 
high in beginning the loop. Unless there is ample power, the pilot 
starts his loop by diving to gain speed. 

The loop is rarely a perfect circle, for the airspeed will decrease 
as the plane climbs up preparatory to going over on its back, and 
after the top of the loop has been passed, the airspeed will increase 
again. 

The exact load factors in a loop cannot be calculated unless the 
exact pattern of the loop is known, but an approximate idea may be 
gained if it is assumed that the path of the loop is a vertical circle. 
If R is the radius of this circle, Vi the velocity and Li the lift at 
the bottom of the loop, and V2 the velocity and L2 the lift at the 
top of the loop, then 

The work done in bringing the airplane from the bottom of the 
loop to the top of the loop is the weight times the vertical distance 
or 2 RW. In climbing, the energy lost is the difference in kinetic 
energy. 



264 TURNS 

Therefore 

ARg = 7i2 - V2' 

Assuming that at the top of the loop the Hft is zero, which is 
approximately true in most cases, 



But 



W 72- 



L. = o=— ^-W 



W=^^ 



g R 

Rg 



Then since Fi^ - Fa' = 4.Rg 

Fi2 -Rg^ 4.Rg 



Fi2 = 5Rg 



W Fi2 
g K 
_W5Rg , ^ 



9 R 
= QW 



or the load factor is 6^. 



CHAPTER XIII 



THE CONTROL SURFACES 

Axes. An airplane may rotate about three axes. The longi- 
tudinal or fore-and-aft axis is an imaginary line through the center 
of gravity of the airplane parallel to the thrust line of the propeller. 
It is called the X axis. The lateral axis is an imaginary line 
through the center of gravity of the airplane, perpendicular to the 
X axis, and horizontal when the airplane is on an even keel. The 
lateral axis is called the Y axis. The vertical or Z axis passes 

through the center of gravity 
perpendicular to the X and Y 
axes and is vertical when the 
airplane is on even keel; see 
Fig. 73. 

Rotational Motions of the 
Airplane. Rotation about the 
X axis is called roll. By tacit 
agreement among designers, it 
is termed positive roll if, when 
viewed from the rear, the air- 
plane rotates in a clockwise 
direction. A rising left wing is the beginning of a positive roll. 
Rotation about the Y axis is called pitch. If, when viewed from 
the left wing tip, the airplane rotates clockwise, it is positive pitch. 
Rotations are caused by moments. The moment which causes the 
airplane to rotate in a positive pitch is called a stalling moment; 
that which causes it to rotate in a negative pitch is called a diving 
moment. 

Rotation about the Z axis is called turn or yaw. If when viewed 
from above the airplane rotates clockwise, it is positive yaw. A 
moment tending to cause a right turn is a positive yawing moment; 
that tending to cause a left turn is a negative yawing moment. 

Angle of Attack of Airplane. Earher in this book, the angle of 
attack of the wing was stated to be the angle between the direction 
of the relative wind and the wing chord. The wing chord is not 

265 




Fig. 73. Axes of airplane. 



266 THE CONTROL SURFACES 

usually parallel to the longitudinal axis of the airplane. The 
angle of attack of the airplane is the angle between the direction 
of the relative wind and the longitudinal axis of the airplane. 

The angle between the chord of the wing and the longitudinal 
axis of the airplane is the angle of incidence. The angle of inci- 
dence is therefore the numerical difference between the angle of 
attack of the wing and the angle of attack of the airplane. Usu- 
ally airplanes are so rigged that the angle of incidence is equal to 
the angle of maximum L/Z) of the airfoil section used. 

Control Surfaces. It is necessary that means be provided to 
cause or stop rotation about any of the three axes, in order either 
to maintain the airplane in straight level flight or to execute vari- 
ous maneuvers. It is universal practice nowadays to make 
each control surface movable with respect to a fixed surface. 

In the Canard type the horizontal and vertical rudders are in 
front of the main wing; in the conventional type, they are at the 
rear of the airplane. 

The control surfaces for producing or regulating turns are the 
vertical tail surfaces. The fixed part is the vertical fin ; the movable 
part is the rudder. Movement of the rudder is by the feet of 
the pilot. Pushing forward with the right foot on the rudder 
bar moves the rudder to the right and causes a right turn. 

The control surfaces for producing or regulating pitch are the 
horizontal tail surfaces. The movable part is the elevator or 
" flipper "; the " fixed " part is the stabilizer. Though termed 
" fixed," the stabilizer except on small light planes can be ad- 
justed during flight from the cockpit. Movement of the elevator 
is made by means of a control stick or post. Pushing forward on 
the stick causes the nose to go down; pulling back on it causes 
the nose to rise. 

The control surfaces for producing or regulating roll are the aile- 
rons. They are located at the rear of each wing tip. They are 
controlled by the control stick on a small plane and by a wheel 
(Deperdessin or " Dep " control) on large planes; moving the 
stick to the left makes the left aileron go up and the right aileron 
go down, causing negative roll. Moving the stick to the right 
causes the left aileron to go down and the right aileron to go up, 
producing positive roll. 

Ailerons. Earher attempts at fljang were thwarted because 
the experimenters did not realize the need for lateral control. 



CONTROL SURFACES 267 

The Wright brothers attained complete control by providing lat- 
eral control. Their first wings had simply upper surfaces, the 
ribs being strips of hard wood which had been steamed and bent 
to the proper curvature. Lateral control was effected by warping ; 
that is, wires connected to the rear corners of the rectangular wings 
could be pulled, forcing the wing to be bent to a greater curvature. 
Upon release of the wire, the natural springiness of the wooden rib 
would cause it to return to its former shape. 

Putting a lower surface on the wing precluded warping. The 
Wrights then used flat surfaces fastened onto the outer forward 
struts of their biplane. To produce a positive roll, the surface at the 
left side was tilted so as to slope backward and downward, on the 
right side the surface sloped backward and upward. It was direct 
impact pressure on these surfaces that caused the wing to move 
up or down. 

The conventional type of lateral control is the aileron. The 
outer rear portion of each wing is movable, the hinged edge being 
more or less parallel to the leading edge. In normal flight these 
ailerons form part of the wing surface. Rarely they are rigged to 
droop slightly to give greater lateral stability. 

With both ailerons in normal position, the lift on the left wing 
is the same as the lift on the right wing. When an aileron is 
moved down, it changes the airfoil section of the outer portion of 
the wing to an airfoil section of greater curvature and conse- 
quently greater lift. Moving an aileron upward changes the air- 
foil section to one of less curvature or of reverse curvature causing 
less lift. 

Ailerons are designed so that moving an aileron down on one 
side causes the wing on that side to increase in lift by the same 
amount that the lift of the other wing is decreased by its aileron 
being moved up. Moving the control stick should change, not 
the total lift, but merely the proportion carried by each side. 
Unfortunately with the simplest ailerons this movement produces 
more drag on the side of the " down " aileron than on the side with 
the " up " one. 

Moving an aileron down not only changes the airfoil to one of 
greater camber but also changes the chord so that the angle of 
attack is greater. 

All control surfaces, whether they be ailerons, elevators, or 
rudders, are more effective at high speeds, since the forces acting 



268 THE CONTROL SURFACES 

on the surfaces vary as the square of the air velocity. The 
controls are said to be sluggish at low airspeeds. At high speeds 
considerable force is required to move a control surface away 
from its neutral position. At low airspeeds there is little re- 
sistance to the movement of a control surface. Even with slight 
experience a pilot knows when he is getting up near a stalled 
position by the looseness (or ease of movement) of his controls 
and by lack of response of the airplane to movements of the control 
surfaces. 

Ailerons are very ineffective at high angles of attack. Exami- 
nation of most of the common airfoil sections shows that the 
angle of maximum lift is about the same for them all. Then 
changing the shape of the airfoil by dropping an aileron will not 
increase lift ; the change in chord puts the angle of attack beyond 
the burble point. The yawing effect of ailerons at high angles of 
attack is very pronounced, and it has been facetiously stated that, 
^' at high angles, ailerons are good rudders." 

To remedy the lack of effectiveness of ailerons for lateral con- 
trol, slots, spoilers, or floating ailerons may be used. 

Slots are primarily devices for giving higher lift, but since at 
high angles of attack they direct the airflow so that it follows di- 
rectly the contour of the wing instead of burbling, slots add to the 
effectiveness of the ailerons. 




Closed 




Open 

Fig. 74. Spoilers. 

Spoilers. Though used very Kttle, spoilers are quite effective 
for lateral control at high angles of attack. A spoiler consists 
of a long metal strip located on the top of the wing above the front 
spar. The strip can be quite narrow (see Fig. 74). 

It is hinged at the front edge. In normal flight, this strip lies 



FLOATING AILERONS 269 

flat against the upper surface of the wing. By moving the stick 
to one side, the spoiler on the wing on that side stands erect. 

The action of the spoiler is to destroy the smooth flow of air over 
the wing so that burbUng takes place. The lift is consequently 
greatly reduced. 

The outstanding objection to spoilers is that, unlike ailerons, 
their use reduces the total lift. When the spoiler is raised on one 
wing, the lift on the other wing remains as before. The reduction 
of lift on the wing with the raised spoiler means the same reduction 
in the total lift, and the airplane loses altitude. Though im- 
material at moderate altitudes, this is exceedingly dangerous 
close to the ground as a crash might result. 

Floating Ailerons. The floating aileron was installed on the 
Curtiss Tanager which won the Guggenheim Safety Contest in 
1929. The floating aileron is placed outside the tip of each wing. 
The aileron is a symmetrical airfoil. Its axis is slightly behind 
the leading edge. Weight is added to the leading edge so that 
the aileron is balanced statically. 




Fig. 75. Floating ailerons. 

The ailerons are so rigged that they turn freely up and down 
provided that the ailerons on opposite wing tips rotate together. 
They will set themselves in the plane of the relative wind, much 
as a weather vane points into the wind. The pilot has no control 
over this action. 

The pilot can displace one aileron with respect to the other. 
No matter at what angle the floating ailerons may have adjusted 
themselves with respect to the wing chord, the pilot by his con- 
trol can move one aileron up and the other down by the same angle. 



270 



THE CONTROL SURFACES 



Since the airfoils are symmetrical, and the positive angle on one 
side the same as the negative angle on the other, the plus lift equals 
the minus lift and the total lift remains the same. In addition, the 
drag caused by the up aileron is the same as the drag of the down 
one, so that there is no yawing or turning tendency. 

Balanced Control Surfaces. Since moving a surface against the 
pressure of a high wind requires considerable force, control sur- 




Paddle Type 




Handley-Page Type 



C 



Overhang Type 
Fig. 76. Balanced ailerons. 

faces are quite often of the balanced type. The axis about which 
the control surface rotates is not immediately at the front edge 
but back a distance so that a portion of the surface is ahead of the 
hinge. As soon as the surface is displaced from its neutral posi- 
tion, the air strikes the front portion and reduces the effort the 
pilot must exert. 

Although balanced controls enable the pilot to move large 
control surfaces easily, care must always be observed that they are 
not too well balanced. A pilot might then with Httle effort move 



SIZES OF AILERONS 



271 



the surface quickly to an extreme position, thus changing the 
motion of the plane suddenly, which causes high dynamic 
loads. 

Differential Ailerons. Because the angular movement of an 
aileron downward causes more drag than the same angular move- 
ment upward, ailerons are sometimes rigged so that, when the 
stick is moved to one side, the aileron that moves upward has 
approximately twice the angular movement of the aileron that 
moves downward. This is done so that the drag on each wing tip 
shall be the same in order to eliminate yawing moment. 

Frise Aileron. Another method of reducing the yawing moment 
is by use of the Frise type of aileron. Bearing a resemblance to 
the balanced type, the aileron is hinged about 20 per cent of the 
chord back from the leading edge and so designed that, when in its 




Down Position ^ Up Position 

Fig. 77. Frise aileron. 



^' down " position, the upper surface of the aileron is a smooth 
prolongation of the curved upper surface of the wing. In its 
^' up " position, the leading edge of the aileron projects below the 
continuation of the curved lower surface of the wing, so that addi- 
tional drag is arbitrarily introduced at this point. When properly 
designed, the drag of both ailerons is the same, and there is no 
yawing moment. 

Sizes of Ailerons. Ailerons are for the production of a rolling 
moment. This rolling moment, in foot-pounds, is the difference 
in lift between the two wings multiplied by the distance in feet 
between the centers of pressure on the two wings. This moment- 
arm is a function of the span. The rolling moment is usually 
expressed as 



272 THE CONTROL SURFACES 

Le = coeflficient of rolling mo- 
ment, dimensionless (N.B. : 
not to be confused with 
Rolling moment = Le^SwV% lift coefficient) 

Sw = wing area, square feet 
V = airspeed, feet per second 
b = span, feet 

In practice, it has been found that the coefficient Le in the 
above equation should be at least 0.03 for all angles of attack. 

To achieve this rolling moment it has been found that the total 
area of the ailerons should be about 10 per cent of the total wing 
area; 3 per cent below or above this value is permissible. The 
chord of the aileron is usually decided by the position of the rear 
spar of the wing, since the aileron hinges cannot be forward of the 
rear spar. This means that the aileron chord may be from 20 
to 33 per cent of the wing chord. The spans of ailerons are 
between 40 and 65 per cent of the semi-spans of the wing. 

Horizontal Tail Surfaces. The horizontal tail surface is very 
important as with it the airplane is " trimmed " so that with vary- 
ing loads the plane may be flown at any angle of attack up to the 
maximum. It is common to use thin symmetrical or nearly sym- 
metrical airfoil contours for tail surfaces. The size of the total 
horizontal surface and the distance behind the wing will be dis- 
cussed in the following chapter on stability. The elevator is 
usually about 45 per cent of the total area although variations of 
plus or minus 10 per cent from this value have worked satisfactorily 
in practice. The aspect ratio of the entire tail surface is usually 
in the neighborhood of 3. 

A large sudden movement of the elevator will cause a violent 
maneuver, putting severe strains on the airplane structure. For 
this reason, the elevators on commercial airplanes are not usually 
permitted to move more than 20° above or below their neutral 
position. On military planes greater movement is permitted, but 
it rarely exceeds 45°. 

Occasionally an airplane is advertised as incapable of spinning. 
Spins usually start from stalls, and it is sometimes true that these 
airplanes have insufficient tail area to force them into a stalled 
position. This improper amount of tail surface is usually in- 
dicative of lack of controllabihty. 



EFFECT OF PROPELLER ACTION 273 

The stabilizer or " fixed " part of the horizontal tail surface has 
its rear edge fastened to a cross-member; the front edge may be 
moved up or down to change the stabilizer setting. Usually a 
vertical screw is used to adjust the height of the front of the stabi- 
lizer; turning the screw feeds the leading edge up or down, an 
arrangement being provided so that the screw may be rotated 
by a crank or handwheel in the pilot's cockpit. The screw by its 
leverage action permits easy adjustment by the pilot while it 
resists any changes of setting which might be caused by air forces 
acting on the tail. 

The rear end of a stabilizer can be braced and thus made quite 
rigid, but the front end of an adjustable stabilizer cannot be braced 
and is held only at its central part. Even with a very sturdy 
structure the outer forward ends are apt to flutter in the slip- 
stream. There appears to be a tendency at the present time to 
make the stabilizer fixed and non-adjustable. Necessary trim of 
the airplane longitudinally is attained by the use of tabs or bungees. 

A tab is an auxiliary control surface, hinged to the rear edge of 
the elevator. When set at a definite angle to the plane of the 
elevator by means of a control wheel in the pilot's cockpit, the 
tab will remain at that angle with respect to the elevator regard- 
less of any movement of the elevator. Stability can be attained 
by use of tabs in much the same way as with an adjustable stabi- 
lizer. They have the further advantage that their drag is much 
smaller than that of an adjustable stabilizer when the latter is 
meeting air at a large angle of attack. 

Vertical Tail Surface. The vertical tail surfaces are to aid in 
maintaining directional stability and to cause turns. The area 
of the fin and rudder combined is usually between 5 and 6 per 
cent of the total wing area and between 40 and 45 per cent of the 
horizontal tail area. The movable rudder is usually 60 or 70 
per cent of the total vertical area. The average maximum per- 
missible movement of the rudder to either side of the center fine 
of the airplane is 30°. 

Effect of Propeller Action. If an ideal propeller could be de- 
vised, the air, after being acted on by this propeller, would pass 
directly to the rear. In practice, however, air is dragged around 
slightly with the propeller so that the slipstream has a helical 
motion; the rotation is clockwise viewed from the rear. The 
vertical tail surfaces receive therefore slightly more pressure on 



274 THE CONTROL SURFACES 

their left than on their right side, this tending to turn the airplane 
to the left. 

For every action there is an equal and opposite reaction. Since 
a rotation is imparted to the propeller, the reaction of the propeller 
on the engine tends to turn the engine in the opposite direction. 
This causes greater pressure on the engine mount on the left side 
than the right side. As the flat surface of the wings would offer 
great resistance to a rapid rotation of the airplane in the opposite 
direction of rotation to that of the propeller, it is quite common to 
give more lift to the left wing than to the right wing to oppose 
this action. The left wing may be set at a very slightly greater 
angle of incidence, or in a biplane by tightening the rigging the 
left wing may be warped or twisted to have slightly greater cam- 
ber. The latter is called '' wash in." As a result of giving the 
left wing greater lift, the left wing has greater drag tending to 
turn the airplane to the left. 

This left-turning tendency can be corrected by the pilot's 
using a small amount of right rudder, or the vertical fin can be set 
at a slight angle to the plane of symmetry. 

The slipstream action and torque reaction are of course evident 
only when the engine is turning over at normal speed. If the en- 
gine stops and the airplane is in gliding flight, the slipstream and 
torque effects disappear, and if the fin has been offset the pilot 
will have to apply left rudder to prevent turning to the right. 

It is to be noted that with twin engines, one mounted on each 
wing, the torque reaction is doubled, not neutralized. 



CHAPTER XIV 
STABILITY 

Definition of Stability. Stability is the property of a body 
which, when the body is disturbed from a condition of equi- 
librium, causes forces or moments which act to restore it to its 
original condition. The greater the disturbance or change from 
its equilibrium position, the greater will be the magnitude of the 
forces or moments tending to return the body to its original 
attitude. 

Stability of an airplane means that the airplane tends to re- 
main at the same attitude with respect to the relative wind. It 
does not imply that the airplane is steady or that the airplane does 
not wobble with respect to the ground or to a fixed reference point 
in space. If the air is rough, the airplane may be constantly 
changing its attitude with respect to the ground. 

The factors which make for a stable airplane are factors which 
preclude maneuverability. For stability, whenever attempt is 
made to change the attitude of the plane, forces resist this change. 
In maneuvering, these forces oppose any alteration in the flight 
path. Racing or pursuit planes should have little or no stability. 

A plane is statically in equilibrium, if, when in flight, the sum 
of all forces acting in all directions equals zero and the sum of all 
moments about any point equals zero. The first part of the fore- 
going statement may be expressed as : the body is in equiUbrium 
when the sum of the vertical forces is zero and the sum of the 
horizontal forces is zero. 

27 = 
S^ = 
SM = 

If the body is disturbed from its equilibrium position, for 
stability there must be a restoring moment, and this moment must 
be larger for larger displacements from equilibrium position. 

To be statically stable, the airplane must have the character- 
istic that a restoring moment or force acts in a direction to move 
the airplane back to the attitude from which it was disturbed. 
In discussing static stability, no thought is given to the magnitude 

275 



276 



STABILITY 



of the restoring moment. This moment which acts to return the 
airplane to its equilibrium position may cause the airplane to 
acquire angular momentum so that it will swing past that position. 
Owing to the stability characteristic, another restoring moment 
will then act in the opposite direction, so that even with static 
stability there may be oscillation. 

For dynamic stability, in addition to the requirement for static 
stability that there shall be a restoring moment, there is the 
further requirement that the moments created shall be of such 
character that the amplitudes of any displacement shall be of de- 
creasing size so that the airplane will cease to oscillate and come 
to rest in its equilibrium position. D3rQamic instability would 
mean that the restoring moment is so strong that each successive 
oscillation would have a bigger amplitude; such an action would 
mean impossibility of control, and disaster. The calculation of 
dynamic stability is an involved process and will not be treated 
in this book. 

As there are three axes of rotation, so there are three classes of 
stability — longitudinal or fore- 
and-aft stability, lateral sta- 
bility, and directional stability. 
They are interrelated, as roll- 
ing may produce turning, and 
vice versa. 

Mean Aerodynamic Chord. 
Any discussion of longitudinal 
stability involves a consider- 
ation of the mean aerodynamic 
chord of the wing or wings, 
usually abbreviated to M.A.C. 
The mean aerodynamic chord 
of a wing cellule is the chord 
of an imaginary airfoil which 
throughout the flight range 
will have the same force vectors as those of the wing cellule. 
This imaginary airfoil does not need to resemble any known air- 
foil. When the mean aerodynamic chord is obtained from, wind- 
tunnel data of tests on a model of the airplane itself its location 
and length are definitely known. When such tests have not been 
made the following rules apply. 




78. Mean aerodynamic chord 
of tapered wing panel. 



MEAN AERODYNAMIC CHORD 



277 



For a rectangular monoplane wing, the mean aerodynamic chord 
is identical with the chord of the wing section. 

For a tapered monoplane, the mean aerodynamic chord of each 
wing panel passes through the centroid of the plan view of the 
wing panel. The leading edge of the M.A.C. is on a line connect- 
ing the leading edges of the root and tip sections, and the trailing 
edge of the M.A.C. is on a line connecting the trailing edges of the 
root and tip sections (see Fig. 78), which shows a swept-back, 
tapered wing. If d is the perpendicular distance from the root 
section to the M.A.C. 

h = perpendicular distance root 
h{a -\- 26) section to tip section 

S{a -\- h) a = length of root chord 

h = length of tip chord 



d = 



The length of the mean aerodynamic chord is 



M.A.C. 



= i[a 



a + 



The distance, m, in a swept-back wing, of the leading edge of the 
M.A.C. to the rear of the leading edge of the root chord, is 



_ s{a + 26) 
"^ ~ 3(a + h) 




Fig. 79. Mean aerodynamic chord of 
wing with straight center section and 
swept-back outer panel. 



s = total sweepback 

In Fig. 78 is shown a 
graphical method of find- 
ing the M.A.C. of a wing 
panel. 

For a monoplane with a 
straight center section and 
swept-back wing panels, the 
M.A.C. of the center section 
and the M.A.C. of the outer 
panel are found separately. 
The distance (e) of the M.A.C. 
of the entire semi-wing from 



the center line of the airplane (see Fig. 79) will be 



Sc + Sf, 



h' 



Sc 

Sp 



= distance from center line of airplane 
to M.A.C. of semi-center-section 

= distance from center line of airplane 
to M.A.C. of wing panel 

= total area of center section 

= total area of both wing panels 



278 STABILITY 

The length of the mean aerodynamic chord is 

Cp = length of M.A.C. wing 
M A P -r I (^c-Cp)(h''-e) panel 

• ^"^ h^'-h' Cc = length of M.A.C. center 

section 

The rearward distance (n) of the leading edge of the M.A.C. 
behind the leading edge of the center section is 

^ m(e — h') m = distance of L.E. of M.A.C. of wing 

h" — h' panel behind L.E. of center section 

For a monoplane with dihedral, 
the M.A.C. of the center section and 
the M.A.C. of the outer panel are 
found separately. The distance 
in') (see Fig. 80) by which the 
M.A.C. is elevated is 

.,(e-^') 



m 



h" 




of outer panel is raised 



Fig. 80. Mean aerodynamic chord 
of wing with dihedral. 



M.A.C. = 



For biplanes, the M.A.C. 's for the upper and lower wings are 
found separately. The length of the M.A.C. of the biplane is 

Su = area upper wing 
Sl = area lower wing 
eCuSu + ClSl Cu = M.A.C. upper wing 
eSu + aSl Cl = M.A.C. lower wing 

e = relative efficiency of upper 
wing (see Fig. 82) 

The relative loading e of the upper wing as given in Fig. 82 is 
only approximate but may be used for all angles of attack. For 
accurate work N.A.C.A. report 458 should be used. 

The vertical distance of the leading edge of the M.A.C. of the 
biplane above the leading edge of the M.A.C. of the lower wing is 
g, where 



MEAN AERODYNAMIC CHORD 



279 



The horizontal distance of the leading edge of the M.A.C. of the 
biplane ahead of the leading edge of the lower M.A.C. is s', where 
, sg s = horizontal distance between upper 

^ ~ G and lower leading edges 




1.50 

1.40 

' 1.30 

1 1.20 



o ° 1.10 



Sl.OO 



.90 



.80 







Gap „ „„ 


/ 




n-^n 


1 / 


/ 




Chore 


r'oo^ 


^/ 


y 






; 


^ 




^ 




/. 


^ 


<is-- 


^^. 


A 










/ 













-30 -20 -10 10 
Stagger in Degrees 



20 30 



Fig. 81. Mean aerodynamic chord 
of biplane. 



Fig. 82. Relative wing loading 
in a biplane. 



Problems 

1. A certain monoplane had a rectangular wing with 5-ft. chord and 
32-ft. span, the maximum forward position of the center of pressure 
being at 30 per cent of the chord from the leading edge. It was found 
that the center of gravity of the airplane was to the rear of the maxi- 
mum forward position of the center of pressure. To remedy this 
situation, it was proposed to leave unchanged the center section of 
the wing, the width of which was 4 ft., while the outer portions of the 
wing were to remain parallelograms of the same area but were to be 
given a sweepback of 2 ft. at each tip. (a) Find distance from 
longitudinal center line of airplane to M.A.C. of the altered wing. 
(6) Find distance that leading edge of M.A.C. has been moved back, 
(c) Find distance that maximum forward position of the C.P. of the 
M.A.C. of the wing has been moved back. 

2. Solve problem 1 when sweepback at each tip is 1 ft., wing area 
being the same. 

3. Solve problem 1 when trailing edge is left perpendicular to the 
longitudinal axis of the airplane while leading edge is given a sweep- 
back beginning at the center so that the sweepback at each tip is 2 ft., 
the span being lengthened so that the wing area remains unchanged. 

4. Solve problem 3 when sweepback at each tip is 1 ft., the span 
being of such a length that the wing area remains unchanged. 



280 STABILITY 

5. Solve problem 1 when the center section of 4-ft. width remains 
unchanged, the trailing edge remains perpendicular to the longitudinal 
axis of the airplane, the leading edge outside of the center section is 
given a sweepback so that the sweepback at each tip is 2 ft., the length 
of the span being such that the wing area remains unchanged. 

6. Solve problem 5 when the sweepback at each tip is 1 ft., the span 
being of such a length that the wing area remains unchanged. 

7. The center of gravity of a monoplane with rectangular wing of 
4-ft. chord and 30-ft. span was not far enough below the wing. The 
proposed remedy was to leave unchanged the center section, the width 
of which was 4 ft., while the outer portions of the wing were given a 
dihedral of 2 ft. at each tip. (a) Find the horizontal distance from 
longitudinal center line of airplane to M.A.C. of altered wing. (6) 
Find the distance that M.A.C. of the wing has been moved up. 

8. Solve problem 7 when the dihedral at each tip is 1 ft. 

9. Solve problem 7 when the center section of the wing is left un- 
changed, the outer portions being given a dihedral of 2 ft. at each tip, 
the leading edge outside of the center section being given a sweepback 
of 2 ft., while the trailing edge remains perpendicular to the longi- 
tudinal axis of the airplane, the span being of such a length that the 
wing area remains unchanged. 

10. Solve problem 9 when the dihedral at each tip is 1 ft. and the 
sweepback at each tip is also 1 ft., the span being of such a length that 
the wing area remains unchanged. 

Longitudinal Balance. By using the convention of a mean aero- 
dynamic chord, the forces resulting from biplane wings or from a 
monoplane wing having taper, dihedral, or sweepback can be con- 
sidered as the forces acting on a rectangular monoplane wing. 
In the following discussion, though a simple wing will be alluded 
to, it may be considered as the (imaginary) rectangular monoplane 
wing whose chord is the M.A.C. of the actual wing or wings. 

In level flight, the forces which must be considered are the 
weight, acting downward; the propeller thrust, acting forward; 
the lift, acting upward ; the total drag, acting backward ; and the 
tail load, which may be either upward or downward. 

In the conventional high-wing monoplane shown in Fig. 83, 
the thrust and drag both act to produce stalling or positive pitch- 
ing moment. The lift must then produce a negative or diving 
moment. The moment of the tail load must be such as to be equal 
in magnitude to the difference of the plus and minus moments 
and of the same sign as the smaller. In order to ensure that the 



LONGITUDINAL BALANCE 



281 



lift always produces a diving moment, the center of gravity must 
be ahead of the most forward position of the center of pressure of 
the airplane. At high angles of attack, for some airfoils the center 
of pressure moves forward to a position 25 per cent of the chord 
back of the leading edge, so that the center of gravity must be in 
front of that. The tail is a symmetrical airfoil with constant 
center of pressure position. 




Fig. 83. Forces on an airplane in flight. 



In Fig. 83, the distance of the center of gravity above ths 
thrust line is a, its distance below the line of action of the total 
drag is h, its distance ahead of the line of action of lift is e, and the 
distance from the center of gravity to the center of pressure of the 
tail is d. Then, 

T Xa-\- D Xh - LXedtz Tail load X d = 



Example. A monoplane, weighing 3,000 lb., having rectangular 
Clark Y wing of 48-ft. span and 8-ft. chord and 5j sq. ft. of equivalent 
flat plate area, is flying at 100 miles per hour. The center of gravity 
of the airplane is 20 in. back of the leading edge of the wing, 12 in. 
above the thrust line, and 8 in. below line of action of total drag. 
It is 25 ft. from the center of gravity of the airplane to the center 
of pressure of the tail. What should be the tail load? 



Solution. 



Cl = 



W 



0.00256 >S 72 
3.000 



0.00256 X 384 X 100' 
= 0.306 



282 STABILITY 

From Fig. 17 when Cl = 0.306, Cd = 0.015 
Aotai = (cd + ^^^') 0.00256 XSV^ 

= ('0.015 + ^'^^3^ ^•^ ) 0.00256 X 384 X lOO" 

= 322 lb. 
Since Sfl" = 0, Drag = Thrust 

From Fig. 17, when Cl = 0.306, C.P. is at 46 per cent of chord, 
that is, 0.46 X 96 in. = 44 in. back of leading edge. 

Moment arm of lift force = e = 44 in. — 20 in. = 24 in. 

TXa + DXb-LXezL Tail load X 25 X 12 = 

^ .. . , 72,000 - 6,440 
Tail load = ^ttt: 

= 218 lb. 

Tail load must act to give positive moment, i.e., must be down. 

Assumptions for Balance Computations. It is desirable to 
make measurements of pitching moments in the wind tunnel using 
a model of the airplane built exactly to scale. It is impracticable 
to have the propeller turning over in the miniature airplane, so 
that the airspeed is the same over the tail as over the wings. This 
is the condition of gliding flight. When engine failure occurs it 
should be possible to keep the airplane in balance, and if there is 
sufficient tail surface to balance under this condition, there will 
be ample surface for balancing under ordinary flight conditions. 

In computing stability, because of the difficulty of predicting 
where the resultant of all the drag forces will be acting it is custom- 
ary to neglect this moment also. Making the tail moment of such 
size as to be able to balance the lift moment is in the nature of a 
safety factor, for using this as criterion the designer will know 
that there will be sufficient horizontal tail area. 

Tail Angle. Even if the adjustable stabilizer is set parallel to 
the wing, the angle of attack of the tail will not be the same as the 
angle of attack of the main wing because of downwash. The 
air flowing around the wing is given a downward velocity. As the 
air moves backward towards the tail this downward deflection 
becomes less. The direction of the air is different at different 
points above and below the plane of the wing. 

The tail is usually located about three chord lengths back of 
the wing, but there is considerable variation in different airplanes. 



TAIL ANGLE 283 

In seaplanes, the horizontal tail surfaces are high above the plane 
of the wing. Several elaborate formulas have been devised giving 
the angle of downwash as a function of the distance back of the 
wing and of the distance above or below the wing. For very- 
accurate work, these should be investigated and the most desirable 
one applied. 

The Army assumes that the angle of downwash at the tail is 
one-half the angle of attack of the wing measured from the angle 
of zero lift. This appears to be a good representative value of 
downwash. 

It should be borne in mind that the angle of downwash is the 
angle between the relative direction of the undisturbed air in front 
of the wing and the direction of the air after it has passed the wing. 
The angle of attack at the tail surface is called the tail angle. The 
tail angle will depend on the stabilizer setting as well as the angle 
of downwash, which in turn depends on the angle of attack of the 
wing. 

Example. What is the tail angle when stabilizer is set at +10° 
to chord of main wing, if angle of attack of airplane is +5°, angle of 
incidence is +11°; and downwash is — 4°? 

Solution. (Assume relative wind as horizontal.) 
Angle of attack of wing = 5° + li° = 6i° 
Angle of tail with horizontal = 6^ + 10 = 16j° 
Angle of tail with downwashed air = 16j — 4 = +12|° 

Example. Airplane has a Clark Y wing, aspect ratio 6, set at +li° 
incidence. Airplane is at +2° angle of attack. Stabilizer set at 
— 5° to chord of wing. What is tail angle? 

Solution. 

Angle of attack of wing = 2° + 1^° = 3j° 

Same measured from angle of zero lift = 8|° 

Angle of downwash 8i° X i = 4^° 

Angle of tail with horizontal = 3i° - 5° = - li° 

Angle of tail with downwashed air — lj° — 4j = — 5|° 

Problems 

1. Airplane has aGottingen 398 wing, aspect ratio 6, set at +J° 
incidence. Angle of attack of airplane is +5°. What is tail angle 
(a) if stabilizer is set at — 10° to wing chord? (6) if stabilizer is set 
at +10° to wing chord? 



284 



STABILITY 



2. Airplane has a C-80 wing, aspect ratio 6, set at +3° incidence. 
Angle of attack of airplane is — 1°. What is tail angle {a) if stabilizer 
is parallel to main wing chord? (b) if set at — 10° to wing? 

3. Airplane has a Clark Y wing, aspect ratio of 6, with 1|° in- 
cidence. Airplane is at —4° angle of attack. What is tail angle if 
stabilizer is set at 5° to wing? 

4. Airplane has an M-6 wing, aspect ratio of 6, with 0° incidence. 
Airplane is at 8° angle of attack. What is tail angle if stabilizer is 
set at — 10° to wing? 

5. Airplane has a Gottingen 398 wing, aspect ratio of 6 with 1° 
incidence. Airplane is at —4° angle of attack. What is tail angle if 
stabilizer is set at —5° to wing? 

When the angle of attack is not given but the lift coefficient is 
known, for the part of the lift coefficient against angle of attack 
curve which is straight, the slope is constant. Assuming that this 
slope is 0.0718, which is approximately true for all airfoils having 
an aspect ratio of 6, then 

Cl 
'^^•^- 0.0718 

If the wing, fictitious or actual, does not have an aspect ratio of 
6, the angle of attack used in calculating downwash needs to be 
corrected. From Chapter V 

«z.L. = angle of attack measured from zero lift 

chord for aspect ratio of 6 
«'z.L. = angle of attack measured from zero lift 

chord for aspect ratio of problem 
^A.R. = aspect ratio factor from Fig. 39 



« Z.L. 



«zx. 

^AR. 



Then angle of downwash, 



e = Wz.L. 



2 X 0.0718 X Fa.r. 

Example. Airplane, weighing 2,000 lb. has 250 sq. ft. of wing area, 
aspect ratio of 8, and is flying at 100 miles per hour. Angle of attack 
of wing is +7°. Stabilizer set at —10° to wing chord. What is tail 
angle? 

W 
^^ "" 0.00256 ,SF2 
2,000 



0.00256 X 250 X 100' 
= 0.314 



TAIL MOMENT 285 

From Fig. 39, for aspect ratio of 8, Fa.r. - 1.06 

Cl 

^ ~ 2 X 0.0718 X i^A.R. 

0.314 

~ 2 X 0.0718 X 1.06 
= 2.08° 

Angle of tail with horizontal = +7 - lO'' = -3° 
Angle of tail with down washed air = —3 — 2.1 = —5. I*' 

Problems 

1. Airplane, weighing 4,000 lb., has 300 sq. ft. of wing area, aspect 
ratio of 9, and is flying at 125 miles per hour. Angle of attack of air- 
plane is +2°. Angle of incidence is 1°. Stabilizer is set at +5°. 
What is tail angle? 

2. A Lockheed airplane, weighing 4,900 lb., has 293 sq. ft. of wing 
area, aspect ratio is 6.24, and it is flying at 150 miles per hour. If 
stabilizer is set at — 10° to wing chord and angle of attack of wing is 
+2°, what is tail angle? 

3. A Fokker monoplane weighing 11,000 lb. has 748 sq. ft. of wing 
area with aspect ratio of 9.1, and is flying at 90 miles per hour. If 
angle of attack of wing is +4° and stabilizer is set at —5° to wing 
chord, what is tail angle? 

4. A Douglas observation plane weighs 4,800 lb., has 376 sq. ft. of 
wing area with E.M.A.R. of 5.2, and is flying at 120 miles per hour. 
If angle of attack of wings is +5° and stabilizer is set at +5° to wing 
chord, what is tail angle? 

5. A training biplane weighs 2,400 lb., has 280 sq. ft. of wing area 
with E.M.A.R. of 4.6, and is flying at 85 miles per hour. If angle of 
attack of wing is +7° and stabilizer is set at —5°, what is tail angle? 

Tail Moment. To find the moment about the center of gravity 
due to the tail load, the lift (either positive or negative) acting on 
the tail must be multiplied by the distance from the center of 
gravity to the center of pressure of the tail surface. The tail 
surface, being usually a symmetrical airfoil, has a constant center 
of pressure position. 

The lift force on the tail is found in the usual way. The tail 
surface is usually of a smaller aspect ratio than 6. To correct for 
this, the angle of attack may be corrected for the smaller aspect 
ratio, or what is the exact equivalent the tail area may be multi- 



286 STABILITY 

plied by the aspect ratio factor, Fa.-r., to give an effective horizontal 
surface area, S\. 

S't = St X Fa.r. 

The airfoil characteristics of the tail airfoil surface are usually- 
different from those for the main wing, but the slope of all airfoil 
lift curves is approximately the same and equal to 0.0718. If 
at is the tail angle measured from the zero lift chord, the lift 
coefficient of the tail surface is 0.0718 X a^. 

Example. Find tail moment, when horizontal surface is 40 sq. ft. 
in area, span of tail is 11 ft., symmetrical airfoil section stabilizer is 
set at —10° to main wing. Distance from center of gravity of air- 
plane to center of pressure of stabilizer is 14 ft. Main wing is Clark 
Y, angle of attack of wing is 4°, airspeed is 100 miles per hour. 

Solution. For Clark Y wing, angle of zero lift is — 5°. 

4° _|_ 5° 
Downwash angle e = ^ 

= 4.5° 
Tail angle = -10.5° 

Aspect ratio of tail = ttt 

= 3.07 

From Fig. 39, for aspect ratio of 3.07, Fa.r. = 0.828 
Lift force on tail (down) 

= 0.0718 XatX p/2 X St X FA.R.taU X V^ 
= 0.0718 X 10.5 X 0.00119 X 40 X 0.828 X T46?7^ 
= 639 lb. 
Mt = 639 X 14 
= +8940 ft-lb. 

Problems 

1. A tail surface, 30 sq. ft. in area, has a span of 11 ft.; its airfoil 
section is symmetrical. Stabilizer is set at —5° to main wing which is 
Clark Y section and is at 2° angle of attack. Distance from center of 
gravity of airplane to center of pressure of stabilizer is 15 ft. What is 
tail moment if airspeed is 120 miles per hour? 

2. A tail surface 45 sq. ft. in area has a span of 13 ft., and its air- 
foil section is symmetrical. Stabilizer is set at 0° to main wing which 
is Clark Y section and is at 1° angle of attack. Distance from center 
of gravity to center of pressure of stabilizer is 16 ft. What is tail 
moment if airspeed is 90 miles per hour? 



MOMENT CURVES 287 

3. A tail surface 47 sq. ft. in area has a span of 15 ft., and its airfoil 
section is symmetrical. Stabilizer is set at — 10° to main wing which 
is Clark Y section and is at 6° angle of attack. Distance from center 
of gravity of airplane to center of pressure of stabilizer is 17 ft. What 
is tail moment if airspeed is 110 miles per hour? 

