Aerodynamic investigation of a multiplace fighter

s%, INST.

by 17 SEP 1935

OSRARl~

Leslie E. Ri chardeon,Lieut.Jjg.

_(CC)

U.S.N.

Submitted in Partial Fulfillment of the Requirements for the degree of Master of Science from the Massachusetts Institute of Technology,

1935

Course XIII-A

Signature of

AuthoQ-Signature

redacted

Certification of Department _. Chairman of Graduate Committee Professor in Charge

-Head of the Department

Cambridge, Mass. Ilay 16, 19:35.

Professor George W. Swett,

Secretary of the Faculty,

Ylass. Inst. of Technology,

Cambridge, Mass.

Dear Sir:

I submit herewith a thesis entitled

"AERODYNAYIIC IINVESTIGATION OF A ITULTIPLACE FIGHTER" in partial fulfillment of the reouirements of the degree of Master of Science from the Massachusetts Institute of Technology.

Respectfully,

Signature redacted

( ' Leslie E. Richardson

The author wishes to express his

ap-preciation to Proflessor 0. C. Koppen of the Aeronautical Engineering Course for his

interest in, and for the many valuable

sug-gestions given during, the preparation of

this thesis. He is also indebted to Professor Shatswell Ober for Counsel and assistance.

TABLE OF CONTENTS

Page Purpose . . . . . . . . . . . . . . . . . .

Weight estimates . . . . ... .... Dimensions, areas, etc.

Estimate of horsepower required . . .

Take off . . . ... Propeller selection . ... 1 2 4 5 6 Horsepower calculations ... Longitudinal stability . . . . . . . Lateral stability . . . . . . . . . . Velocity in vertical dive . . . . . . Lateral stability in vertical dive.

Aileron rolling moment and angular velocity of roll . . . . . . Rate of roll due to rudder . .

Design of elevator . . . . . . . . . Trimming of tabs on elevator in verti

7 12 15 18 19 . . . 21 . . .

23

. . . 24 cal dive 26

Rudder tab angle necessary at spped of

best climb with one engine cut out . . . . .

Discussion and conclusions . . . . . . . . . 27 28

PURPOSE

The purpose of this thesis is to in-vestigate certain aerodynamic characteristics of a low wing fighter monoplane, multiplace (3),

with two engines.

In the following investigation it is not to be assumed that the characteristics as determined are deemed to be satisfactory in every instance although some revision was made from time to time to try and improve certain features.

For the time available, it was considered best to determine, in a general way, what might be

expected from an airplane designed to fulfill certain requirements rather than to be constantly revising the design to achieve the best results in certain particular characteristics.

The requirements referred to were

ar-bitrarily chosen to include a landing speed of sixty miles per hour in still air and an overall wing span of less than sixty feet. These would be requisite for operation aboard a carrier.

IVEIGHT ESTIMATES

Useful Load

Crew of 3 with parachutes

Gasoline (3 x 2 x 550 x .50)

Oil (31 x 2 x 550 x .035 + 8 x 7.75)

Ordnance

3 machine guns with mounts

2400 rounds ammunition bombs and bomb rack Equipment

Radio (two way)

Fire extinguishers (3)

Safety Belts & Live Preservers Total useful load

Weight Empty

1. Structure

A. Wings complete (2 x 533) B. Tqil group complete (110 x 1) C. Body group Fuselage (43 x 12 x 1.5) Landing Gear Tail Skid 600 1924 197 192 726 650 50 24 21 4384 lbs. 1066 110 774 400 20

A. Engine group 2370

& Motors dry 1950

2 Propellers and hubs 220

Exhaust system 40

Controls ₃₀

B. Fuel System (Z21 x 1.0) 321

C. Lubricating System 50

D. Starting and ignition

2 Starters 60

2 Generators 40

Battery and box 50

III. Fixed Equipment and miscellaneous

Instrument board & instruments 30

Surface controls 60

Flooring ₃₀

Firewall ₁₀

Miscellaneous ₂₇₅

Total weight empty 5536 + total useful load 4384 Total weight 9920

-4-Dimensions, areas, etc.

Length = 516" = 431.0 Span = 678" = 56'.5

For landing speed of 60 m.p.h at sea level with C0 (flaps) = 2.0, (flap angle 450)

Wing area, S 2.0 x 9.3 - 533 sq. ft.

(56.5)2_

Aspect ratio, wing, 533 6.00

Mean chord, wing, = 113"' = 9.42

L = length from c.g to rudder post = 318" = 26t.5

Vertical tail surface = 31.9 sq. ft. Horizontal tail surface = 75.0 sq. ft.