The sum of the moment due to the wing and the moment due to 
the tail at various angles of attack is the criterion of the airplane's 
longitudinal stabihty. Not only must it be possible for the air- 
plane to balance at all normal flying conditions, but for stability, 
when the angle of attack is increased there must be a diving mo- 
ment, and vice versa. 

For stability, the slope of the curve of moments plotted against 
angle of attack must be negative, so that with larger angle of 
attack there must be less positive moment. If the slope of the 
moment curve is positive the airplane will be unstable. 

Moment Curves. In a first approximation, the velocities over 
the wing and over the tail may be considered the same. If the 
airplane is balanced for a given angle of attack for one velocity, it 
will be balanced at any other velocity. If it is unbalanced at one 
velocity, it will be unbalanced for all velocities. For simplicity of 
calculation, therefore, one velocity may be assumed for all angles 
of attack. 

As the horizontal tail surface is a symmetrical airfoil, its lift 
coefficient is assumed to vary directly with angle of attack. In the 
following computation, the elevator is in line with the stabilizer 
in all conditions; this is termed " fixed elevator condition." 

The moment of the wing, drag being neglected, is the product 
of the lift, which varies with angle of attack, times its moment 
arm, which also varies with angle of attack. The moment of the 
tail is the product of its lifting force, which varies with its angle of 
attack and may be up or down, times its moment arm. Suice the 
tail surface is a symmetrical airfoil, its center of pressure does not 
move and the moment arm of the tail, which is the distance from 
the center of gravity of the airplane to the center of pressure of the 
tail, is constant . Upward forces are positive as are stalling moments . 

Example. A monoplane has a rectangular Clark Y wing, 36 ft. by 
6 ft. The rectangular horizontal tail surface has 25 sq. ft. area with 
an aspect ratio of 6. The center of gravity of the airplane lies on 
the chord of the wing and is 27 per cent of the chord back of the lead- 
ing edge. The distance from the center of gravity to the center of 



288 



STABILITY 



pressure of the tail is 18 ft. Consider velocity constant at 100 ft. per 
sec. Find the moment curve {a) when chord of the stabilizer is paral- 
lel to wing chord, (&) when stabilizer is set at —5° to wing chord. 
Solution. 

0° Stabilizer Setting 



(1) 



(2) 



(3) 



(4) 



(5) 



(6) 



(7) 



(8) 



(9) 



(10) 



(11) 



(12) 



Cl 
0.360 
0.645 
0.930 
1.190 
1.435 



C.P. 

0.424 
0.347 
0.316 
0.300 
0.296 



C.P.-27 
0.154 
0.077 
0.046 
0.030 
0.026 



Mw 
-853 
-766 
-660 
-547 
-573 



e 

2.5 
4.5 
6.5 
8.5 
10.5 



-2. 
-0. 

+1. 
+3. 
+5. 



Cli 
-0.180 
-0.036 
+0.108 
+0.252 
+0.395 



t 

- 53.5 

- 10.7 
+ 32.1 
+ 75.0 
+117.2 



Mt 
+ 963 
+ 192 
- 577 
-1350 
-2110 



M 
+ 110 
- 574 
-1237 
-1897 
-2683 





— 5° Stabilizer Setting 








(1) 


(2) (3) (4) (5) (6) 


(7) 


(8) 


(9) 


(10) 


(11) 


(12) 


a 


Cl C.P. C.P.-27 Mw oco 


e 


«/ 


Cit 


t 


Mt 


M 









-7.5 


-0.539 


-160.0 


+2880 


+2027 


4 






-5.5 


-0.395 


-117.2 


+2110 


+1344 


8 


same as 




-3.5 


-0.252 


- 75.0 


+1350 


+ 690 


12 


above 




-1.5 


-0.108 


- 32.1 


+ 577 


+ 30 


16 






+0.5 


+0.036 


+ 14.8 


- 260 


- 833 



Explanation of Table 

Column 2 obtained from Fig. 17. 
Column 3 obtained from Fig. 17. 
Column 4 obtained by subtracting 0.27 from items in column 3. 

Column 5 obtained by multiplying (2) X (4) X ^SV^c. 

Column 6 obtained by subtracting angle of zero Hft (—5*) from (1). 

Column 7 obtained by halving items in column 6. 

Column 8 obtained by inspecting columns 1 and 7. 

Column 9 obtained by multiplying items in column 8 by 0.0718. 

Column 10 obtained by multiplying (9) X -^t^'^. 

A 

Column 11 obtained by multiplying (10) X 18. 
Column 12 obtained by adding (5) and (11). 

The results from the above tables are plotted in Fig. 84. While 
the curve for wing moment has a slightly positive slope, the curve for 
combined wing and tail moment has a negative slope. With 0° 



MOMENT COEFFICIENT CURVES 



289 



stabilizer angle the airplane is balanced at approximately 1° angle of 
attack. With —5° stabilizer angle, the plane is balanced at 12|° 
angle of attack. In either case, when the angle of attack is increased 
there is a diving moment tending to bring the nose down, while if the 
angle of attack is decreased there is a stalling moment tending to bring 
the nose up. 

It will be noticed that a change in the stabilizer angular setting does 
not appreciably change the slope of the total moment curve but merely 
shifts it to the left or right. 

3.000 



2.000 



1.000 



-1.000 



-2.000 



-3.000 



Fig. 84. 



N 


\. 








\ 


> 












x 
X 






\, 


N 




M.. 






\, 




X?^ 


f 


\ 


\ 


^ 


"l 






\ 


\. 








^ 


^^^ 


-^c 


^ 










\ 


^ \ 


^0 








^\* 


"-^^ 


















N 



12' 



16' 



Angle of Attack 
Moment curves for illustrative example. 



Moment Coefficient Curves. It is usual to plot moment coeffi- 
cients instead of the moments themselves. 

The v^ing moment coefficient, Cmo, as shown in Chapter IV, is 
such that 

c = chord in feet 

p = distance in percentage of chord back 
from L.E. of point on chord about 
which moment is desired 
C.P. = center of pressure location in per- 
centage of chord back from L.E. 



M 

and 



= C^(|>^T^^c) 



Cm = Cl(p - C.P.) 



290 



STABILITY 




Fig. 85a. Moment coefficients about center of gravity. 

.10 
.09 
.08 
.07 
.06 
.05 
.04 
.03 
.02 



E.Ol 
o 

.02 
.03 
.04 
.05 
.06 
.07 





\ 


s. 






















\ 
























\ 






















> 


\ 


















V 




\ 


f 
















\l 


s. 




\ 


s. 
















N 






\ 


















\ 






\ 


















\ 






\ 
















\ 


V 




\ 


\, 
















\ 






\ 


\ 














> 


\ 






\ 
















\ 
























\ 






^^^^ 


■^^^^ 














"c;:;^ 


\ 


\ 






















\ 


L 






















> 


\ 






















\ 



6 8 10 12 14 16 18 
Angle of Attack 



Fig. 856. Moment coefficients versus angle of attack (illustrative example). 



MOMENT COEFFICIENT CURVES 291 

If the center of gravity is on the M.A.C., the above expressions 
are exactly right for moment and moment coefficient, when the 
distance of the center of gravity backward from the leading edge is 
substituted for p. When the center of gravity is above or below 
the wing chord, the wing drag will produce a moment, so that it 
is necessary to consider the resultant force, instead of merely 
lift component, and the moment arm should be the perpendicular 
distance from the center of gravity to the vector of the resultant. 

Referring to Fig. 85a, if the location of the center of gravity is 
known, R and 6 either are known or can be found. The moment 
arm of the lift is the perpendicular distance from the center of 
gravity to the line of direction of lift. From an inspection of 
Fig, 85a, this distance is C.P. cos a — R sin (6 — a), expressed in 
percentage of chord. The moment arm of the wing drag is the per- 
pendicular distance from the center of gravity to the line of action 
of the drag. From Fig. 85a, this is R cos {6 — a) — C.P. sin a. 
Finding the sum of the moments of the lift and of the drag is 
exactly the same as finding the moment of the resultant itself. 

While earlier in this chapter it was assumed that the velocity of 
the air past the tail was the same as the velocity of the air past the 
wing, it has been found that the air stream is slowed up by friction 
with the fuselage so that the air velocity over the tail surface is less 
than the airspeed. It is usual to correct for this by assuming that 
the speed at the tail is 90 per cent of the airspeed. Then the 
velocity squared at the tail is 81 per cent of the airspeed squared. 

M = Mwing + Mtail 

= Cowing (^ SV'C^ + Clu^ I >Sftail7^aiia' 

= ^Mwing (^ SV'CJ + 0.0718 X atail X i^A.R. tail X 

1^X0.81- 

Cm = Cjfwing + 0.0718 X atail X i^A.R.tail X 1^ X 0.81 - 

^wing 

It is quite common to use a tail coefficient, Cr, where 

Cf = 0.0718 X 1^ X FA.R.taii X 0.81 X - 
Substituting in above 

Cm = — Cowing + Cfata.il 

These moments may be plotted for various angles of attack. 



(iSV^c) 



292 



STABILITY 



Example. Find and plot wing moment coefficient, tail moment, and 
total moment coefficient for the following high-wing monoplane. 
The wing is rectangular, Clark Y section, 48-ft. span, 6.5-ft. chord. 
Total horizontal tail area is 43 sq. ft.; span of tail is 12.5 ft. Stabi- 
lizer is set at — 5° to main wing. The center of gravity of the airplane 
is 30 per cent of the chord back of the leading edge of the M.A.C. 
and 25 per cent of the chord below the M.A.C. Distance from center 
of gravity to center of pressure of tail is 12 ft. 

Solution, Area of main wing = 48 X 6.5 = 312 sq. ft. 

48 
Aspect ratio of wing = ^ = 7.39 

From Fig. 39, for aspect ratio of 7.39, Fa.r. = 1.043 

30 



25 



d (see Fig. 85a) = tan-i 

= 50.2° 

25 
R (see Fig. 85a) = ^ g 5Q 2° = ^^'^ P^^ ^^^* M.A.C. 



TABLE XVI 



(1) 


(2) 


(3) 


(4) 


(5) 


(6) 


(7) 


^ARe 


Cl 


OiARQ - «Ai27 39 


«Aie7.39 


^''ARe 


^^ARr^^AR,^,, 


^^^^7.39 





0.36 


0.21 


- 0.21 


0.011 


0.001 


0.010 


4 


0.65 


0.37 


+ 3.63 


0.017 


0.004 


0.013 


8 


0.94 


0.54 


+ 7.46 


0.033 


0.009 


0.024 


12 


1.19 


0.68 


+11.32 


0.060 


0.014 


0.046 


16 


1.44 


0.82 


+15.18 


0.139 


0.021 


0.118 



(8) 


(9) 


(10) 


(11) 


(12) 


(13) 


(14) 


(e-cx) 


sin(0 - a) 


COS(0 — a) 


R sin(0 — a) 


R cos(^ - a) 


C.P. 


C.P. sin a 


50.41 


0.771 


0.637 


0.301 


0.249 


0.424 


-0.002 


46.57 


0.726 


0.688 


0.283 


0.268 


0.347 


+0.022 


42.74 


0.679 


0.734 


0.265 


0.286 


0.316 


+0.041 


38.88 


0.628 


0.779 


0.245 


0.304 


0.300 


+0.059 


35.02 


0.574 


0.819 


0.224 


0.319 


0.296 


+0.078 



MOMENT COEFFICIENT CURVES 



293 



TABLE XVI {Continued) 



(15) 


(16) 


(17) 


(18) 


C.P. cos a 


Cl(C.P. cos a 

- R sin {d - a)) 


Cd{R COs(0 - a) 

- C.P. sin a) 


Cmw 


0.424 
0.346 
0.313 
0.294 
0.286 


-0.0445 
-0.0410 
-0.0457 
-0.0588 
-0.0891 


+0.0025 
+0.0032 
+0.0059 
+0.0113 
+0.0285 


-0.0420 
-0.0378 
-0.0398 
-0.0475 
-0.0606 



Explanation of Table 

Columns 2 and 5 obtained from Fig. 17. 

Column 3 obtained by multiplying items in column 2 by 18.24 (1/6 
- 1/7.39). 

Column 4 obtained by subtracting items in column 3 from items in column 1. 

Column 6 obtained by squaring items in column 2 and multiplying by 
l/7r(l/6 - 1/7.39). 

Column 7 obtained by subtracting items in column 6 from items in column 5. 
12^32 

Tail aspect ratio = ~To~ = 3.64 

From Fig. 39, for A.R. of 3.64, Fa.r. = 0.874 



43 X 0.874 



12 



^/ = ^-^^1^ X 312 X 1.043 X 0-81 X ^ = 0.0124 



TABLE XVII 



(1) 


(2) 


(3) 


(4) 


aZ.L.(A.R.7.39) 


e 


«< 


Cut 


4.79 


2.395 


-7.61 


+0.0944 


8.63 


4.315 


-5.68 


+0 . 0704 


12.46 


6.230 


-3.77 


+0.0468 


16.32 


8.160 


-1.84 


+0.0228 


20.18 


10.090 


+0.09 


-0.0011 



Explanation of Table 

Column 1 obtained by subtracting —5° (angle of zero lift for Clark Y) from 
items in column 4, Table XVI. 

Column 2 obtained by dividing items in column 1 by 2. 

Column 3 obtained by adding stabilizer angle (—5°) to items in column 4, 
Table XVI and from this sum subtracting items in column 2. 

Column 4 obtained by multiplying items in column 3 by C/(= 0.0124). 



294 STABILITY 

As shown by Fig. 856 this gives balance or zero moment at about 
8.5° angle of attack. This is a stable condition for locked elevator, 
i.e., elevator held in same plane with stabilizer; since a greater angle 
of attack means a diving moment, at a lesser angle of attack, there is a 
stalling moment. 

A change of stabilizer setting does not change the wing moment 
coefficient, nor does it change slope of tail moment coefficient curve. 
The curve of total moment coefficient is moved to right or left, so that 
equilibrium is found at a different angle of attack. 

Problems 

1. Plot moment coefficient curve for airplane in illustrative ex- 
ample with stabilizer set at — 10° to wing chord. 

2. Plot moment coefficients for an airplane similar to that described 
in the illustrative example, except that it is a low- wing monoplane and 
the center of gravity is 25 per cent above M.A.C., stabilizer set at 
— 5° to wing chord. 

Free Elevator. This condition is where the pilot releases the 
control stick entirely. The elevator being hinged at its forward 
edge would tend to sag down from its own weight. The air pass- 
ing under the stabilizer impinges on the under side of the elevator, 
so that the elevator assumes a position where the moment about 
its hinge due to its weight just balances the moment due to the air 
pressure on the under side. 

Lateral Stability. Because the airplane is symmetrical about a 
vertical plane through the longitudinal axis, lateral stabiHty does 
not present as difficult a problem as does longitudinal stability. 
A low center of gravity position as in flying boats is a great aid in 
lateral stability. 

When flying straight and level, the air will meet the wing (or 
wings) on the left side of the airplane at the same angle of attack 
as on the right side and therefore there will be the same lift on 
one side of the airplane as on the other side. If for any reason the 
airplane tips, the wing on the side which is going down will meet 
air, so that relative to the wing air not only is coming backward 
at the wing (due to the forward motion of the airplane), but also 
is coming upward (due to the roll). The relative wind is then not 
horizontal but backward and upward, so that temporarily the 
descending wing has a greater angle of attack. The other, rising, 
wing is meeting air on its upper surface, that is, momentarily the 



DIHEDRAL 295 

rising wing has a relative wind, which is backward and downward, 
giving a lesser angle of attack. The descending wing has a greater 
angle of attack and consequently greater lift; the ascending wing 
will have less lift, tending to restore the plane to its original atti- 
tude. The descending wing will also have more drag, the rising 
wing less drag, tending to turn the airplane. 

This action is effective only while the airplane is actually rolling. 
If an airplane has ceased rolling and is flying level with one wing 
low, there is the same angle of attack on both the left and the 
right wings and there is no tendency on the part of the airplane to 
right itself. 

Dihedral. One of the most effective ways of securing lateral 
stability is with dihedral. Instead of the wings being straight 
across the span, they slope outward and upward from the center. 
Dihedral angle is the angle which the wings slope upward from the 
horizontal. A small dihedral angle of 1J° or 2° is sufficient to give 
ample lateral stability. 

When an airplane tips sideways, it will sideslip. If the wings 
are tipped up to a vertical position, there will be a great deal of 
sideslip; but if the plane is only slightly tilted, there will still be 
some amount of slip as the resultant of lift and weight will have a 
side component. As soon as any slip takes place, the relative wind, 
instead of coming directly in front of the wing, will come from a 
direction which is to one side of dead ahead. 

With a straight wing, even when the relative wind comes from 
the side instead of from dead ahead, the angle of attack will still 
be the same on the right wing as the left wing. With dihedral, 
if the right wing drops, the plane will slip to the right. The 
relative wind is then coming at the airplane from the right of 
dead ahead. Owing to the dihedral angle, the right wing will 
have a greater and the left wing a smaller angle of attack. This 
will give more lift on the low wing and less on the high, tending to 
restore the airplane to an even keel. 

The first Wright airplane happened to have great inherent 
stabiUty so the upper wings were given negative dihedral, called 
cathedral, to decrease it. This is never done in modern plane 
construction. Too much lateral stability makes a cross-wind 
landing extremely difficult. 

An explanation sometimes given of the action of dihedral in 
aiding lateral stability is that the wing going down has a greater 



296 



STABILITY 



horizontal projected area than the wing going up. As the usual 
dihedral angle is 14° or 2°, the additional projected area of a wing 
dropping from normal to a horizontal position would be (1 — 
cos 2°) X area or (1 — 0.9994) X area, which would be a very- 
small part of the area. 

Some tapered-wing monoplanes have a straight upper surface, 
the decrease in thickness coming entirely on the under surface. 
This helps lateral stability, and is known as effective dihedral. 



Normal Flight 




Flight at or 

above burble 

point 



-4 4 8 12 16 20 24 28 
Angle of Attack 

Fig. 86. Autorotation. 

Autorotation. As explained in a previous paragraph, when an 
airplane tips, the descending wing has a greater and the rising wing 
a lesser angle of attack. At low or medium angles of attack, a 
wing with greater angle of attack has greater lift. The curve of 
lift coefficient versus angle of attack is a straight line with a 
positive slope to near the stalling angle, where it rounds off to a 
maximum. Beyond the angle of maximum lift, the lift coeffi- 
cient has negative slope. 

When an airplane is flying at or near the angle of maximum Hft, 
if the airplane tips, the descending wing will have a greater angle 
of attack than the rising; but since beyond the burble point the 
slope of the lift curve is negative, a greater angle of attack will 
mean less lift. Since the descending wing has the less lift and the 
ascending wing has the more lift, the rising wing will rise more and 



SPINS 



297 



the wing which is going down will go down more. This rotation 
will continue indefinitely unless controls are applied to stop the 
maneuver. Rotation about the longitudinal axis with wing at 
or above the angle of maximum lift is called autorotation. 

Spins. In a spin, the airplane wing is at a high angle of attack. 
At any instant, the path of the airplane is vertically downward 
so that the relative wind is vertically upward. Even though the 
nose of the airplane is down, the angle of attack is beyond the 
burble point. With the nose be- 
tween 20° and 40° below the hori- 
zontal, the spin is termed a flat spin; 
with the nose at a greater angle than 
40° below the horizontal, it is a 
normal spin. In a normal spin the 
angle of attack is about 35°, in a flat 
spin the angle of attack may be as 
high as 70°. 

Combined with the downward 
motion is autorotation and sideslip. 
In a steady spin, the actual path 
is a vertical spiral, the axis of this 
spiral being termed the axis of spin. 
Lift, being perpendicular to the 
relative wind, is horizontal and it 
balances centrifugal force. Drag, 
which is vertically upward, is oppos- 
ing the downward effect of weight. 
Air striking the under side of the 
stabiHzer tends to throw the nose 
downward; this is balanced by 
centrifugal inertia moments which 
tend to make the airplane assume a more horizontal position. 

A spin is usually started from horizontal flight by pulling back 
the stick till the airplane is in a stall; the rudder is then kicked, 
causing the airplane to sideslip and autorotate. To come out of a 
spin, it is first necessary to stop the autorotation. In order to do 
this, the angle of attack must be reduced to below the stalling 
angle. This may be accomplished by shoving forward on the 
stick. This action may be hastened by opening the throttle to 
send a blast of air against the tail surface. While this will cause 




Fig. 87. Forces in a spin. 



298 STABILITY 

the airplane to assume a more vertical position, the autorotation 
will stop and the plane will be in a simple dive from which recovery- 
is made by merely pulling back on the stick. When the airplane 
is coming down in a spin, the air is pressing against the under side 
of the elevator forcing it into an " up " position; with a properly 
designed plane, bringing the elevator into its mid-position should 
be sufficient to stop the spin. The Department of Commerce 
requires that after a six-turn spin, the airplane shall recover in no 
more than one and a half additional turns after the controls are 
put in neutral, without the use of the engine. 

Directional Stability. Because of the symmetry of the airplane, 
usually little difficulty is experienced in achieving stability in yaw 
or directional stability. A deep fuselage will aid in directional 
stability. 

In pursuit planes, which have shallow well-rounded fuselage, 
use is often made of sweepback. Sweepback or the sloping of the 
wings backward from the center section aids in the following 
manner. When an airplane turns to the left, the swept-back right 
wing will become more nearly at right angles to the direction of 
flight, its effective span will be greater and its drag will increase, 
while the drag on the left wing will decrease. A right turn will 
show an opposite reaction. In each case, the change in relative 
drag of the two wings will be such as to cause the airplane to 
return to its original heading. 



CHAPTER XV 
AUXILIARY LIFT DEVICES 

Speed Range. The ratio of maximum velocity to minimum 
velocity is called the speed range. With retractable landing gear 
and general clean design, racing planes can have a speed range of 
3.3; airplanes not so carefully streamlined or those so heavily 
loaded that they cannot fly at the angle of attack of minimum 
drag will have a smaller speed range. Unless design is very poor, 
or the load is exceedingly heavy, the speed range should be at 
least 2.5. 

The above is based on fixed wings. The landing speed is de- 
termined by the maximum lift coefficient and the wing loading. 
If the maximum lift coefficient can be increased in any way, the 
wing area can be decreased without changing the minimum speed. 

Fixed Slot. G. Lachmann in Germany and F. Handley-Page in 
England appear to have developed the slotted wing at about the 
same time. The slot is a narrow opening near the leading edge 
and parallel to the span. The small section in front of the slot 
may be considered as a miniature airfoil. When the main wing 
is at a high angle of attack, the small airfoil in front of the slot 
is at a small angle of attack. Whereas with a simple wing burb- 
ling takes place at 18° to 20° angle of attack, because air is unable 
to change direction so as to follow the upper surface of the wing, 
with the slot, the air is given a downward deflection in passing 
over the small auxiliary wing section, so that it can follow closely 
the upper surface of the main wing. 

By using a slot, the maximum lift angle is increased to 28° or 
30°. The slope of the lift curve remains constant, so that the 
maximum lift coefficient is increased about 50 per cent by the use 
of a slot. 

With a fixed slot, at low angles of attack, there being less pressure 
on the upper than at the lower end of the slot, air will travel up- 
ward through it. This will divert the main air stream and cause 
burbHng so that the drag at low angles of attack is much greater 
with the fixed slot than with the simple wing. 

299 



300 



AUXILIARY LIFT DEVICES 



Automatic Slot. To F. Handley-Page should go the credit for 
obviating the increased minimum drag of the fixed slot. The 
auxiliary airfoil is held by a linkage mechanism or a series of studs 
working in pairs of rollers in the main wing, so that it will move 
freely from a closed position where the auxihary is butted against 
the leading edge of the main wing to an open position where there 
is a gap of an inch or more between the auxiliary airfoil and the 
main wing. 




Fixed Slot at High Angle of Attack 




Fixed Slot at Low Angle of Attack 




Automatic Slot at High Angle of Attack 



Automatic Slot at Low Angle of Attack 
Fig. 88a. Wing slots. 

By examining Figs. 14, 15, and 16, it will be seen that at low 
angles of attack there is a pressure on the nose which is utilized to 
press the auxiliary airfoil tightly against the main wing, while at 
high angles of attack there is a negative pressure to draw the auxil- 
iary wing to its *' open " position. The automatic slot requires 
no manipulation on the part of the pilot. At low angles of attack, 
the drag is only very slightly greater than with the ordinary wing. 



FLAPS 



301 



Flaps. Flaps, in the simplest form, merely mean that the rear 
part of the main wing is hinged so that it can be swung downward. 
In appearance, they resemble ailerons, except that ailerons extend 
only a small portion of the span and are so linked that, when the 
aileron on one side goes up, that on the other side goes down. 
Flaps extend across the span, except for the small portion which is 
aileron, and the flaps on both sides go down together. Flaps are 
moved by a control mechanism in the pilot's cockpit. 

The effect of depressing the flaps is to increase the effective 
camber of the upper surface of the wing as well as the concavity 
of the under surface. This has the effect of increasing the lift 
coefficient, partly because of the action on the under side, where 
the depressed rear edge acts to hinder the smooth flow of air; 
pressure will build up at this point, which will cause an increase in 
lift. It will also cause the center of pressure to move rearward. 













/ 


/^ 


Wing 
slot ai 


with 
id flap 










/ 


A 














o 


// 




/ 


yi 


Wing 
with 
slot 








i 




\ 


/ 




1 








r 




/ 


V. 




\ 






w 




/ 


f 




\ 






7/ 




. 


f/ 










1 


f 




1 


7 










1 






/ 












// 




/ 


Y 












// 




/ 















Cl 



■10^ 



0" 10° 

Angle of Attack 



Fig. 88&. Effect on lift coefficient of slots and flaps. 

This rearward center of pressure movement, when the flaps 
are swung downward, means that it is usually necessary to inter- 
connect the flaps with the stabilizer; otherwise the airplane will 
be thrown out of balance longitudinally. 

With flaps, the maximum lift coefficient can be increased 50 



302 AUXILIARY LIFT DEVICES 

per cent. At low angles of attack, the flaps are in their normal 
position so that they do not detract from high speed. Both slots 
and flaps may be installed on the same wings, and by so doing, 
maximum lift coefficient may be increased 100 per cent. 

Special forms of flaps have been devised and several have proved 
to be very practical. Among these are the '' Zap " flap, in which 
the lower surface of the rear portion of the wing swings down while 
the upper surface remains intact. Another type is the Fowler 
wing; the rear portion of the lower surface slides backward and 
downward, so that not only the camber but also the wing area 
are increased. 



CHAPTER XVI 

UNCONVENTIONAL TYPES OF AIRCRAFT 

Autogiro. In the autogiro, lift is derived, not from fixed wings 
as in airplanes, but from wings rotating in a horizontal plane. 
In the early models, a small fixed wing gave part of the Uft; in 
the more recent types all the lift is obtained from the rotating 
vanes. The essential features of the autogiro are shown in Fig. 89. 



Vertical Pivot and 
Friction Damper 



Vertical Pivot and 
Friction Damper 




(b) 

Fig. 89. Autogiro. 

An autogiro is not a helicopter. In a helicopter, the rotating 
vanes are driven by the engine, and in case of engine failure dis- 
aster would result. In the autogiro, the vanes are caused to 
rotate by aerodynamic forces produced by the vanes themselves. 
To hasten the start of rotation at the beginning of a flight the 
vanes are connected through a clutch with the engine, but imme- 
diately as the vanes rotate at the proper speed the engine is dis- 
connected and the vanes rotate freely. 

303 



304 



UNCONVENTIONAL TYPES OF AIRCRAFT 




(a) Side View 




(b) Top View 
Fig. 90. Autogiro with blades folded. 



The rotating vanes are four in the older, three in the newer, 
types. The latest type, without a fixed wing, permits, by the 
removal of one bolt on each vane, the swinging together of all 
three vanes over the fuselage, and in this position the autogiro 
may be taxied along a road 
or trundled into an ordinary- 
sized garage (see Fig. 90). 

The airfoil section most 
favored for the vanes is the 
Gottingen 429 profile. The 
normal position of three blades 
would be 120° apart, but 
during rotation a blade ex- 
periences more drag when 
moving forward than when 
moving backward, so that if 
the blades were fastened 
rigidly there would be re- 
peated variation in the 
stresses. For this reason, each blade is hinged about a vertical 
pin, its horizontal movement being restricted by friction dampers 
so that only a slight relative motion is permitted. 

Each blade is also hinged about a horizontal pin, so as to per- 
mit flapping or movement in a plane through the axis of the rotor 
head. The rotor head supports the inner ends of the three vanes. 

In horizontal flight, the advancing blade has a greater relative 
velocity than the retreating blade; consequently the advancing 
blade would have greater lift. Rigidity of the blades would tilt 
the entire gyro, and it is for this reason that the blades are hinged 
about a horizontal axis. There is no restriction to their upward 
sweep; care is taken, however, that when rotation is slow the 
blades cannot sink down so as to strike the propeller. Folding 
up is impossible, owing to centrifugal force. The angle which the 
span of any blade makes with the horizontal is due to the resultant 
of centrifugal force and the lift on that blade. This means that 
the advancing blade is at a higher angle from the horizontal than 
the retreating blade. When there is no forward motion and the 
autogiro is parachuting down, the relative wind is verticaUy 
upward and all blades are at the same coning angle. 

In vertical descent, there will always be rotation. Even if it 



AUTOGIRO 



305 



could be conceived that the blades were stationary, for the airfoil 
section used, as well as for most airfoils, zero lift occurs at 93° or 
94° angle of attack. At 90° angle of attack there is always a 
small lift, which acting perpendicular to relative wind, see Fig. 
91a, will cause rotation. As rotational speed becomes greater, 
the velocity of the relative wind becomes greater, and since the 
relative wind is the resultant of the vertical descent velocity and 




^:S^ 



Resultant 




(a) 



Resultant 
Velocity 
(b) 



Forces in Vertical Descent 



Resultant ^ 
Lift-^ 


n 




A 


1 


<"' 


Relative Wind " V /"^^ 


/ Component 
/ causing 
' Rotation 


(c) 


Forces 


in 


Flight 



Fig. 91. Forces on autogiro blade. 

the rotational velocity, the angle of attack becomes less. The 
resultant force has a forward component which causes rotation 
^\ t> 1 ^ (Fig. 916). As the speed of rotation increases, the direction of the 
relative wind becomes more and more inclined to the vertical and 
the angle of attack of the airfoil becomes less. At some one angle 
of attack the resultant force will be vertical. Under this condition 
there will be no forward-acting component of force, and without this 
accelerating force there will be no increase in rotational velocity. 



306 UNCONVENTIONAL TYPES OF AIRCRAFT 

The speed of rotation will be constant at about 100 r.p.m. If, for 
any reason, the resultant should act backwards from the vertical, 
it would immediately have a component tending to retard ro- 
tation. The decrease in rotational velocity would change the 
direction of the relative wind to give a larger angle of attack, 
which would cause the line of action of the resultant to move for- 
ward again. 

When the autogiro is moving forward in horizontal flight, the 
main axis of rotation of the rotor head is tilted slightly backward. 
Rotation of the rotor head is counter-clockwise viewed from above, 
and this rotation is always in effect before take-off. Then a 
blade on the right side has greater speed than a blade on the left. 
If there were no flapping the advancing right blade would have 
greater lift than the retreating left blade. Being hinged to permit 
upward movement, the forward-moving blade is rising, giving a 
smaller angle of attack, while the backward-moving blade is being 
moved downward by the effect of centrifugal force. These 
effects tend to make the lifts on each side more nearly equal. 

When the axis of rotation is tilted backward to the resultant 
force produced by the action of the relative wind so that there is a 
component producing rotation as shown in Fig. 91c, the rotor will 
accelerate. When the resultant force is parallel to the axis of 
rotation there will be no acceleration, the rotational speed will be 
constant, and the system will be in equilibrium. If the resultant 
acts in a direction to give a component decelerating the rotor, the 
autogiro will start to descend, speeding up the rotor. 

The rotor system is supported on a sturdy tripod, the head being 
mounted on a universal joint, so that it may be tilted in any di- 
rection by the control column. This control column hangs down 
instead of protruding up from the floor as in a conventional air- 
plane. The movement of the pilot's hand is the same as with the 
ordinary control stick. Pulling back on the stick causes the nose 
of the autogiro to go up; pushing forward causes it to go down. 
Moving the stick to the left causes the fuselage to tilt in a manner 
similar to the way an airplane fuselage would be tilted in banking 
for a left turn. The rudder on the rear of the fuselage is operated 
by pedals or rudder bar as in the conventional airplane. 

The landing gear resembles that of the airplane except that a 
much wider tread is given to the wheels. The shock-absorbers 
have greater travel to withstand harder landings. 



GYROPLANE 



307 



Gyroplane. In the autogiro the blade airfoils have a fixed angle 
with respect to the rotor disk (an imaginary plane perpendicular 
to the axis of rotation). In the gyroplane this angle can be 
changed (see Fig. 92). 




Fig. 92. Gyroplane. 



Four blades are used, set 90° apart. The opposite blades are 
rigidly connected, but each pair is supported in bearings in such a 
manner that the blades may be " feathered " or have the angle 
with the rotor disk altered as the blades proceed around the disk. 
This movement is caused by a central cylindrical cam at the rotor 
hub. 

The gyroplane has a fixed wing as well as the rotating vanes. 
At high speeds the small fixed wing furnishes sufficient lift, and the 
rotating blades are turned to the zero lift angle offering a minimum 
of drag. As the speed is decreased, more of the load is taken by the 
rotor. 

Another feature of the gyroplane is its lateral control. The 
control stick is connected with the feathering cam by linkage. 
A movement of the stick to the left will cause the blades on the 
right side to have an increased angle of attack, and the blades as 
they proceed to the left side will have a decreased angle of attack, 



308 



UNCONVENTIONAL TYPES OF AIRCRAFT 



resulting in more lift on the right than on the left. Conversely 
a stick movement to the left causes the blades on the left side to 
increase their angle, and on the right side to decrease their angle. 
The control stick is also connected to conventional ailerons on the 
fixed wing for control at high speed. 

Although the gyroplane is still in the developmental stage, it is 
aerodjoiamically sound. It should be as safe as other rotating- 
wing aircraft and holds possibilities of a greater speed range. 

Cyclogiro. The cyclogiro, sometimes termed the '' paddle- 
wheel " airplane, is of the rotating- wing type, having several 
blades on each side rotating about a horizontal axis perpendicular 
to the direction of normal flight (see Fig. 93). By means of cams 




Fig. 93. Cyclogiro. 

or linkages, the angle which an element makes with the tangent 
to its circular path may be changed as the blade rotates. Chang- 
ing the setting of the cam operating this feathering can be made to 
produce a resultant in any desired direction. 

The combination of wings, i.e., the '' paddle-wheels," is driven 
by the engine, but a clutch mechanism is provided so that in case 
of engine stoppage the airfoil systems are free to rotate. They 
autorotate, thus permitting hovering or slow descent. With the 
engine connected, a reasonable forward speed and climb are ob- 
tainable. 

The autorotation, with the clutch out, may be understood by 
examining Fig. 94. The three symmetrical airfoils A, B, and C 
are pivoted at their centers of pressure at three points 120° apart 
on a circular framework which rotates about its center 0. From 
eccentric point 0' links are connected to each airfoil. The posi- 
tion of 0' with respect to 0, both in direction and distance, may 
be controlled by the pilot. In Fig. 94a, it is assumed that the 



CYCLOGIRO 



309 



system is stationary. The relative wind will then be vertically 
upward. The resultant forces on each wing will be as shown. 
Each of these resultant forces has a tangential component tending 
to cause the system to rotate in a counter-clockwise manner about 
center 0. After rotation has started, conditions are as represented 



Resultant force 
A 





■j^- — >J I / wind 
^Peripheral/ 

velocity/ Resultant' 




y<- Upward 
' wind 



Upward 
wind 



(a) Vertical Descent 



Peripheral 
velocity 

(b) Vertical Descent 



Resultant force 
A 




(c) Forward Flight 
Fig. 94. Operation of cyclogiro. 

in Fig. 946. The relative wind (Vr) at each wing is the resultant 
of that due to the downward motion and to the rotary motion of 
each individual airfoil. The resultant forces on each wing are 
shown approximately in the diagram. All these forces have an 
upward component tending to retard descent. In the diagram, 
the resultant force on the wing at A has a component tending to 



310 UNCONVENTIONAL TYPES OF AIRCRAFT 

retard rotation, but both wings at B and C have components tend- 
ing to increase rotation. As long as the components acting to 
cause counter-clockwise rotation are greater than those tending 
to cause clockwise rotation, the speed of rotation will increase. 
At some speed of rotation, equilibrium will be found. Figure 94 
shows conditions on the left side of the cyclogiro; on the right 
side the conditions will be reversed. 

To obtain forward motion the position of the eccentric center 
0' is shifted by the pilot as is shown in Fig. 94c. The relative 
wind at each airfoil is the resultant of the relative wind due to 
forward motion and that due to the rotation. The resultant 
force on each airfoil can be resolved into a component tangent 
to the circular path and radial to the path. The tangential 
components either aid or retard rotation, the algebraic sum of these 
components usually retard rotation in forward flight, and it is the 
function of the engine to furnish torque to overcome this re- 
sistance in the same way that the engine furnishes force to over- 
come drag of the wings in the ordinary airplane. The vectorial 
sum of the radial components of the resultant forces on the in- 
dividual airfoils should have an upward component, which is the 
lift. 

Although the cyclogiro is still in experimental stages, it has 
been proved that the device is capable of slow hovering descent in 
case of engine failure; that a steep descent, not unlike that of an 
autogiro, can be made; and that horizontal flight at a fair rate of 
speed can be maintained. 



CHAPTER XVII 
MATERIALS AND CONSTRUCTION 

Load Factors. In th^ preceding chapters, it has been shown 
that, while in flight, external forces are acting on the airplane struc- 
ture. Some of these forces, lift and weight, are dependent on the 
size and design; other forces, inertia forces, are dependent on the 
suddenness with which maneuvers are executed. In one kind of 
maneuver a certain part of the airplane is stressed highly, and in 
another kind of maneuver a different part of the airplane may have 
its greatest stress. In designing parts of an airplane, each part 
must be planned to withstand the greatest stress that that part 
may receive in any ordinary maneuver. The term *' applied 
loads " means the actual forces produced on a structure by the 
accelerations during a maneuver. The load factor is the accelera- 
tion expressed in terms of g. 

For safety, parts are made of greater ultimate strength than 
just enough to stand the greatest loading they will undergo. The 
applied load multiplied by the factor of safety gives the design 
load. 

The load factors in maneuvers are based to a great extent on 
past experience and are therefore semi-empirical. Many calcu- 
lated accelerations have been checked by accelerometer readings 
obtained in airplanes in actual flight. The gust load factors are 
based on an arbitrarily assumed sharp-edged vertical gust of 30-ft.- 
per-sec. velocity. 

Materials. The materials used in the construction of airplanes 
cannot be unduly heavy or bulky in giving the required strength, 
as unnecessary size or weight detracts from the performance. 
The cost of raw material is small compared with the labor cost, but 
a material which is extremely high in price would probably not be 
desirable. The material should, if possible, be adapted to use by 
workmen of ordinary skill. 

The quantity of airplanes of one particular design will to some 
extent dictate the choice of material. Metal stampings are very 

311 



312 MATERIALS AND CONSTRUCTION 

satisfactory and cheap on quantity production, but in small lots 
the expense of dies is not justified. 

Wood. Wood was the chief material used in early airplane 
construction, but it is employed to only a limited extent at the 
present time. First-quality selected spruce is hard to obtain and 
is consequently expensive, and other woods are not entirely satis- 
factory. 

Plywood, which is made of several thin sheets of veneer, some- 
times serves for flooring and other parts of the fuselage. In one 
or two airplanes, plywood has been used for wing covering. Ply- 
wood is very unsatisfactory in the tropics as the combination of 
heat and moisture causes the glue to lose its strength. 

Probably the main reason that wood has been largely superseded 
is its lack of uniformity. It must be aged or seasoned before 
fabrication. It must be protected by varnish from absorbing 
moisture. There is usually a large waste. Because of its non- 
uniformity and the ever-present dangers of defects not discernible 
on the surface, a high safety factor must always be applied in the 
design of wooden members. 

Cloth. The first airplanes had cloth for wing covering, and the 
first fuselages were cloth-covered. The standard cloth was un- 
bleached mercerized Grade-A cotton, having a minimum tensile 
strength of 30 lb. per in. After the cloth covering has been 
sewed in place, it is treated with dope — a colloidal solution of 
either cellulose acetate or cellulose nitrate. Doping the fabric 
makes it weatherproof, tightens the fabric, and produces a rigid 
surface. Without the dope, damp air would make the fabric 
slack. Because clear dope is transparent, and because sunlight 
is the chief cause of the deterioration of fabric, it is customary to 
impregnate the dope with pigment. Even when protected with 
pigmented dope, cloth loses its life and strength after about a 
year. Because of this need for frequent renewals, fabric is being 
used less and less. 

Steel. Low-carbon steel is used to some extent in modern air- 
planes for less important parts of the structure. Its tensile 
strength is about 55,000 lb. per sq. in. It is used either in the 
form of tubes or in sheets. It is assembled either by welding or 
riveting. 

Chrome-molybdenum steel, usually termed " Chrome-Moly," 
is used to a great extent in the fabrication of the modern airplane. 



ALUMINUM ALLOYS 313 

This steel, unheat-treated, has an ultimate tensile strength of 
95,000 lb. per sq. in., with a yield point of 60,000 lb. per sq. in. 
When properly heat-treated it may develop an ultimate tensile 
strength of 200,000 lb. per sq. in. It is slightly more difficult to 
machine than the low-carbon steel but it may be welded just as 
easily. 

Aluminum Alloys. Absolutely pure aluminum does not corrode 
as does impure aluminum, but the addition of other metals greatly 
increases its strength, so that aluminum alloys are used for air- 
craft rather than the pure aluminum. The alloys are treated in 
various ways to minimize the danger of corrosion. Originally the 
term duralumin was applied to one specific composition alloy; 
now it is common practice to use the term duralumin loosely to 
cover all strong aluminum alloys. The most popular alloy has 
4 per cent copper, | per cent manganese, and J per cent magne- 
sium; it is used for tubing and sheets. Changes in the composi- 
tion are made to give special properties. Alloys suitable for 
castings usually have silicon added. 

Duralumin has a specific gravity of 2.8, while chrome-molyb- 
denum steel has a specific gravity of 7.9, so that, volume for 
volume, duralumin has 35.4 per cent of the weight of steel. The 
tensile strength of 17ST duralumin is 58,000 lb. per sq. in., while 
that of unheat-treated steel is around 100,000 lb. per sq. in. 
Then a piece of duralumin in tension has 35.4 per cent of the 
weight and 58.0 per cent of the strength of a piece of steel of the 
same cross-section. If the duralumin piece had a larger cross- 
section so the weight were the same as that of the steel piece, the 
duralumin would be 1.6 times stronger in tension than the steel. 
In bending such as would be experienced in a beam the extra 
depth of the duralumin beam for the same weight would give 
much greater stiffness than that of a steel beam. 

Duralumin can be welded or riveted. Sheet duralumin is 
being used a great deal for wing and fuselage covering, and spot 
welding appears to be superseding riveting for joining sheets. 

Great care must be exercised in heat-treating duralumin, 
especially if the material is worked cold during fabrication. 
Aluminum alloy rivets should invariably be driven within a half- 
hour of being heat-treated. If a longer time has elapsed, they 
should not be used until after they have been reheat-treated. 

To protect aluminum alloys from corrosion, the surface must be 



314 MATERIALS AND CONSTRUCTION 

protected by a thin coating of oxide. The most popular method 
is called the '' anodic " process. This consists in passing an elec- 
tric current through the piece of duralumin while it is immersed 
in a 3 per cent solution of chromic acid. The duralumin is the 
anode of this electrolytic bath. 

Stainless Steel. An alloy of steel containing 18 per cent 
chromium, 8 per cent nickel, and a httle less than 0.18 per cent 
carbon is known as stainless steel because of its remarkable 
resistance to corrosion. It has a high tensile strength and is very 
tough and ductile. By cold rolHng, tensile strengths of 400,000 
lb. per sq. in. have been obtained, but in this state it is very 
difficult to fabricate. Ordinarily 200,000 lb. per sq. in. is a fair 
value of its tensile strength. 

The maximum resistance to corrosion is obtained after heat- 
treatment. The heat of the ordinary welding process causes 
stainless steel to lose some of its properties. A special method of 
spot- welding, called '^ shot- welding," has been developed which 
does not harm the steel. 

In 1932, an amphibian airplane was built in this country en- 
tirely of stainless steel. In Russia, stainless steel is used a great 
deal in airplane construction, and it may be presumed that it will 
be employed more widely in the United States when its prop- 
erties and method of handling are better understood. 

Construction. The early form of construction of an airplane 
was to make the fuselage of oblong cross-section, with four length- 
wise members called longerons and suitable cross-pieces and 
bracing. The wings had two wooden spars with ribs or formers 
spaced at equal intervals to give the desired airfoil shape. 

When metal was first introduced as an airplane material, every 
part was reproduced in metal as an exact replica of the wooden 
piece. Because of the nature of wood, it absorbs vibrations better 
than metal, so that troubles due to vibration that had not been 
experienced with the wooden construction were encountered 
in the metal construction. More important, however, was the 
item that wood had Httle tensile or shear strength, so that the 
design of the structure had to be extremely simple. Metal can 
be rolled or worked into innumerable shapes and can be used in 
very thin sheets still retaining requisite thickness. The entire 
design of aircraft has therefore been modified to take advantage 
of all the special properties of metal not possessed by wood. 



MILITARY PLANES 315 

At higher speeds, parasite resistance becomes of paramount 
importance, and as fuselages are built of better streamline form, the 
monocoque form of construction becomes more desirable. The 
monocoque or stressed-skin construction relies entirely on the 
skin to carry all the bending moments and shear. The pure 
monocoque has only vertical bulkheads to reinforce the skin. 
A type called semi-monocoque has longerons in addition to the 
bulkheads. 

Military Planes. Airplanes for military uses are divided into 
various types depending on the use to which they are put. Train- 
ing airplanes should have a fairly low landing speed, and this will 
mean a moderate maximum speed. The training plane will be a 
two-seater for the student and the instructor, with dual-control, 
a small gas supply, and no equipment. The observation plane is 
a two-seater with fairly high speed. It is essential that there be 
excellent visibility for the observer. The equipment would in- 
clude radio, camera, and machine guns for both pilot and observer. 
The bombing plane is primarily to carry heavy loads of bombs at 
as high a speed as is consistent with the load carried. It has a 
crew of four or five. For defense it is equipped with machine 
guns. Some maneuverabihty is usually sacrificed in order to 
obtain load-carrying ability. The pursuit plane is the fast single- 
seated fighter. It must have high maximum speed, big rate of 
climb, and high degree of maneuverability. Its engine must be 
supercharged so that the plane may operate at high altitude. 
The equipment consists of two fixed synchronized machine guns, 
a bomb rack for carrying small demolition bombs, radio, and oxy- 
gen tank. On some airplanes, fixed machine guns are installed in 
the wings to be operated from the cockpit. 

Figure 95 shows an Army observation airplane. It is a high- 
wing braced monoplane, designed to give the observer extremely 
good visibility of the ground. 

Figure 96 shows an army attack airplane. It is a low- wing canti- 
lever monoplane. Special attention should be drawn to the clean- 
ness of design, the absence of wire bracing, and the streamlining 
of the landing gear. 

Figure 97 shows an army pursuit airplane. It has an extremely 
high speed and high degree of maneuverability with full military 
load. 

Figure 98 shows an army two-seater pursuit airplane. Although 



:i 



316 



MATERIALS AND CONSTRUCTION 




Fig. 95. Douglas Observation Airplane. 




Fig. 96. Northrop Attack Airplane. 




Fig. 97. Boeing Pursuit Airplane. 



MILITARY PLANES 



317 

1 




f 



Fig. 98. Consolidated Two-seater Pursuit Airplane. 




k „,j?Sslr^- 



Fig. 99. Douglas Observation Amphibian. 




Fig. 100. Martin Bombing Airplane. 



318 



MATERIALS AND CONSTRUCTION 



the top speed is not as great as that of the single-seater, it is well 
over 200 miles per hour. The supercharger which enables the 
engine to give full horsepower at high altitude is shown in this 
photograph. 

Figure 99 shows a twin-motored amphibian observation air- 
plane, adapted for alighting or taking-off from either land or 
water. 

Figure 100 shows a twin-motored bombing airplane. It is a 
mid-wing cantilever monoplane with retractible landing gear. 
By reason of the clean design, this plane can carry a heavy bomb 
load at a high speed. 







Fig. 101. Sikorsky "Brazilian Clipper. 



Non-Military Airplanes. The design of non-military airplanes 
is influenced by the purpose of the plane whether a sport or a 
commercial airplane. If it is a commercial plane, it must be decided 
whether it is to carry a moderate load at high speed or a greater 
load at a lesser speed. The following paragraphs describe briefly 
a few commercial airplanes of different types. 

Figure 101 shows the Sikorsky S-42, Brazilian Clipper, high- 
wing cabin monoplane seaplane which was used on the pioneer- 
ing flights from the United States to the Hawaiian, Midway, and 
Wake Islands. It is equipped with four Pratt and Whitney 
Hornet engines giving 700 hp. each. The wings are two spar 



NON-MILITARY AIRPLANES 319 

construction with metal skin on both sides of wing forward of the 
rear spar and with fabric skin rearward of the rear spar. The 
boat hull is metal-covered over a metal framework, divided for 
safety into nine watertight compartments. It is designed for 
thirty-two passengers. The wing span is 114 ft., the wing area 
is 1,330 sq. ft. Flaps are located between the ailerons and are 
hydrauHcally operated. The weight empty is 21,950 lb.; the 
useful load is 16,050 lb. ; the maximum speed at sea-level is 180 
miles per hour; the landing speed is 65 miles per hour. 

Figure 102 shows the Seversky low- wing twin float, two- or 
three-seater monoplane amphibian. It is equipped with a 420- 
hp. Wright Whirlwind engine. The fuselage is all-metal mono- 
coque construction; the wings are multi-box all-metal construc- 
tion. Flaps are of the split trailing-edge type and extend between 
the ailerons. The wing span is 26 ft. ; the wing area is 209 sq. ft. 
The weight empty is 2,550 lb., and the useful load, 1,650 lb.; 
the maximum speed at sea-level is 185 miles per hour, and the 
landing speed 55 miles per hour. 

Figure 103 shows the Lockheed Electra, a twelve-place low-wing 
cantilever monoplane. It is equipped with two Pratt and Whitney 
Wasp Junior 400-hp. engines. The fuselage is all-metal mono- 
coque with longitudinal reinforcements. The wings are all-metal 
stressed skin, with trailing-edge flaps. The wing span is 55 ft.; 
the wing area is 458 sq. ft. The weight empty is 6,200 lb. ; the 
useful load, 3,550 lb. ; the maximum speed is 210 miles per hour, 
and the landing speed 63 miles per hour. The landing gear is 
retractable, and the entire airplane is a splendid example of good 
streamlining. 

Figure 104 shows the Fairchild Cargo Carrier, a high-wing 
braced monoplane. It is equipped with one 750-hp. Wright 
Cyclone engine. The fuselage is of welded chrpme-molybdenum 
steel tubing, fabric-covered. The wing spars are single solid web 
type with extruded angles as flanges all of chrome-molybdenum 
steel; the ribs are built up of aluminum framework; the wing is 
fabric-covered. Special attention should be drawn to the Zap- 
type flaps in the illustration in their " down " position. The 
wing span is 84 ft. ; the wing area, 800 sq. ft. The weight empty 
is 7,320 lb.; the useful load, 5,680 lb. ; the maximum speed is 167 
miles per hour, and the landing speed 52 miles per hour. 

Figure 105 shows the Monocoupe, a high-wing braced cabin 



320 



MATERIALS AND CONSTRUCTION 




Fig. 102. Seversky Amphibian. 




Fig. 103. Lockheed "Electra" Airplane. 




Fig. 104. Fairchild Cargo Transport, 



NON-MILITARY AIRPLANES 



321 



monoplane. This is a typical sport plane for a private owner. 
It is equipped with a 90-hp. Lambert radial engine. The fuselage 
is of welded steel tubing, fabric-covered. The wings have two 
spruce spars, the ribs have basswood webs with spruce cap-strips; 




Fig. 105. Monocoupe Sport Airplane. 

the wing is fabric-covered except that sheet aluminum is used on 
the leading edge. The flaps are fabric-covered. The wing span 
is 32 ft.; the wing area, 132 sq. ft. The weight empty is 935 
lb. ; the useful load, 650 lb. ; the maximum speed is 135 miles per 
hour, and the landing speed 40 miles per hour. 



CHAPTER XVIII 
INSTRUMENTS 

Introduction. Instruments are practically indispensable for 
safe flying. Although on short hops around an airport the 
necessity for instruments is slight, on cross-country flights, safety 
demands that every possible instrumental aid be given the pilot. 

On the Wright brothers' first flights, their only guide was a long 
streamer of cloth tied to one of the struts. It served as airspeed 
indicator: at low speeds the rag hung limply down; at higher 
speeds its position became more nearly horizontal. It also in- 
dicated slipping by swinging away from the center of turn and 
skidding by swinging towards the center of turn. 

Many instruments have been added since the days of the early 
Wright flights, but each has been the result of a distinct need for 
that instrument. All instruments have a definite function, and 
they are placed on the instrument board because at some time 
the occasion will arise when the information given by that in- 
strument will be vital. 

It is of primary importance that aircraft instruments be accurate 
at all times. With only the owner flying the airplane, if one in- 
strument reads consistently low or high, the pilot can of course 
make a mental correction; but this is a very poor practice and is 
extremely dangerous. The Army Air Corps has a fixed policy of 
" either dead-right or stopped. '' Interpreted, this means that it is 
more desirable to have an instrument completely fail to function 
than to have it function imperfectly. If an instrument is not 
working at all, the pilot will disregard it ; if it apparently is func- 
tioning the pilot will rely on its indications and, if they are in- 
correct, disaster may result. 

All aircraft instruments are classed under two headings, engine 
instruments and avigation instruments. All instruments whose 
readings do not pertain to the operation of the engine are termed 
avigation instruments. Since time should not be lost while the 
pilot's eye is searching about the instrument board, the engine 
instruments should be grouped together and the avigation instru- 
ments should be together. 

322 



ENGINE INSTRUMJ^NTS 323 

In considering instruments, care must be taken to differentiate 
between accuracy and sensitivity. Sensitivity is the ability to 
detect very small differences in the amount of the quantity being 
measured. It is not the same as accuracy or correctness. An 
instrument may be sensitive and at the same time very inaccurate. 
For example, a tachometer might be extremely sensitive in that it 
would show a very small change in the number of revolutions per 
minute, but the actually indicated revolutions per minute might 
be greatly in error. 

Another important feature in instruments is absence of lag. 
The instrument should respond instantly to any change in the 
magnitude of the quantity being measured. For example, an 
altimeter would be of little value if the airplane dropped a thou- 
sand feet and the altimeter did not show this decrease for several 
seconds. A properly designed instrument will have the parts so 
balanced that the time lag will be reduced to a negligible amount. 

Engine Instruments. Engine instruments are either to enable 
the pilot to operate the engine most efficiently or to warn him of 
impending trouble. Some engine troubles cannot be alleviated 
in the air, but the instrument performs the very useful service of 
indicating wherein the trouble lies, so that, the cause of stoppage 
being known, the fault can be remedied after landing. Much 
time can be saved the mechanic in repairing a stopped engine if 
he knows what first caused the trouble. 

Airplane engines that have passed the rigid tests of the Depart- 
ment of Commerce if they have no defective parts and are prop- 
erly mounted should run for long periods provided that they are 
oiled properly, proper and sufficient fuel is suppHed, etc. The 
throttle of an airplane engine is a hand throttle. When it is set 
at a certain throttle-opening to give a certain engine speed, the 
engine will continue to turn over at that speed unless some part 
of the engine or its auxiliaries is not functioning properly. 

Engines do not stop without cause, nor do they stop instantly. 
Whatever ultimately causes the engine to quit entirely, first causes 
it to slow down. The malfunctioning of some part of the operation 
causes a loss of power which results in a drop in engine speed. 
If the reason for this malfunctioning is not removed, the loss in 
power will increase and finally result in stoppage. For example, 
if a bearing is getting insufficient lubrication, it will get hot and 
expand. The added friction will slow up the revolutions per min- 



324 INSTRUMENTS 

ute as well as increase the heating to cause more expansion till 
finally the bearing and journal '' freeze." For another example, 
clean spark plugs do not suddenly become fouled. There is a 
gradual accumulation which causes poor firing, lessening the power 
and decreasing the speed. As the accumulation increases the 
firing becomes poorer until the spark fails to jump and the cylinder 
misses completely. 

Probably the only reasons for instantaneous failure of an engine 
are breakage of the fuel-line or breakage of an electric wire, both 
of which would be due to vibration from improper mounting. 
Since, on a properly mounted engine, complete failure is always 
preceded by a more or less brief period of poor functioning, a care- 
ful watch should be kept of the engine speed; a slight decrease in 
revolutions per minute is a warning of impending trouble. 

The tachometer is essentially a cautioning instrument; if it 
shows a decrease in engine speed, the pilot should scan the other 
engine instruments to locate, if possible, the source of the trouble. 
If the oil or water is too cool, the pilot can remedy this in the air 
by closing shutters. Most of the other troubles cannot be recti- 
fied in the air, but with the advance warning the pilot can often 
land before his engine has quit entirely. 

Tachometers. The tachometer indicates the revolutions per 
minute of the engine. Although its most valuable function is to 
warn of impending engine trouble, it is at all times useful in aiding 
the pilot to set the engine at the most economical cruising speed, 
etc. On the line, it helps the mechanic in judging to what extent 
an engine is " tuned-up." If an airplane engine can be throttled 
down to 250 r.p.m. and it turns over without missing, it is in very 
good condition; if it can be throttled down to 200 r.p.m. without 
missing, it is in excellent condition. 

Several types of tachometers are in use, the most popular being 
the chronometric and the centrifugal types. The chronometric 
type has more working parts and therefore the production cost 
should be greater than for the centrifugal type. With wear of the 
working parts, the centrifugal type loses in accuracy while still 
functioning; the chronometric type is " either dead-right (within 
instrumental accuracy) or stopped." 

Chronometric Tachometer. This type actually counts the 
revolutions in 1-sec. periods, but shows on the dial the revolutions 
per second multiphed by 60 (i.e., the revolutions per minute). 



CHRONOMETRIC TACHOMETER 



325 



The pointer remains stationary for 1 sec. If the engine speed is 
constant, the pointer remains fixed ; but if the engine speed is vary- 
ing, at the end of the 1-sec. counting period, the pointer jumps to a 
new position. This does not mean that the engine speed is erratic 
or jumpy, but merely that the average speed for 1 sec. is different 
from that of the previous second. 

The mechanism is roughly as follows. A flexible shaft from the 
engine turns a gear in the tachometer at a fixed ratio to engine 
speed. This driving gear does two things. First, it winds the 
mainspring of a clockwork, through a friction clutch, so that the 
spring will not be wound too tight ; as the tachometer operates the 
clockwork tends to run down but the driving gear keeps it wound 
up. The other function of the driving gear is to rotate a counting 
gear. This counting gear, marked G in Fig. 106, is in mesh with 



P5^^ 




Fig. 106. Chronometric tachometer mechanism. 



a cylindrical rack or fine-tooth worm. This worm is split longi- 
tudinally into three segments. A, B, and C. The three segments 
encase a shaft and are so keyed to the shaft that they may move 
independently along it. On one end (left end in Fig. 106) is a 
pinion with an escapement operated by the clockwork so that by 
an intermittent mechanism, at the end of each second, the shaft 
is rotated through one-third of a revolution. On the other end of 
the shaft is a loose collar S, which by a spring is held against the 



326 INSTRUMENTS 

end of the segmented worm. The collar S moves the pointer over 
a scale. 

Assuming segment A in mesh with counting gear G, at the be- 
ginning of the 1-sec. counting period, the rotation of G starts 
moving segment A to the right. At the end of the second the 
linear movement of A has been proportional to the speed of gear 
G. Synchronously with the end of the second, the clock escape- 
ment rotates the shaft and segmented worm through 120°, so that 
segment A occupies the position previously occupied by segment 
C, segment C has moved to replace segment B, and segment B has 
moved to engage with counting gear G. The linear displacement 
of segment A is maintained by a locking pawl. Collar S, linked 
with the pointer, is resting against the end of segment A. 

During the next second, segment B is being fed to the right by 
the counting gear. At the end of the second, segment B moves 
down to the position of A, and segment A slides back along the 
shaft to its zero position, actuated by a spring not shown. 

The above-described type is the ^' Tel " mechanism; other 
types use gears instead of the segmented worm. 

Even a very cheap clock can be regulated so as not to gain or 
lose more than half a minute in a day; in 1 sec. the inaccuracy 
is very small indeed, and the only error should be due to the mesh- 
ing of the teeth of the counting gear which, on a good instrument, 
should not exceed plus or minus 10 r.p.m. If a chronometric 
tachometer is functioning at all, it should be correct within these 
limits. Sometimes when very cold the oil in the clockwork be- 
comes thick and gummy, stopping the clockwork. Wear of teeth 
will cause a jam, stopping the operation. 

In addition to the merit of either being correct or stopped, the 
chronometric type is usable when the engine is turning over very 
slowly. The centrifugal type described in the next paragraph 
does not give a readable indication much below 500 r.p.m. 

Centrifugal Tachometer. The centrifugal type depends on the 
centrifugal action of a pair of weights disposed about a shaft 
which is rotated by the engine through a flexible shaft. The 
weights, see Fig. 107, are connected by links to fixed collar A and 
movable collar B. As the speed of rotation of the shaft is in- 
creased, the weights tend to move outward raising collar B, this 
motion being resisted by a spring. Centrifugal force varies as 
the square of the speed, so the upward movement of collar B is 



THERMOMETER 



327 



Weight 




Centrifugal tachometer 
mechanism. 



not proportional to the changes in revolutions per minute. Usu- 
ally, since the actuating force is very small at low speeds, no 
attempt is made to read the instrument at low speed. As the 

parts wear, the lost motion 
causes the instrument to be 
inaccurate, even though ap- 
parently functioning. The 
inertia of the moving parts 
produces a time lag which 
may delay the instrument 
several seconds in indicating 
a change in rate of revolution. 
Pressure Gages. Pressure 
gages are used to measure the 
gasoline pressure near the car- 
buretor in force-feed systems 
and also to measure the pres- 
sure of the oil system. Because loss of pressure means engine 
failure, the pressure gages should be watched closely, especially 
before take-off. Overpressure indicates a stoppage in the line; 
underpressure indicates lack of oil, failure of pump, or broken 
line. 

The gages are usually of the Bourdon type, the expansion mem- 
ber being a seamless drawn bronze tube of elliptic cross-section 
bent into an arc of more than 180°. One end of this tube is closed, 
and when internal pressure is applied through the open end, the 
tube tends to straighten out. The movement of the free end is 
communicated to a pointer. 

Thermometers. Thermometers are used to find the tempera- 
ture of the cooling liquid, the lubricating oil, or the cylinder walls 
or heads. The thermometer in the cooling system warns of engine 
trouble due to overheating, aids in operating the engine at maxi- 
mum efficiency, warns if the cooling medium is near its boiling 
or freezing point, and warns if the engine has become too cool in 
a glide to pick-up readily. 

There are two general types of aircraft thermometers: the 
pressure types and the electric types. All thermometers must be 
distant-reading; that is, there must be an indication on the in- 
strument board in front of the pilot of the temperature at some 
point several feet distant. The pressure types are especially 



328 INSTRUMENTS 

suited to finding the temperature of a liquid such as the oil or the 
cooling medium; the electric type is suited best for finding the 
temperature of parts of the engine and is used on air-cooled 
engines. 

The pressure-type thermometer has three parts: a bulb, which 
is immersed in the oil or water ; a pressure gage on the instrument 
board; and a length of tubing which connects the bulb to the 
gage. There are two kinds of pressure thermometers — the 
liquid-filled and the vapor-pressure thermometers. 

In the liquid-filled type, the bulb, the tubing, and the pressure 
chamber in the gage are all filled completely with a liquid. When 
the bulb is heated, the liquid in the bulb expands and causes in- 
ternal pressure in the system. For an increase of 100° C, the 
increase in pressure is several hundred pounds per square inch. 
This pressure is indicated on the pressure gage, the scale, however, 
reading directly in degrees of temperature. Since only a part of 
the liquid is heated to the temperature which it is desired to meas- 
ure, there is liable to be an inaccuracy due to the liquid in the 
connecting tubing and the gage not being at this temperature. 
This error is made negligible by having the connecting tubing of 
very fine bore and by putting a bimetallic compensation on the 
pressure gage. 

In the vapor-pressure type, the bulb contains a volatile liquid, 
usually ether. The quantity of liquid must be such as not to 
fill the bulb completely. The remainder of the bulb, the capillary 
connecting tubing, and the pressure chamber of the gage are filled 
with the vapor from the liquid. It is essential that the free surface 
of the liquid be inside the bulb, since in a closed system containing 
a liquid and its vapor the vapor pressure depends on the tem- 
perature of the liquid surface in contact with the vapor. 

The liquid-pressure type depends on the expansion of the 
liquid with heat, so the pressure changes will be lineal with tem- 
perature; with the vapor-pressure type, the pressure change will 
not be linear with temperature but will be greater at higher 
temperatures, so that the scale will be more open there than at 
low temperatures. The pressures developed with the liquid- 
filled type will be much greater at the same temperature than with 
the vapor-pressure type so that the gage is much more rugged for 
the liquid-filled type. Owing to these high pressures in the 
liquid-filled system, the changes in atmospheric pressure due to 



ALTIMETERS/ 329 

altitude are insignificant; with the vapor-pressure type there 
may be an error of several degrees resulting from this. Both types 
use a fine capillary tube connecting the bulb to the gage, and care 
must be taken in dismantling an airplane that a careless mechanic 
does not cut the tubing under the impression that it is a copper 
wire. 

The electric-type thermometer is customarily a thermocouple 
connected to a millivoltmeter. One wire is iron, the other con- 
stantin. The " hot junction " may be a true joint, or the same 
effect is achieved if the two wires are inserted into holes a short 
distance apart in the cylinder wall and peined securely in place. 
In a thermocouple, the difference in potential indicated by the 
millivoltmeter on the instrument board depends on the difference 
in temperature between the " hot " and " cold " junctions. 
This means that some form of temperature compensation must be 
used or the thermometer will merely indicate the difference be- 
tween engine temperature and cockpit temperature. 

Ice-warning indicators are merely thermometers with special 
markings for the temperatures between 32° and 29° F., which is 
the temperature range that is especially dangerous from the 
standpoint of ice formation. 

Avigation Instruments. Avigation instruments are those in- 
struments which aid the pilot in flying the airplane safely and in 
guiding him to his destination. They tell him his altitude, his 
speed, his direction, and the attitude of the airplane with respect to 
the ground or some other reference. 

Altimeters. The usual altimeter is a barometer or pressure gage 
which measures the atmospheric pressure. Other types have 
been proposed but none have passed beyond the experimental 
stage. These types are electrical or sonic. 

For bombing, it is desirable to know the altitude very accurately 
at heights of 12,000 to 15,000 ft. in order that the trajectory may 
be calculated correctly. For photographic mapping, usually done 
at about 12,000 ft., it is primarily essential that all the photo- 
graphs be taken at exactly the same altitude in order that each 
part of the mosaic be the same scale. It is very desirable that 
this altitude be known correctly to check ground distances. 

In flying over mountain ranges in cloudy weather, flight should 
be made at sufficient altitude to have ample clearance above the 
mountain tops. The readings of the ordinary airspeed indicator 



330 INSTRUMENTS 

are subject to an altitude correction, but an error of a hundred 
feet in altitude makes a very small change in the altitude correction 
to the indicated airspeed. 

Therefore, except for bombing and photographic work and the 
somewhat rare attempts for altitude record, the need for accuracy 
at medium or high altitudes is not very great. For landing in 
fog, however, the altimeter should be extremely accurate. 

Aneroid Altimeters. The standard type of altimeter is a 
pressure gage, but since the pressure differences to be observed are 
very small, the gage must be exceedingly sensitive. For this 
reason an aneroid is used. An aneroid is a thin metallic box 
shaped like a poker chip. The flat sides are very thin and flexible. 
Formerly these sides were corrugated to give flexibility, but 
better results are obtained with flat sides. When pressure on 
the outside is removed, the aneroid expands. As the flat sides 
move outward, this motion is magnified and transmitted to the 
pointer. 

The dial instead of being graduated in pressure units is marked 
in units of altitude. Since the expansion of the aneroid depends 
on pressure, a simple Hnkage would result in an approximately 
uniform pressure scale. The pressure difference for 1-ft. altitude 
difference is much greater close to the ground than at altitude, so 
that if the pointer movement were proportional to pressure differ- 
ence, the altitude scale would be non-uniform; that is, the gradu- 
ations would be far apart near the ground and closer together as 
altitude is increased. To have a uniform altitude scale, a very 
ingenious arrangement of links and chain and roller is used. 

Having a uniform altitude scale makes it possible to have the 
scale adjustable. With the usual circular dial, a knob is provided 
outside the case whereby the dial may be rotated irrespective of 
the position of the pointer. 

Formerly there was no fixed barometric scale on the altimeter. 
Before taking off, the pilot would turn the dial till the zero of his 
altitude scale was directly back of the pointer. After fljdng 
around, in coming down to land he would assume that the alti- 
meter would read zero when his wheels touched the ground. 
This would be true only if the atmospheric pressure had not 
changed while he was in the air. If, while he had been flying, the 
atmospheric pressure at the airport had decreased by, say, 0.2 in., 
his altimeter would read 200 ft. when he landed. Conversely if 



ANEROID ALTIMETERS 331 

the barometric pressure on the ground had risen while he was 
aloft, his altimeter might indicate zero while he was still several 
hundred feet in the air. 

Modern altimeters have a fixed barometric scale in addition 
to the usual adjustable altitude scale. This is a desirable adjunct 
to landing at an airport in fog. While the airplane is still in the 
air, an operator on the ground, hearing the motor noise, carefully 
reads the barometer at the airport and radios the reading to the 
pilot. The pilot thereupon sets the zero of the altitude scale op- 
posite this reading on the fixed barometric scale on his altimeter. 
He then comes down to land knowing that when the pointer 
reaches the zero on the altitude scale, his wheels should be touch- 
ing the ground. 

The altitude-pressure relation for altimeter calibration is given 
in N.A.C.A. Report 246 and is practically identical with the 
standard atmosphere given in Table I of this text. This relation 
is based on an arbitrary assumption of the temperature change 
with altitude. This assumption checks very closely with the 
average observed variations of temperature with altitude at 
latitude 40° in the United States. These observations were taken 
throughout the year. On any one day, the temperature at some 
altitude may be above or below the average yearly temperature 
at that altitude. 

For an aneroid altimeter to be absolutely correct, the tempera- 
ture not only must be 15° C. (59° F.) at sea-level but it also must 
vary in an orthodox manner as the altitude is increased. When 
these conditions do not exist the altimeter will not read correctly 
at altitude. A correction to the altimeter reading can be made if 
the average temperature of the air between the plane and the 
ground is known. A very close approximation of this mean 
temperature can be made by reading the temperature every few 
hundred feet during the climb and averaging these readings. 
An approximation sufficiently close for most photographic work 
is made by taking the arithmetic mean between the ground air 
temperature and the air temperature at the altitude at which the 
flying is to be done. 

Since air inside the aneroid would expand with an increase in 
temperature and build up a pressure, and vice versa, it is custom- 
ary to pump out most of the air before sealing the aneroid to de- 
crease the error from this source. The pressure inside the altimeter 



332 INSTRUMENTS , 

case, outside the aneroid, should be atmospheric pressure; there- 
fore, there should always be an opening in the case. Ignorant 
mechanics, seeing a hole in the case, sometimes, with mistaken 
zeal, plug it up. This should be guarded against. 

Frequently the pressure in the cockpit is not exactly the same 
as the atmospheric pressure immediately around the airplane. 
This is especially true in cabin planes. For accurate readings a 
tubing is connected to the altimeter case, the outer open end of the 
tubing being outside the cabin at some point on the structure 
where there is only static pressure. 

To make the altimeter sensitive, the side of the aneroid is of 
large area. A very slight difference of pressure will then make 
an appreciable pointer movement. 

It is important that an altimeter have very little lag. Inertia 
varies with mass, so that, if the aneroid or other moving parts 
have considerable mass, they will have more inertia or resistance 
to sudden movement. Sensitive altimeters are built with large 
aneroids, but sufficient care is taken in balancing the moving parts 
so that lag is reduced to a minimum. 

It should be noted that the word " sensitive " is not synony- 
mous with accurate. A sensitive altimeter is one in which the 
pointer moves a detectable amount for very small changes in 
altitude. A " sensitive " altimeter is subject to the same errors 
as other aneroid altimeters if the atmosphere at the time of flight 
is not standard, i.e., if the pressure and temperatures do not 
correspond with the standard pressure-temperature-altitude 
relation. 

Altimeters, Miscellaneous Types. At sea, sonic depth-finders 
are used, the time interval being measured for a sound to travel 
to the bottom of the ocean and echo back. Boehm in Germany 
and Jenkins in the United States have attempted to adapt this 
scheme to aircraft but with only moderate success. 

A capacity altimeter depends on the principle that the mutual 
electrical capacity of two adjacent plates increases with the 
approach of a third plate. Two plates are fastened to the bot- 
tom of the fuselage and their capacity increases as the plane nears 
the earth, which acts as the third plate. This device acts ad- 
mirably as a landing altimeter because the sensitivity increases 
greatly as the altitude becomes small. Owing to weight and cost, 
this type is rarely used. 



PITOT-STATIC AIRSPEED INDICATORS 333 

At night, flying over water, a searchlight with a diverging beam 
can be aimed directly downward. If alongside the searchlight 
proper is mounted a telescope equipped with stadia wires, it is 
possible to measure the diameter of the light circle on the water 
and compute the altitude. Such a device was used on the Graf 
Zeppelin. By comparing the altitudes obtained by this method 
with the readings of the aneroid altimeter, the meteorologist 
ascertained whether the Graf was in a low- or high-pressure area 
to aid him in his weather predictions. Another scheme used on 
the Graf was to drop plaster eggs and time the interval till they 
shattered on the water. 

Airspeed Indicators, The airspeed indicator shows the speed 
of the airplane with respect to the surrounding air, i.e., the speed 
of the relative wind. In still air, the airspeed is the ground speed, 
but if any wind is blowing, the airspeed and ground speed are 
not the same. The airspeed indicator is a valuable aid to the 
pilot. It warns him when the plane is near its stalling speed. By 
variations of its indications, he knows if he is nosing-up or diving. 
In taking-off, the airspeed indicator tells him when he has attained 
sufficient speed. He can know when he is at best climbing speed, 
speed corresponding to L/Aotai maximum, etc. In dives he is 
warned when the speed is becoming excessive for structural 
strains. 

Pitot-Static Airspeed Indicators. The standard airspeed in- 
dicator makes use of a Bitot tube, which is a tube parallel to the 
airplane axis, its open end pointing in the direction of flight. 
The pressure at the open end of a correctly designed Bitot tube 
will be 

Pt is Bitot pressure in pounds per square foot 
p.2 Ps is static (atmospheric) pressure in pounds 

Pt = Ps + ^o- per square foot 

p is mass density of air in slugs per cubic foot 
V is airspeed in feet per second 

If the airspeed, V, is expressed in miles per hour instead of feet 
per second this becomes 



= P, + 1.075 p72 



334 INSTRUMENTS 

If the Pitot and static pressures are in pounds per square inch, 

= Ps + 0.00747 pV^ 

With air at standard density (p = 0.002378 slug per cubic foot), 
this becomes 

Pt= Ps + 0.00001776 72 

In cahbrating airspeed indicators against pressure gages, the 
pressure gage shows the difference between the Pitot pressure and 
the static pressure, so that 

Pt- Ps = 0.00001776 72 
or 7 = 237.3 Vp^^HP^ 

If the cahbration of the airspeed indicator is against a U-tube of 
water, since 1 in. of water is equivalent to a pressure of 0.0362 lb. 
per sq. in., 

h = 0.000491 V^ h is head of water in inches 

or 7 = 45.2 Vh V is in miles per hour 

Because cockpit pressure is not exactly the same as atmospheric 
pressure, a static tube is mounted immediately adjacent to the 
Pitot tube, and the other end of the static tube is connected to 
the inside of the case of the pressure gage. The pressure gage will 
then be operating on the dynamic pressure, or difference between 
Pitot and static pressure, at the point where the Pitot and static 
tubes are mounted. It is important that the tubes be in a proper 
location where the air flow is not too greatly disturbed. For a 
biplane, the best location is on a forward interplane strut about 
midway of the gap. For a monoplane, the preferable location 
is one-third the semi-span in from a tip and about half a chord 
length forward of the leading edge in the plane of the under sur- 
face of the wing. Mounting the Pitot and static tubes on a rod 
projecting this far in front of the wing, though it gives more 
accurate airspeed readings, is undesirable from the liability of 
their being damaged as the airplane is moved around in the hangar. 
For this reason it is customary to mount the tubes about a foot 
below the under surface of the wing, one-third the semi-span in 
from the tip and vertically below the leading edge. 



ALTITUDE CORRECTION FOR AIRSPEED INDICATORS 335 

The tubing connecting the Pitot-static heads to the indicator 
in the cockpit must be absolutely airtight. All connections should 
be soldered or a metal union should be used. The former practice 
of using rubber tubing to make joints is to be condemned as with 
age the rubber cracks open. 

Under certain weather conditions, ice forms on the open for- 
ward-projecting end of the Pitot tube. Because it is pressure, 
not force, that is being measured, the actual size of the opening 
makes very little difference. Ice partially blocking the opening 
of the Pitot tube will not affect the readings of the airspeed in- 
dicator as long as there is even a pinhole opening. With the 
opening totally sealed over by ice, the airspeed indicator fails to 
function. Some Pitot tubes have means for being electrically 
heated to prevent ice formation. It should be borne in mind 
that the same weather conditions that would cause stoppage by 
ice of the Pitot tube would cause ice to form on the propeller and 
leading edge of the wing, and a pilot should avoid this. 

Pitot-Venturi Airspeed Indicator. At one time, it was almost 
universal practice to mount a Venturi tube alongside a Pitot 
tube and connect these by two hues of tubing to a pressure gage 
airspeed indicator in the cockpit, so that one side or compartment 
of the pressure gage had the Pitot pressure, the other had the suc- 
tion from the throat of the Venturi tube. The suction (or nega- 
tive pressure) at the throat of a properly designed Venturi tube is 
numerically several times the (positive) pressure of a Pitot tube, 
the air velocity being the same. The pressure gage operating on 
the pressure differential between a Pitot tube and a Venturi tube 
could be much more rugged and sturdy than one operating on the 
pressure differences between a Pitot tube and a static tube. 

The reason for the discontinuance of the manufacture and use 
of the Pitot-Venturi type of airspeed indicator is that any deposit 
of ice or mud on the forward lip of the Venturi nozzle destroyed 
the smooth flow of air through the throat of the Venturi tube. 
Unless the airflow was devoid of eddies and burbling, the suction 
would not be proportional to airspeed and the instrument would 
read incorrectly. 

Altitude Correction for Airspeed Indicators. The Pitot-static 
as well as the obsolete Pitot-Venturi airspeed indicators are termed 
'' dynamic," since the pressures on which the indicators operate 
vary with the dynamic pressures, i.e., they vary with (p/2)V^. 



336 INSTRUMENTS 

These instruments, when the airplane is travehng at the same air- 
speed at altitude as at the ground, will be subjected to less pressure 
difference at altitude since the density (p) is less. These airspeed 
indicators will therefore read low at altitudes. 

The mechanism of the indicator is designed to give a uniform 
airspeed scale, so that, while the pressure varies as the square of 
the velocity, a linkage is used that gives a pointer movement 
varying as the square root of the pressure. Then as the density 
(p) decreases with altitude, the pressure decreasing in the same 
ratio, the indicated airspeed will decrease as the square root of the 
density. 