9920 = 18.6

S 533

- 9920 = 9.01 H.P. ₁₁₀₀

Vertical tail surface required : (Diehl)

S"= 6 x 10-5 X 9920 x 56.5 - 31.9 sq. ft. 0.0015 (26.5);

Estimate of Horsepower required

Assume Vmax = 210 m.p.h at sea level

CL = = 9920

Sq _{53 x 113.5} = 0.164 Cdprof = 0.0105

Estimated parasite drag coefficient = 18.6 X .008 = .0149

Cdi = (0.164)2 = .0014 10 d-x 6.0 H.P. required = 210 x 533 x 11Z.5 x .0268 375 = 908 H.P. available = 1100 x .80 880

Vmax at sea level will, therefore, be about 210 m.p.h

Take off

Take off distance =

X 200 x 18.6 - 3720 -75-1 7.43 8 .9 200W 75 _-1 P 501 ft. (Warmer)

L =

Cj

,

e

s v

2 9920 _{= 12.40 = q} 1.5 x 533 Wind speed Take off speed

.. Take off speed = 69.0 m.p.h = 3 = 0.435

69

From Diehlits chart, Ground run into wind = 0.32 Ground wind into calm

-6-Propeller Selection

The selection of a satisfactory propeller to absorb the power from an engine supercharged to 20000 ft., with a plane velocity of about 260 m.p.h is ex-tremely difficult. Little published information is available for propellers operating at the resulting high values of CS and Q

Engine delivers 550 h.p. @ 2200 r.p.m. at 20000 ft.

A geared propeller must be used to avoid tip speed

losses. For 3 blades:; .70 x 550 = 385 gearing 3:2 gives 1467 R.P.M. Assume Vmax = 260 m.p.h at 20000ft. =S 0.638 x 260 0.638 x 260 =24 C 85 - /5 X (1467)2/5 3.73 X 18.54 _2.40

Select a propeller with - 370 giving a~n- = 1.50 D - 88 x 260 10.4 ft.

1467 X 1.5

NOTE* For the charts used (unpublished N.A.C.A.) a

correction is necessary to convert the propeller efficiencies given to propulsive efficiencies. This correction was worked out to be

.89 - .76 = 14.6%

.89

Diameter correction

₌

.0194 -m

The propeller selected was a standard peller (Navy drawing 4412) which, for high speed pro-pellers has a low constant geometrical pitch. It would be better to have a constant geometrical pitch

of say about 250 and then turn the propeller from that

point.

Horsepower available at sea level

By the use of a "constant speed" controllable

pitch propeller, the r.p.m remains constant.

CS . Ei'V 2/ .638 V -004 Cs =(385)-.//D (1467) 3.29 x 18.54 .01045 V V 88V _{- .00577 V} ND 1467 X 10.4 C v .01045 .00577 1 Cs (1) 2.20 (2) 1.90 (3) 1.60 (4) 1.30 (5) 0.90 ST 1.216 1.050 0.884 0.719 0.497 28.7 26.5 24.5 22.3 21.5 Propeller Efficiency .872 .899 .908 .890 .770 Propulsive Corrected(for Efficiency n dia. amd 3 .745 .741 blades) .768 .764 .775 .770 .760 .756 .658 .657

-8-(1) V a JND = 1.216 x 1467 x 10.4

₌

210.5 T.H.Pa e 1100XO.741:815 88 88 (2) V

=

173.1 x 1.050 = 181.9 T.H.Pa = 1100-.764=841 (3) V = 173.1 x 0.884 153.1 T.H.Pa = 11OOxO.770.848 (4) V 173.1 x 0.719 124.6 T.H.Pa - 11OOxO.756=831 (5) V 173.1 x 0.497 86.1 T.H.Pa 1100x0.657=723

Horsepower available at 20000 feet

Cs .638 x V .638 V 0.00922 V (38511/5(1467)2/5 3.73 x 18.54 J - 88 V - 0.00577 V 1467 x 10.4

CS

J___ ____ 0.00922 0.00577 00

J

C 1.598 Propeller (3 efficiency Propulsive efficiency Corrected n 2.40 2.10 1.80 1.50 1.10 1.502 1.314 1.128 0.939 0.689 V = 173.1 x V 173.1 x V 173.1 x V = 173.1 x V = 173.1 x 37.0 35.3 33.5 32.0 31.6 1.502 1.314 1.128 0.939 0.689 0.880 0.892 0.880 0.848 0.650 0.752 0.762 0.752 0.724 0.555 = 260.0 T.H.Pa= 1100 x = 227.5 T.H.Pa * 1100 x 195.1 T.H.Pa 1100 x 162.7 T.H.Pa: 1100 x = 119.2 T.H.Pa 1100 x 0.748 0.758 0.748 0.721 0.557 0.748 - 823 0.758 835 0.748 . 823 0.721 = 793 0.557 613 C J (1) (2) (3) (4) (5) (1) (2) (3) (4) (5) .. N.W