A rough rule is to add 2 per cent of the indicated airspeed for 
every thousand feet of altitude to get the true airspeed. More 
accurate results are obtained by multiplying the indicated air- 
speed by a/po/p, using values of p/po from Table I in this text. 

The true airspeed is of Httle use in flying except, with wind 
velocity known or by using a wind-star, to obtain groundspeed. 
For handling the airplane the pilot desires to know not the true 
airspeed but the dynamic airspeed or airspeed indicated on a 
dynamic-type airspeed indicator. Maximum rate of climb, 
stalling, stresses, etc., are all dependent on dynamic airspeed and 
not on true airspeed. For example, if an airplane has a certain 
stalling speed at sea-level, the stalling speed a t any altitude will be 
the sea-level stalling speed multiplied by V po/p. The dynamic 
airspeed indicator at any altit ude in dicates a speed which is the 
true airspeed multiplied by Vp/po, or conver sely, the true air- 
speed is the indicated airspeed multipHed by Vpo/p. If the stall- 
ing speed is known at sea-level, whatever the altitude, when the 
airspeed indicator reads that speed, the airplane is on the point 
of stalling. In other words, if at some high altitude the airplane 
is put in a stall and the airspeed indicator read, at any other 
altitude when the airspeed indicator shows the same reading, the 
airplane is stalled. 

True Airspeed Indicator. Any freely revolving device such as 
an anemometer or spinner (miniature propeller) will rotate at a 
rate proportional to the true airspeed provided that there is little 
or no friction. When friction is not negligible so that the rotor 
must do some work, its speed is no longer independent of density. 

Compasses, Magnetic-Needle Type. A magnetic compass is an 
instrument which tells direction by means of the earth's mag- 



COMPASSES, MAGNETIC-NEEDLE TYPE 337 

netism. The earth's magnetic field is represented by magnetic 
lines of force which, in general, run from the north magnetic pole 
(located near 70° N., 96° W.) to the south magnetic pole (near 
71° S., 148° E.) but with erratic diversions due to local deposits 
of iron and other causes. These lines slope downward, except at 
the magnetic equator, so that at the latitude of San Antonio the 
lines slope 50° from the horizontal; farther north at the latitude 
of New York the dip is 72°. 

A magnetic needle, unaffected by nearby iron or steel, tends to 
lie along the magnetic lines of force. In the United States, it is 
customary in making compasses to suspend the needles from a 
point considerably above the needles so that there will be little 
tendency for the needles to dip. They will be free to swing into 
the vertical plane of the magnetic lines of force. At only a few 
places on the earth are these lines running true north and south. 
Compasses therefore do not indicate true north, but when they 
are in the plane of the magnetic lines of force, they are said to be 
indicating magnetic north. The difference between true north 
and magnetic north is the magnetic variation (or magnetic declina- 
tion). Variation is said to be plus or westerly when the compass 
indicates a north which is west of the true north; variation is 
easterly or negative when the compass indicates a north which is 
east of the true north. 

Charts or tables tell the variation at various places; a sample 
chart for the United States is given in Fig. 108. The variation 
changes from year to year so that the most recent chart should 
always be used. 

The pull of magnetic force is in the direction of the magnetic 
lines of force; it is only the horizontal component of this force 
that has a directional effect on the compass needles. Near the 
earth's magnetic poles the direction of the force is nearly vertical 
so that the compass needles have little tendency to point in any 
direction. 

Besides variation, compass needles are subject to the additional 
error of deviation, which is due to the nearby presence of steel or 
iron. Direct electric currents also affect the compass. The 
deviating effect of these disturbing elements varies as the mass and 
inversely as the square of the distance ; a small piece of steel close 
to the compass may have more effect than a large piece of steel at a 
considerable distance. When deviation is such that the compass 



338 



INSTRUMENTS 



indicates a north which is west of magnetic north, deviation is called 
westerly or plus; when the compass north is east of the magnetic 
north, deviation is easterly or minus. 

The compass needles point in a general northerly direction. 
As an airplane's heading is changed, the absolute direction with 




Fig. 108. Magnetic variation for the United States, 1935. 



respect to the compass of the engine, fuselage members, and other 
steel parts is changed. The deviating effect of the engine, etc., is 
therefore different with each heading of the airplane. 

Variation depends on the geographic location of the airplane; 
deviation depends on the heading. 

To ensure that there shall not be large deviations, a compass on 
an airplane should be compensated, which means placing small 
magnets close to the compass needles to counteract as much as 
possible the deviating influence of steel objects on the airplane. 
It is usually not possible to correct all deviation; the most that is 
attempted is to compensate on north heading and on east heading 
so that there shall be zero deviation on these headings. The 
airplane is then set on other headings (every 30° from north) and 
the deviation recorded on a card which is fastened on or near the 
compass. By consulting this deviation card, interpolating if 
necessary, the pilot can correct his compass reading to obtain his 
true heading. 



COMPASSES, MAGNETIC-NEEDLE TYPE 339 

Directions are given in degrees clockwise from north, so east is 
90°, south is 180°, west is 270°, etc. Headings measured from true 
north are true headings; measured from magnetic north they are 
magnetic headings; and measured from the north indicated by the 
compass they are compass headings. 

A course is the direction an airplane should head to reach its 
destination provided there is no cross-wind; if there is a cross- 
wind, the proper heading is not the same as the course. 

On airway strip maps, the line between airports is marked with 
the magnetic course (M.C.) at intervals, wherever the variation 
has changed by as much as 1°. Before these magnetic courses are 
used by the pilot, they must be changed to compass courses by 
applying deviation. If a pilot is using maps not so marked but 
obtains a true course by any means, the true course must be 
changed to a compass course by applying both variation and 
deviation. 

To find a magnetic course knowing the true course, the variation 
is added to the true course if the variation is westerly, and sub- 
tracted from the true course if variation is easterly. To find the 
true course knowing the magnetic course, the variation is sub- 
tracted from the magnetic course if the variation is westerly, 
added to the magnetic course if variation is easterly. 

To find a compass course knowing the magnetic course, the 
deviation is added to the magnetic course if the deviation is 
westerly, and subtracted from the magnetic course if the deviation 
is easterly. To find a magnetic course knowing the compass 
course, the deviation is subtracted from the compass course if 
deviation is westerly, and added to the compass course if deviation 
is easterly. 

Example. It is desired to head due east. Variation (from map) 
is 14° W. Deviation (from card on instrument board) is 10° E. What 
should compass read? 

Solution. True heading should be 90°; since magnetic N. is 14° west 
of true N., magnetic heading should be 104°. Since compass N. is 10° 
east of magnetic N., compass heading should be 94°. 

Example. The compass heading is 260°. Variation is 21° E. 
Deviation is 5° E. What is true heading? 

Solution. Compass heading is 260°; since compass N. is 5° E. of 
magnetic N., magnetic heading is 265°. Since magnetic N. is 21° B, 
of true N., true heading is 286°. 



340 INSTRUMENTS 

Problems 

1. Compass heading is 80°. Variation 15° W., deviation 10° W. 
What is true heading? 

2. It is desired to head 195° true. Variation 15° W., deviation 
10° E. What should compass read? 

3. True heading is 290°. Variation 25° E., deviation 5° W. What 
is compass heading? 

4. Magnetic heading is 340°. Deviation is 12° E. What is com- 
pass heading? 

5. Compass heading is 355°. Variation 12° E., deviation 10° E. 
What is true heading? 

6. Magnetic heading is 5°. Deviation is 20° E. What is compass 
heading? 

7. True course is 160°. Variation is 10° W.; deviation is 15° E.; 
there is no wind. What should be the compass heading? 

8. True course is 15°. Variation is 20° W.; deviation is 8° E.; 
there is no wind. What should be the compass heading? 

9. Magnetic course is 274°. Deviation is 5° W. ; there is no wind. 
What should be compass heading? 

10. Magnetic course is 129°. Deviation is 10° E.; there is no wind. 
What should be compass heading? 

Airplane compasses have two parallel bar magnets of tungsten 
steel fastened to a spider of non-magnetic material. Affixed to the 
spider is the '' card " or scale, a cylindrical strip of celluloid or 
thin brass, on which the graduations are marked. The compass 
being mounted so that the pilot is virtually looking backwards at 
the compass, the point on the card corresponding with the south- 
seeking end of the magnets is marked N (North). The card is 
usually marked with 5° graduations. 

From the center of the card, a stellite pivot projects downward, 
resting in a jewel-cup on the top of a post extending up from the 
bottom of the bowl. A retaining device prevents the pivot from 
jumping out of the cup (see Fig. 109). 

All airplane compasses are of the liquid type. The compass 
magnets and card are inside a spherical metal bowl which has a 
glass window on one side, and the entire bowl is full of mineral 
spirits. The liquid serves two purposes; it buoys up the card so 
that there is less friction at the pivot, and it damps the swingings 
of the card. 

Oscillations of the card are caused either by the airplane's nose 
swinging to one side which by friction with the interior of the 



COMPASSES, MAGNETIC-NEEDLE TYPE 



341 



bowl causes the liquid to swirl, moving the card; by a wing 
dropping which makes the compass card pivot no longer vertical 
permitting the magnets to dip ; or by vibrations transmitted from 
the engine. To prevent the last, the entire compass is mounted 
on felt washers. 

Drawer for Compensating 
Magnets 



Sylphon- 




• Magnetic Needles 



'Glass Window 



Expansion Chamber— i ^ ^Retaining Claws 

Fig. 109. Compass, magnetic needle type. 

Ordinarily compass cards have a regular period for an oscillation 
and are termed periodic. The period is the time required to swing 
from an extreme position on one side to an extreme position on the 
other side and return. Whether the amount of swing is large or 
small, the period is practically the same. V/ith excessive damp- 
ing, the periods are not of the same duration ; such compasses are 
termed aperiodic. An aperiodic compass, if disturbed, will re- 
quire a longer time to return to north than the more common type 
of compass, but the aperiodic card will not swing so far past north. 
In a very few swings it will be steady enough to read, whereas an 
ordinary compass will swing several times before the amplitude 
has decreased to such a small amount that it may be read. As 
yet, aperiodic compasses have not been so built as to be adaptable 
for instrument-board mounting; they serve well as master com- 
passes to be placed back in the cabin, away from magnetic ma- 
terial, to check the pilot's steering compass. 

Steering entirely by compass is very tiring to the pilot, and for 
this reason the compass should be so mounted as to be as nearly 
as possible in the pilot's ordinary line of vision. The compass 
should be placed where it will be affected least by local magnetic 
material such as steel fuselage members, voltage regulators, and 
guns. Care should be taken when mounting to ensure that the 
bolts, washers, and nuts are brass or aluminum. 



342 INSTRUMENTS 

Compasses, Induction Type. The ordinary compass is a direct- 
reading instrument; the induction-type compass is a distant- 
reading instrument, that is, the part affected by the earth's mag- 
netism is a considerable distance from the part the pilot reads. 
The induction-type compass consists of three main parts, a gen- 
erator, a control-dial, and an indicator. The generator is located 
in a part of the airplane as free as possible from steel, usually in 
the extreme rear of the fuselage. The indicator is placed in the 
cockpit where it can be easily seen by the pilot. The control dial 
is located conveniently for the avigator, or for the pilot if he is 
functioning also as avigator. 

The induction compass depends on the principle that, if mag- 
netic lines of force are cut by a moving wire, a difference in electric 
potential is created at the ends of the wire. If the wire moves 
parallel to the magnetic lines of force, no lines being cut, there 
will be no difference of potential. If a coil of wire is connected 
to a galvanometer and the coil is rotated about a vertical axis, at 
the instant that the coil is in the magnetic north-south plane, the 
galvanometer will indicate a maximum current, but at the instant 
that the coil is in the magnetic east-west plane the galvanometer 
will indicate zero current. Instead of a single coil, one may con- 
ceive of a number of coils in different planes all wrapped around 
a cylindrical core with the ends of each separate coil terminating 
in commutator segments, and a pair of brushes connecting with 
the galvanometer. As long as the brushes are in the magnetic 
east-west plane, they are connected through the commutator 
with a coil just as it is not cutting any lines of force. If the brushes 
are not in the east- west plane, they will be connecting with a coil 
when it is cutting the magnetic lines of force and therefore a cur- 
rent will be indicated on the galvanometer. A zero reading on the 
galvanometer would show that the brushes were in the magnetic 
east-west plane. 

The brushes are mounted on a turntable which may be rotated 
by the control dial. With the elementary type described above, 
when it is desired to fly magnetic north, the brushes are set 
exactly thwartship (perpendicular to the longitudinal axis of the 
airplane). As long as the airplane is headed due magnetic north, 
the galvanometer will read zero. If the airplane's heading changes 
by even a slight amount to one side, the brushes will no longer be in 
the east-west plane and the galvanometer will be deflected to one 



COMPASSES, INDUCTION TYPE 



343 



side. If it is desired to fly, say, a 30° magnetic heading, the 
brushes are turned through a 30° angle counter-clockwise viewed 
from above. With the brushes in this position, right rudder is 
given till the galvanometer reads zero, which will be when the 
airplane is headed 30° magnetic. Veering to one side or other from 
this heading will cause corresponding deflections of the needle. 

In the induction compass as actually constructed, instead of a 
number of separate coils, the armature is drum- wound, greatly 
adding to the sensitivity of the instrument. The functioning is 
the same as described above except that the brush positions are 
shifted through an angle of 90°. The vertical armature shaft 
projects up through the top of the fuselage, and on the upper end 
is a paddle-wheel or impeller which is caused to revolve by the 
slipstream. The axis of the armature coils must be truly vertical 
at all times; therefore, since the airplane may be flying at different 
angles of attack, a universal joint is necessary between the upper 
and lower ends of the armature shaft (see Fig. 110). 

lU Propeller mounted 

M — .^ in slipstrearri 




Fig. 110. Compass, induction type. 

The induction compass is a magnetic compass, and corrections 
must be made for variation in order to obtain true directions. 
The rotor or armature can usually be so located that deviation is 
negligible, but if deviation is appreciable, compensation can be 
made by strapping permanent magnets to the vertical shaft- 
housing of the rotor. 

The induction-type compass wifl read zero either when the 
airplane is on the desired heading or when it is 180° off from the 
desired heading. When the airplane is on its correct heading a 
shift of the heading to the right will make the galvanometer needle 
deflect to the right, and a shift of heading to the left wfll make 



344 i INSTRUMENTS 

the needle deflect to the left. When the airplane is 180° off from 
correct heading, a shift of heading to the right will make the needle 
deflect to the left, and vice versa. 

One of the chief advantages of the induction-type compass 
over the ordinary type is that with the ordinary type on long flights 
the pilot must constantly bear in mind the heading that he must 
fly. With the induction type, after the control dial has once been 
set, the pilot has only to fly so as to keep the needle of the gal- 
vanometer always on zero. 

Compass, Magneto T3rpe. Because of the weight of the in- 
duction compass (14 to 15 lb. for total installation) and because of 
trouble with failure of the universal joint, the General Electric 
Company has brought out a modification called the magneto 
compass, weighing less than 9 lb. In line, one on each side of the 
armature are two bars or pole pieces of PermaUoy, a special aUoy 
of iron which is extraordinarily permeable to magnetic lines of 
force. If the bars were placed in a north-south plane there would 
be a very strong magnetic flux across the poles. The poles, how- 
ever, are so placed by the control dial that, whenever the airplane 
is on its proper heading, the poles are in the east-west plane. 
There is then no magnetic flux across the space between the poles 
where the armature is turning. If the heading is changed even 
slightly, there will be a magnetic flux cut by the armature coils 
and a current will be shown on the galvanometer. Slip-rings are 
used instead of an armature. The armature shaft is rigid through 
its entire length, no universal joint being necessary. The pole 
pieces are mounted on a heavy pendulum so that the poles are at 
all times in a horizontal position. 

Directional G3rro. A true gyroscopic compass such as installed 
on marine vessels is inapplicable to airplane use. Necessarily the 
rotating mass must be heavy, making the total installation very 
weighty. On the Graf Zeppelin, the only aircraft on which a 
gyroscopic compass has been tried, the installation was more than 
300 lb. in weight. A gjn-oscopic compass is a north-seeking device 
in that a weight is suspended by a yoke on the gyroscope shaft 
so that, if the shaft is ever off from a true north-south line, the 
weight, in conjunction with the earth's rotation, wiU cause pre- 
cession back to a true north indication. 

The directional gyro is not a gyroscopic compass in that it has 
no directive force to make it seek a north heading. It is set by 



DIRECTIONAL GYRO 



345 



reference to a magnetic or radio compass. After being set, the 
directional gyro will remain on heading for several minutes, the 
exact time depending on the bumpiness of the air and the amount 
of maneuvering of the airplane. The ordinary compass is very 
difficult to steer by. The card of the magnetic-needle type is 
constantly swinging from side to side; in extreme cases the card 
may spin completely around. With the induction type, the 
pointer of the indicator, except on extremely calm days, is flicker- 
ing from side to side. The indications of the directional gyro are 
remarkably steady and consequently easy to stear by. 

The operation of the directional gyro may be said to depend on 
the law of gyroscopic inertia. The energy stored up by the rota- 
tion of the gyro wheel tends to keep the axis of rotation in its 
original position or in a position parallel to its original position. 

The directional gyro is in a case which is airtight except for two 
openings. One opening is connected either to the throat of a 
Venturi tube located in the sHpstream or to an engine-driven pump, 



Direction of Flight >■ 

Lr— Vertical Axis 



Rotor 
Horizontal Ring 



Caging Surface 




Caging 
Knob 



Synchronizer Pinion 

Fig. Ill 



Vertical Ring 



Card 



Horizontal Axis 

\^ 

Caging Arm 
Synchronizer Gear 
Directional gyro. 



causing a partial vacuum (3 J in.) inside the case. The other 
opening is to the outer air but it is so arranged that entering air 
passes through nozzles to strike against buckets cut in the rim of 
the gyro rotor, causing the rotor to spin at a speed of 10,000 to 
12,000 r.p.m. The rotor wheel weighs 11 oz. Its horizontal 
shaft is held in bearings in a horizontal gimbal ring. This hori- 
zontal ring is pivoted and free to turn about a horizontal axis in 
the plane of rotation of the gyro rotor. 
A view of the directional gyro is shown in Fig. 111. Ordinarily 



346 



INSTRUMENTS 



the synchronizer pinion is not in mesh with the synchronizer gear. 
To effect resetting, the caging knob, which projects outside the 
front of the case, is pushed inward, accomphshing two things. 
It engages the bevel pinion with the synchronizer bevel gear so 
that, by turning the caging knob, the vertical ring is rotated about 
its vertical axis. Pushing in the caging knob also operates a cam, 
not shown in the figure, which raises the caging arms. The free 
ends of the caging arms are united by a cross-bar, the upper sur- 
face of which bears against the lower surface of the horizontal 
gimbal ring, locking it in a horizontal position when the caging 
arms are raised. 

The nozzles, not shown in the figure, are two in number. They 
are pointed upward, shooting the air up in two parallel, vertical 
jets. They aid in keeping the rotor shaft horizontal in that, if the 
shaft is tipped, the nozzle on the high side strikes the side of the 
buckets, while the nozzle on the low side strikes the face of the 
buckets in an off-center position, causing a righting moment. 

Turn Indicators. The turn indicator was introduced shortly 
after the World War to overcome the difficulty in flying straight 
due to the inabihty of the magnetic compass to detect small turns 
or turns of short duration. The turn indicator does not indicate 
any particular direction or 
heading. If the pointer is 
on zero it shows that the 
airplane is not turning 
about a vertical axis; if the 
pointer is off zero it shows 
that the airplane is turning. 
After the turn has stopped 
the pointer returns to its 
zero position. 

The turn indicator is gy- 
rostatic in principle. The 
rotating wheel turns about 
an axis which is perpen- 
dicular to the fore-and-aft 

axis of the airplane and is ordinarily horizontal. The shaft of 
the rotating wheel is supported at each end in a bearing in the 
horizontal gimbal ring. This horizontal ring is suspended by 
pivots at the extremities of the fore-and-aft diameter of the ring; 



Air Nozzle 



Gimbal Ring 



Rotor Wheel 




Pointer 



Turn indicator. 



BANK INDICATORS 347 

see Fig. 112. The gyrostat of the turn indicator has therefore 
only one degree of freedom, the axle may turn about an axis which 
is parallel to the longitudinal axis of the airplane. This is different 
from the directional gyro in which the gyroscope has the further 
freedom of rotation about a vertical axis so that the axle may 
point in any direction and at any vertical angle. In the turn 
indicator, the gimbal ring is ordinarily held in a horizontal posi- 
tion by a centrallizing spring. The gyrostat wheel has buckets on 
its rim. Air is removed from the case by a Venturi tube in the 
slipstream or by an engine-driven pump, the atmospheric pressure 
forcing air to enter through a nozzle. The entering air strikes the 
buckets causing the wheel to spin at approximately 9,000 r.p.m. 
The direction of rotation is such that the upper rim of the wheel 
goes in the direction towards the nose of the airplane. 

In operation, a horizontal turn of the airplane turns the axle 
of the gyro wheel. The combination of this turning force and the 
inertia force of rotation causes the gyro to precess. A right turn 
of the aircraft causes the right end of the gyro axle to move up or 
to make the gimbal ring rotate in a counter-clockwise manner as 
viewed from the pilot's seat. A left turn of the airplane would 
cause precession to move the right end of the gjn^o axle downward. 
The gimbal ring has a projecting lug which moves in a slotted arm 
connected to the pointer, so that the pointer moves to the same side 
as does the bottom of the gyro wheel in precessing. 

It must be emphasized that the turn indicator shows turns 
about the apparent vertical or Z axis of the airplane and not 
about the true vertical in space. It shows both properly banked 
or skidded turns. In a banked turn, since the tipping of the 
spin axis of the gyro is opposite to the bank, the instrument is very 
sensitive. In a sideslip, where the airplane is banked without 
turning, the turn indicator registers zero. Likewise, in a verti- 
cally banked turn where the apparent vertical or Z axis of the air- 
plane is actually horizontal, the turn indicator shows zero. The 
name "turn indicator^' is incorrect; the instrument should be 
called a yaw indicator since it shows yaw, not actual turn. 

Bank Indicators. The banking indicator is a lateral incli- 
nometer. Its purpose is in straight flight to indicate if the air- 
plane is laterally level and in turning flight to indicate if the air- 
plane is properly banked. 

In a turn, centrifugal force combines with gravity to form a 



348 INSTRUMENTS 

resultant force which acts on every part of the airplane in the 
same direction. A pendulum or spirit level indicates an apparent 
vertical which is the direction of this resultant force. A spirit 
level mounted on the instrument board, laterally with respect to 
the airplane, may serve as a bank indicator. In straight flight 
with wings level, the bubble is in the central highest part of the 
glass tube; if one wing is low the bubble goes to the high side. 
In a properly banked turn, the bubble is in the center of the tube. 
In skidding, where the outer wing is too low, centrifugal force 
throws the liquid to the outer end of the tube, making the bubble 
appear at the inner (or too high) end. In slipping, where the 
inner wing is too low, the liquid goes to the inner or low end, the 
bubble appears at the high end. 

Because with a spirit level a bump will cause the fluid to surge 
back and forth, and there being no way of damping this bubble 
movement, the preferred bank indicator is the ball-in-glass-tube 
type. In this type the glass tube is curved downward at the 
middle; the steel ball has almost as large a diameter as the inner 
diameter of the tube and thus has a dashpot action with respect 
to the liquid in the tube. The fluid is usually a mixture of alcohol 
and glycerine. By painting the outer back surface of the tube 
with luminous paint, this type works nicely in night-flying, the 
ball standing out as a black disk against the luminescent back- 
ground. 

With the ball-in-tube type, the ball goes towards the end of the 
tube that is too low. If in straight flight the wings are level, or 
if in turning flight the wings are banked the correct angle, the 
ball remains in the center of the tube. To correct the airplane, 
the pilot needs to remember only the simple rule, to move the stick 
in the same direction that he would move the ball in bringing it to 
the center of the tube. 

Usually the bank indicator and turn indicator are in the same 
case, the combination instrument being called the bank-and- 
turn indicator. 

Pitch Indicators. The pitch indicator is a fore-and-aft in- 
clinometer to tell whether or not the aircraft is flying level longi- 
tudinally. With constant revolutions per minute the airspeed 
indicator can be used for this purpose, a decreased airspeed mean- 
ing climb, an increased airspeed meaning dive, and an unchanged 
airspeed denoting level flight. The increased airspeed comes, 



PITCH INDICATORS 349 

however, after the nose has gone down, and the decreased air- 
speed comes after the nose has gone up. To ehminate this sUght 
lag which tends to cause overcontrol, and also because in cold 
weather there is a danger of the Pitot tube becoming clogged with 
ice, it is desirable to have an instrument that gives directly the 
angular attitude of the airplane with respect to the horizontal. 

The problem of finding longitudinal inclination is extremely- 
difficult because of the effects of accelerations. A spirit level 
resting on a horizontal table top shows the bubble in the center of 
the glass. If the spirit level is shoved along the table as the move- 
ment is started, owing to inertia the liquid surges backward and the 
bubble moves forward in the direction the spirit level is being 
moved. When constant motion is attained, the bubble becomes 
centered again. When the movement is stopped, the liquid 
surges onward so that the bubble goes to the rear of the glass. 
A pendulum hanging in a moving vehicle hangs vertical when the 
vehicle is not moving or is at a constant speed, swings backward 
when the vehicle is accelerated and forward under the effect of a 
deceleration. 

If a plumb-bob is suspended at the side of an airplane against a 
circular scale, it will hang vertically when the speed of the airplane 
is constant. Suddenly opening the engine throttle to increase the 
airspeed would cause the plumb-bob to swing backward, and 
momentarily at least the pilot looking at the reference scale would 
be led to believe that he had nosed up. If the nose of the airplane 
drops 10° after the airplane has attained a constant speed con- 
sistent with his throttle opening and this angle of dive, the plumb- 
bob on the side of the cockpit will indicate the 10° downward 
angle. As the nose goes down, however, the speed increases, and 
this increase of speed may be at such a rate that the plumb-bob 
may actually be thrown backward, indicating to the pilot a cHmb. 
If the nose of the airplane goes up, the decrease in speed may cause 
the plumb-bob to swing forward and it will remain forward of its 
correct position till the lessened speed is constant. A spirit level 
mounted on the side of an airplane will give wrong indications 
similar to that of the plumb-bob. Neither a simple pendulum nor 
a simple spirit level will serve as a longitudinal inclinometer. 

A solution that works very effectively is a gyro-controlled pen- 
dulum, as shown in Fig. 113. A short-period pendulum is con- 
nected by a link to the gimbal ring of a gyrostat. The gyro wheel 



350 



INSTRUMENTS 



rotates about an axis ordinarily parallel to the longitudinal axis 
of the airplane in a direction which is counter-clockwise viewed 
from the pilot's seat. The axle of the wheel is supported at each 
end in bearings in a vertical gimbal ring. This ring ordinarily is 
in a vertical plane parallel to the longitudinal axis of the aircraft, 
but it may rotate about a vertical axis. A dashpot, not shown in 
the figure, prevents the mechanism from oscillating back and forth. 



■< Direction of Flight 

rAxis of 
Gimbal Ring 

i 



Plvolt 




Fig. 113. Pitch indicator. 



If the airplane noses up, during the time that the nosing-up 
is taking place, the clockwise turning of the airplane about its 
Y axis combined with the inertia force of rotation of the rotor 
causes the gyro to precess in a counter-clockwise direction viewed 
from above. This motion causes the link connecting the gimbal 
with the pendulum to move forward away from the pilot. This 
tends to move the pendulum backward towards the pilot and show 
an " up " indication on the dial. During the time that nosing- 
up is taking place, the speed of the airplane is being slowed down. 
This deceleration tends to throw the pendulum forward. The 
precessional force is greater than the decelerating force so that the 
pointer is moved in the proper direction. When the airplane has 
assumed a steady upward angle, there are no decelerating or pre- 
cessional forces and the pendulum determines the setting of the 
pointer. 

In nosing-down, exactly the opposite reactions take place. 



GYRO-HORIZON 



351 



The speed acceleration tends to throw the pendulum backward, 
but the greater precessional force moves the pendulum forward. 

When the pitch indicator, bank indicator, and turn indicator 
are all mounted in one case, the combination is called the flight 
indicator. 

Gyro-Horizon. Although the flight indicator gives the desired 
information, it is generally conceded that it is easier for the pilot 




Right Bank 



Left Bank 



Direction of Flight 



Rotor Housing 



Horizon Bar 



Gimbal Ring 



Pointer Guide Pin 




Pendulum Assembly 
Airport 





Fig. 114. 



(c) 

Gyro-horizon. 



to visualize the attitude of the airplane from a miniature replica 
of the airplane and the horizon than from reading two separate 
and distinct arbitrary scales. The gyro-horizon gives the angle 
of pitch and the true angle of bank. On the face of the instrument 
is a small representation of the airplane, and back of this moves a 
bar representing the horizon (see Fig. 114). 



352 INSTRUMENTS 

The rotor in the gyro-horizon rotates about what is ordinarily a 
vertical axis, in a counter-clockwise direction viewed from above. 
The rotor is encased in a housing which is swung in a gimbal ring, 
so that the housing may rotate in a vertical plane parallel with the 
longitudinal axis of the airplane. The gimbal ring is swung on 
pivots so that the gimbal may rotate on an axis parallel with the 
longitudinal axis of the airplane. The front pivot of the gimbal 
and the left pivot of the rotor housing are so constructed that 
outside air may enter through the gimbal pivot and pass through 
a small channel inside the gimbal, then through the gimbal pivot 
to the interior of the rotor housing. With this construction, no 
matter what the angular position of the gimbal or the housing, 
outside air will be able to enter the housing. After entering the 
housing, the air divides and goes through two passages to nozzle 
openings on diametrically opposite sides of the rotor. After doing 
the work of causing the rotor to spin, the air leaves by four ports 
at the bottom of the rotor housing. 

A long arm hinged to the front of the gimbal ring terminates in 
a '' horizon " bar across the face of the instrument. This bar is 
always parallel to the plane of the gimbal ring. A pin fastened to 
the rotor housing projects through a slot in the gimbal to actuate 
the arm of the horizon bar, moving the bar up when the upper part 
of the rotor housing is moved backward in the case towards the 
pilot and down when the rotor housing has an opposite relative 
motion. 

At the bottom of the rotor housing are four air-ports or openings, 
set 90° apart, so that two are on the fore-and-aft axis of the air- 
plane and two on the lateral axis. At each opening is a vane, hung 
pendulously, so that, when the rotor housing is exactly vertical, 
each vane covers half of its corresponding opening. An equal 
quantity of air then issues from each opening. If the gyro de- 
parts from its upright position, gravity holding the vanes vertical, 
one vane completely closes its port as shown in the lower left-hand 
diagram in figure, while the opposite vane completely opens its 
port as shown in the lower right-hand diagram. Air streaming 
out in a jet from an opening when none is issuing from the opposite 
side causes a reaction tending to swing the bottom of the gyro 
housing in the opposite direction from the side from which the jet 
is issuing. This force combined with the inertia force of the gyro 
rotation causes precession in a direction at right angles to the air 



RATE-OF-CLIMB INDICATORS 



353 



reaction which would be in the direction shown by the arrow in the 
lower right-hand diagram. This corrective movement brings the 
gyro back to its normal position. Any tendency of the gyro to 
depart from its true vertical position, caused by acceleration forces 
or by friction in the bearings, is thus corrected. Although the 
pilot apparently sees a movement of the horizon bar over the 
face of the instrument, the horizon bar is actually stationary, and 
it is the miniature airplane, as well as the real plane, that is moving 
relatively to the bar. 

Rate-of-Climb Indicators. Although pitch indicators and 
artificial horizons show the attitude of the plane with respect to the 
horizontal, an airplane can have a constant angle with the hori- 
zontal and be either climbing, settling, or flying level, depending 
on whether the plane has excess speed, insufficient speed, or the 
correct speed corresponding to the angle of attack. A sensitive 
altimeter will tell a pilot whether he is going up or down but will 
not tell him directly the rate at which he is gaining or losing 
altitude. 

In climbing, there is one angle of attack and one airspeed at 
which the rate of climb is greatest; the airspeed for maximum 
rate of climb varies with the load on the airplane, the angle of 
attack bears no simple relation to the angle of the longitudinal 
axis with the horizontal plane. In gliding, there is one angle of 
attack and one airspeed for the flattest angle of glide, and this 
airspeed varies with the load. 



Capillary 
Tubev 



^ v> ^ ^^ — J 



Scale • 



Fixed 
Pivot 



^Pin C*^^'"- 



3J 



-Pointer 



Fig. 115. Rate-of -climb indicator. 



For these reasons an instrument giving directly the rate of 
climb, either positive or negative, is very useful in fog flying and 
in flight testing. The most satisfactory rate-of-climb indicator is 
the capillary-leak type, sometimes called the '* leaky altimeter '^ 
type. An aneroid is mounted in a case with a large opening from 
the interior of the aneroid box to the outside air. In the case, 
outside of the aneroid box, is an opening to the outside air through 



354 INSTRUMENTS 

a long, very fine capillary glass tube (see Fig. 115). As long as the 
aircraft stays at the same altitude, the air in the case outside the 
aneroid is at the same pressure as that in the interior of the aneroid. 
When the altitude is changed, the air inside the aneroid because 
of the large opening instantly assumes the air pressure corre- 
sponding to the new altitude, but some time must elapse before 
sufficient air can escape or enter by the capillary tube for the air 
in the case outside the aneroid to attain the pressure correspond- 
ing to the new altitude. 

In cHmbing, the air inside the aneroid is at less pressure than 
the air outside, and so the side walls contract. This motion is 
transmitted and magnified by a lever arm and chain to a spring 
roller to which the pointer is attached. In a glide the air inside 
the aneroid is at greater pressure than the air outside, so that the 
side walls expand, giving a reverse motion to the pointer. Flow 
through a simple orifice depends on density, but flow through a 
capillary tube is within wide limits practically independent of 
density and depends only on the viscosity. An instrument, such 
as described above, can therefore be graduated in rate of climb, 
and it will retain its calibration at all altitudes. 

Drift Indicators. Whenever there is a cross-wind, an aircraft 
does not travel over the ground in the same direction as the air- 
craft is headed. The actual path of the aircraft over the ground 
is called the track. The angle between the aircraft heading and 
the track is called the drift angle or simply the drift. 

To ascertain the drift, it is necessary to see objects on the 
ground. If the ground objects that are seen directly ahead of the 
aircraft pass directly below the aircraft and then appear directly 
astern there is no drift. If the objects as they are passing to the 
rear appear to be moving over to one side, the airplane is said to 
drift. To be able to estimate the angle of drift, devices are 
provided through which the operator sights on objects on the 
ground. The basic principle of all drift indicators is essentially 
the same. A bar or wire on the outside of the cockpit or a glass 
plate with ruled parallel lines set in the floor of the cockpit is so 
arranged that the bar, wire, or glass plate is rotatable in a hori- 
zontal plane. The wire or lines are turned until, to the observer 
in the airplane, objects on the ground appear to be moving back- 
ward along the wire. If the airplane is rolling or pitching, th3 
objects will describe an erratic path as viewed from the airplane. 



GYROPILOT 355 

The observer must mentally average out these deviations from a 
straight path and so set his wire that the object appears to move 
equally on either side of the wire. 

For over-water flights, smoke bombs by day and flares by 
night are dropped overboard from the airplane. On the hori- 
zontal tail surface are painted lines radiating from a sighting point 
in the roof of the cabin, where the observer can place his eye. 
The central line runs along the center line of the fuselage; the 
other lines are 5° apart. After dropping over the bomb or flare, 
the observer waits till he sights the smoke or hght over the tail. 

Gyropilot. The gyropilot is sometimes called the automatic 
pilot or robot pilot. Thus relieved of the burden of constantly 
flying the airplane, the human pilot may attend to navigation, 
radio, and engine control. The human pilot can engage or disen- 
gage the automatic device instantly for landing, take-off, or other 
maneuvers. 

The control mechanism of the gyropilot is a combination of the 
directional gyro and the gyro-horizon, both of which are described 
earlier in this chapter. By attaching a pick-up mechanism to these 
instruments, any deviation from the desired path of the airplane 
will cause the controls of the airplane to be so operated as to bring 
the airplane back to its predetermined path. 

There are three sets of pick-up devices, one on the directional 
gyro and two on the gyro-horizon. The device is essentially as 
follows. Leading from the air-relay to the air-tight box containing 
the gyroscope are two tubes whose ends are normally open, as 
shown in Fig. 116a. When the position of the airplane is disturbed 
as in Fig. 1166, the gyroscope will tend to remain in its original 
position and the end of one tube will be closed while the other will 
be left open. 

The air-relay is a small box divided into halves by a thin dia- 
phragm. On each side is a small air inlet and a tube connecting 
with the gyroscope box. The air inlet openings are smaller than 
the tubes to the gyro box. 

When the gyroscope is in its central position there is equal suc- 
tion on each side of the diaphragm and it stays in its mid-position. 
When the airplane tilts or turns, so that there is suction through 
only one of the two tubes, the diaphragm of the air-relay is moved 
over to one side as shown in Fig. 1166. 

Fastened to the center of the air-relay diaphragm is a piston-rod 



356 



INSTRUMENTS 



which has on its other end a balanced oil valve. This slide valve 
is on an oil circulating system. When the valve is in its mid- 
position, the oil-pump forces oil into the valve and it returns 
through pipes a and b to the oil sump. 




Servo-piston-^ Servo-piston- 

{a) (6) 

Fig. 116. Mechanism of a gyropilot. 

When the valve is to one side as in Fig. 1166, the oil from the 
pump passes through the valve to pipe c leading to one side of the 
servo-piston. From the other side of the servo-piston, oil passes 
through pipe d, the oil valve, and pipe b to the sump. 

There are three air-relays, three oil valves, and three servo- 
pistons. The movement of the servo-piston is transmitted to the 
rudder, elevator, or ailerons. 



BIBLIOGRAPHY 

N.A.C.A. Reports 125-128. 
U. S. Army, T.R. 1440-50. 
Sperry Gyroscope Co., The Sperryscope. 



CHAPTER XIX 
METEOROLOGY 

Introduction. The ancients believed that the presence of 
meteors in our atmosphere was largely responsible for weather 
changes and therefore gave the name of meteorology to the art 
or science of weather forecasting. This conception has long since 
been proved false, but the old name of meteorology is still retained. 
The United States Navy has adopted the term aerology. It is 
more appropriate and less tongue-twisting and is being used more 
and more by aviators. 

Meteorology is an important adjunct to aviation. From a 
standpoint of safety, a study of the meteorological conditions 
should give warning of flying hazards such as storms or fog. 
When flying over clouds, consideration of certain meteorological 
factors should enable the pilot to estimate, at least in a general 
way, his drift. Pilots of free balloons can judge at what altitude 
to travel to secure favorable winds. Dirigibles can alter their 
course in order to secure suitable weather conditions. 

The United States Weather Bureau was started many years 
ago as a division of the Department of Agriculture. Its pre- 
dictions were primarily for the farmer, to warn him in a general 
way of impending freezing weather or of rain. In the early days, 
the observations were taken only at ground level or on the roofs of 
tall buildings. 

With increasing knowledge of the technique of weather fore- 
casting and the growth of flying demanding a more exact and 
thorough prediction, in recent years there has been improvement 
in the precision and minuteness of weather prediction. Today, 
in addition to ground observations made twice daily at some 250 
stations in the United States, upper-air measurements of tem- 
perature, pressure, and humidity are made daily by the Weather 
Bureau at seven stations, by the Naval Air Service at seven, and 
by the Army Air Corps at eight. One important airUne has its own 
staff of meteorologists to supplement the governmental agencies. 
All these agencies interchange data, and in addition to the Weather 

357 



358 METEOROLOGY 

Bureau system of intercommunication, the Department of Com- 
merce maintains a teletype system at the principal airports. 

The upper-air measurements are made by means of a meteoro- 
graph carried aloft by an airplane or by small unmanned free 
balloons called sounding balloons. The meteorograph is a small 
combination instrument containing a clockwork-operated smoked 
drum on which are recorded temperature, pressure, and humidity. 
The temperature-indicating unit is a bimetalHc strip of dissimilar 
metals which bends on heating. The pressure-indicating unit 
is a tiny aneroid. The humidity indicator consists of a number 
of animal hairs stretched over an elastic framework. 

Temperature. The heat radiating from the sun passes first 
through our atmosphere before reaching the earth. Dry air 
absorbs practically no heat from the sun, but moist air does 
absorb it. As half of the air is below 22,000 ft. and the air above 
that altitude has scarcely any moisture content, practically all the 
heat absorption by the air occurs at the lower altitudes. 

The sun's radiation not absorbed by the atmosphere strikes the 
earth's surface. Part of the radiation is absorbed and part re- 
flected or re-radiated, the exact proportion depending on whether 
the surface is rock, field, trees, ice, etc. Dark surfaces, like plowed 
fields, absorb more heat and are at a slightly greater temperature 
than reflecting surfaces such as green fields. The air in contact 
with the ground is heated by direct conduction, and this air heats 
air with which it is in contact by direct conduction. Also warm 
air rises, so that warm air may be found at considerable altitude. 
These rising currents of warm air are called convection currents, 
and the presence of warm air at altitude from this cause is said to 
be due to convection. 

When a mass of air at a given temperature is compressed adia- 
batically, that is, without the addition or subtraction of heat, the 
temperature of the air increases. Conversely, when a mass of 
air is allowed to expand without heat being added or heat being 
allowed to escape, the temperature drops. The heat conduc- 
tivity of atmospheric air being very low, if a mass of dry air moves 
from one level to a higher level, since the pressure is less at the 
higher altitude, the air will expand very nearly adiabatically and 
consequently drop in temperature. This drop will be approxi- 
mately 1° F. for every 185-ft. increase in altitude. On the other 
hand, air moving downward 185 ft. should increase 1° in tempera- 



PRESSURE 359 

ture. Actually other factors affect the heating and cooling so that 
the change in temperature is usually less than 1° for 185-ft. 
altitude difference. 

Both for the reason explained in the last paragraph and because 
the heating effect of the earth is less with higher altitude, the 
temperature decreases as one rises above the earth. The rate at 
which temperature decreases with altitude is called the tempera- 
ture gradient. The temperature continually decreases until the 
stratosphere is reached. In the stratosphere, the temperature is 
presumed to be constant. 

At night, the ground cools quickly so that the air adjacent to the 
ground drops in temperature. When this occurs, the air gets 
warmer as one ascends a short distance. This is termed reverse 
temperature gradient. At a comparatively low altitude, this 
reversal stops and the temperature decreases as further ascent is 
made. 

On weather maps, lines, usually dotted ones, are drawn connect- 
ing geographic points that have the same ground temperature. 
These lines are called isotherms. 

Humidity. Warm air can hold more water vapor in suspension 
than cold air can. The amount of saturation of air with water 
vapor is called the humidity. Absolute humidity is the actual 
amount of water vapor present in a cubic inch or cubic centimeter 
of air. Relative humidity is the ratio of the mass of moisture 
present in the air to the amount required for complete satu- 
ration. 

When warm air with high relative humidity is chilled, the abso- 
lute humidity may remain the same, but since at lower tempera- 
ture less moisture is required to saturate the air completely, in 
lowering the temperature, the relative humidity increases. If 
the relative humidity increases to more than 100 per cent, pre- 
cipitation will take place. 

Pressure. Atmospheric pressure is due to the weight of the 
column of air extending from the point of measurement upward 
to the limit of our atmosphere. Atmospheric pressure is usually 
given in terms of the height of a mercury column that would be 
supported by the atmosphere. Even though glass tubes con- 
taining mercury are rarely used now except as basic standards, 
aneroid barometers are marked off in units of height of mercury 
columns. Standard atmospheric pressure is 29.92 in, or 760 mm. 



360 METEOROLOGY 

A unit only used in meteorological work is the millibar (1,000 
millibars = 750 mm. of mercury = 29.53 in. of mercury). 

On weather maps, lines are drawn connecting places that have 
the same equivalent sea-level barometric pressure. These lines 
are called isobars. 

Warm air expands and its density decreases. This would tend 
to make the barometric pressure less over heated areas and 
greater over surrounding cooler areas. 

Considering the earth as a whole, as the equatorial regions re- 
ceive nearly three times the amount of heat received by the polar 
regions, it might be expected that there would be a region of low 
barometric pressure in the tropics and of high barometric pressure 
near the poles. This is true in so far that there is low pressure in 
the tropics, but owing to the earth's rotation, centrifugal force 
draws the air away from the poles. The resulting effect is to have 
in general a comparatively high-pressure band around the earth 
at about 30° latitude and low-pressure areas at the equator and 
around the poles. 

Winds. Air, like any other fluid, travels from a region of high 
pressure to one of low pressure. A large difference in barometric 
pressure at two places a short distance apart geographically will 
cause a strong air movement or wind. A small pressure difference 
will cause a correspondingly weaker wind. On American weather 
maps, isobars are drawn for 0.1-in. pressure difference. Where 
these isobars are close together, there are strong winds; where 
there are big spaces between the isobars, the winds are very mild. 

Because air travels from high-pressure to low-pressure areas, it 
might be expected that, in the northern hemisphere, at the south- 
ern edge of the temperate zone there would be north winds and at 
the northern edge of the temperate zone there would be south 
winds. These north and south winds are modified by the earth's 
rotation as follows. The earth is rotating from west to east. 
*' Stationary " air on the equator actually is moving eastward 
with a velocity greater than 1,000 miles per hour. Air farther 
from the equator has less eastward velocity. At any latitude, the 
ground and " stationary " air has less eastward velocity than the 
ground and " stationary " air at less latitude. Air tending to move 
directly northward in the northern hemisphere, owing to its 
greater eastward velocity, becomes a southwest wind. Air 
tending to move southward is moving into a region which has a 



WEATHER MAPS 



361 



greater eastward velocity, so that the north wind becomes a north- 
east wind. This is true whether the air movement is on a large or 
small scale. 

When, from local heating or other causes, there is a low-pressure 
area, usually called simply a " low," in the northern hemisphere, 
air instead of moving radially inward deviates to the right, so 
that there is a counter-clockwise rotation of wind about a low, as 




(a) Wind Direction in Low, Northern Hemisphere 




(b) Wind Direction in High, Northern Hemisphere 
Fig. 117. Wind directions (a) in low, (6) in high. 

shown in Fig. 117a. About a high-pressure area, or high, the air 
instead of moving radially outward at right angles to the isobars 
veers to the right as shown in Fig. 1176. 

Weather Maps. Areas of high and low barometric pressure are 
constantly moving across the United States from west to east. 
A rapid movement and constant succession of highs and lows 
impHes frequent changes in the weather. A slow movement 
portends a continuation of the present weather. 



362 



METEOROLOGY 



Although a study of a daily weather map gives a certain amount 
of information, it is always advisable to study, in conjunction with 
it, the maps of the two previous days. By doing this, the paths 
of the disturbances can be visualized and the probable future 
positions of the disturbances predicted. For special reasons dis- 
turbances sometimes appear seemingly spontaneously; but the 
highs and the lows are usually noted first on the west coast and 
they follow more or less well-defined paths across the United States. 

Lows. In meteorology, the term " cyclone " is synonymous 
with '' low." In some localities, owing to distinct temperature 
differences from adjacent regions which maintain themselves for 
long periods, there are lows which are more or less stationary or 
semi-permanent. Mostly, however, the lows move eastward at 

Cold air ^ ^^^^ ^^ about 25 milcs 



Clear weather 



Cold air 




Warm air 



Fig. 117c. Weather in a low. 



per hour in the summer 
and about 40 miles per 
hour in the winter. 

Examining Fig. 117c, it 
will be seen that, if a line 
is drawn through the 
center of the low in a 
northeast to southwest di- 
rection as shown by the 
dotted line, winds on the 
southeast side of this fine 
come from a southerly direction while those on the other side 
have a component from the north. Winds from the south bring 
warm air northward. Winds from the north bring cold air 
southward. 

When the warm air meets the cooler air, approximately in the 
area marked A in Fig. 117c, the warm air being of lighter density 
rises and flows over the cold air. The mass of cold air meeting 
the warmer air at B, Fig. 117c, forces its way under the warm air. 
The leading surface of a mass of air in motion is called a '' front '^; 
that at ^ is a warm front, that at 5 a cold front. Fronts are not 
sharp planes of discontinuity between warm and cold air. Wher- 
ever a mass of air at one temperature meets a mass at a different 
temperature, there is always turbulence and a mixing of warm and 
cold air. A front may be many miles in length. 

In the northeast section of the low, the rising warm front de- 



HIGHS 363 

creases in temperature both because of expansion and because of 
mixture with the colder air. If the warm air contained moisture, 
the cooHng will result in the water vapor condensing out to form 
clouds. A continuation of the cooling will cause precipitation in 
the form of rain or snow. The ascension of warm air is gradual 
as the air moves northward, so the rainy area may be quite ex- 
tensive. 

In the southwest section of the low, the cold front is much more 
sharply defined. The change in temperature is quite sudden, and 
the precipitation is much more intense. The turbulence along the 
cold front is very intense, and the air is quite " bumpy '^ to the 
pilot. 

On the northwest side of a low, the air is cold. Cold air holds 
little moisture in suspension. Any increase in temperature per- 
mits the air to absorb more water vapor, so that in this sector are 
clearing weather and blue skies. 

By watching the way in which smoke drifts away from chim- 
neys or flags flutter, a pilot may often be able to ascertain the 
wind direction on the ground during the day. Knowing the direc- 
tion from which the wind blows enables the pilot to estimate the 
direction of the center of the low, and he can head his airplane so 
as to avoid the extremely bad weather. A very valuable rule to 
remember is known as Buys-Ballot's law. It is, briefly, to 
imagine oneself facing into the wind; the barometric pressure on 
the right hand is lower than the left, and the center of the low will 
be on the right-hand about 135° from dead ahead. 

Highs. A region of high barometric pressure moves eastward 
at less speed than a low-pressure area, and a large high-pressure 
area may be stationary for several days. An exception to this 
is the case where the high is situated between two lows; when this 
occurs the high has the same velocity as the lows. As a rule, 
when a high sets in, settled weather may be expected to last for 
two or three days. Regions of high pressure are sometimes called 
anticyclones. 

In highs, the isobars have the same oval shape as in lows, but 
the isobars are farther apart. The pressure gradient being less, 
the winds are less in strength than those found in a low. A high 
usually covers more territory, never being as concentrated as the 
lows. The weather in a high is generally clear, but fog sometimes 
occurs in the autumn. Rains are always of short duration. 



364 METEOROLOGY 

Since a high usually follows a low in progression across the 
United States, when a high comes into a region, the rain area of 
the preceding low has traveled onward. With the rise in barome- 
ter the weather clears and the temperature drops. At the center 
of the high a perfect calm is usually found with a cloudless sky or 
at most a few scattered clouds. Temperature is low at night but 
rises rapidly during the day. 

Winds Aloft. While winds at low altitudes cross the isobars at 
an angle which is somewhat towards the right from going straight 
across, as altitudes are increased the angle from a perpendicular 
crossing becomes greater. At about 8,000-ft. altitude, the winds 
are generally practically parallel to the isobars. Continuing 
higher the winds change direction so as to become more nearly 
west winds. If the wind at lower altitude is southeast it will 
probably veer through south and southwest to west; if the wind 
is northeast at lower altitude it will probably back through north 
and northwest to west. 

The velocity increases with altitude up to a height of 20,000 ft. 
Above this, the wind usually decreases in strength. Roughly, 
the wind at 5,000-ft. altitude has twice as great velocity as at the 
ground. 

Clouds. The suspension of water vapor in the air is determined 
mainly by the temperature. When the temperature is high, a large 
amount of water vapor may exist; when the temperature is low 
only a small amount of water vapor can be held in suspension. 
Consequently, if a mass of warm air carrying a high percentage of 
moisture comes in contact with or moves completely into a cooler 
region, the air is chilled and the water vapor is condensed to liquid 
water in the form of minute drops. Particles of dust usually act 
as nuclei about which this condensation collects, but sometimes a 
drop of water acts as nucleus. These tiny drops form clouds, and 
if the condensation continues, these drops grow in size until 
precipitation as rain or snow results. 

Clouds may be formed in several ways, such as when : 

(a) Warm saturated air blows over cold sea or land, as when 
warm currents pass over mountains especially if the mountains 
are snow-covered, or in winter, warm south winds. Clouds thus 
produced are usually of the stratus type; see below. 

(h) Rising air expands and in expanding it is cooled, condensing 
out its moisture. This usually produces cumulus clouds. 



CLOUDS 365 

(c) Air is forced to rise by encountering mountain slopes. 

(d) Air may part with its heat by radiation on cold nights and 
in cooling cannot contain so much moisture. 

(e) Air is forced to rise when a warm front meets the cooler air 
on the northeast sector of a low. 

There are many different types of clouds but, all may be classi- 
fied under or are a combination of four main subdivisions. 

Cirrus. These are thin and wispy. They are composed of ice 
particles and are found at altitudes of 25,000 to 30,000 ft. Cirrus 
clouds appear like giant, curling feathery plumes and are some- 
times called " mares' tails." 

Cumulus. These are detached, fleecy clouds. They appear 
like big lumps of absorbent cotton or shaving-soap suds. They 
have a fiat base. They are composed of water particles and 
may be found at altitudes from 5,000 up to 25,000 ft. 

Stratus. These are low, flat, spread-out clouds, resembling 
fog but not resting on the ground, and are in distinct layers. 
This type often merges into nimbus. 

Nimbus. These are thick, extensive layer of formless clouds, 
black on the under side from which rain or snow is falling. 

Fog is cloud resting on or close to the ground. Fog is divided 
into two classes, radiation fog and advection fog. 

Radiation fog forms along rivers, creeks, etc., during still 
cloudless nights of the summer or autumn. During a calm, warm 
day, water evaporates into the lower atmosphere of such regions, 
where it remains since there is no wind. During the night the 
ground cools; the lower, moisture-laden air drops in temperature, 
and condensation takes place. 

Advection fog forms when warm air drifts slowly over a cold 
surface. This occurs in winter in front of an advancing low. 
Advection fog is also formed when a cold wind passes under a 
mass of warm, damp air such as in winter in front of a high. 

BIBLIOGRAPHY 

HuMPHEEY, Physics of the Air. 
Gregg, Aeronautical Meteorology. 
Maguire, Aerology. 



CHAPTER XX 
AVIGATION 

Introduction. Avigation is the guidance of aircraft along a 
desired path and the ascertainment of the aircraft's geographic 
position. The term ''avigation" has superseded the older ex- 
pression '' air navigation," which is anomalous since the word 
navigation refers to travel by water. 

The chief advantage of air travel over other means of trans- 
portation is the speed in reaching the destination. The two 
reasons for this speed are: first, aircraft are faster than other 
means of travel; and second, aircraft can travel the shortest 
possible path. When aircraft follow railroads, highways, or 
winding rivers, the only gain is due to the superior speed of the 
aircraft. By cutting across mountain ranges, etc., to follow the 
most direct route, the distance to be flown is reduced considerably. 

Flying by following a definite landmark, such as a railroad, 
river, or highway, or by sighting recognizable landmarks ahead 
immediately after a known landmark has passed astern is a form 
of avigation; but the term avigation is usually applied to the 
conducting of an aircraft from place to place and finding position 
when recognizable landmarks are not continuously in sight. 
Dead reckoning (contracted from deduced reckoning) is the term 
applied to the method of finding position by means of the direction 
and distance flown from the last recognized landmark. It is 
sometimes called compass flying. When long distances are flown, 
dead reckoning is liable to considerable error, owing to the effect 
of the wind not being known accurately, and the position should be 
checked if possible by astronomic means or by the aid of radio. 

Maps and Charts. The earth is an oblate spheroid, the polar 
diameter of which is 7,900 miles and the equatorial diameter 
7,926 miles. For all avigation work the earth is assumed to be a 
sphere. The earth rotates daily about an axis, the ends of which 
are the north and south poles. The equator is the circle formed by 
a plane cutting the earth perpendicularly to the axis and equi- 
distant from the poles. The equator is everywhere 90° distant 

366 



MAPS AND CHARTS 367 

from each pole. Circles passing through both poles are meridians. 
Their direction is everywhere north and south. Circles formed by 
planes cutting through the earth perpendicular to the axis are 
called parallels of latitude. 

The geographic position of any place on the earth may be de- 
noted by its latitude and longitude. The latitude of a place is the 
length of the arc of the meridian through the place, between the 
equator and the place. It is measured in degrees and minutes, 
and is designated north or south according as the place is north 
or south of the equator. The longitude of a place is the angle 
between the plane of the meridian through the place and the plane 
of the meridian through Greenwich, England. The angle is 
measured in degrees and minutes, and longitude is designated 
west or east according as the place is west or east of Greenwich. 

A circle formed by any plane passing through the center of the 
earth is called a great circle. The shortest path between two 
places on the earth is along the great circle passing through the 
two points. 

A map or chart represents in miniature, upon a flat surface and 
according to a definite system of projection, a portion of the earth. 
Although the two names are used indiscriminately, a chart pic- 
tures a portion of the earth the greater part of which is water, 
whereas a map depicts a portion of the earth's surface which is 
mostly land. 

No spherical surface can be depicted correctly on a flat surface. 
There can be no accurate maps. 

There are four uses for maps: (a) to find distances, (b) to find 
directions, (c) to recognize natural features, and (d) to compare 
areas. No map is ideal for all these purposes ; a map may furnish 
correct information for one of these uses and approximately cor- 
rect information for one or two of the others, but it cannot be 
correct for all four. 

The scheme or arrangement on which the parallels of latitude 
and meridians are laid out is called the system of projection. An 
infinite number of systems of projection may be thought of. At 
one time or another several hundred have been used, and at 
present over three dozen are actually in use in the maps made by 
various nations and by various agencies and for various purposes. 
The three most common projections which are widely used are the 
mercator, the polyconic, and the great circle projections. For 



368' AVIGATION 

areas under 200 square miles the earth may be considered flat; 
for larger areas the curvature of the earth becomes important. 
It is important to know the system of projection of the map being 
used and the limitations and inaccuracies of that system. 

Polyconic Projection. The polyconic projection is used in the 
United States Coast and Geodetic Survey maps, United States 
Post Office maps, and the airway strip maps issued by the Depart- 
ment of Commerce and Army Air Corps. This projection can 
usually be identified by being the only one of the three common 
projections having a scale of miles printed on it. This scale of 
miles is correct in the central portion of the map but is likely to be 
slightly inaccurate near the sides of a map of such size as to cover 
the area of a continent. 

It is an ideal projection for areas the size of a single state. It is 
best suited to maps which are long in the north-south dimension 
and narrow in the east-west dimension, thus making it a satis- 
factory projection for maps of the Atlantic or Pacific coast. For 
maps of the entire United States there is considerable distortion 
near the seacoasts. 

The parallels of latitude are circles of different radii. If the 
map is large scale, short portions of the circular parallels may be 
considered as being straight lines. When longer distances are 
being dealt with, the curvature enters into the problem. On the 
earth, the parallels of latitude are due east-west lines. On the 
polyconic maps, since the parallels are represented by circles, 
east-west lines have a different inclination in different parts of the 
map ; therefore, if a line is drawn between two places on the map 
a considerable distance apart, the direction of this line cannot be 
ascertained. The polyconic projection depicts more accurately 
than other projections the actual surface of the ground so that 
landmarks, rivers, and shorehnes can be recognized readily. 

Mercator Projection. The mercator projection is used in the 
maps issued by the United States Navy Hydrographic Office and 
is the standard projection for ocean-sailing charts. This pro- 
jection can be identified by being the only one of the three common 
projection having the meridians as straight vertical equidistant 
lines and the parallels of latitude as straight horizontal lines, the 
degrees of latitude increasing in length farther from the equator. 
It is the only one of these three projections on which the entire 
inhabited portion of the earth can be represented on one map* 



GREAT-CIRCLE CHARTS 369 

Since its limit of construction is about 70° latitude, polar regions 
cannot be shown on the ordinary mercator chart. 

On the mercator projection, the meridians and parallels intersect 
at right angles; north is the same direction in all parts of the map. 
It is therefore very easy to find the direction of a line drawn on a 
mercator chart by measuring the angle at which the line crosses 
a meridian. A straight line drawn on a mercator chart is called a 
rhumb line. 

On the sides of mercator charts is a scale of latitude. This scale 
is not constant. To measure short distances on a mercator chart, 
the length of a degree of latitude on the side scale, at the average 
latitude between the two places, is used as measuring unit. This 
gives the distance in degrees and fractions of a degree of latitude. 
A degree of latitude is 60 nautical miles in length, a nautical mile 
being 6,080 ft. long. 

A rhumb line is not the'shortest path between two points on the 
earth's surface; it is merely a Hne which has a constant direction. 
The shortest path is a great circle, which is shown as a curve on a 
mercator chart. 

Great-Circle Charts. For the use of seamen, the Hydrographic 
Office has published five great-circle charts, namely, of the North 
Atlantic, South Atlantic, North Pacific, South Pacific, and Indian 
Oceans. These charts have the property that a straight line 
between two points on the chart represents the great circle through 
these points on the earth, and shows all the localities through which 
this most direct route passes. The meridians are straight lines, 
either parallel as shown in Fig. 118, or converging towards the 
pole. The parallels of latitude are represented as conic sections. 

It is very comphcated to measure directions on this projection. 
The usual procedure, after drawing a straight line on this chart as 
a great-circle path between two points, is to spot a number of 
points on this line. By reading the latitude and longitude of 
each point, they may be transferred to a mercator chart. On 
the mercator chart, a smooth curve is then drawn through these 
points. On the great circle, direction is constantly changing. 
In practice, instead of actually following the great circle, a series of 
straight tracks are flown approximating the great circle, the head- 
ing of the aircraft being changed whenever the direction of the 
great circle changes a degree. 

The rules of the Federation Aeronautique International state 



370 



AVIGATION 



that, in making a record for distance flown, the distance from 
starting to landing point shall be measured along the great-circle 
route. Directions given by radio direction-finders, such as the 




(a) Polyconic Projection 




(b) Mercator Projection 




(c) A Great Circle Projection 

Fig. 118. Common map projections. 

radio beacon or the radio compass, are along the great-circle 
course. For short hops of only a few hundred miles the mileage 
saved is inappreciable, but for transcontinental or transoceanic 
flights the distance along the great-circle path is many miles shorter 
than along the rhumb-line path. 



GREAT-CIRCLE DISTANCE 371 

Distances cannot be scaled off directly on a great-circle chart, 
and though an auxiliary graph is provided, it is quite complicated 
so that the great-circle distance is usually computed. 

Great-Circle Distance. The distance along the arc of the great 
circle between two points A and B may be found by the following 
formula : 

cos D = sin La sin Lb + cos La cos Lb cos LOab 

Z) is distance LOab is longitude difference 

La is latitude of point A between points A and B 

Lb is latitude of point B 

Example. Find the great circle distance from Curtiss Field, Long 
Island (40° 45' N., 73° 37' W.), to Le Bourget Field, Paris (48° 50' N., 
2° 20' E.). 

Solution, 

La = 40° 45',N. 
Lb = 48° 50' 
LOab = 73° 37' + 2° 20' = 75° 57' 

(Note: Since longitudes are of opposite names, the difference is 
found by adding.) 

sin 40° 45' sin 48° 50' = 0.6528 X 0.7528 
= 0.4914 
cos 40° 45' cos 48° 50' cos 75° 57' = 0.7576 X 0.6582 X 0.2428 

= 0.1211 
cos D = 0.4914 + 0.1211 
= 0.6125 
D = 52° 14' 

On a great circle, a degree is 60 nautical miles and a minute is 1 
nautical mile in length. 

Distance (D) = 52 X 60 + 14 

= 3,134 nautical miles 

= 3,134 X Ifg^or 3,609 land miles 

Problems 

1. Find the distance from Rome (41° 54' N., 12° 29' E.) to Natal, 
Brazil (5° 37' S., 35° 13' W.). (Del Prete's flight.) 

2. Find the distance from San Francisco (37° 47' N., 122° 28' W.) 
to Honolulu (21° 22' N., 157° 48' W.). (Hegenberger's flight.) 



372 



AVIGATION 



3. Find the distance from Cranwell, England (53° 05' N., 0° 25' W.), 
to Walvis Bay, South Africa (22° 43' S., 14° 20' W.). (Flight of 
Gayford and Nicholetts.) 

4. Find the distance from Burbank, California (34° 12' N., 118^ 
18' W.), to Floyd Bennett Field, N. Y. (40° 40' N., 73° 50' W.). 
(Flight of Haizlip and others.) 

5. Find the distance from Harbor Grace, Newfoundland (47° 43' N., 
53° 08' W.), to Clifden, Ireland (53° 30' N., 10° 04' W.). (Flight of 
Alcock and Brown.) 

Velocity Triangle. The velocity of an aircraft with respect to 
the ground is the resultant of two velocities: that of the aircraft 
with respect to the air, and that of the air with respect to the 
ground, the latter being the wind velocity. Velocity has two 
characteristics, speed and direction. 

The velocity of the aircraft with respect to the air has a speed 
which is the reading of a properly corrected airspeed indicator 
and a direction which is the true heading of the aircraft determined 
from a properly compensated compass reading corrected for 

variation. The velocity of the aircraft 
with respect to the ground has a speed 
which is the ground speed and a direc- 
tion which is the track or actual path of 
the aircraft over the ground. Drift is 
the angle between the heading and the 
track. 

To find the resultant of the air veloc- 
ity and the wind velocity, use is made 
of the parallelogram of velocities, the 
two component velocities being drawn to the same scale and in 
the proper respective directions as two sides of a parallelogram, 
and the resulting velocity being its diagonal. By omitting two 
sides of the parallelogram, the remaining velocity triangle is 
sufficient to obtain a solution. Care must be taken that the 
arrows indicating direction of velocity are correctly marked to 
give continuity. The use of the velocity triangle is illustrated 
by the following example in conjunction with Fig. 119. 

Example. An airplane is heading due northeast with a corrected 
airspeed of 150 miles per hour. There is a 40-mile-per-hour west wind. 
What are the ground speed, track, and drift? 

Solution (see Fig. 119). Draw line AB, 150 units in length; let 




Fig. 119. Velocity triangle 
finding ground velocity. 



VELOCITY TRIANGLE 373 

direction of ABhe northeast. From point B, draw line BC, 40 units 
in length, direction oi BC being east (since a west wind blows towards 
east). Connect points A and C. 

Then the line AC represents ground velocity. 

The length of AC being 181 units, the ground speed is 181 miles per 
hour. 

The line AC makes an angle of 54° from north, so that track is 54°. 

Angle BAC is 9°, which is drift of 9° right. 

Problems 

1. An airplane is headed east with an airspeed of 125 miles per hour. 
There is a 30-mile-per-hour northeast wind. What are ground speed, 
track, and drift? 

2. An airplane is headed 30° with an airspeed of 150 miles per hour. 
There is a south wind of 25 miles per hour. What are ground speed, 
track, and drift? 

3. An airplane is headed 340° with an airspeed of 140 miles per hour. 
There is a northeast wind of 20 miles per hour. What are ground 
speed, track, and drift? 

4. An airplane is headed 265° with an airspeed of 100 miles per hour. 
There is a 35-mile-per-hour wind from 320°. What are ground speed, 
track, and drift? 

5. An airplane is headed 220° with an airspeed of 120 miles per 
hour. There is a south wind of 30 miles per hour. What are ground 
speed, track, and drift? 

The desired track being known, the aircraft must be headed 
into the wind from this track in order that the wind in combination 
with the aircraft's own motion shall carry the aircraft along the 
desired track. The angle by which the heading is altered from the 
desired track into the wind is called the angle of crab. 

In taking-off on a cross-country flight, if the aircraft is first 
headed on the desired track and the drift measured, the heading 
should not be crabbed by the drift angle since when the heading 
is changed the wind will be at a different angle and consequently 
there will be a different drift angle. The heading can, of course, 
be altered in this way to an approximately correct heading, but 
the drift on the new heading should be measured and the heading 
re-altered for the difference between this drift and the drift on the 
former heading. This is called the cut-and-try method. 

When the wind speed and direction are known, the proper 
heading can be found by a velocity triangle. In this form of 




374 AVIGATION 

problem, the airspeed is known but not its direction; the track is 
known but not the ground speed. Various devices have been 
made for solving this problem mechanically. The graphic 
solution is illustrated by the following example. 

Example. It is desired to fly northeast. Airspeed is 150 miles 
per hour. There is a 40-mile-per-hour west wind. What should be 
the proper heading, and what will be the 
ground speed? 

Solution (see Fig. 120). Draw line AW, 
40 units in length, in a direction representing 
due east. Through point A draw line AX 
of indefinite length, in a direction represent- 
ing northeast. With W as center and 150 
■^;o — ^^ units as radius, strike an arc, intersecting 

Fig. 120. Velocity tri- line AX at point 5. 

angle, finding heading. Then triangle ABW is the velocity tri- 

angle. The length of AB is the ground 
speed, 176 miles per hour. The direction of line WB^ 34°, is the 
heading. 

Problems ^ 

1. It is desired to fly due north. Airspeed is 150 miles per hour. 
The wind is from 80°; wind speed is 40 miles per hour. What should 
be proper heading, and what will be ground speed? 

2. It is desired to fly northeast. Airspeed is 135 miles per hour. 
The wind is from west; wind speed is 30 miles per hour. What should 
be heading, and what will be ground speed? 

3. It is desired to fly 5°. Airspeed is 100 miles per hour. Wind is 
from 300°; wind speed is 40 miles per hour. What should be heading, 
and what will be ground speed? 

4. It is desired to fly 160°. Airspeed is 125 miles per hour. Wind 
is from 70°; wind speed is 35 miles per hour. What should be head- 
ing, and what will be ground speed? 

5. It is desired to fly 230°. Airspeed is 1 10 mfles per hour. Wind 
is from 170°; wind speed is 35 miles per hour. What should be head- 
ing, and what will be ground speed? 

Wind-Star, Although, at many airports, information is avail- 
able as to winds at altitude, there is no certainty that winds will 
not have changed after the observations were taken. The wind 
conditions may be entirely different a few miles away from the 
airport. 



WIND-STAR 375 

Drift indicators installed on the aircraft can be used to measure 
drift angle with a fair degree of accuracy. Finding ground speed 
by use of two cross-hairs on the drift indicator presupposes knowl- 
edge of the actual height of the aircraft above the ground. Alti- 
meters give barometric height, i.e., altitude above sea-level, not 
height above the ground. When the location of the aircraft is 
known, the height above sea-level of the ground below the air- 
craft can be found from contour maps; but when the geographic 
location of the aircraft is recognizable on a map, the pilot resorts 
to landmark flying, not dead reckoning. 

Ground-speed indicators have been devised on the principle of 
synchronizing the speed of an endless celluloid belt on the aircraft 
with the apparent rearward speed of the ground. These devices 
have not yet passed out of their developmental status. 

A single drift measurement does not give ground speed, nor 
does it give sufficient information to obtain wind speed and 
direction. Measurements of drift on two headings will give ground 
speed and wind velocity by utilizing the principle of the wind-star. 

For example — see Fig. 121a — if an airplane is headed north 
with an airspeed of 150 miles per hour and a 15° right drift is 
measured, this may be due to a 40-mile-per-hour west wind or to a 
45-mile-per-hour northwest wind or to a 78-mile-per-hour south- 
west wind. There^^may be an infinite number of winds which will 
produce a''15° right drift on a north heading. 

If the heading is changed to east and the drift is found to be, 
say, 5° left, the wind is then determined, because there can be only 
one wind that will give a 15° right drift on a north heading and a 
5° left drift on an east heading. By drawing line AB — see Fig. 
1216 — 150 units long (airspeed is 150 miles per hour) in a direction 
representing north and drawing line AX oi indefinite length at an 
angle of 15° to the right of AB, the track on the first heading is 
represented. From point B, a line BA' is drawn in a direction 
opposite to the second heading (i.e., westward); the length BA' 
is 150 units. Then A 'B represents the airspeed and heading after 
heading has been changed. Through A' a line A'X' is drawn of 
indefinite length making an angle of 5° left of A'B. Call the 
intersection oi AX and A'X' the point W, and connect points B 
and W. Then the vector BW represents the wind speed of 48 
miles per hour and wind direction of 249°, this being the only wind 
velocity that will give a drift of 15° right on a north heading and a 



376 



AVIGATION 



drift of 5° left on an east heading. The length of ATF is the 
ground speed 173 miles per hour on first heading, and the length 
of A'TFis the ground speed 195.5 miles per hour on second heading. 
After the drifts on two headings are found so that the wind is 
known, the drift can be found on any heading by drawing a line 
representing the heading into point B, and from the other or 
starting end of the vector drawing a line through point W. 




Fig. 121. Wind-star. 



Figure 121c shows the finding of drift on south, southwest, and 
west headings, AiB, A^B, and AzB representing the respective 
headings and AiWj A2W, and A^W representing the respective 
tracks and ground speeds. When headings and tracks are drawn 
for several different headings as in Fig. 121c, the drawing bears a 
crude resemblance to the conventional star-form, which is the 
reason for giving this process the name of the wind-star. 

After the wind velocity has been found by measuring the drift 



ASTRONOMIC AVIGATION 377 

on two headings, it is possible to find the proper heading for the 
aircraft to follow a desired track by the following procedure. 
From the arrow end of the wind-vector draw a line of indefinite 
length in a direction opposite to that of the desired track. From 
the non-arrow end of the wind- vector as center and with the air- 
speed as radius strike an arc intersecting the line just drawn. 
The radius from this point of intersection gives the proper head- 
ing. For example, the wind having been found on a north head- 
ing and an east heading as Fig. 121?), it is desired to find the 
heading which will give a true northeast track. Through point 
W draw a line of indefinite length WX" in a southwest direction; 
see Fig. 121c?. With B as center and 150 units as radius, inter- 
sect line WX" at point A". Then direction of A'' B is the proper 
heading 37.5°, and length of A"W is the ground speed, 192 miles 
per hour. 

In using this method, it is customary to head 30° to the right of 
the desired track and 30° to the left, since by so doing a good inter- 
section is obtained. The wind-star method does not involve 
knowledge of altitude. 

Problems 

1. On a 30° heading there is a 10° left drift; on a 330° heading there 
is a 15° left drift. If airspeed is 125 miles per hour: (a) what is wind 
speed and (6) direction; (c) what heading should be flown to obtain a 
due north track; (d) what will be ground speed? 

2. On a 15° heading there is a 10° left drift; on a 75° heading there 
is a 20° left drift. Airspeed is 160 miles per hour, (a) What is wind 
speed and (b) direction? (c) What heading should be flown to obtain 
a northeast track? (d) What will be ground speed? 

3. On a 60° heading there is a 5° right drift; on a 120° heading 
there is a 10° left drift. Airspeed is 140 mile per hour, (a) What 
is wind speed and (b) direction? (c) What should heading be to fly 
east, (d) What will be ground speed? 

Astronomic Avigation. For centuries sailboats and more re- 
cently steamships have navigated by means of observations on the 
sun, moon, or stars. The angular height of a celestial body above 
the horizontal plane being measured by a sextant, an oblique 
spherical triangle is solved and a line partially determining the 
ship's position is obtained. Two such fines exactly fix the position. 

On marine vessels, the horizon, where sky and water meet, is 



378 AVIGATION 

used to determine the horizontal plane. Unless an aircraft is at 
an altitude less than 1,000 ft., the observer is unable to see the 
horizon because of haze. For this reason, and also to enable 
observations to be taken at night, some form of artificial horizon 
is used; generally a spirit level is utilized. In bumpy air the 
bubble dances about badly, but by taking the average of a number 
of observations the error can be reduced to a minimum. 

Computation of the position from the sextant observations was 
formerly a long, tedious matter. Short methods have been 
evolved which permit the calculations to be made in a minute or 
less. 

Astronomic avigation is used on long overwater flights, such as 
trans-Atlantic or trans-Pacific hops. 

Radio Avigation. Radio is an important adjunct to flying. 
A simple receiving set enables the pilot to receive storm-warning 
and other weather information as well as to use the radio beacon. 
A two-way set enables the pilot to report his progress and other- 
wise keep completely in touch with the ground. 

To make use of radio a battery is required, so that an airplane 
whose engine operates from a magneto must add a battery. This 
is necessary for reception; to transmit, a source of high-voltage 
current is needed in addition. Apparatus have been developed 
which are very compact and fight weight. 

The ordinary aircraft engine has a spark gap in each cylinder for 
each ignition. This sparking sets up radio oscillations which 
would interfere with the reception of radio messages if the engine 
were not shielded. This means that every electric wire must be 
covered with a coating of braided metal, the plugs themselves 
shielded with a metal cup, and every other part of the electrical 
equipment surrounded with a metal covering. 

The metal structure of the airplane is made to serve as a ground, 
which means that every metallic part of the airplane must be 
connected electrically to every other metallic part. This is called 
bonding. Proper bonding prevents trouble caused by intermove- 
ment of parts and eliminates possible sparking. 

Radio is used in avigation in several ways. If two or more 
stations on the ground are equipped with radio compasses, 
the pilot can request them to get his bearings, then sending out a 
series of dots from the plane enables the ground operators to get 
the direction of the airplane each from his own station. With 



RADIO BEACON 



379 



the bearings from two or more ground stations, either the pilot or 
one of the ground operators can draw corresponding Unes on a map 
and the intersection gives the geographic location of the airplane. 
If the ground operator does the plotting he promptly radios the 
position to the airplane. 

Radio Beacon. Another form of radio avigation is the utiHza- 
tion of the radio beacon. The Department of Commerce has 




(a) Intensity from Directional Antenna 



PureV 




Pure B 
(b) Signal Zones 
Fig. 122. Radio beacon. 

installed radio beacons at many airports, and it has been very 
successful in guiding airplanes from one airport to another. No 
special equipment is needed on the airplane; a receiving set is 
required, but a transmitter is not necessary. No special skill is 
required by the pilot. 

The radio beacon transmitter is shown diagrammatically in 
Fig. 122. In broadcasting, the transmitter is designed to send 
signals in all directions with equal strength, but it is relatively 
simple to erect an antenna which will send stronger signals in one 



380 AVIGATION 

direction than another. The directional antenna usually has 
" figure 8 " characteristics as shown in Fig. 122a. This diagram 
means that, if A 5 is the antenna, the strongest signals will be 
heard in a vertical plane through AB. At any point not in this 
plane the signal heard on the plane will not be as strong as it would 
be if the airplane were the same distance away from AB but in the 
plane AB. The intensity of the signal received on the airplane, 
which is a measure of the audibihty, is proportional to the length 
of the arrow on the diagram in the direction of the airplane. 

When two directional antennas, AB and CD, are crossed as in 
Fig. 1226, but not connected electrically in any way, signals may 
be sent out simultaneously from each. Two code letters are used, 
which are easily recognizable but are so selected that, if they are 
both started at the same time, the dash of one code letter is sounded 
at the same instant as the dot or dots of the other letter. The 

letters A (• — ) and N ( ) answer this requirement, as do 

D ( ) and U ( ), or B ( ) and V ( ). If the radi- 
ations from each antenna are received with equal strength, the 
receiver will respond to both equally, and the signal heard by the 

pilot will be a T or long dash ( ). If the reception of one 

signal is of greater intensity than the other, the powerful signal 
will be heard while the weaker signal will be either faint or in- 
audible. 

For example, if antenna A 5 is emitting a characteristic signal 
B and antenna CD is sending out V, in the two planes, which 
bisect the two angles made by A 5 and CD, both signals would be 
received with equal strength, and the T signal would be heard. 
If the airplane's position is nearer the plane through AB than the 
plane through CD, though the V signal might be heard faintly, the 
B signal would predominate. 

Because it is somewhat difficult to distinguish the difference in 
signals when the pilot is in or close to the T zone, a visual-type 
radio beacon is frequently used. One antenna transmits with a 
frequency of 66 cycles, the other with a frequency of 86| cycles. 
Located on the instrument board is a glass-faced box containing 
two springy metal strips or " reeds." One responds to a frequency 
of 66 cycles ; the other to 86| cycles. When receiving the modula- 
tion frequencies, the reed vibrates rapidly in vertical plane. The 
greater the intensity received, the greater is the amplitude of 
vibration. The pilot looks at the ends of the reeds, which are 



RADIO LANDINGS 381 

painted white; when vibrating rapidly they appear to be two 
white vertical lines. The greater the ampHtude of vibration, the 
longer is the line. When the airplane is in the T zone, the circuits 
for reeds are receiving the same intensity, and the two white lines 
are the same length. When the airplane is off to one side of the 
T zone, the circuit for one reed is receiving greater intensity so that 
one white Hne is longer and the other is shorter. The visual beacon 
is less tiring for the pilot, but since the receiving hook-up is slightly 
more complicated there is more danger of failure. 