Horseiower required at sea level = 125 (I)2

1

V m.p.h Vm.ph x 1.47 - () 2 X550 b 1 X ZVmp.h (9920)2 X 1 - 10290 H.P-ind 56.5 X3 ~ v V H.Pind 212 48.5 200 51.4 180 57.1 160 64.3 140 73.5 120 85.8 100 102.9 80 128.6 CL= W 18.6 Sq

_s

CLU, at Vmax Parasite dra = 18.6 = 0.161 115.4

g coefficient at Vmax _max - 18.6 X 0.0076 ₁₀ = 0.0142 H. P - ind

-10-CL CD prof 0.161 0.0085 0.181 0.0085 0.224 0.0087 0.282 0.0089 0.372 0.0090 0.503 0.0100 0.727 0.0110 1.128 0.0167 H 212 x 533 x 115.4 x 0.0227 375 200 x 533 x 103.0 x 0.0227 . 375 180 x 533 x 83.1 x 0.0230 375 160 x 533 x 66.0 x 0.0235 375 140 x 533 x 50.0 x 0.0238 375 120 x 533 x 37.0 x 0.0249 375 100 x 533 x 25.6 x 0.0262 375 80 x 533 x 16.5 x 0.0342 375 .P.required

par+ prof. H.P.req.

791 840 665 716 490 353 237 157 95 64 547 417 311 243 198 193 V 212 200 180 160 140 120 100 80 q 115.4 103.0 83.1 66.0 50.0 37.0 25.6 16.5 B 1.00 1.00 1.01 1.03 1.04 1.05 1.07 1.23 CD par 0.0142 0.0142 0.0143 0.0146 0.0148 0.0149 0.0152 0.0175 prof. V 212 200 180 160 140 120 100 80 CDpar + 0.0227 0.0227 0.0230 0..0235 0.0238 0.0249 0.0262 0.0342

Horsepower required at 20000 feet 0' = 0.533 212 200 180 160 140 120 100 80 X 1.37 I! IT tl it Vt

1

0.730 yO~.5338 0.730 290.6 _{H. P.reCL} 274.0 " 246.5 " 219.2 " 191.8 " 164.5 "V 137.0 " 109.7 " = 840 X = 716 X = 547 X = 417 X = 311 X = 243 X = 198 X = 193 X = 1.37 1.37 it IT " " 1150 981 750 571 426 333 272 264 V = V = V = V = V = V = v = v =

-12-Longitudinal Stability*

T

2 U h 6. 00

At maximum speed, horizontal flight, and 20000f t. altitude 9920 32.2 X 0.0006Z4 X 533 X 377 Mean chord = 9.42 C-L = - 9920 _{= - 0.206} 0.000634 X 533 X (Z77)2 q = 0.000634 X (Z77)2 = 90.20 CD = - (0.0227 - C) = - 0.0227 - (-0.206)2 0.0250 1TR 6 T Zu =2 (-.206) =-0.412 d CL=7 -5.70 xlTx 6 = -107.6 do W 6TT - (-5.70) 24.58 - 4.38 L = - 26.5 a = + 35" d = + 15" h = 108" 18.6 0.000634 x 32.2 x 26.5 =34.4 dCD _{-_2 (-0.206) (-4.38)} = - 0.0956 61IT = d = d C. - CL = - 0.0956 - (-0.206) + 0.1104 CD = Ct at o< = 0

dCD

= d Ct at CL= 0 X= 2 ( -. 0250) - 0.62 (0.0250) = -. 0500 - 0.0155 X(= - 0.0655

* For definition of symbols used, see notes on "Airplane

Stability and Control" by Professor 0.C.Koppen.