The radio beacon guides a pilot in a proper direction. With a 
cross-wind, the pilot will probably zigzag a bit till he finds experi- 
mentally a heading which does not direct him from the B zone 
across the T zone into the V zone, or vice versa. Although there 
is a zone of silence directly over the beacon, a pilot flying in fog 
cannot rely on silence meaning he has reached his destination; 
his receiving set might have developed trouble. The beacon 
gives no indication of the distance to the transmitting station. 
In mountainous districts, the radio waves suffer peculiar dis- 
tortions and echoes which affect the accuracy of the beacon. 
The beacon requites a special transmitting installation. 

Radio Compass. By constructing the receiving antenna in the 
form of a coil or loop, and by suitably designing the tuned circuits 
in rotating this loop about a vertical axis till a signal fades 
out completely, the direction of the transmitter of that signal 
can be ascertained. Such an arrangement is called a radio 
compass. 

A radio compass can be used on any broadcasting transmission; 
the sending set need have no directional effect. The radio com- 
pass is useful in fog flying, since the pilot can take bearings on two 
or more broadcasting stations and plot these bearings on his map; 
the intersection gives the plane's position. The radio compass 
can also be used as a " homing " device. The pilot can find the 
direction of a broadcasting station located near his destination 
and then head his airplane in that direction. 

Radio Landings. The following method of making a landing 
in fog, etc., was devised by Captain A. F. Hegenberger, A.C., 
and is termed the Army method. 

Two transmitters are positioned near the landing field. These 
transmitters are portable so that the line connecting them can be 
put in the direction of the prevailing wind. Transmitter A is lo- 



382 AVIGATION 

cated outside the landing area but not more than 1,000 ft. outside 
the boundary of the field. Transmitter B is located farther from 
the field at a distance of a mile to a mile and a half from A. If an 
airplane flies over B headed for A, continuing his flight, without 
changing his heading, should bring the airplane directly over the 
landing area. The transmission from A and B are on different 
frequencies. 

The transmitters at A and B are not especially powerful. It 
is sufficient if signals can be picked up at a distance of 25 miles. 
The antenna for this transmission is non-directional; that is, the 
signals are broadcast with equal intensity in all directions. 

At each station, A and J5, is also a marker beacon. This beacon 
uses a low-power ultra-high-frequency transmitter, the wave length 
of the oscillator being 4 meters. A very short horizontal doublet 
antenna is used, located one-quarter wavelength above the ground. 
This form of antenna is very directional. The antenna is placed 
so that it is in line with the approach to the field, i.e., along the 
fine AB. Then the radiation is a fan-shaped wave in a plane 
through A or through B perpendicular to the line AB. 

The procedure for utilizing the above-described equipment is 
as foUows. The pilot, by making use of radio beacon, radio com- 
pass, or dead reckoning, arrives within receiving distance of 
station A. He then tunes in on B with his radio compass, and 
flies directly towards it, setting his directional gyro on his head- 
ing. He is then flying directly away from the field. After passing 
over B, the pilot makes a gradual flat turn of 180°, using his di- 
rectional gyro to tell him when he has exactly reversed his heading. 
Altitude should be gradually lost so that the altitude is approxi- 
mately 700 ft. above the ground when the airplane passes the 
outer station B. The exact time of passing B is told by the 
signal denoting passing the standing wave of the marker beacon. 
The receiver of the radio compass is then tuned on A and the 
airplane headed in that direction. In this, the pilot is aided by his 
directional gyro which he had previously set on the reverse head- 
ing. In proceeding from B to A the throttle is set back approxi- 
mately at 1,000 r.p.m. and the plane is put in a steady glide at an 
airspeed about 30 per cent above stalling speed. If this procedure 
has been followed the plane should arrive at station A at an alti- 
tude of 200 ft., and continuing the steady glide the plane should 
pass over all obstacles and reach the landing area properly. 



RADIO LANDINGS 383 

As soon as the wheels touch the ground, the pilot cuts the throttle, 
permitting the plane to come to rest. 

The pilot knows when he is passing over station A either by the 
" zone of silence " or from the signal from the marker beacon 
located at A. If his altitude at that time is less than 200 ft., the 
pilot merely opens the engine, putting the airplane in level flight 
for a brief interval of time, and then permits the plane to settle 
into its glide again. If the plane is higher than 200 ft. in passing 
over A, the pilot may either momentarily glide more steeply 
before resuming his normal glide or he may return to B and start 
his glide again at a slightly steeper angle. The altitudes men- 
tioned in this paragraph are read from a sensitive altimeter set 
at correct ground barometer reading from information radioed 
up to the pilot by the ground operator. 

This method of landing without seeing the ground has been 
tested by many hundred landings either in hooded cockpits or in 
actual fog. No accidents have ever been reported in using this 
method. 



CHAPTER XXI 
AEROSTATICS 

Introduction. Aerodynamics treats of air in motion; aero- 
statics deals with air at rest. In aviation, aerostatics deals with 
the problems of lighter-than-air craft. 

All lighter-than-air craft are balloons, but present-day usage 
is to employ the term balloon only for craft which has no motive 
power and the term dirigible balloon or simply dirigible for a 
balloon supphed with motive power. The term airship is 
synonymous with dirigible; an airplane should not be called an 
airship. 

Balloons are classified in two types, captive and free balloons. 
Captive balloons are moored to the ground and are used for 
observing artillery fire, etc. Captive balloons, if spherical in 
shape, have a strong tendency to rotate, so that they are usually 
elongated in shape with protuberances so designed that the bal- 
loon lies in the direction of the wind. This shape is called a kite- 
balloon. Free balloons are usually the so-called spherical shape. 
The upper part is a true hemisphere; the lower half is hemi- 
spherical except that at the extreme lower part the skin cones 
down to a long narrow tube called the appendix. In the earlier 
forms a rope netting was arranged over the upper half of the bag, 
the ends coming down just below the appendix where they were 
fastened to a metal concentration-ring from which the basket for 
passengers was hung. Because of the weight of the rope and the 
criss-cross of the netting forming many tiny pockets to retain 
rainwater, the netting arrangement was discarded for the suspen- 
sion band or " bellyband," a strip of fabric around the equator 
of the bag, ropes from the concentration-ring being attached to 
this band. The most modern method is to fasten each rope from 
the concentration-ring to a piece of fabric, called a finger-patch, 
cemented to the bag at the point where the curve of the sus- 
pension rope is just tangent to the sphere of the envelope. A 
valve, normally held shut by springs, is opened by pulling the 

384 



INTRODUCTION 



385 



valve-rope which hangs down inside the bag, coming through 
the appendix, the lower end being at the basket. 

Dirigibles are classified according to their method of construc- 
tion as rigid, semi-rigid, and non-rigid. The rigid ones are fre- 



Valve 



■Valve 




Netting 



Envelope 



Appendix 



Concentration 
Ring 



Basket 

Free Balloon with 
Netting Suspension 

<a» 




Basket 

Free Balloon with 
Band Suspension 



-Valve 



Mooring Bands 




Free Balloon with 
.Patch Suspension 

(c) 



Fig. 123. Free and captive balloons. 



Vertical 

Stabilizing 

Lobe 



quently called Zeppelins after Ferdinand Count Zeppelin who was 
the first successful designer of this type. The rigid type has a 
cloth-covered metal framework which gives the airship its shape, 
the gas being contained in a number of individual cells. The 
semi-rigid type has a metal keel the entire length of the ship, taking 



386 AEROSTATICS 

care of the transverse loads, the gas being in a single large envelope. 
The non-rigid is a single rubberized-cloth gas-tight bag containing 
the gas. Since, in non-rigid and semi-rigid types, the pressure of 
the gas inside the bag is relied upon to give the bag its shape, both 
these types are termed pressure airships. 

Gases. Hot-air, coal-gas, hydrogen, and helium are the prin- 
cipal gases used for lighter-than-air work. Heated aif is used 
sometimes for parachute jumps at carnivals; as soon as the air 
cools lift is lost. Coal-gas or ordinary illuminating gas is used 
sometimes for free balloons, the hydrogen contained in these gases 
being chiefly responsible for the lifting power. Hydrogen has 
been used for more than a century and is best from the point of 
lift but has the disadvantage that, if it mixes with air, it is highly 
inflammable and explosive. Helium though furnishing less lift 
than hydrogen is perfectly inert so that in its use there is not the 
fire hazard always present with hydrogen. 

Previous to 1915, helium cost approximately $2,000 per cubic 
foot. During the war the cost to the United States government 
was reduced to $400 per 1,000 cu. ft. At present, owing to im- 
provements in the method of extraction from natural gas, the cost 
is about $10 per 1,000 cu. ft. The cost of hydrogen is about $5 
per l,000;'cu. ft. 

Under standard conditions of 59° F. temperature and 29.92 
in. of mercury pressure, the weight per cubic foot of the gases 
important in aerostatics is as follows 

Air 0.07651 lb. 0^ ^ 

Hydrogen 0.00532 lb. \(S^ 

Helium 0.01056 lb. \o^ 

Average illuminating gas 0.0323 lb. 

Laws of Aerostatics. Six physical laws find application in 
aerostatic work. Full explanations are given in any standard 
textbook on physics. Briefly these laws are as follows. 

Archimedes' law: the buoyant or upward force exerted upon a 
body immersed in a fluid is equal to the weight of the fluid dis- 
placed. 

Boyle's law: at a constant temperature, the volume of a gas 
varies inversely as the pressure. 

Charles' law : at a constant pressure, the volume of a gas varies 
directly as the absolute temperature. 



LIFT 387 

Dalton's law: the pressure of a mixture of several gases in a 
given space is equal to the sum of the pressures which each gas 
would exert by itself if confined in that space. 

Joule's law: gases in expanding do no interior work. 

Pascal's law: the fluid pressure due to external pressure on 
the walls of the containing vessel is the same at all points through- 
out the fluid. 

Lift. A body completely immersed in a fluid displaces its own 
volume of fluid. If the weight of the fluid displaced equals the 
weight of the body, the body is in equihbrium. If the weight of 
the fluid displaced is greater than the weight of the body, the 
body rises. If the weight of the fluid displaced is less than the 
weight of the body, the body falls. 

The operation of a balloon and a submarine is somewhat similar 
in that both are entirely submerged in a fluid. The submarine is 
entirely sealed when under water, and the skin must be of suffi- 
cient strength to withstand stresses due to difference in pressure 
between inside and outside the hull. Balloons either are open to 
the air or have other arrangements so that there will be little or 
no pressure difference. 

Lift is obtained directly from Archimedes' principle. Unit 
lift is the difference between the weight of a cubic foot of air and 
the weight of a cubic foot of the gas. The gross lift or buoyancy 
of a balloon or airship is expressed by the following equation: 



L = V{Ba - A) 



L = gross Hft in pounds 
V = volume of gas in cubic feet 
Da = weight of a cubic foot of air 
Dg = weight of a cubic foot of gas 



The net lift or useful load is the difference between the gross lift 
and the dead weight of the bag, ropes, basket, etc. 

From the laws of Boyle and Charles, volume varies inversely 
as pressure and directly as absolute temperature. If Pq is the 
standard pressure. To the standard temperature, and Vo the 
standard volume, then Ti, the volume when the pressure is Pi 
and the absolute temperature is Ti, may be found by the fol- 
lowing, 



388 AEROSTATICS 

The foregoing is true only if the gas is in a perfectly elastic 
container. 

Example. A cloth bag contains 1,000 cubic feet of air, the tempera- 
ture being 59° F. and the pressure 29.92 in. What is the volume if the 
pressure is increased to 40.0 in. and the temperature decreased to 
0°F.? 

Solution. 

i -r 1 
, ^^^460 29.92 
= 1'^00 519X40:0 
= 663 cu. ft. 

In practice the balloon bag is made of rubberized cloth. If 
there is not sufficient volume of gas to fill the bag, the cloth sides 
will fold in and the bag will be flabby. When the volume of the 
gas is just equal to the cubical content the bag will be fully in- 
flated. Any further increase in volume of the gas means that 
gas will escape through the appendix. If the appendix is closed, 
a decrease in the air pressure outside or an increase of tempera- 
ture of the gas inside will cause the bag to burst. For this reason, 
though the appendix is usually tied shut while the balloon is being 
handled on the ground, the tie-off is broken as soon as the balloon 
leaves the ground. 

Density varies inversely as volume, so that if Do is the density 
under normal pressure Po and normal absolute temperature To, 
the density Di under pressure Pi and temperature Ti is found by 
the following relation : 

Di^To Pi 
Do Ti ^ Po 

Under normal conditions, the temperature of the gas inside the 
envelope is the same as the temperature of the adjacent air. The 
open appendix ensures that the pressure of the gas and air are the 
same. 

Considering the pressure of the gas the same as the pressure of 
the surrounding air and the temperature of the gas the same as 
that of the air, the expression for lift under other than standard 
conditions becomes 



ASCENSION OF A FREE BALLOON 389 

L = gross lift in pounds 

V = volume of bag in cubic feet 

■Dflo = weight of a cubic foot of 

air under standard con- 

/ T P T P\ ditions 

L = V[Dao^^ - D,, ^ ^ D,o = weight of a cubic foot of 

\ i 1 /^o J^i ^0/ gas under standard con- 

= V(B — n ^ — — ditions 

y yi^a, ^go) rp^ p^ p^ ^ Standard pressure 

Pi = actual pressure 
Tq = standard absolute temper- 
ature 
Ti = actual absolute tempera- 
ture 

Example. A 10,000-cu.-ft. free balloon is filled with pure hydrogen. 
Air and gas are at a temperature of 32° F. and a pressure of 28 in. 
What is the lift? 

Solution. 

Lift = V{Dao- Dgo)~X^ 

= 10,000 (0.07651 - 0.00532) ^ X 2II2 
= 703 lb. 

Problems 

1. A 30,000-cu.-ft. free balloon is full of pure hydrogen; air and 
gas are at a temperature of 80° F. and a pressure of 26.3 in. What 
is the lift? 

2. A 10,000-cu.-ft. free balloon is full of pure helium; air and gas 
are at a temperature of 75° F. and a pressure of 27.5 in. What is the 
lift? 

3. A 20,000-cu.-ft. free balloon is full of pure hydrogen; air and 
gas are at a temperature of —10° F. and a pressure of 21.3 in. What 
is the lift? 

4. A 20,000-cu.-ft. free balloon is full of pure helium; air and gas are 
at a temperature of —10° F. and a pressure of 21.3 in. What is the 
lift? 

5. A 30,000-cu.-ft. free balloon is full of pure hydrogen; air and gas 
are at a temperature of 20° F. and a pressure of 23.7 in. What is the 
lift? 

Ascension of a Free Balloon. When a free balloon, full of gas 
on the ground, has lift greater than its weight, it rises, and be- 
cause the atmospheric pressure decreases with altitude, gas will 
expand and tend to occupy greater volume. Since the fabric 
does not stretch, the volume is fixed and gas is forced out of the 



390 AEROSTATICS 

appendix. As the altitude increases, the density of the air de- 
creases; but the density of the hfting gas decreases at the same 
rate, so that the difference between the two weight densities, 
which is the unit Hft, also decreases at the same rate. The 
gross lift which is the constant volume multipHed by the unit lift 
decreases. Ascension will continue until the gross lift just equals 
the weight. 

If the bag is only partially inflated at the ground, on rising, 
the atmospheric pressure being less, the gas will expand, round- 
ing out the bag more fully. There will be no loss of gas till the 
bag is fully rounded out. The altitude where the bag is completely 
full and where any further increase in altitude will cause gas to 
escape is called the pressure height. 

A free balloon which is fully inflated at the ground will start to 
lose lift immediately on leaving the ground and there will be a 
continuous loss of lift as the balloon rises. A free balloon which 
is partially inflated at the ground will have a constant Hft until 
the pressure height is reached; above that altitude, Hft will de- 
crease with altitude. 

While inflating a balloon, sufficient sandbags are placed in the 
basket or hung on the netting to ensure that the balloon will not 
leave the ground. When ready to start, the balloon is weighed-o£f. 
This process is the removing of ballast until the total weight of the 
balloon just equals the total lift. In this condition, a man stand- 
ing on the ground can with practically no exertion move the 
balloon up in the air a foot or so, where it will stay in equihbrium. 
Adding a fraction of an ounce of sand will make the balloon sink 
slowly to the ground. Tossing overboard a little ballast will 
cause the balloon to rise slowly; heaving over a lot of sand will 
cause the bag to rise swiftly. 

If the bag is fuUy inflated at the ground, after weighing-off, 
weight equals Hft. If ballast is then jettisoned, the weight is 
lessened. The balloon will rise. As it rises, gas will escape, so 
that the Hft will decrease. At the start, the accelerating force will 
be equal to the difference between the weight and the lift; i.e., 
the force in pounds will equal the pounds of sand put over. In 
ascending, since the difference between the weight of the balloon 
and the lift will be less owing to the decreasing lift, the upward 
accelerating force diminishes. At the height where the Hft has 
decreased till it equals the weight, there will be no unbalanced 



ASCENSION OF A FREE BALLOON 391 

upward accelerating force and the balloon will be in equilibrium. 
To ascend further, more ballast must be put overboard. The 
ascent to a high altitude can be made by easy stages, by first 
dropping a little ballast, then a little more, and a little more till 
the desired height is reached. 

A bag, which is only partially inflated at the ground is weighed- 
off in the same manner as a fully inflated bag. When a little 
ballast is dropped, however, the balloon will immediately rise to 
its pressure height, the upward accelerating force being constant 
to that altitude. On reaching the pressure height gas will begin 
to escape, and with decreasing lift, the difference between lift and 
weight will get smaller, till equilibrium is reached. 

A partially inflated balloon will be at pressure height when its 
volume has increased to equal the capacity of the balloon. As 
the volume of a gas varies inversely with density, the reciprocal of 
the ratio of the new volume to the old is the ratio of new density 
to the old. Using Table I, interpolating if necessary, the altitude 
may be found corresponding to this density. 

If it is desired to ascend to some predetermined moderately 
high altitude, there is no special merit in fully inflating the bag for 
gas will start to escape immediately as the balloon rises. A par- 
tially inflated balloon on reaching its pressure height is exactly in 
the same condition as if it had been fully inflated on the ground. 
It should be noticed that there is a difference in maneuvering. 
On the ground, the fully inflated bag has more lift so that more 
ballast can be carried. In ascending with the full bag, ballast is 
gradually dropped so that, when the altitude is reached correspond- 
ing to the pressure height of a partially inflated bag, the amount 
of ballast remaining in each case would be the same. The bag 
which was fully inflated on the ground can be brought up to any 
desired altitude slowly. A bag only partially inflated on the 
ground will have a constant upward accelerating force; there- 
fore there will be a constantly increasing upward velocity. Ar- 
riving at the pressure height, the balloon will have considerable 
upward momentum, the product of its mass times its velocity, 
and this momentum will tend to carry the balloon on upward 
beyond its equilibrium point. At the equilibrium point, the lift 
equals the weight. If momentum carries the balloon above its 
pressure height, the gas having already expanded to fill the bag, 
gas will be forced out and lift will decrease. Lift being less than 



392 AEROSTATICS 

the weight, there is a downward accelerating force which acts 
first to decelerate the upward velocity, and, when this has been 
reduced to zero, the downward force will cause a downward ve- 
locity with ever-increasing speed. A skilful pilot will time his 
actions so that, just when the balloon has reached its highest point 
and is about to start on its downward plunge, just enough ballast 
is thrown over so that the weight remaining equals the then-exist- 
ing lift. 

Example. A 10,000-cu.-ft. free balloon is inflated with 8,000 cu. ft. 
of pure hydrogen at the ground, temperature 59° F., pressure 29.92 in. 
What is lift at 5,000-ft. altitude (temperature 45° F., pressure 24.8 in.)? 
At 10,000-ft. altitude (temperature 15° F., pressure 21.1 in.)? What 
is the pressure height? 

Solution. 
At ground 

L = 8,000 (0.07651 - 0.00532) 

= 570 lb. Y 

At 5,000-ft. altitude: To determine if bag is full 

_8 000??^'x^ 
-8,U0U 24.8,^519 

= 9,391 cu. ft. (volume at 5,000-ft. alt.) 

L = 9,391 (0.07651 - 0.00532) ^ X ^q^ 
= 570 lb. 
At 10,000-ft. altitude: To determine if bag is full 

o r.r.^ 475 29.92 
= 8,000X5^9X20- 

= 10,951 cu. ft. = volume of original gas at 10,000-ft. 

altitude; i.e., 951 cu. ft. has escaped. 

519 21 1 
■ L = 10,000 (0.07651 - 0.00532) ^ X ^g^ 

- 549 lb. 
To find pressure height: 
Pi _ 8,000 
Po ~ 10,000 
= 0.8 
From Table I by interpolation Pi/Po is 0.8 at 7,075-ft. altitude. 



DESCENT OF A FREE BALLOON 393 

Problems 
(See Table I for pressures and temperatures at altitude.) 

1. A 10,000-cu.-ft. free balloon is inflated at the ground with 7,000 
cu. ft. of pure hydrogen under standard atmospheric conditions. 
(a) What is the lift at the ground? (6) What is the lift at 5,000-ft. 
altitude? 

2. A 25,000-cu.-ft. free balloon is inflated at the ground with 
15,000 cu. ft. of pure helium under standard atmospheric conditions. 
(a) What is the lift at the ground? (6) What is the lift at 5,000-ft. 
altitude? (c) What is the lift at 10,000-ft. altitude? 

3. The total weight of a 20,000-cu.-ft. balloon, including bag, 
basket, crew, instruments, and 300 lb. ballast, but less gas, is 1,450 lb. 
The bag is filled with pure hydrogen on a day when the temperature is 
65° F. and the barometer is 29.6 in. (a) How much ballast must be 
dropped off in weighing-off, i.e., having weight just equal lift at the 
ground? (6) How much more ballast must be dropped in order for the 
bag to be in equilibrium at 5,000-ft. altitude? 

4. The balloon described in problem 3 is inflated with only 18,000 
cu. ft. of hydrogen under the same atmospheric conditions as in 3. 
(a) How much ballast must be dropped in weighing-off? (b) What is 
the pressure height? (c) How much more ballast must be dropped in 
order to be in equilibrium at 5,000-ft. altitude? 

Descent of a Free Balloon. In descending, the atmospheric 
pressure increases. Any gas that has escaped from the appendix 
or that has been valved is gone. When the balloon is in equilib- 
rium, if gas is valved, the lift becomes less than the weight, so that 
balloon starts downward under an accelerating force which is the 
excess of the weight over the lift. The effect of the increase in 
atmospheric pressure while descending will be to decrease the 
volume of the gas in the bag. The same weight of gas as at the 
beginning of the descent will displace less and less volume of air, 
but the density of the gas and air will increase as the volume 
decreases. The lift will remain the same all the way down, and 
there will therefore be a constant accelerating force downward 
which will cause ever-increasing downward velocity. 

The only way to check descent will be to introduce a decelerat- 
ing force, that is, an upward force. This can be accomplished 
only by making the weight less than lift, which is done by dropping 
ballast. The upward excess of lift will decrease the downward 
speed to zero and then cause an upward acceleration. 



394 AEROSTATICS 

If it is desired to descend to a definite altitude and remain there 
in equilibrium, the pilot, after valving a shght amount of gas to 
start descent, must drop ballast as he nears the desired altitude. 
At the instant that his downward progress is checked completely 
and before he starts to rise again, he must valve just enough gas 
to gain equihbrium. 

The faster the balloon is descending, the greater is the amount of 
ballast needed to check the speed. A good pilot valves only a 
Uttle gas at a time, so that only a little ballast must be sacrificed in 
maneuvering. 

In landing, the pilot comes down towards the ground, and at the 
proper height, which he has learned by experience, the pilot tosses 
over ballast so that his downward speed is zero when he is just a 
few feet off the ground. At that instant, he is pulling the rip- 
cord, which entirely opens one seam of the bag, completely empty- 
ing it of gas. 

Example. A 10,000-cu.-ft. bag is full of hydrogen at 12,000-ft. 
altitude (temperature 16.2° F., pressure 19.03 in.) and is in equilibrium, 
(a) What is the total weight of the bag including ballast? (6) If 100 
*cu. ft. of gas is valved, what is the loss of lift at this altitude? (c) At 
7,000-ft. altitude, what is the volume of the gas in the bag? (d) What 
is lift at 7,000-ft. altitude? (e) What ballast must be dropped at 
7,000-ft. altitude to secure equilibrium? 

Solviion. 

(a) Li2,ooo = 10,000 (0.07651 - 0.00532) ^^ X |^ 

= 494 lb. 

(6) L = 100 (0.07651 - 0.00532) ^^ X ^ 

= 5 1b. 

19 03 494 
(c) 77,000 = 9,900 X 23^0^ X ^ 

= 8,468 cu. ft. 

23 09 519 

■ (d) L7,ooo = 8468 (0.07651 - 0.00532) ^^ X ^ 

= 489 lb. 
(e) Ballast to be dropped W — Ly.ooo = 5 

Problems 

1. A 30,000-cu.-ft. balloon is only partially inflated with 20,000 
cu. ft. of pure hydrogen under standard atmospheric conditions at the 



SUPERHEAT 395 

ground, (a) What is the lift at the ground? (b) What is the lift at 
15,000-ft. altitude? (c) If balloon is in equilibrium at 15,000-ft. 
altitude, what is excess weight after 150 cu. ft. of gas are valved? 

(d) What is volume of bag after descending to 10,000-ft. altitude? 

(e) What is lift at 10,000-ft. altitude? 

2. A 20,000-cu.-ft. balloon is filled with pure helium at the ground 
under standard atmospheric conditions, (a) What is the lift at the 
ground? (b) What weight of ballast must be tossed over to rise to 
5,000-ft. altitude; (c) to 10,000-ft. altitude? (d) If 200 cu. ft. of 
helium are valved at 10,000-ft. altitude, what is the remaining lift? 
(e) When balloon has descended to 5,000-ft. altitude, 200 cu. ft. addi- 
tional of helium are valved; what is the remaining lift? 

Superheat. The sun's radiation consists not only of the long 
heat waves but also of the shorter light and electric waves. It 
is the heat radiation that warms the earth and to a smaller extent 
the atmosphere. 

Light radiation may change into heat radiation in passing 
through surfaces. This effect may be noticed in a tent which has 
the flaps down; the confined air is at a higher temperature than 
the outside air because the light in passing through the tent fabric 
is changed to heat. In greenhouses there is no ventilation, and 
the air inside is hotter than outside, as the result of light changing 
to heat in passing through the glass. Naturally the foregoing is 
true only on sunny days. In free balloons, the sun shining on the 
bag heats up the gas inside the bag to a higher temperature than 
the surrounding air. This is called superheat. 

Free balloons are suspended in air. If the air is moving, the 
balloon travels with it. Air does not travel past the balloon; 
the only heat lost is by direct conduction to the surrounding 
air. 

Kite balloons are moored to the ground for observational pur- 
poses. In perfectly still air, the same conditions apply as in a 
free balloon. If any breeze is blowing, the air passing the bal- 
loon takes up heat, so that any superheat is quickly lost, the 
temperature of the gas being reduced to that of the surrounding 
air. Dirigibles in motion are forcing their way through air, and 
this air acts to cool off any superheat. 

Superheat give additional lift to a balloon. It is a false, 
treacherous lift, since if the sun goes behind a cloud or the sun 
sets, the additional lift is quickly lost. 



396 AEROSTATICS 

Black surfaces absorb heat and light; shiny surfaces reflect 
heat and light. Balloons whose envelopes are of dark material 
have been found to have in the center of the bag a temperature 
70° hotter than the outer air. With lighter-colored surface the 
superheat will be less. 

The increase in temperature due to superheat will cause the gas 
to expand. Below the pressure height, this will increase the full- 
ness of the bag; for a fully inflated bag, superheat will force out 
gas; in either case, lift is increased. 

If a balloon or airship is brought out of a hangar, the gas is 
presumably at the same pressure and temperature as the atmos- 
phere. If the balloon is weighed-off immediately, the unit lift 
is due to the difference in weight density of the air and gas at the 
same temperature. Should the balloon stay on the ground with 
the sun shining brightly, the gas will receive superheat and will 
expand. The additional lift involves adding ballast to prevent the 
balloon or airship from rising. The weight of ballast that must 
be added exactly represents the additional lift due to superheat. 
When rising, the balloon moves upward through the air, and this 
air passing the envelope tends to cool it slightly. 

The additional Hft can be calculated if the number of degrees of 
superheat is known. If the balloon is initially inflated, the in- 
creased temperature will cause expansion, reducing the density, 
the volume being the same. If the balloon is only partially in- 
flated, the expansion of the gas will cause a greater displacement 
of the bag until the bag is fully inflated. Further expansion will 
cause gas to escape. For the partially inflated bag, gas expanding 
but not escaping will not change the weight of the gas contained 
in the envelope; more air will be displaced, however, so that the 
lift will be increased. 

It will be noted that, when a fully inflated bag is superheated, 
the increase in lift is exactly equal to the weight of the gas forced 
out of the bag by the expansion. 

The finding of the increased lift from superheat is illustrated 
by the following two examples. 

Example. An 8,000-cu-ft. balloon is fully inflated with pure hydro- 
gen under standard conditions. What is the additional lift due to 
superheat of 40°? 



SUPERHEAT ' 397 

Solution. 

Lift without superheat = 8,000 (0.07651 - 0.00532) 

= 569.5 lb. 
Weight of displaced air= 8,000 X 0.07651 
= 612.1 lb. 

^ 519 

Weight of superheated gas = 8,000 X 0.00532 X ^ 

= 39.5 lb. 
Lift with superheat = 612.1 - 39.5 
= 572.6 lb. 
Gain = 572.6 - 569.5 
= 3.1 lb. 

Check. 

559 
Volume of superheated gas = 8,000 X kTq 

= 8,616 cu. ft. 
Volume of escaping gas = 8,616 — 8,000 
= 616 cu. ft. 

519 

Weight of escaping gas = 616 X 0.00532 X ^ 

= 3.1 lb. 

Example. A 10,000-cu.-ft. balloon is inflated with 8,000 cu. ft. 
of pure hydrogen under standard conditions. What is the additional 
lift due to superheat of 40°? 

Solution, 

Lift without superheat = 8,000 (0.07651 - 0.00532) 

= 569.5 lb. 

559 
Volume of superheated gas = 8,000 X cTq 

= 8,616 cu. ft. 
Weight of displaced air = 8,616 X 0.07651 
= 659.2 lb. 
Weight of superheated gas = 8,000 X 0.00532 

= 42.6 lb. 
Lift with superheat = 659.2 - 42.6 
= 616.6 lb. 
Gain=47.11b. 

Problems 

1. A 10,000-cu-ft. hydrogen-filled balloon is brought out of the 
hangar. Both air and hydrogen are at 29.9 in. pressure and 45° F. 
temperature, (a) What is the lift? (b) Owing to sun's rays, the 
hydrogen experiences 30° superheat; what is then the lift? 



398 ' AEROSTATICS 

2. On a cold day a 25,000-cu.-ft. balloon is filled with pure hydrogen. 
Air and gas are at 30.1 in. pressure and 18° F. temperature, (a) What 
is the lift? (6) If the hydrogen is superheated 45° F., what is the lift? 

3. A 10,000-cu.-ft. balloon contains 7,000 cu. ft. of pure hydrogen. 
Air and gas are at 29.9 in. pressure and 70° F. temperature, (a) What 
is lift? (6) What is lift if the hydrogen is superheated 35° F.? 

4. A 10,000-cu.-ft. balloon contains 9,000 cu. ft. of pure helium. 
Air and gas are at 29.9 in. pressure and 32° F. temperature, (a) What 
is lift? (b) What is the lift if the helium is superheated 50° F.? 

5. A 10,000-cu.-ft. balloon is full of hydrogen at 75° F. temperature. 
Atmospheric pressure is 29.9 in. and atmospheric temperature is 40° F. 
(a) What is lift? (6) What is the lift if the sun goes behind clouds and 
the gas loses all its superheat? 

Purity. Any gas which remains inside a balloon for any length 
of time contains impurities owing to air seeping in through the 
appendix and mixing with the gas and to other causes. Although 
the impurities may be dry air, water vapor, carbon dioxide, or 
other substances, only a slight error is involved if all impurities 
are considered as being dry air. 

The purity of a gas is defined as the ratio of the volume of pure 
gas in the mixture to the total volume of impure gas. Considering 
all impurities as being dry air, they merely support themselves and 
furnish no lift. The impurities merely subtract from the total 
volume of lifting gas. A volume of gas of x per cent purity is x 
per cent of the volume of pure gas, giving lift, and (100 — x) per 
cent of the volume of dry air giving no lift. 

Example. A 10,000-cu.-ft. balloon is inflated with hydrogen of 95 
per cent purity under standard conditions. What is the lift? 
Solution. 

Gas is 95 per cent pure hydrogen 
5 per cent dry air 
Lift = 0.95 X 10,000 X (0.07651 - 0.00532) 
= 676.3 lb. 
Check. 

Weightof 9,500 cu. ft. of pure hydrogen = 9,500 X 0.00532 

= 50.5 lb. 
Weight of 500 cu. ft. of dry air = 500 X 0.07651 

= 38.3 lb. 
Weight of 10,000 cu. ft. of impure gas = 88.8 lb. 
Weight of 10,000 cu. ft. of displaced air = 10,000 X 0.07651 = 765.1 lb. 
Difference in weight of displaced air and gas = 676.3 lb. 



NON-RIGID AIRSHIP 399 

Non-Rigid Airship. A non-rigid airship is propelled through the 
air by one or more engines. To reduce the drag resistance of this 
motion, the gas bag is made of streamline shape. The rubberized 
cloth is tailored to the correct shape, and the pressure inside the 
bag is relied upon to fill out the bag properly. Use is made of two 
or more ballonets to preserve the shape of the bag. 

Without ballonets, a non-rigid airship, which might have been 
fully inflated on the ground, on ascending would need to have gas 
valved to prevent bursting, on descending would be flabby. In 
free balloons there is no special objection to flabbiness; but in 
dirigibles, flabbiness will destroy the streamline contour of the 
envelope. Partial inflation means that the nose will be cupped or 
dished in, increasing the drag enormously. 

A ballonet is a bag or compartment in the main gas bag that is 
formed by a diaphragm made of the same kind of gas-tight rub- 
berized cloth as the outer skin. The ballonet compartment is to 
hold air, and it has a valve opening to the atmosphere. The 
main gas bag is sealed to the outer air except for a gas valve used 
only in emergency. 

During inflation the ballonet is filled with air while the main 
bag is being filled with gas. On rising, the expansion of the gas 
causes the flexible wall of the ballonet chamber to collapse, expel- 
ling air. When the ballonet wall has entirely flattened out 
against the skin of the envelope so that all the air has been ex- 
pelled, the non-rigid airship is at its pressure height. On descend- 
ing, air is introduced into the ballonet, thus maintaining the 
rigidity of shape of the main envelope. To force air into the 
ballonet a scoop hangs down from the ballonet with its open end 
in the slipstream of the propeller. Sometimes a small auxiliary 
blower is used to pump air into the ballonet. 

If the non-rigid airship ever ascends above its pressure height, 
gas win have to be valved, and on descending, even with full 
ballonets, the main envelope will still be flabby. For this reason 
great care needs to be exercised that pressure airships do not rise 
to too great altitudes. 

The aerostatics of pressure airships does not vary much from 
that of free balloons below the pressure height, bearing in mind 
that the air in the ballonets contributes no lift. 

Superheat affects the operation of an non-rigid airship in the 
following manner. Bringing the airship out of the hangar into the 



400 AEROSTATICS 

bright sunshine, the pilot will notice immediately by his pressure 
gage or manometer that the pressure inside the gas chamber is 
increasing. The valve releasing air from the ballonet must be 
opened at once to prevent the gas bag from bursting. Each 
pound of air valved increases the lift by 1 lb. 

When the airship starts to move forward, the superheat is 
reduced by the air circulating past the outside of the envelope. 
The gas contracts, air is forced into the ballonet, lift is reduced, and 
therefore ballast must be discarded. 

The ballonet capacity is small compared with the gas capacity 
of an airship. Because air is heavier than the gas, the ballonets 
are always located on the lower side of the hull. In this location 
the sun's rays do not shine on the ballonets but on the upper side 
of the hull; any heating of the air would be by direct conduction 
from the warm gas through the separating wall to the air. Any 
heating of the air would add to the lift in the same way that lift 
was obtained in the old-style hot-air balloons; ordinarily this 
additional lift is so small that it is neglected. 

Example. A non-rigid airship of 200,000 cu.-ft.-capacity is brought 
out of the hangar. Atmospheric temperature is 50° F. and pressure 
is 30.2 in. Ballonets have 40,000-cu.-ft. capacity. How much added 
lift will the dirigible pick up if gas is superheated 30° F. before take- 
off? 

Solution. 
Volume of unsuperheated gas = 200,000 - 40,000 

= 160,000 cu. ft. 

Original lift = 160,000 (0.07651 -0.00532) ^x|^ 

= 11,7001b. 

540 
Volume of superheated gas = 160,000 Xttk 

= 169,410 cu. ft. 

Volume of displaced air = 169,410 - 160,000 

= 9,410 cu. ft. 

519 30 2 
Weight of displaced air = 9,410 X 0.07651 X ^ X 29^ 

= 738 lb. gain in lift 

Check. 

Kin 3Q 2 

Original lift = 160,000 (0.07651 -0.00532) ^Jq X 29^ 
= 11,700 lb. 



RIGID AIRSHIPS 401 

519 30 2 
Weight of air displaced after superheat = 169,410 X0.07651 X^ X29^ 

= 13,314 lb. 

519 30 2 
Weight of gas after superheat = 160,000 X 0.00532 X^ X 29^ 

= 874 lb. 
Lift after superheat = 13,314 - 874 
= 12,440 lb. 
Gain in lift = 12,440 - 11,700 
= 740 lb. 

Problems 

1. A non-rigid airship of 300,000-cu.-ft. capacity has its 50,000- 
cu.-ft. ballonets full of air; the remaining space is filled with hydrogen 
at standard conditions. What is the pressure height? 

2. A non-rigid airship of 225,000-cu.-ft. capacity has its 30,000- 
cu.-ft. ballonets full of air. The airship is inflated with pure helium. 
Air and gas are at 60° F. temperature and 30.1 in. pressure, (a) What 
is the lift if there is no superheat? (6) What is the lift with 20° F. 
superheat? 

3. A non-rigid airship of 200,000-cu.-ft. capacity is filled with 
hydrogen of 95 per cent purity; ballonets are 25,000-cu.-ft. capacity; 
superheat is 35° F. (a) What is the total lift? (6) How much ballast 
must be tossed over when all superheat is lost? 

Rigid Airships. A rigid airship has a metal framework which 
consists of a series of huge rings of varying diameter connected by 
longitudinal girders extending from the nose to the tail. During 
the war, for ease and cheapness in construction, the rings, except 
for those at the nose and tail, were of the same diameter. This 
gave the " pencil-shape " to the war-time Zeppelins. The modern 
Zepps have a nicely streamlined contour even though this requires 
separate design of each ring and more intricate tailoring of the 
envelope. 

There are three keels running lengthwise of the modern rigid 
ship. The framework of these keels is triangular in cross-section, 
permitting their use as passageways from stem to stern. In 
addition to these keels and other longitudinal members, the struc- 
ture is braced by a multiplicity of wires and cross-pieces. 

The gas is contained in ten to twenty individual cells instead of 
one large cell as in the pressure-type airship. This gives pro- 
tection similar to the bulkheads of marine vessels: if one or two 



402 AEROSTATICS 

cells become punctured, the remaining cells will have sufficient 
buoyancy to keep the ship in the air. A network of wires around 
each cell prevents undue expansion. 

Over the entire framework is stretched the outer cover, which is 
a doped fabric designed to reflect heat and give a smooth flying 
surface. The air space between the outer cover and the walls of 
the gas cells acts as ventilation space. 

The cells are only partiafly inflated on the ground so that 
expansion of the gas may take place when the airship rises. 
Flabbiness of the gas cells is permissible in the rigid type, since the 
streamline shape is preserved by the framework and outer cover. 

The dimensions of three famous rigid airships are given below. 

Los Graf ^^^^^ 

Angeles Zeppelin 

Nominal gas volume (cu, ft.) 2,470,000 3,700,000 6,500,000 

Over-aU length (ft.) 658 776 785 

Maximum diameter (ft.) 91 100 132 

Slenderness ratio 7.2 7.7 6.0 

Gross lift (lb.) 153,000 258,000 403,000 

Useful lift (lb.) 60,000 182,000 

Maximum velocity (miles per hour).. 73 80 84 

BIBLIOGRAPHY 

Chandler and Diehl, Balloon and Airship Gases. 

Warner, Aerostatics. 

Blakemore and Pagon, Pressure Airships. 

Burgess, Airship Design. 

U. S. Army Air Corps, TR 1170-295. 



APPENDIX A 
NOMENCLATURE 

(From N.A.C.A.-T.R. 474.) 

accelerometer — An instrument that measures the accelerations of an aircraft 

in a defined direction. 
acrobatics — Evolutions voluntarily performed with an aircraft other than 

those required for normal flight. 
adjustable propeller — See propeller, adjustable. 
aerodynamic center, wing section — A point located on or near the chord of 

the mean line approximately one-quarter of the chord length aft of the 

leading edge and about which the moment coefficient is practically 

constant. 
aerodynamics — The branch of dynamics that treats of the motion of air and 

other gaseous fluids and of the forces acting on solids in motion relative 

to such fluids. 
aerostatics — The science that treats of the equilibrium of gaseous fluids and 

of bodies immersed in them. 
aileron — A hinged or movable portion of an airplane wing, the primary fimc- 

tion of which is to impress a rolling motion on the airplane. It is usually 

part of the trailing edge of a wing. 
Frise aileron — An aileron having the nose portion projecting ahead of 

the hinge axis, the lower surface being in line with the lower surface of 

the wing. When the trailing edge of the aileron is raised, the nose por- 
tion protrudes below the lower surface of the wing, increasing the drag. 
airfoil — Any surface, such as an airplane wing, aileron, or rudder,' designed 

to obtain reaction from the air through which it moves. 
airfoil profile. — The outline of an airfoil section. 
airfoil section — A cross-section of an airfoil parallel to the plane of symmetry 

or to a specified reference plane. 
airplane — A mechanically driven fixed-wing aircraft, heavier than air, which 

is supported by the djrnamic reaction of the air against its wings. 
canard airplane — A type of airplane having the horizontal stabilizing and 

control surfaces in front of the main supporting surfaces. 
pusher airplane — An airplane with the propeller or propellers aft of the 

main supporting surfaces. 
tailless airplane — An airplane in which the devices used to obtain stabihty 

and control are incorporated in the wing, 
tractor airplane — An airplane with the propeller or propellers forward 

of the main supporting surfaces. 
airship — An aerostat provided with a propelling system and with means of 

controlling the direction of motion. A dirigible balloon; lighter than the 

airship. 

403 



404 



APPENDIX A 



non-rigid airship — An airship whose form is maintained by the internal 

pressure in the gas bags and ballonets. 
rigid airship — An airship whose form is maintained by a rigid struc- 
ture, 
semi-rigid airship — An airship whose shape is maintained by means of 
a rigid or jointed keel in conjunction with internal pressure in the gas 
containers and ballonets. 
air speed — The speed of an aircraft relative to the air. 

airspeed head — An instrument which, in combination with a gage, is used 
to measure the speed of an aircraft relative to the air. It usually con- 
sists of a Pitot-static tube or a Pitot-Venturi tube. 
airway — An air route along which aids to air navigation, such as landing 
fields, beacon lights, radio direction-finding facilities, intermediate fields, 
etc., are maintained. 
airworthiness — The quality of an aircraft denoting its fitness and safety for 

operation in the air under normal flying conditions. 
altimeter — An instrument that measures the elevation of an aircraft above a 

given datum plane, 
altitude : 

absolute altitude — The height of an aircraft above the earth. 

density altitude — The altitude corresponding to a given density in a 

standard atmosphere. 
pressure altitude — (1) The altitude corresponding to a given pressure in a 
standard atmosphere. (2) The altitude at which the gas bags of an 
airship become full. 
amphibian — An airplane designed to rise from and alight on either land or 

water. 
angle: 

aileron angle — The angular displacement of an aileron from its neutral 
position. It is positive when the trailing edge of the aileron is below the 
neutral position. 
blade angle — The acute angle between the chord of a section of a propeller 

and a plane perpendicular to the axis of rotation. 
dihedral angle — The acute angle between a line perpendicular to the plane 
of symmetry and the projection of the wing axis on a plane perpendicular 
to the longitudinal axis of the airplane. If the wing axis is not approxi- 
mately a straight line, the angle is measured from the projection of a line 
joining the intersection of the wing axis with the plane of symmetry and 
the aerodynamic center of the half-wing on either side of the plane of 
symmetry. 
downwash angle — The angle through which an air stream is deflected by 
any lifting surface. It is measured in a plane parallel to the plane of 
symmetry. 
drift angle — The horizontal angle between the longitudinal axis of an 

aircraft and its path relative to the'ground. 
elevator angle — The angular displacement of the elevator from its neutral 
position. It is positive when the trailing edge of the elevator is below 
the neutral position. 



APPENDIX A 405 

gliding angle — The angle between the flight path during a glide and a 

horizontal axis fixed relative to the air. 
zero-lift angle — The angle of attack of an airfoil when its lift is zero, 
angle of attack — The acute angle between a reference line in a body and the 
line of the relative wind direction projected on a plane containing the 
reference hne and parallel to the plane of symmetry. 
absolute angle of attack — The angle of attack of an airfoil, measured from 

the attitude of zero lift. 
critical angle of attack — The angle of attack at which the flow about an 
airfoil changes abruptly as shown by corresponding abrupt changes in the 
lift and drag, 
effective angle of attack — See angle of attack for infinite aspect ratio. 
induced angle of attack — The difference between the actual angle of attack 
and the angle of attack for infinite aspect ratio of an airfoil for the same 
Hft coefficient. 
angle of attack for infinite aspect ratio — The angle of attack at which an 
airfoil produces a given lift coefficient in a two-dimensional flow. Also 
called " effective angle of attack." 
area, equivalent flat-plate — The area of a square flat plate, normal to the 
direction of motion, which offers the same amount of resistance to motion 
as the body or combination of bodies under consideration. 
area, measurement of (performance calculations) : 

horizontal tail area — The horizontal tail area is measured in the same 
manner as the wing area, that is, with no deduction for the area blanketed 
by the fuselage, such blanketed area being bounded within the fuselage 
by lateral straight lines that connect the intersections of the leading and 
trailing edges of the stabilizer with the sides of the fuselage, the fairings 
and fillets being ignored. 
vertical tail area — The area of the actual outline of the rudder and the fin 

projected in the vertical plane, the fairings and fillets being ignored, 
wing area — Wing area is measured from the projection of the actual outline 
on the plane of the chords, without deduction for area blanketed by fuse- 
lage or nacelles. That part of the area, so determined, which lies within 
the fuselage or nacelles is bounded by two lateral lines that connect the 
intersections of the leading and trailing edges with the fuselage or na- 
celle, ignoring fairings and fillets. For the purpose of calculating area, a 
wing is considered to extend without interruption through the fuselage 
and nacelles. Unless otherwise stated, wing area always refers to total 
area including ailerons. 
area, propeller-disk — The total area swept by a propeller; i.e., the area of a 

circle having the same diameter as the propeller. 
aspect ratio — The ratio of the span to the mean chord of an airfoil; i.e., the 
ratio of the square of the span to the total area of an airfoil. 
effective aspect ratio — The aspect ratio of an airfoil of elliptical plan 
form that, for the same lift coefficient, has the same induced-drag coeffi- 
cient as the airfoil, or the combination of airfoils, in question. 
atmosphere: 

standard atmosphere — An arbitrary atmosphere used in comparing the 



406 APPENDIX A 

performance of aircraft. The standard atmosphere in use in the United 
States at present represents very nearly the average conditions found at 
latitude 40° and is completely defined in N.A.C.A. Report 218. 

autogiro — A type of rotor plane whose support in the air is chiefly derived 
from airfoils rotated about an approximately vertical axis by aerody- 
namic forces, and in which the lift on opposite sides of the plane of sym- 
metry is equalized by the vertical oscillation of the blades. 

automatic pilot — An automatic control mechanism for keeping an aircraft in 
level flight and on a set course. Sometimes called " gyro pilot," " me- 
chanical pflot," or " robot pilot." 

axes of an aircraft — Three fixed lines of reference, usually centroidal and 
mutually perpendicular. The horizontal axis in the plane of symmetry is 
called the longitudinal axis; the axis perpendicular to this in the plane of 
symmetry is called the normal axis; and the third axis perpendicular to 
the other two is called the lateral axis. In mathematical discussions, the 
first of these axes, drawn from rear to front, is called the X axis; the 
second, drawn downward, the Z axis; and the third, running from left 
to right, the Y axis. 

axis, wing — The locus of the aerodynamic centers of all the wing sec- 
tions. 

balance — A condition of steady flight in which the resultant force and mo- 
ment on the airplane are zero. 

bank — The position of an airplane when its lateral axis is incHned to the 
horizontal. A right bank is the position with the lateral axis inclined 
downward to the right. 

bank — To incline an airplane laterally; i.e., to rotate it about its longi- 
tudinal axis. 

biplane — An airplane with two main supporting surfaces placed one above 
the other. 

blade element — A portion of a propeller blade contained between the sur- 
faces of two cylinders coaxial with the propeller cutting the propefler 
blades. 

blade face — The surface of a propeller blade that corresponds to the lower 
surface of an airfoil. Sometimes called " thrust face " or " driving face." 

blade section — A cross-section of a propeller blade made at any point by a 
plane parallel to the axis of rotation of the propeller and tangent at the 
centroid of the section to an arc drawn with the axis of rotation as its 
center. 

boundary layer — A layer of fluid, close to the surface of a body placed in a 
moving stream, in which the impact pressure is reduced as a result of the 
viscosity of the fluid. 

camber — The rise of the curve of an airfoil section, usually expressed as the 
ratio of the departure of the curve from a straight Une joining the ex- 
tremities of the curve to the length of this straight line; " Upper camber" 
refers to the upper surface; " lower camber " to the lower surface; and 
" mean camber " to the mean line of the section. 

ceiling : 
absolute ceiling — The maximum height above sea level at which a given 



APPENDIX A 407 

airplane would be able to maintain horizontal flight under standard air 
conditions. 
service ceiling — The height above sea level, under standard air conditions, 
at which a given airplane is unable to climb faster than a small specified 
rate (100 ft. per min. in the United States and England). This specified 
rate may differ in different countries. 

cellule (or cell) — In an airplane, the entire structure of the wings and wing 
trussing of the whole airplane on one side of the fuselage, or between 
fuselages or nacelles if there are more than one. 

center of pressure of an airfoil — The point in the chord of an airfoil, pro- 
longed if necessary, which is at the intersection of the chord and the line 
of action of the resultant air force. 

center-of-pressure coefficient — The ratio of the distance of the center of 
pressure from the leading edge to the chord length. 

center section — The central panel of a wing; in the case of a continuous wing 
or any wing having no central panel, the limits of the center section are 
arbitrarily defined by the location of points of attachment to the cabane 
struts or fuselage. 

chord — An arbitrary datum line from which the ordinates and angles of an 
airfoil are measured. It is usually the straight line tangent to the lower 
surface at two points, the straight line joining the ends of the mean line, 
or the straight line between the leading and trailing edges. 

chord, mean aerodynamic — The chord of an imaginary airfoil which would 
have force vectors throughout the flight range identical with those of 
the actual wing or wings. 

chord length — The length of the projection of the airfoil profile on its chord. 

cockpit — An open space in an airplane for the accommodation of pilots or 
passengers. When completely enclosed, such a space is usually called 
a cabin. 

control column — A lever having a rotatable wheel mounted at its upper end 
for operating the longitudinal and lateral control surfaces of an airplane. 
This type of control is called " wheel control." 

controllability — The quality of an aircraft that determines the ease of operat- 
ing its controls and/or the effectiveness of displacement of the controls 
in producing change in its attitude in flight. 

decalage — The difference between the angular settings of the wings of a bi- 
plane or multiplane. The decalage is measured by the acute angle 
between the chords in a plane parallel to the plane of symmetry. The 
decalage is considered positive if the upper wing is set at the larger angle. 

dive — ■ A steep descent, with or without power, in which the airspeed is greater 
than the maximum speed in horizontal flight. 

downwash — The air deflected perpendicular to the direction of motion of an 
airfoil. 

drag — The component of the total air force on a body parallel to the relative 
wind. 
induced drag — That part of the drag induced by the lift; 
parasite drag — That portion of the drag of an aircraft exclusive of the 
induced drag of the wings. 



408 APPENDIX A 

profile drag — The difference between the total wing drag and the induced 

drag. 
profile drag, effective — The difference between the total wing drag and the 
induced drag of a wing with the same geometric aspect ratio but eUip- 
tically loaded. 

drag force or component (stress analysis) — A force or component, in the 
drag direction, i.e., parallel to the relative wing. 

drag strut — A fore-and-aft compression member of the internal bracing 
system of an aircraft. 

dynamic pressure — The product ipF^ where p is the density of the air 
and V is the relative speed of the air. 

engine: 

compression-ignition engine — A type of engine in which the fuel is sprayed 
into the cylinder and ignited by the heat of compression of the air charge, 
right-hand engine — An engine whose propeller shaft, to an observer facing 
the propeller from the engine end of the shaft, rotates in a clockwise 
direction. 
supercharged engine — An engine with a compressor for increasing the air 
or mixture charge taken into the cylinder beyond that inducted normally 
at the existing atmospheric pressure. 

engine, dry weight of — The weight of an engine exclusive of fuel, oil, and 
liquid coolant. 

engine weight per horsepower — The dry weight of an engine divided by the 
rated horsepower. 

equivalent monoplane — A monoplane wing equivalent as to its lift and drag 
properties to any combination of two or more wings. 

fairing — An auxiliary member or structure whose primary function is to re- 
duce the drag of the part to which it is fitted. 

fin — A fixed or adjustable airfoil, attached to an aircraft approximately paral- 
lel to the plane of symmetry, to afford directional stability; for example, 
tail fin, skid fin, etc. 

fineness ratio — The ratio of the length to the maximum diameter of a stream- 
line body, as an airship hull. 

fishtail — A colloquial term describing the motion made when the tail of an 
airplane is swung -from side to side to reduce speed in approaching the 
ground for a landing. 

flap — A hinged or pivoted airfoil forming the rear portion of an airfoil, 
used to vary the effective camber. 
split flap — A hinged plate forming the rear upper or lower portion of an 
airfoil. The lower portion may be deflected downward to give increased 
lift and drag; the upper portion may be raised over a portion of the wing 
for the purpose of lateral control. 

flight path — The flight path of the center of gravity of an aircraft with refer- 
ence to a frame fixed relative to the air or with reference to the earth. 

flow: 

laminar flow — A particular type of streamline flow. The term is usually 
applied to the flow of a viscous liquid near solid boundaries, when the 
flow is not turbulent. 



APPENDIX A 409 

streamline flow — A fluid flow in which the streamlines, except those very 

near a body and in a narrow wake, do not change with time. 
turbulent flow — Any part of a fluid flow in which the velocity at a given 
point varies more or less rapidly in magnitude and direction with time. 

flutter — An oscillation of definite period but unstable character set up in any 
part of an aircraft by a momentary disturbance, and maintained by a 
combination of the aerodynamic, inertial, and elastic characteristics of 
the member itself. 

fuselage — The body, of approximately streamline form, to which the wings 
and tail unit of an airplane are attached. 
monocoque fuselage — A fuselage construction which relies on the strength 
of the skin or shell to carry either the shear or the load due to bending 
moments. Monocoques may be divided into three classes (reinforced 
shell, semi-monocoque, and monocoque), and different portions of the 
same fuselage may belong to any one of these classes. The reinforced 
shell has the skin reinforced by a complete framework of structural 
members. The semi-monocoque has the skin reinforced by longerons and 
vertical bulkheads, but has no diagonal web members. The monocoque 
has as its only reinforcement vertical bulkheads formed of structural 
members. 

gap — The distance separating two adjacent wings of a multiplane. 

glide — To descend at a normal angle of attack with little or no thrust. 

ground loop — An uncontrollable violent turn of an airplane while taxying, or 
during the landing or take-ofif run. 

horn — A short lever attached to a control surface of an aircraft, to which 
the operating wire or rod is connected. 

horsepower of an engine, rated — The average horsepower developed by a 
given type of engine at the rated speed when operating at full throttle 
or at a specified altitude or manifold pressure. 

Immelman turn, normal — A maneuver made by completing the first half of a 
normal loop, then, from the inverted position at the top of the loop, half- 
rolling the airplane to the level position, thus obtaining a 180° change 
in direction simultaneously with a gain in altitude. 

impact pressure — The pressure acting at the forward stagnation point of a 
body, such as a Pitot tube, placed in an air current. Impact pressure 
may be measured from an arbitrary datum pressure. 

inclinometer — An instrument that measures the attitude of an aircraft with 
respect to the horizontal. 

induction system, rotary — A carburetor induction system used on radial 
engines, in which a rotary fan assists in distributing the fuel charge to the 
cylinders. 

instability, spiral — A type of instability, inherent in certain airplanes, which 
becomes evident when the airplane assumes too great a bank and side- 
slips; the bank continues to increase and the radius of the turn to decrease. 

instrument flying — The art of controlling an aircraft solely by the use of 
instruments; sometimes called " blind flying." 

Interceptor — A lateral-control device consisting of a small plate placed just 
back of a wing slot to spoil the effect of the slot at high angles of attack. 



410 APPENDIX A 

interference — The aerodynamic influence of two or more bodies on one 

another. 
landing gear — The understructure which supports the weight of an aircraft 
when in contact with the land or water and which usually contains a mech- 
anism for reducing the shock of landing. Also called " undercarriage." 
retractable landing gear — A type of landing gear which may be withdrawn 
into the body or wings of an airplane while it is in flight, in order to re- 
duce the parasite drag. 
leading edge — The foremost edge of an airfoil or propeller blade; 
level-off — To make the flight path of an airplane horizontal after a climb, 

glide, or dive. 
lift: dynamic — The component of the total aerodynamic force on a body per- 
pendicular to the relative wind. 
lift/drag ratio — The ratio of the lift to the drag of any body. 
load: 

full load — Weight empty plus useful load; also called gross weight, 
payjoad — That part of the useful load from which revenue is derived, viz., 

passengers and freight, 
useful load — The crew and passengers, oil and fuel, ballast other than 
emergency, ordnance, and portable equipment. 
loading: 

power loading — The gross weight of an airplane divided by the rated horse- 
power of the engine computed for air of standard density, unless otherwise 
stated. 
Span loading — The ratio of the weight of an airplane to its equivalent 

monoplane span. 
tinsymmetrical loading (stress analysis) — A design loading condition for 
the wings and connecting members, representing the conditions as in a 
roll, 
wing loading — The gross weight of an airplane divided by the wing area. 
longeron — A principal longitudinal member of the framing of an airplane 
fuselage or nacelle, usually continuous across a number of points of 
support. 
loop — A maneuver executed in such a manner that the airplane follows a 

closed curve approximately in a vertical plane. 
maneuverability — That quality in an aircraft which determines the rate at 

which its altitude and direction of flight can be changed. 
mean line (of an airfoil profile) — An intermediate line between the upper and 

lower contours of the profile. 
mixture control, altitude — A device on the carburetor for regulating the 
weight proportions of air and fuel supplied to the engine at different 
altitudes. 
monoplane — An airplane with but one main supporting surface, sometimes 
divided into two parts by the fuselage. 
high-wing monoplane — A monoplane in which the wing is located at, or 

near, the top of the fuselage. 
low-wing monoplane — A monoplane in which the wing is located at, or 
near, the bottom of the fuselage. 



APPENDIX A 411 

xnidwing monoplane — A monoplane in which the wing is located approxi- 
mately midway between the top and bottom of the fuselage. 
parasol monoplane — A monoplane in which the wing is above the fuselage. 

multiplane — An airplane with two or more main supporting surfaces placed 
one above another. 

nacelle — An enclosed shelter for personnel or for a power plant. A nacelle 
is usually shorter than a fuselage, and does not carry the tail unit. 

nose-down — To depress the nose of an airplane in flight. 

nose-over — A colloquial expression referring to the accidental turning over 
of an airplane on its nose when landing. 

oleo gear — A type of oil-damping device that depends on the flow of oil 
through an orifice for its shock-absorbing effect in a landing gear. 

oscillation: 
phugoid oscillation — A long-period oscillation characteristic of the dis- 
turbed longitudinal motion of an aircraft. 
stable oscillation — An oscillation whose amplitude does not increase, 
unstable oscillation — An oscillation whose amplitude increases continu- 
ously until an attitude is reached from which there is no tendency to 
return toward the original attitude, the motion becoming a steady 
divergence. 

over-all length — The distance from the extreme front to the extreme rear of 
an aircraft, including the propeller and tail unit. 

overhang — (1) One-half the difference in span of any two main supporting 
surfaces of an airplane. The overhang is positive when the upper of the 
two main supporting surfaces has the larger span. (2) The distance from 
the outer strut attachment to the tip of a wing. 

overshoot — To fly beyond a designated mark or area, such as a landing field, 
while attempting to land on the mark or within the area. 

panel (airplane) — A portion of an airplane wing constructed separately from 
the rest of the wing to which it is attached. 

pitch — An angular displacement about an axis parallel to the lateral axis of 
an aircraft. 

pitch of a propeller: 

effective pitch — The distance an aircraft advances along its flight path for 

one revolution of the propeller, 
geometrical pitch — The distance an element of a propeller would advance 
in one revolution if it were moving along a helix having an angle equal to 
its blade angle, 
zero-thrust pitch — The distance a propeller would have to advance in 
one revolution to give no thrust. Also called " experimental mean pitch.'* 

pitch ratio (propeller) — The ratio of the pitch to the diameter. 

pitot-static tube — A parallel or coaxial combination of a Pitot and a static 
tube. The difference between the impact pressure and the static pressure 
is a function of the velocity of flow past the tube. 

Pitot tube — A cylindrical tube with an open end pointed upstream, used in 
measuring impact pressure. 

Pitot-Venturi tube — A combination of a Pitot and a Venturi tube. 

plane (or hydroplane) — To move through the water at such a speed that the 



412 APPENDIX A 

support derived is due to hydrodynamic and aerodynamic rather than to 
hydrostatic forces. 
plan form, developed — The plan of an airfoil as drawn with the chord lines 
at each section rotated about the airfoil axis into a plane parallel to the 
plane of projection and with the airfoil axis rotated and developed and 
projected into the plane of projection. 
plan form, projected — The contour as viewed from above. 
profile thickness — The maximum distance between the upper and lower 
contours of an airfoil, measured perpendicularly to the mean line of the 
profile. 
propeller — Any device for propelling a craft through a fluid, such as water 
or air; especially a de\'ice having blades which, when mounted on a 
power-driven shaft, produce a thrust by their action on the fluid, 
adjustable propeller — A propeller whose blades are so attached to the 

hub that the pitch may be changed while the propeller is at rest. 
automatic propeller — A propeller whose blades are attached to a mechanism 
that automatically sets them at their optimum pitch for various flight 
conditions. 
controllable propeller — A propeller whose blades are so moimted that 

the pitch may be changed while the propeller is rotating. 
geared propeller — A propeller driven through gearing, generally at some 

speed other than the engine speed. 
pusher propeller — A propeller mounted on the rear end of the engine or 

propeller shaft. 
tractor propeller — A propeller mounted on the forward end of the engine 
or propeller shaft. 
propeller eflBciency — The ratio of the thrust power to the input power of a 

propeller. 
propeller rake — The mean angle which the line joining the centroids of the 
sections of a propeller blade makes -^ith a plane perpendicular to the axis. 
propeller root — That part of the propeller blade near the hub. 
propeller thrust — The component of the total air force on the propeller which 

is parallel to the direction of advance, 
propeller thrust, effective — The net driving force developed by a propeller 
when mounted on an aircraft, i.e., the actual thrust exerted by the 
propeller, as mounted on an airplane, minus any increase in the resistance 
of the airplane due to the action of the propeller, 
propeller thrust, static — The thrust developed by a propeller when rotating 

without translation. 
propeller tipping — A protective covering of the blade of a propeller near the 

tip. 
propulsive efl5ciency — The ratio of the product of the effective thrust and 
flight speed to the actual power input into the propeller as mounted on 
the airplane. 
pull-out — The maneuver of transition from a dive to horizontal flight. 
pull-up — A maneuver, in the vertical plane, in which the airplane is forced 
into a short chmb, usually from approximately level flight (c/. zoom). 
sudden pull-up (or sudden pull-out) (stress analysis) — A loading con- 



APPENDIX A 413 

\ 

dition for the tail surfaces resulting from a sudden application of up- 

elevator (c/. dive). 
purity (of gas) — The ratio of the partial pressure of the aerostatic gas in the 

container to the total pressure of all the contained gases. 
range, maximum — The maximum distance a given aircraft can cover under 

given conditions, by flying at the economical speed and altitude at all 

stages of the flight. 
range at maximum speed — The maximum distance a given aircraft can fly 

at full speed at the altitude for maximum speed under given conditions. 
rate-of-climb indicator — An instrument that indicates the rate of ascent or 

descent of an aircraft. 
Reynolds number — A non-dimensional coefficient used as a measure of the 

dynamic scale of a flow. Its usual form is the fraction Vl/v in which V 

is the velocity of the fluid, Z is a linear dimension of a body in the fluid, 

and V is the kinematic viscosity of the fluid (c/. scale effect). 
righting or restoring moment — A moment that tends to restore an aircraft 

to its previous attitude after any small rotational displacement. 
ring cowling — A ring-shaped cowling placed around a radial air-cooled engine 

to reduce its drag and improve cooling. 
roll — A maneuver in which a complete revolution about the longitudinal 

axis is made, the horizontal direction of flight being approximately 

maintained. 
aileron roll — A roll in which the motion is largely maintained by forces 

arising from the displacement of the aileron. 
outside roll — A roll executed while flying in the negative angle-of -attack 

range, 
snap roll — A roll executed by a quick movement of the controls, in which 

the motion is largely maintained by autorotational couples on the wings. 
rudder — A hinged or movable auxiliary airfoil on an aircraft, the function of 

which is to impress a yawing moment on the aircraft. 
rudder bar — The foot bar by means of which the control cables leading to 

the rudder are operated. 
scale effect — The change in any force coefficient, such as the drag coefficient, 

due to a change in the value of Reynolds number. 
seaplane — An airplane designed to rise from and alight on the water, 
sesquiplane — A form of biplane in which the area of one wing is less than half 

the area of the other. 
sideslipping — Motion of an aircraft relative to the air, in which the lateral 

axis is inclined and the airplane has a velocity component along the lateral 

axis. When it occurs in connection with a turn, it is the opposite of 

skidding, 
sinking speed — The vertical downward component of velocity that an air- 
craft would have while descending in still air under given conditions of 

equilibrium, 
slip — The difference between the geometrical pitch and the effective pitch 

of a propeller. Slip may be expressed as a percentage of the mean geo- 
metrical pitch, or as a linear dimension. 
slip function — The ratio of the speed of advance through the undisturbed 



414 APPENDIX A 

air to the product of the propeller diameter and the number of revolutions 
per unit time: V /nD. 
slipstream — The current of air driven astern by a propeller. 
slot — The nozzle-shaped passage through a wing whose primary object is to 
improve the flow conditions at high angles of attack. It is usually near 
the leading edge and formed by a main and an auxihary airfoil. 
span — The maximum distance, measured parallel to the lateral axis, from 
tip to tip of an airfoil, of an airplane wing inclusive of ailerons, or of a 
stabilizer inclusive of elevator, 
effective span — The true span of a wing less corrections for tip loss, 
speed: 
ground speed — The horizontal component of the velocity of an aircraft 

relative to the ground. 
landing speed — The minimum speed of an airplane at the instant of con- 
tact with the landing area in a normal landing. 
minimum flying speed — The lowest steady speed that can be maintained, 
with any throttle setting whatsoever, by an airplane in level flight at 
an altitude above the ground greater than the span of the wings. 
rated engine speed — The rotative speed of an engine at which its reliabihty 

has been determined for continuous performance. 
stalling speed — The speed of an airplane in steady flight at its maximum 

coefficient of lift. 
take-off speed — The airspeed at which an airplane becomes entirely air- 
borne. 
spin — A maneuver in which an airplane descends along a helical path of large 
pitch and small radius while flying at a mean angle of attack greater than 
the angle of attack at maximum hft {cf. spiral). 
flat spin — A spin in which the longitudinal axis is less than 45° from the 

horizontal. 
inverted spin — A maneuver having the characteristics of a normal spin 

except that the airplane is in an inverted attitude. 
normal spin — A spin which is continued by reason of the voluntary po- 
sition of the control surfaces, recovery from which can be effected within 
two turns by neutralizing or reversing aU the controls. Sometimes 
called " controlled spin." 
uncontrolled spin — A spin in which the controls are of little or no use 
in effecting a recovery, 
spinner — A fairing of approximately conical or paraboloidal shape, which is 
fitted coaxially with the propeller hub and revolves with the propeller. 
spiral — A maneuver in which an airplane descends in a helix of small pitch 
and large radius, the angle of attack being within the normal range of 
flight angles. 
split S — A maneuver consisting of a half snap roll followed by a pull-out 
to normal flight, thus obtaining a 180° change in direction accompanied 
by a loss of altitude. 
Spoiler — A small plate arranged to project above the upper surface of a wing 
to disturb the smooth air flow, with consequent loss of lift and increase of 
drag. 



APPENDIX A 415 

stability — That property of a body which causes it, when its equihbrium is 
disturbed, to develop forces or moments tending to restore the original 
condition, 
automatic stability — Stability dependent upon movable control surfaces 

automatically operated by mechanical means. 
directional stability — Stability with reference to disturbances about the 
normal axis of an aircraft, i.e., disturbances which tend to cause yawing, 
dynamic stability — That property of an aircraft which causes it, when its 
state of steady flight is disturbed, to damp the oscillations set up by the 
restoring forces and moments and gradually return to its original state. 
inherent stability — Stability of an aircraft due solely to the disposition 
and arrangement of its fixed parts; i.e., that property which causes it, 
when disturbed, to return to its normal attitude of flight without the use 
of the controls or the interposition of any mechanical device, 
lateral stability — Stability with reference to disturbances about the longi- 
tudinal axis. 
longitudinal stability — Stability with reference to disturbances in the plane 
of symmetry; i.e., disturbances involving pitching and variation of the 
longitudinal and normal velocities, 
static stability — That property of an aircraft which causes it, when its 
state of steady flight is disturbed, to develop forces and moments tending 
to restore its original condition. 

stabilizer (airplane) — Any airfoil whose primary function is to increase the 
stability of an aircraft. It usually refers to the fixed horizontal surface. 

stagger — A term referring to the longitudinal position of the axes of two 
wings of an airplane. Stagger of any section is measured by the acute 
angle between a line joining the wing axes and a line perpendicular to the 
upper wing chord, both lines lying in a plane parallel to the plane of sym- 
metry. The stagger is positive when the upper wing is in advance of the 
lower. 

stall — The condition of an airfoil or airplane in which it is operating at an 
angle of attack greater than the angle of attack of maximum lift. 

static pressure — The force per unit area exerted by a fluid on a surface at 
rest relative to the fluid. 

streamline — The path of a particle of a fluid, supposedly continuous, com- 
monly taken relative to a solid body past which the fluid is moving; 
generally used only of such flows as are not eddying. 

streamline form — The form of a body so shaped that the flow about it tends 
to be a streamline flow. 

strut — A compression member of a truss frame. 

supercharge — To supply an engine with more air or mixture than would be 
inducted normally at the prevailing atmospheric pressure. The term 
supercharged is generally used to refer to conditions at altitudes where the 
pressure in the intake manifold is partly or completely restored to that 
existing under normal operation at sea-level. 

supercharger — A pump for supplying the engine with a greater weight of air 
or mixture than would normally be inducted at the prevailing atmos- 
pheric pressure. 



416 APPENDIX A 

centrifugal-type supercharger — A high-speed rotary blower equipped with 
one or more multiblade impellers which, through centrifugal action, 
compress the air or mixture in the induction system, 
positive-driven-type supercharger — A supercharger driven at a fixed ratio 

from the engine shaft by gears or other positive means. 
Roots-type supercharger — A positive-displacement rotary blower con- 
sisting of two double-lobed impellers turning in opposite directions on 
parallel shafts within a housing, the impellers rolling together except for 
a small clearance, meanwhile alternately trapping incoming air or mix- 
ture in the ends of the housing and sweeping it through to the outlet. 

sweepback — The acute angle between a line perpendicular to the plane of 
symmetry and the plan projection of a reference line in the wing. 

tachometer — An instrument that measures in revolutions per minute the 
rate at which the crankshaft of an engine turns. 

tail, airplane — The rear part of an airplane, usually consisting of a group 
of stabilizing planes, or fins, to which are attached certain controUing 
surfaces such as elevators and rudders; also called " empennage." 

tail boom — A spar or outrigger connecting the tail surfaces and the main 
supporting surfaces. 

tailheavy — The condition of an airplane in which the tail tends to sink when 
the longitudinal control is released in any given attitude of normal 
flight. 

tail skid — A skid for supporting the tail of an airplane on the ground. 

take-ofE — The act of beginning flight in which an airplane is accelerated from 
a state of rest to that of normal flight. In a more restricted sense, the 
final breaking of contact with the land or water. 

taper in plan only — A gradual change (usually a decrease) in the chord 
length along the wing span from the root to the tip, with the wing sections 
remaining geometrically similar. 

taper in thickness ratio only — A gradual change in the thickness ratio along 
the wing span with the chord remaining constant. 

taxi — To operate an airplane under its own power, either on land or on 
water, except as necessarily involved in take-off or landing. 

thickness ratio — The ratio of the maximum thickness of an airfoil section to 
its chord. 

trailing edge — The rearmost edge of an airfoil or of a propeller blade. 

turn indicator — An instrument for indicating the existence and approxi- 
mate magnitude of angular velocity about the normal axis of an air- 
craft. 

velocity, terminal — The hypothetical maximum speed that an airplane could 
attain along a specified straight flight path under given conditions of 
weight and propeller operation, if diving an unlimited distance in air of 
specified uniform density. If the term is not qualified, a vertical path 
angle, normal gross weight, zero thrust, and standard sea-level air 
density are assumed. 

Venturi tube (or Venturi) — A short tube of varying cross-section. The flow 
through the Venturi causes a pressure drop in the smallest section, the 
amount of the drop being a function of the velocity of flow. 



APPENDIX A 417 

visibility — The greatest distance at which conspicuous objects can be seen 
and identified. 

warp — To change the form of a wing by twisting it. Warping was formerly 
used to perform the function now performed by ailerons. 

wash — The disturbance in the air produced by the passage of an airfoil. 
Also called the " wake " in the general case for any solid body. 

washin — A warp of an airplane wing giving an increase of the angle of attack 
toward the tip. 

washout — A warp of an airplane wing giving a decrease of the angle of attack 
toward the tip. 

weight: 

empty weight — The structure, power plant, and fixed equipment of an 
aircraft. Included in this fixed equipment are the water in the radiator 
and cooling system, all essential instruments and furnishings, fixed 
electric wiring for fighting, heating, etc. In the case of an aerostat, it 
also includes the amount of ballast that must be carried to assist in mak- 
ing a safe landing. 
fixed power plant weight for a given airplane weight — The weight of the 
power plant and its accessories, exclusive of fuel and oil and their tanks. 
gross weight (airplane) — The total weight of an airplane when fully 

loaded (c/. load, full). 
net weight (stress analysis) — The gross weight, less some specific partial 
weight. Very often the partial weight is the dead weight of the wings, 
but it may be the useful load. The partial weight in question should 
always be clearly indicated by the context. 

\irheel, tail — A wheel used to support the tail of an airplane when on the 
ground. It may be steerable or non-steerable, fixed or swivehng. 

wind, relative — The velocity of the air with reference to a body in it. It is 
usually determined from measurements made at such a distance from 
the body that the disturbing effect of the body upon the air is negligible. 

window, inspection — A small transparent window fitted in the envelope of a 
balloon or airship, or in the wing or fuselage of an airplane, to allow in- 
spection of the interior. 

wind tunnel — An apparatus producing an artificial wind or air stream, in 
which objects are placed for investigating the air flow about them and 
the aerodynamic forces exerted on them. 

wing — A general term applied to the airfoil, or one of the airfoils, designed to 
develop a major part of the lift of a heavier-than-air craft. 
equivalent wing (stress analysis) — A wing of the same span as the actual 
wing, but with the chord at each section reduced in proportion to the 
ratio of the average beam load at that section to the average beam load 
at the section taken as the standard. 

wingheavy, right or left — The condition of an airplane whose right or left 
wing tends to sink when the lateral control is released in any given attitude 
of normal flight. 

wing-over — A maneuver in which the airplane is put into a climbing turn un- 
til nearly stalled, at which point the nose is allowed to fall while continu- 
ing the turn, then returned to normal flight from the ensuing dive or 



418 APPENDIX A 

glide in a direction approximately 180° from that at the start of the 
evolution. 
wing rib — A chord-wise member of the wing structure of an airplane, used to 
give the wing section its form and to transmit the load from the fabric 
to the spars. 
compression wing rib — A heavy rib designed to perform the function of an 
ordinary wing rib and also to act as a strut opposing the pull of the wires 
in the internal drag truss. 
former (or false) wing rib — An incomplete rib, frequently consisting only 
of a strip of wood extending from the leading edge to the front spar, which 
is used to assist in maintaining the form of the wing where the curvature 
of the airfoil section is sharpest. 
wing section — A cross-section of a wing parallel to the plane of symmetry or 

to a specified reference plane, 
wing skid — A skid placed near the wing tip to protect the wing from contact 

with the ground. 
wing spar — A principal span-wise member of the wing structure of an airplane, 
wing-tip rake — A term referring to the shape of the tip of the wing when the 
tip edge is sensibly straight in plan but is not parallel to the plane of 
symmetry. The amount of rake is measured by the acute angle between 
the straight portion of the wing tip and the plane of symmetry. The 
rake is positive when the trailing edge is longer than the leading edge, 
wire (airplane) : 

antidrag wire — A wire intended primarily to resist the forces acting for- 
ward in the chord direction. It is generally enclosed in the wing. 
drag wire — A wire intended primarily to resist the forces acting backward 

in the chord direction. It is generally enclosed in the wing. 
landing wire — A wire or cable which braces the wing against the forces 

opposite to the normal direction of the lift. 
lift wire — A wire or cable which braces the wings against the lift force; 

sometimes called " flying wire." 
stagger wire — A wire connecting the upper and lower wings of an airplane 
and lying in a plane substantially parallel to the plane of symmetry; 
also called " incidence wire." 
wire (airship): 

antiflutter wire — A wire in the plane of the outer cover for local reinforce- 
ment and for reducing flutter due to variations in air pressure or propeller 
wash. 
chord wire — A wire joining the vertices of a main transverse frame. 
yaw — An angular displacement about an axis parallel to the normal axis of 

an aircraft. 
zoom — To climb for a short time at an angle greater than the normal climb- 
ing angle, the airplane being carried upward at the expense of kinetic 
energy. 



APPENDIX B 
ANSWERS TO PROBLEMS 



Chap. I, page 3 



(1) 0.00175 slug 

(2) 0.00166 slug 



per cu. ft. 
per cu. ft. 



(3) 0.0659 lb. per cu. ft. 



(1) 0.001225 



Chap. I, page 6 
(2) 0.000992 



(3) 0.000810 



(1) 0.000634 



Chap. I, page 7 
(2) 0.000472 



(3) 0.000376 



(1) 4190 lb. 



Chap. II, page 16 

(2) 3.8 hp. (3) 2.4 hp. 



(4) 1.8 hp. 



(1) o. 79.8 lb. 
6. 15.2 1b. 

c. 81.4 1b. 

d. 81.4 lb. 