703 (dCL _{- 5.7 X 77 X 4 3.92} doc Rq-4 4

r

+ 5.7 r - 2 X 533 2 X (- 0.206) (-0.0250) 1.62 - 0.62

]

gX (10.4) -0.206 r - 1 = + 0.0785 r = 1.0785 (f = 1.038

Q

=.95 -1 = 0.038 75 -0.141 S 533

Assume tail efficiency of .85

mq (3.92) (0.141)

(0.098)

_{+ 0.95 x 0.0383} mq = - 6.16 Udr -2 x 533 (-0.206) (-0.0250 0.62 -

3.24

du (104)-0206 0.6 L.24 = 0.205 U .0989 0.06 -(-0.206)1 3-0.9.)-.0205 x 1 2 ( 25) M"770989 13 13-265"1U.0946

(assume Ct = CD for _{very small angles of.}

₎

mu, - 0.0508 9.42 x 0.95(-0.205) 0.0989 -26.5 1.075 MU - 0.0331 mw .1 9.42 1.25 0._{0 9}891-4.38 x-26.5(.31-.25)-o.ll4 --26.5-3.92(0.141)(0.85) _( -4.3 8 m =10.1 + 0.094 + 0.005 - 0.298J mw = 10.1 x (-0.199) = - 2.01 10 = 0

-14-P = - (-0.o06) (34.4) (-2.01) (-0.412) - (-0.206) (34.4) (-0.33?(+4.38) E = 5.88 - 1.03 = 4.85 C = (-0.0655)(-4.38)+(-0.0655 - 4.38) (-6.16)-34.4(-2.0l)-(-0.412) ,(O.1104) C = + 0.287 + 27.27 + 69.24 + 0.04 C = 96.94 4.85 96.94 2 1' 0. 0500 2 7r = 28.0 units 0.224 period = 28.0 x 2.418 = 67.6 seconds p

Lateral Stability

At maximum speed, horizontal flight, and 20000 ft. altitude Unit of length =

I

= 28.25 18.6 0.000634 x 32.2 x 28.25 T = 2.418 (same as long.) = 32.4

For oscillation in roll and yaw = 0 CC E ( --1) .( T- ) -4 B D B;&-C period of oscillation,= 2Vf

assume "a" = 0.06 and "C" = 0.115 (weight distribution factors)

W

0 U= ₀

i =

-

CL]

x

-0.0956 -

(0.206)]

L do KC x

1

4 x 0.115

"k" is taken as 4 for low angles of attack (elliptical distribution) P =- 0.1104 x 1 -0.240

4 x 0.115

1 =

cz.

X

1

=

-

4.38 x _{= --} 18.25

p d< Ka 4 x 0.06

-16--2 CL = - 2(-0.206) ak 4 x 0.06 assume _y S it _31.9 = 533 = + 1.718 = - 0.4 -= 0.060 (dCL = - 5.7 x

i7

x 1.5 26.9 = - 2.58 e)1.5 1.5 Ir + 5.7 10.4 S=( " 2.58 X 2(-26.5) x 0.06 ₊ 1.262 56.5 x 0.115 nr (tail) X (.2 - 2.58 x _0.06 x( -53.0 2 = - 1.183 0.115 56.5 nr (wing) Z CD X 1 - 2(-0.025) 0 -1088 KC 4 x 0.115 .'. nr (total) = - 1.292

Now to make E = 0, i.e. Ulane is neutral as regards spiral

stability.

xlr IV K nr

(1.262) (1.718) = lv (-1.292) ny

A =0 B = - (-18.25 - 1.292 - 0.400) + 19.94 C = 18.25 (- 0.400 - 1.29) + (-0.4) (-1.29) + 32.4xl.262-(-0.24) (1.72 C = + 30.84 + 0.52 + 40.85 + 0.41 = + 72.62 D = 32.4(-1.679()-0.240-0.206)-(-18.25) [-(0.4)(-l.292)+32.4(-1.262 +(-0 .240) (1.718) (-a.4) D = + 24.2 + 18.25(41.37) + .16 D = 779.4 I E= 0 2 B or D B D - C2 = 19.94 x 779.4 - (72.62)2 = 15660 - 5270 = 10390 1 C2 5 2 70 -264 20 20

.*. the factors are not sufficiently accurate Taking the equation

Ad4 + B d3 + 0d2 + Dd + E = 0, and setting E =0, we get Ad3 + Bd2 + Cd + D = 0

d3 + 19-.94 d2 + 72.62d + 779.4 = 0

one real root about _{- 18, i.e. approximately 1P}

a f(d)

-18 + 96 by graph, one real - 19 - 259 root = - 18.3

-18-d = - 18.3 factoring, (d + 18.3) (dZ + 1.64d + 42.55) = 0 -1.64 * V2.64 - 4 (42.55) :. d = 2 d - 0.82 \'- 41.88 2 IT _17 _eod period-6.4 6 x 2.418 = 2.37 seconds

time to damp to 1/2 period = 9.69 x 2.418

0.82

= 2.03 seconds

Velocity in a vertical drive (i.e. path is vertical CL₁ = 0 CD prof CDpar CD flaps CD total 9920 = s2 0<= -100 = 0.0100 = 0.0142 = 0.0400 = 0.0642 0 (split flaps, 0.25c, tapered, tip section,

removed, flap angle 45*)