Chap. II, page 17 

(2) 1.2 1b. 

(3) a. 7.7 lb. 
h. 1.1 lb. 
c. 87° 28' 



(4) a. 125.5 lb. 
6. 11801b. 
(6) 840 lb. 



(1) 11.9 lb. 



Chap, n, page 21 
(2) 47.6 lb. 



(3) 6.1 lb. 



(1) 4,480,000 

(2) 5,900,000 



Chap. Ill, page 27 

(3) 3,400,000 

(4) 1,200 mi. per hr. 



(5) 26.6 atmos. 



(1) 4,570 lb. 

(2) 3,260 lb. 

(3) 1.9'' 

(4) 4.6° 



Chap. IV, page 37 

(5) 8.4° 

(6) 606 sq. ft. 

(7) 6,420 lb. 

(8) 4,760 lb. 

419 



(9) 56.0 mi. per hr. 

(10) 65.0 mi. per hr. 

(11) 45 lb. per sq. ft. 

(12) 33 lb. per sq. ft. 



420 APPENDIX B 

Chap. IV, page 38 

(1) 56.9 mi. per hr. (4) 4.9 lb. per sq. ft. (7) 7.4 lb. per sq. ft. 

(2) 55.0 mi. per hr. (5) 2,700 lb. (8) a. 44.8 mi. per hr. 

(3) 418 sq. ft. (6) 69.0 mi. per hr. b. 52.0 mi. per hr. 

Chap. IV, page 40 

(1) 308 lb. (3) 237 hp. (5) a. 126 lb. 

(2) 664 lb. (4) 15.4 hp. ' b. 21.7 hp. 

Chap. IV, page 42 

(1) a. 2,2601b. (3) 2,320 1b. (6) 3,2501b. 
b. 116 lb. (4) 55.8 mi. per hr. (7) 2,380 lb. 

(2) 8661b. (5) 65.1 mi. per hr. 

Chap. IV, page 51 j 

(1) a. 176 lb., 70.5 hp. (3) a. 259 lb., 73.0 hp. 

b. 150 lb., 60.3 hp. b. 259 lb., 92 hp. 

c. 148 lb., 59.6 hp. (4) a. 96.5 hp. 

(2) a. 299 lb., 100 hp. b. 77 A hp. 

b. 400 1b., 133 hp. (5) a. -1.5° 

c. 548 lb., 183 hp. b. 1.7° 

Chap. rV, page 54 

(1) b. 189 lb. (2) b. 87 lb. (3) b. 83 lb. 

c. 250 lb. c. 5° c. 3° 

d. 236 lb. d. 107 lb. 

Chap. IV, page 58 

(1) 40.3 hp. (5) 244 lb. (9) a. 11,900 lb. 

(2) 350 sq. ft. (6) 63.5 hp. b. 909 lb. 

(3) 194 lb. (7) 34.9 hp. (10) 118 ft. per sec. or 

(4) 5,760 lb. (8) 18.8 80.4 mi. per hr. 

Chap. IV, page 63 

(1) 1,480 lb. (6) a. 20.45 lb. per sq. ft. (8) 46.5 mi. per hr. 

(2) 52 mi. per hr. b. 15.1 lb. per sq. ft. (9) a. 131 mi. per hr. 

(3) 4° (7) a. 90.9 mi. per hr. 6. 152 mi. per hr. 

(4) a. 4.5° b. 7.8° b. 105.3 mi. per hr. (10) 98.6 sq. ft. 

(5) a. 110.91b. 
b. 88.4 lb. 

Chap. IV, page 64 

(1) 66.5 hp. (4) a. 14.5 hp. (6) o. 24.25 hp. 

(2) 102 mi. per hr. b. 18.4 hp. b. 30.6 hp. 

(3) 286 sq.ft. 



APPENDIX B 421 

Chap. IV, page 66 

(1) 98.7 hp. (3) 76.5 hp. 

(2) 36.8 hp. (4) 23.1 hp. 

Chap. IV, page 69 
(2) 7,890 ft-lb. (3) 13,400 ft-lb. (4) 3,430 ft-lb. (5) 2,220 lb. 

Chap. IV, page 73 

(1) 41 per cent from L.E. (3) a. 37 per cent from L.E. (4) -.008 

(2) a. -5,440 ft-lb. b. 36.9 per cent from L.E. (5) .4 

6. -.114 c. -.078 (6) -.05 



c. -.0326 



Chap. V, page 86 



(1) 1.4°, 0.016 (5) 3.8 hp. (9) a. 88 lb. 

(2) 2.42°, 0.0304 (6) 82 lb. b. 119 lb. 

(3) 125 lb. (7) 72 lb. c. 172 lb. 

(4) 35.4 hp. (8) 9.7 hp. (10) a. 268 lb. 6. 67 lb. 

Chap. V, page 91 

(1) a. 0.027 (4) o. -0.4° (7) a. -0.6° 
b. 25.9 b. 0.015 b. 0.028 

(2) a. 0.0295 (5) a. 8.7° (8) a. 1.9° 
b. 23.7 b. 0.079 6. 0.040 

(3) a. 0.036 (6) a. 7.3° ' (9) a. 7.3° 

b. 19.4 b. 0.036 b. 0.000142 

Chap. V, page 93 

(1) 141.6 lb. (5) 549 lb. (9) 295 lb. 

(2) 166 lb. (6) 258 lb. (10) 388 lb. 

(3) 48 lb. (7) 276 lb. 

(4) 145 lb. (8) 671.3 IK 

Chap. V, page 95 

(1) 0.0452 (3) 0.0308 (5) 0.0375 

(2) 0.0372 (4) 0.0324 

Chap. V, page 99 

(1) 0.0657 (3) 0.56 (6) 0.408 

(2) 0.0974 (4) 0.492 

Chap. VI, page 108 

(1) 4.47 (3) 3.9 (6) 2.93 

(2) 4.6 (4) 3.7 



422 



APPENDIX B 



(1) 0.384, 0.48 

(2) 0.416, 0.493 



(1) 3.58 

(2) 2.88 

(3) 3.98 

(4) 4.25 



(1) 29.4 ft. 

(2) 36.0 ft. 

(3) 38.7 ft. 

(4) 41.5 ft. 



(1) 
(2) 
(3) 
(4) 
(5) 



112 lb. 
114 lb. 

113 lb. 

111 lb. 

112 lb. 



(1) 254 lb. 



(1) 60.6 lb. 

(2) 42.0 lb. 



(1) 1.07 

(2) 1.29 

(3) 0.87 

(4) 0.90 



(4) 0.25 
(6) 0.293 



(9) 6.60 
(10) 6.29 



Chap. VI, page 111 

(3) a. 0.154 
b. 0.257 

Chap. VI, page 115 

(5) 3.7 

(6) 4.36 

(7) 4.50 

(8) 5.15 

Chap. VI, page 116 

(5) 99.0 ft. (9) 34.0 ft. 

(6) 23.1 ft. (10) 38.9 ft. 

(7) 44.5 ft. 

(8) 34.9 ft. 

Chap. VI, page 119 



(6) 1061b. 

(7) 4.5 ft. 
2.19 ft. 
7.13 ft. 
Ill lb. 


(8) 57.4 lb 

(9) 1661b. 
(10) 299 lb. 


Chap. VI, page 123 




(2) 255 lb. 


(3) 183 lb. 


Chap. VII, page 132 




(3) 3.145 lb. 

(4) 20.82 lb. 


(5) 1561b. 


Chap. IX, page 164 




(6) 1.46 

(6) 1.85 

(7) 1.55 

(8) 1.37 


(9) 1.22 
(10) 1.85 




76 per cent, 
73 per cent. 



(1) 27° 

(2) 24' 



Chap. IX, page 171 

(3) 24" 

(4) 18" 



(6) 2V 



APPENDIX B 



423 



(1) 16° 

(2) 2V 



(1) 53.7 mi. per hr. 

(2) 67.5 mi. per hr. 



(1) 5042 miles. 

(2) 231 mi. 



(1) 1280 ft. 

(2) 1373 ft. 



(2) 83.8 mi. per hr. 

(3) 71,600 ft. 



(2) 338.6 mi. per hr. 

(3) 313 mi. per hr. 



(1) 47.6 mi. per hr. 

(2) 46.6 mi. per hr. 



(1) 104 sq. ft. 

(2) 74.8 mi. per hr. 



(1) 122 mi. per hr. 

(2) 161.5 mi. per hr. 



(1) 14,400 ft. 

(2) 15,650 ft. 



(1) a. 7.4 min. 
b. 10.0 min. 



(1) 19,781 ft. 

(2) 20,250 ft. 



Chap. IX, page 174 

(3) 23° 

(4) 22° 

Chap. X, page 200 

(3) 80.7 mi. per hr. 

(4) 61.7 mi. per hr. 

Chap. X, page 203 

(3) 3307 mi. 

(4) 1341 mi. 



(5) 26^ 



(5) 76.3 mi. per hr. 



(5) 1023 mi. 



Chap. X, page 214 

(3) 664 ft. 

(4) 172 ft. 

Chap. X, page 219 

(4) 91.8 mi. per hr. 
(6) 418 mi. per hr. 

Chap. X, page 220 

(4) 287 mi. per hr. 

(5) 6.47 sq. ft. 

Chap. X, page 223 

(3) 82.7 mi. per hr. 

(4) 50.2 mi. per hr. 

Chap. X, page 225 

(3) 94.1 mi. per hr. 

(4) 53.0 mi. per hr. 

Chap. X, page 227 



(5) 587 ft. 



(6) 496 mi. per hr. 



(6) 53.3 mi. per hr. 



(5) 43.4 mi. per hr. 



(3) 270 hp. 

(4) 190 mi. per hr. 

Chap. XI, page 238 

(3) 7.6 hp. 

(4) 473 ft. per min. 

Chap. XI, page 239 

(2) 47.8 min. 

(3) 15.0 min. 

Chap. XI, page 240 

(3) 10,200 ft. 

(4) 6,202 ft. 



(5) 169.5 mi. per hr. 



(5) 500 ft. per min. 



(4) 23.0 min. 

(5) 10.8 min. 



(5) 8,000 ft. 



L24 


APPENDIX B 








Chap. Xll, page 254 






(1) a. 2,000 lb. 

b. 2,8301b. 

c. 672 ft. 


(2) 100 mi. per hr. 

(3) a. 64.6 hp. 
b. 70.5 hp. 


(4) 89.3°, 140,000 lb 

(5) a. 92 hp. 
6. 121 hp. 




Chap. Xil, page 259 






(1) 1.5 

(2) 4.9 


(3) 6.9 

(4) 12.2 

Chap. XII, page 261 


(5) 


17.9 


(1) 226 ft., 8.6 

(2) 554 ft., 8.8 


(3) 768 ft., 14.6 

(4) 96 ft., 4.4 


(5) 202 ft., 6.5 




Chap. XTT, page 263 






(1) 12.7 

(2) 15.3 


(3) 11.1 

(4) 17.7 

Chap. XIV, page 279 


(5) 


14.4 


(1) a. 7.9 ft. 

b. 0.9 ft. 

c. 0.9 ft. 

(2) a. 7.9 ft. 

b. 0.4 ft. 

c. 0.4 ft. 

(3) a. 9.2 ft. 

b. 0.9 ft. 

c. 0.6 ft. 


(4) a. 8.6 ft. 

b. 0.5 ft. 

c. 0.3 ft. 

(5) a. 8.8 ft. 

b. 0.8 ft. 

c. 0.6 ft. 

(6) a. 8.3 ft. 

b. 0.4 ft. 

c. 0.3 ft. 

Chap. XIV, page 283 


(7) 

(8) 

(9) 

(10) 


a. 7.3 ft. 

b. 0.9 ft. 

a. 7.4 ft. 

b. 0.4 ft. 

a. 8.4 ft. 

b. 0.8 ft. 

a. 7.9 ft. 

b. 0.4 ft. 


(1) a. -10.25° 
6. 9.75° 


(2) a. 0.50° 
b. -9.50° 

Chap. XIV, page 285 


(3) 
(4) 
(5) 


1.25° 

-6.00° 

-9.50° 


(1) 5.8° 

(2) -10.0° 


(3) -5.6° 

(4) 7.5° 

Chap. XIV, page 286 


(5) 


-1.5° 


(1) 7,000 ft-lb. 


(2) 1,890 ft-lb. 


(3) 16,020 ft-lb. 




Chap. XV 111, page 340 






(1) 55° 

(2) 200° 

(3) 270° 

(4) 328° 


(5) 17° 

(6) 345° 

(7) 155° 

(8) 27° 


(9) 279° 
(10) 119° 



APPENDIX B 425 

Chap. XX, page 371 

(1) 4,445 mi. (3) 5,305 mi. (6) 1,885 mi. 

(2) 2,387 mi. (4) 2,453 mi. 

Chap. XX, page 373 

(1) 106 mi. per hr., 101°, 11° right. (4) 85 mi. per hr., 245°, 20° left. 

(2) 172 mi. per hr., 26°, 4° right. (5) 99 mi. per hr., 131°, 11° right. 

(3) 133 mi. per hr., 332°, 8° left. 

Chap. XX, page 374 

(1) 15°, 137 mi. per hr. (4) 143°, 120 mi. per hr. 

(2) 35°, 155 mi. per hr. (5) 214°, 88 mi. per hr. 

(3) 343°, 76 mi. per hr. . 

Chap. XX, page 377 

(1) a. 34 mi. per hr. (2) a. 56 mi. per hr. (3) a. 48 mi. per hr. 
h. 60° h. 155° h. 260° 

c. 14° c. 64° c. 94° 

d. 105 mi. per hr. d. 170 mi. per hr. d. 187 mi. per hr. 

Chap. XXI, page 389 

(1) 1,8031b. (3) 1,169 1b. (6) 1,8291b. 

(2) 588 lb. (4) 1,083 lb. 

Chap. XXI, page 393 

(1) a. 498 lb. (2) a. 990 lb. (3) a. 60 lb. (4) a. 198 lb. 

6. 498 lb. h. 990 lb. 6. 188 lb. 6. 3,560 ft. 

c. 9901b. c. 251b. 

Chap. XXI, page 394 

(1) a. 1,423 lb. (2) a. 1,319 lb. 
6. 1,350 1b. h. 186 1b. 

c. 71b. c. 344 1b. 

d. 25,600 cu. ft. d. 965 lb. 

e. 1,3431b. e. 954 1b. 

Chap. XXI, page 397 

(1) a. 730 lb. (3) a. 488 lb. (5) a. 742 lb. 
h. 734 lb. 6. 524 lb. 6. 688 lb. 

(2) a. 1,935 lb. (4) a. 625 lb. 
h. 1,957 lb. 6. 702 lb. 

Chap. XXI, page 401 

(1) 6,099 ft. (2) a. 12,910 lb. (3) a. 12,435 lb. 

h. 13,463 lb. h. 905 lb. 



INDEX 



Absolute ceiling, 237 
Absolute coefficients, 33, 59 
Absolute humidity, 359 
Acceleration of gravity, standard, 2 
Accelerations, 311 

in dives, 261-262 

in loops, 263-264 

in turns, 258-260 
Accuracy, 323 
Adiabatic changes, 358 
Adjustable-pitch propellers, 162 
Advection fog, 365 
Aerodynamic center of pressure, 

70-72, 103 
Aerology, 357-365 
Aerostatics, 384-402 
Ailerons, 266, 268 

balanced, 270 

differential, 271 

floating, 269 

Frise, 271 

size of, 271, 272 
Air, constituents, 1 

standard, 1-3 
Air-cooled engines, 139 
Airfoils, choice of, 40, 41 

definition, 28 

stable, 68 

symmetrical, 35, 41, 68 

thick, 42 

unstable, 67, 68 
Airplane, angle of attack of, 265, 

axes, 265 
Airplanes, commercial, 318-321 

military, 315-318 
Airscrew, 148 
Airship, 384, 399-402 
Airspeed indicators, dynamic, 235, 
333-335 

true, 336 



Akron, 402 
Altimeters, 329-333 
Altitude, standard, 3 

effect on airplane performance, 

230-241 
effect on engine power, 145, 146 
effect on propeller performance, 

171-174 
effect on wing drag and power, 
48-50 
Altitude-pressure relation, 3, 331 
Aluminum alloys, 313, 314 
Amphibian, Seversky, 318, 319 
Aneroid altimeter, 329-332 
Angle of attack of airplane, 265 
of flat plate, 17 
of wings, 31 

correction for aspect ratio, 87-94 
effective, 87, 88 
for zero lift, 91, 97 
geometric, 78 

induced, 78, 81, 83, 87, 88, 96 
of minimum drag, 35 
variation with velocity, 47 
with infinite aspect ratio, 79, 98 
Angle of bank, 252 
Angle of climb, 207, 208 
Angle of crab, 373 
Angle of downwash, 282, 283 
Angle of drift, 373-376 
Angle of glide, 215-217, 257 
Angle of incidence, 54, 266 
Angle of stagger, 110 
Angle, blade, 161, 166 
Angle, dihedral, 278, 296 
Angle, tail, 282, 283 
Anti-cyclones, 363, 364 
Aperiodic compass, 341 
Apparent span, 115, 116 
Appendix, 384 
Applied loads, 311 



427 



428 



INDEX 



Archimedes, principle of, 386, 387 
Area, equivalent flat plate, 125, 126 

swept, 82 

wing, 30, 93, 105 
Ascension of balloons, 389-393 
Aspect ratio, 30, 75, 82, 87-96 

biplane, 104, 105, 112, 117 

definition, 30 

effect on performance, 228, 229 

effect on wing characteristics, 87 

equivalent monoplane, 104, 105, 
112, 117 

factor, 96-100, 284, 286 

infinite aspect ratio, 79, 81, 89 

monoplane, 87, 94 

of models, 75, 92, 121 
Astronomic avigation, 377, 378 
Atmosphere, standard, 1-7 
Atmospheric pressure, 1-3, 359, 360 
Atmospheric temperature, 1, 3, 358, 

359 
Attack, angle of (see angle of attack) 
Autogiro, 303-306 
Automatic slots, 300 
Autorotation, 296, 297 
Avigation, 366-383 

astronomic, 377, 378 

radio, 378, 383 
Avigation instruments, 329-356 
Axes, airplane, 265 



B 

Balance, lateral, 276, 294 

longitudinal, 280-294 
Balanced controls, 270, 271 
Ballonets, 399, 400 
Balloons, free, 358, 384-398 

kite, 384, 385 

sounding, 358 
Balsa, 125 
Banking, 251, 252 
Bank indicator, 347, 348 
Barometric pressure, standard, 2, 359 
Beacon, marker, 382 

radio, 379, 381 
Beam component, 100 
Bernoulh's principle, 149 



Biplane, lift distribution in a, 108- 
111, 279 
mean aerodynamic chord of a, 279 
Blade angle, 161, 166-169 
Blade element theory, 152-158 
BUnd landing, 381-383 
Body interference, 176 
Boeing pursuit airplane, 315, 316 
Boundary layer, 231 
Boyle's law, 2, 386, 387 
Brake horsepower, 144, 145 
Breene, R. G., 136 
Breguet formula, 203 
Bulb, thermometer, 327, 328 
Bungee, 273 
Burble, 23, 28, 31, 299 
Buys-Ballot's law, 363 



C-80 airfoil, 41, 59 
Cable, drag of, 130 
Camber, 29, 41 
Canard type airplane, 266 
Capacity altimeter, 332 
Captive balloon, 384, 385 
Cathedral, 295 
Ceiling, 237, 238 
Celestial avigation, 377, 378 
Center of gravity, 291 
Center of pressure, 36, 66-69 
Characteristics of airfoils, 34 

C-80, 41, 59 

Clark Y, 30, 35, 56, 62 

Gottingen 398, 40 

M-6, 42 

RAF-15, 43 

U.S.A.-35, 57 
Charles' law, 2, 386 
Charts, 337, 338, 366-371 
Chord, definition of, 29, 93 

mean aerodynamic, 93, 276-279 
Chord component, 100 
Cirrus clouds, 365 

Clark-Y airfoil, 30, 35, 56, 62, 90, 153 
Chmb, angle of, 207, 208 
Climb, rate of, at sea-level, 204-207 

rate of, at altitude, 236, 237 



INDEX 



429 



Climb, time to, 238, 240 
Cloth, 312 
Clouds, 364, 365 
Coefficients, absolute, 33, 59 

engineering, 60-64 

flat plate, 17 

fuselage drag, 133, 134 

induced drag, 80, 82, 83, 87, 89, 
117, 126 

lift, 33 

parasite drag, 125-133 

power-speed, 160-163 

profile drag, 81, 89 

propeller, 158 

tail moment, 280, 281, 285-294 

thrust, 158 

torque, 158 

wing moment, 70-73, 287-291 
Cold front, 362, 363 
Commerce, Dept. of, 76, 98, 100, 138, 

181, 182, 323, 358 
Commercial airplanes, 318-321 
Compass, induction, 342-344 

magnetic-needle, 336-341 

magneto, 344 

radio, 381 
Compensation, compass, 338 
Compression ratio, 144 
Construction, airplane, 314, 315 
Control surfaces, horizontal, 266-273 

vertical, 273 
Controllable pitch, 162 
Convection, 364 

Correction, aspect ratio, 87 -89, 92- 
102 

compass, 338 

drift, 372-377 
Course, correction of, 338 

great-circle, 371 

wind effect on, 372-377 
Covering, wing, 312 
Cowling, engine, 136 
Crab, angle of, 373 
Cumulus clouds, 365 
Curtiss Tanager, 269 
Curved plates, 18-21 
Curves, characteristics (see character- 
istics) 



Cyclogiro, 308-310 
Cyclones, 362, 363 
Cylinders, drag of, 129, 130 
Cylindrical plates, 18-21 



Dalton's law, 387 

Dead reckoning, 366 

Density, standard air, 2, 3, 386 

relative, 3, 61 
Dep control, 266 
Department of Agriculture, 357 
Department of Commerce, 76, 98, 

100, 181, 182, 298, 323, 358 
Deperdessin control, 266 
Descent of balloons, 393, 394 
Design load, 311 
Deviation, magnetic, 338, 339 
Diameter of propeller, 166, 170 
Diesel engines, 143 
Dijfferential ailerons, 271 
Dihedral, effect on M.A.C. of, 278 

effect on stability, 295 
Dip, magnetic, 337 
Directional gyro, 344-346 
Directional stability, 298 
Dirigibles, 384, 399-402 
Distance, great-circle, 371 
Distance, take-off, 208 
Distribution of lift across span, 75, 76 
Distribution of lift in biplane, 112- 

115, 279 
Dives, 214-217, 261, 262 
Diving moment, 280 
Dope, 312 

Downwash, 282, 283 
Drag, definition of, 16, 33 

equation of, 39 

fuselage, 133-134 

induced, of biplanes, 104-124 

induced, of monoplanes, 65-103 

parasite, 52, 125-137, 184, 229 

profile, 80, 88 

structural, 125 

tests of, 219 
Drift, 372-374 
Drift indicator, 354, 355 



430 



INDEX 



Drzewiecki, 152 
Duralumin, 373, 314 
Duration of flight, 54, 196-200 
Dynamic airspeed indicator, 235, 333 
Dynamic pressure, 12 

E 

Eddies, 22, 23, 31 
Edges, leading and trailing, 29 
Effect of altitude on airplane per- 
formance, 230-250 

on engine performance, 145, 146 

on propeller performance, 171-174 
Effect of ground, 38 
Effect of power loading, 226, 227 
Effect of span loading, 84, 86, 228, 229 
Effect of weight, 221 
Effect of wing loading, 36, 47, 223- 

225, 227 
Effective angle of attack, 79 
Effective dihedral, 296 
Effective pitch, 162 
Efficacy, 52 

Efficiency of engines, 144 
Efficiency of propellers, 152, 155, 158, 

193 
Efficiency of wings, 52 
Elasticity, Modulus of, 9 
Electric thermometers, 329 
Elevators, 266, 272, 273, 294 

free, 294 
Elliptic loading, 81, 82 
Energy of air in motion, 11, 151 
Engine cowHng, 136 
Engines, airplane, 138-146 
Engineering coefficients, 60-64, 87, 

120 
Envelope, balloon, 384, 386 
Equihbrium in dives, 261 
Equilibrium in level flight, 280-294 
Equilibrium, dynamic, 276 

static, 275 
Equivalent flat plate, 125, 126 
Equivalent monoplane aspect ratio, 

84, 85, 112, 117 
Equivalent monoplane span, 115 
Errors, compass, 337 



Ethylene glycol, 140 
Experimental mean pitch, 160 



Fabrics, airplane, 312 

Factors, aspect ratio, 96-100, 284 

fuselage drag, 133 

load, 311 

Munk's span, 117 

Prandtl's interference, 104-106, 110 
Fairchild monoplane, 319, 320 
Fairing, 125, 127 
Fin, 266, 273 
Fineness, 127, 131, 402 
Finger patch, 384 
Fittings, drag of, 132 
Flaps, 301, 302 
Flat plates, 12-17 
Flight, ghding, 214-218 

level at altitude, 48-50, 230-235 

level at sea-level, 36-39, 183-204 
Flight indicators, 351 
Floating ailerons, 269 
Flow, streamhne, 22, 23 

turbulent, 23 
Fog, 365 

Force, centrifugal, 251 
Forces on airplane, 281 
Fowler wing, 302 
Free balloon, 384-398 
Free elevator, 294 
Frise aileron, 271 
Fronts, 362, 363 
Froude, R. E., 148, 149 
Fuel consumption, 196-200, 235, 236 
Fuselage drag, 133 
Fuselage, monocoque, 315 



Gages, pressure, 327 
Gap, 278, 279 
Gap-chord ratio, 109 
Gap effect on aspect ratio, 104, 117 
Gap effect on induced drag, 106, 109 
Gases, balloon, 386 
Gear, landing, resistance of, 125, 135 
retractable, 136 



INDEX 



431 



Geared propellers, 174r-176 

Glauert, H., 83, 89 

GUde, spiral, 256-258 

Gliding angle, 21^219 

Gliding distance, 217 

Gottingen 429 airfoil, 304 

Gottingen 398 airfoil, 40 

Graf Zeppelin airship, 385, 402 

Graf Zeppelin instruments, 333, 344 

Great circles, 367, 369 

Greene, C., 136 

Ground effect, 38 

Ground speed, 372-377 

Gyro, directional, 344-346 

Gyro-horizon, 351-353 

Gyro-pilot, 355, 356 

Gyroplane, 307 

Gyroscopic horizon, 351-353 

H 

Handley-Page slots, 299 

Headings, compass, 339 

Hegenberger, A. F., 381 

Hehum, 2, 386 

Herman, F., 121 

High angle of attack, 32 

High-lift airfoils, 42 

Highs, 363, 364 

Horizon, gyroscopic, 351-353 

Horizontal tail surface, 266-273 

Horsepower, available, 193-195 

brake, 144 

definition, 14 

minimum, 46, 54, 65, 196-200, 235 

required in level flight at altitude, 
49 

required in level flight at sea-level, 
39, 183 

required in turns, 254, 255 
Humidity, 359 
Humidity Indicator, 358 
Hydrogen, 2, 386 



Ice-Warning Indicator, 329 
Impact Pressure, 12, 13 
Incidence, angle of, 54, 266 



Inclinometers, lateral, 347, 348 

longitudinal, 348-351 
Indicators, airspeed, 333-337 

bank, 347, 348 
Indicators, drift, 354, 355 

flight, 351 

humidity, 358 

ice-warning, 329 

pitch, 348-351 

rate-of-climb, 353, 354 

turn, 346, 347 
Induced angle of attack, 79, 81, 83 
Induced drag. Monoplane, 80, 81, 83 
84, 87, 94 

Biplane, 107 
Induction compass, 342, 344 
Infinite aspect ratio, 79, 81, 89 
Isobars, 360, 363, 364 
Isotherms, 359 
Instruments, 322-356 
Interference, body, 176 
Interference factor, Prandtl's, 104, 

105, 106, 110 
Inverted flight, 262 
Inward flow, 77 



Joukowski streamline, 127, 128 
Joule's law, 387 

K 

Kinematic viscosity, 8 
Kinetic energy, 11, 151 
Kite balloons, 384, 385 



Lachmann, G., 299 

Lag, 323 

Laminar flow, 22, 23 

Landing-gear, resistance of, 125, 135 

retractable, 136 
Landing speed, 38 
Lateral control, 267 
Lateral inclinometer, 347, 348 
Lateral stability, 294 
Latitude, 367 



432 



INDEX 



LID ratio 52, 56, 92, 93, 122 

Leading edge, 29 

Lenticular wire (see streamline wire), 

130 
Level flight at sea-level, 36, 39, 183-201 

at altitude, 48-50, 230-236 
Lift distribution along chord, 32 

along span, 75-77, 81 
Lift distribution in biplane, 108, 279 
Lift of balloons, 387-398 
Lift of inclined plates, 16 
Lift of wings, 33 
Liquid-cooled engines, 139 
Load factor, 256-264, 311 

in dives, 262 

in loops, 263, 264 

in turns, 256-261 
Loading, power, 226, 227 
Loading, span, 84, 86, 228, 229 
Loading, wing, 36, 47, 223-225, 227 
Locked elevator, 281-294 
Lockheed airplane, 319, 320 
Longitude, 367 

Longitudinal balance, 280, 281 
Longitudinal inclinometer, 348-351 
Loops, 263-264 
Los Angeles, 402 
Low angle of attack, 32 
Low-pressure tires, 135 
Lows, 361-363 

M 
Magnesium, 313 
Magnetic compass, 336-344 
Magnetic variation, 337 
Magneto compass, 344 
Maneuverability, 275 
Map projections, gnomonic or great- 
circle, 369 

mercator, 368 

polyconic, 368 
Maps, 366-371 
Maps, weather, 361-363 
Materials, 311-314 
Maximum ordinate, 29, 41 
Maximum speed at sea-level, 196 

at altitude, 235 
Mean aerodynamic chord, 276-279 



Mean effective pressure, 144 
Medium angle of attack, 22 
Mercator projection, 368 
Meridians, 368 
Metal propellers, 181 
Metals, duralumin, 313, 324 

steel, 313, 314 
Meteorograph, 358 
Meterology, 357-365 
Metric units, 59, 60 
Micarta propellers, 180 
MiUibar, 360 
Military airplanes, 315 
Minimum drag, 35 
Minimum fuel consumption, 54, 196- 

200 
Minimum radius of turn, 254-256 
Minimum speed at sea-level, 38, 183, 

186, 191 
Minimum speed at altitude, 235 
Model tests, 133 
Models, aspect ratio of, 75 

correction for biplanes, 121, 122 

correction for monoplanes, 92 
Moment coefficient, 66-69, 289-293 
Moment, diving, 280 

pitching, 265, 280 

rolling, 265, 271, 272 

stalling, 265, 280 

tail, 282, 285, 286 

thrust, 280, 281 

wing, 70-73, 287-289 
Momentum, 11 

Momentum propeller theory, 148-152 
Monocoque, 315 
Monocoupe airplane, 321 
Monoplane, correction for aspect 
ratio, 87-94 

mean aerodynamic chord, 276-279 
Monoplane span, equivalent, 115 
Munk, M., 81 
Munk's span factor, 115 
Mutual Interference of biplanes, 104 



N 

N.A.C.A., 2, 29, 110, 278, 331 
N.A.C.A. airfoils, 44, 45 



INDEX 



433 



N.A.C.A. engine cowling, 136 
N.A.C.A. wind-tunnel, 26 
Navigation, 366-383 
Newton's laws, 11 
Nimbus clouds, 365 

O 

Observation airplane, 315, 316 
Optimum gliding angle, 216, 217 
Oswald method, 243-250 
Outward flow, 77 



Packard Diesel engine, 143 
Paddle-wheel airplane, 308-310 
Parallelogram of velocities, 372 
Parasite drag, 52, 125-137, 184, 229 
Partial inflation of balloons, 389-393 
Pascal's law, 387 

Performance, airplane at sea-level, 
183-229 

airplane at altitude, 230-250 

effect of power loading on, 226, 
227 

effect of weight on, 221-223 
y effect of wing-loading on, 223-225 
Performance, engine at sea-level, 144, 
145 

engine at altitude, 145, 146 
Performance, propeller, 171-180 

at altitude, 171, 172 

at sea-level, 176, 177 
Pitch, adjustable, 162 

controllable, 162 

effective, 162 

experimental mean, 160 

geometric, 161 

variable, 162 
Pitch Indicator, 348-351 
Pitch ratio, 161, 162, 166 
Pitching moment, 265, 280 
Pitot-static tube, 235, 333-335 
Pitot-Venturi tube, 335 
Planform, taper in, 46, 277 
Plates, curved, 18-21 
Plates, flat, 12-17 



Plate area, equivalent flat, 125, 126 
Polar curves, 55, 90, 136, 137, 218 
Poly conic projection, 367, 368 
Power, definition, 14 
Power, available at sea-level, 193- 
196 

available at altitude, 232-235 

required by wings, absolute co- 
efficients, 39 
Power, required by wings, engineering 
coefficients, 64 

required by wings at altitude, 48, 
49, 50, 64 

required by airplane at sea-level, 
183-192 

required by airplane at altitude, 
230-232 
Power-loading, effect on performance, 

226, 227, 228 
Power-speed coefficient, 160, 163 
Prandtl, L., 104-106, 110 
Pressure, atmospheric, 1-3, 359, 360 
Pressure, dynamic, 11 
Pressure, center of, 66-69 

distribution along chord, 32 

distribution along span, 75-77, 81 
Pressure gages, 327 
Pressure on flat plates, 13 
Profile drag, 80, 81, 89 
Projections, map, 366-371 
Propeller, adjustable, 162 

blade angle of, 161, 166, 167 

construction, 180-181 

controllable, 162 

effect on tail, 273 

efficiency, 152, 155, 179, 180, 192 

geared, 174-176 

momentum theory, 148, 149 

pitch, 176-179 

power-speed coefficient, 160, 163- 
167, 170, 176 

thrust of, 53, 150, 154^159 

torque of, 156-158 

variable pitch, 162, 163 
Pull-out, stresses in, 262 
Purity, 398 

Pursuit airplane, 315, 316 
Pusher type airplane, 139 



434 



INDEX 



R 

Radial engines, 141, 142 

Radiation fog, 365 

Radiation, solar, 358 

Radiator, 140 

Radio avigation, 378-385 

Radio beacon, 379-381 

Radio compass, 381 

Radio landing, 381-383 

Radius of turn, 252, 254-256 

RAF-6, 153 

RAF-15 airfoil characteristics, 43 

Range, speed for maximum, 201-204 

Rate of climb at altitude, 236 

at sea-level, 204-207 
Rate-of-climb indicators, 353, 354 
Ratio, aspect, 30, 75, 82, 87, 89, 100 

fineness, 127, 131, 402 

gap /chord, 109 

gap /span, 104-106 

L/D, 52, 56,92,93, 121, 186 

pitch, 161, 162, 166 

slenderness, 402 
Recovery from dive, 262 
Relative humidity, 359 
Relative wind, 31, 78 
Relative wind, true, 78, 80 
Resistance of cables, 130 

of cylinders, 129, 130 

of fittings, 132 

of fuselage, 133, 134 

of landing-gear, 125, 135, 136 

of wheels, 135 
Resultant force on wings, 33 
Retractable landing gear, 136 
Reynolds' number, 24-26 
Ring cowling, 136 
Rolling moment, 265, 271, 272 
Rotary engine, 142 
Rotation, 265 
Rudder, 266, 273 

S 

Scale effect, 26 

Semi-eUiptic distribution, 81, 82 

Sensitivity, 323 

Seversky airplane, 319, 320 



Sextant observations, 377, 378 

Shielding, radio, 378 

Sikorsky airplane, 318, 319 

Similar flow, 25 

Skidding, 251, 252 

Skin friction, 23, 127 

Slenderness ratio, 402 

SHp, 162 

Slipping, 252 

Slipstream, effect on tail, 273, 274 

Slots, 299, 300 

Sound, velocity of, 10 

Span, apparent, 112, 115, 116 

definition, 30, 93 

equivalent monoplane, 112, 115 
116 
Span factor, Munk's, 115-117 
Span loading, 84-86, 228, 229 
Speed, ground, 199, 200, 372-377 

in dives, 218 

landing, 38 

maximum at altitude, 235 

maximum at sea-level, 196 

miminum at altitude, 231 

minimum at sea-level, 38, 183, 186, 
191 

stalling, 38, 183, 186, 191 

take-off, 38 

terminal, 218 
Speed range, 299 
Speed versus angle of attack, 47 
Spin, 272, 297, 298 
Spiral glide, 256-258 
Spoilers, 268, 269 
Stabihty, directional, 276, 295 

dynamic, 276 

lateral 276, 294, 295 

longitudinal, 276, 282-294 

static, 275 
Stabilizer, 266, 273 
Stagger, 278, 279 
Stainless steel, 314 
Stall, 268, 272 
Stalling moment, 265 
Stalling speed, 38, 183, 186, 191 
Star, wind, 374-377 
Static thrust, 210 
Steam engine, 141 



INDEX 



435 



steel, 312-314 
Stick control, 266 
Stratosphere, 4, 359 
Stratospheric flight, 141, 241-243 
Stratus clouds, 366 
Streamline flow, 22 
Streamline struts, 130, 131 
Streamline wire, 130 
Streamlining, 127, 129 
Strip maps, 339, 368 
^ Structural drag, 125 
Structure, 314 

Struts, resistance of, 125, 130, 131 
Surface, control, balance, 270 

horizontal, 266, 272, 273 

vertical, 266, 273 
Suspension band, 384 
Sweepback, effect on M.A.C., 277 
Swept area, 82 
Symmetrical airfoils, 35, 41, 68 



Tabs, 273 

Tachometer, centrifugal, 326, 327 

chronometric, 324-326 
Tail angle, 283, 284 
Tail load, 280 
Tail moment, 285 
Take-off distance, 208 
Take-off speed, 38 
Tanager airplane, 269 
Taper, effect on induced drag, 94, 96 

effect on M.A.C., 277 
Taper, standard, 46 
Temperature, atmospheric, 2, 358, 

359 
Terminal speed, 218 
Thermometer, 327, 328 
Thrust, 53, 150, 154, 210 
Time to climb, 238-240 
Tip, wing, vortices, 77, 78, 104 
Torque, 156-158 
Townend ring, 136 
Track, 372-377 
Tractor airplane, 139 
Trailing edge, 29 
Triangle of velocity, 372-374 
Troposphere, 4 



True airspeed indicator, 336 
Tube, Pitot-static, 235, 333-335 

Pitot-Venturi, 335 
Tubing, resistance of, 129 
Turbulent flow, 23, 129 
Turns, 251-259, 265 
Turn indicator, 346, 347 
Twist, 263, 274 
Two-dimensional flow, 81 



Variable density wind tunnel, 26> 
103 

Variable pitch propeller, 162 

Variation, magnetic, 337 

Velocity in dives, 216-218 

Velocity of sound, 10 

Velocity triangle, 372-374 

Velocity variation with angle of at- 
tack, 47 

Vertical banks, 259-261 

Vertical tail surfaces, 266, 273 

Verville, F., 136 

Viscosity of air, 7, 8, 23 

Visibflity, 109 

Vortices, 77, 78, 104 

W 

Warm front, 361 

Warp, 263, 274 

Washin, 274 

Washout, 274 

Weather Bureau, 357 

Weather maps, 361-363 

Weight, effect on performance, 221- 

223 
Weight, specific, of air, 3, 386 

specific, of gases, 386 
Wheel control, 266 
Wheels, resistance of, 135 
Wind, aloft, 364 

atmospheric, 360, 361 

effect on fuel consumption, 199, 200 

relative, 31, 78, 80 
Wind star, 374-377 
Wind tunnel, 26, 103 



436 



INDEX 



Wing, definition of, 28 

Wing [loading, 36, 37, 47, 48, 110, 

223-225 
Wing moment coefl5cient, 70-73, 

289-294 
Wing structure, 52 
Wing tip vortices, 77, 78, 104 
Wire, resistance of, 129, 130 
Wood, 312 

Wooden propeller, 180 
Work, 14 



Wright biplane, 267, 295 
Wright wing, 28 



Yaw, 265 



Zap flaps, 302 
Zeppelin, Graf, 385, 402 
Zeppelin, Graf, instruments, 333, 344 
Zoom, 208 




AUG 7 



#fi^ 



138159 



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