C D 2 x 9920 533 x 0.0642 x 0.00176 tat 10000ft. V2 = 329000 V = 574 f.p.s = 391 m.p.h

Lateral Stability in Vertical Dive T m 9920 32.2 x 0.0000 x 533 x 574 2 18.6 32.2 x 0.000878 x 28.25 = 23.25 For the quartic eqation:

CD replaces CL in solving fofthe coefficients CD = - 0.0642

w

₀ V U @ = 0 dCD- ₀ np =0 (CL = 0) "af = 0.06 19 p= -18.25

Xas

before) = -0.4 (assumed) (dCL) d r=l.5 nr wing = -2.58 = 2(-0.0642) 4 x 0.115 nv = + 1.262 (as before) = - 0.279 = - 1.183 (as before) = 1.462 1.679 = 1.145 "c It = 0.115 :r tail nrtotal 1_{v =}

- -20-A =-1 B = -

(18.25

- 1.462 - 0.4) = + 20.11 C = - 18.25 (-0.4 - 1.462) + (-0.4) (-1.462) + 23.25 (1.262) C = 34.00 + 0.58 + 29.40 = + 63.98 D = _{23.25(-1.679)(-0.0642)-(-18.25) -0.4(-1.462) + 23.25(1.262)7} D = + 2.50 + 18.25 x 29.98 = 549.7 E = 23.25 (-0.0642) - (-1.679) (-1.462)] E =+ 3.67

The factors are not sufficiently accurate

sinceE

ft

1 B or D and B D - C2

d4 + 20.11 d3 + 63.98 d2 + 549.7 d + 3.67 = 0

To facilitate solving this equation, E can be

assumed = 0, without appreciable error. Actually, this root equals (aDproximately) - 0.0067 instead of the assumed 0.

Now, d3 + 20.11 d2 _{+ 63.98 d + 549.7 = 0}

d f(d) -18 + 85 - 18.2 + 16 - 18.3 - 15

Taking d = - 18.3 as one root and factoring, (d + 18.3) (d2 _{+ 1.81d + 30.0) = 0} d = - 1.81

*

d = - 0.91

V_3.28

- 4 (30) 2 V- 29.18 period = 2 4 x 1.145 5.40 = 1.33 seconds

time to damp to 12= 0.69 x 1.145 = _{0.87 seconds}

0.91

The period is very short, but because of the excellent damping, is satisfactory.

Aileron rolling moment and angular velocity of roll at

Vma,. horizontal flight, 20000 ft.

Ailerons are differential and balanced at 201 of the chord.. Aileron chord 0.25 of wing chord and span = 0.40 b

2

Assume a stick force moment of 100tlbs. For a similar aileron but unbalanced, this would correspond to a moment of 300 ft. lbs.

Hinge moment H = CH q c s

where CH = absolute coefficient q = dynamic pressure

c = wing chord in feet S = wing area, square feet

C300 = 0.000662

Hr 90.25 x 9.42 x 5Z3

The differential aileron used (similar to Fig.3, No. 1 N.A.C.A. Technical Note 449) with aileron angles

of 3%0 will give this coefficient. Each aileron is de-flected 3?0 here because the differential action is not yet apparent at such low angles.

-22-From N.A.C.A. Technical Notes 449, C1 = 0.007 + 0.003 = 0.010

Rolling moment = 90.25 x 56.5 x 533 x 0.01 = 27200 ft. lb.

dL

_{1 (d}

c u

where c = wing chord in feet

b=

wing span in feet

U= speed in ft; sec.

dL

0.000634 x 4.38 x 9.42(56.5)3 377

dL

- 111200 ft. lb. d'P pIL ₌ 27200 dip

- p 27200 0.244 radians per second

111200

Check on the value of "p"

L

l.p

= d L x 5

T dp A

d L -A .1

.'. d p g T

From Technical Note 375,

A= 0.017 x 992 (9.02 + 56,52) = 17150

Also, (to check), A = 0.06 x 9920 56.5)2 14750 32.2 _A

where 0.06 was the assumed value of "at XL

',IL _{= 17150 x 18.25 =}

129700 2.418

P = 27200 = 0.210 radians per second

129700

or using, A 14750 9

p = 27200 = 0.244 radians per second 111300

which is the same value as obtained by the first calculation:

Rate of roll due to rudder, horizontal flight, Vmax, at 20000?. Make t = _{0.65 and assume rudder angle of 50}

C = 1.90 x o.65 - 0.57 (0.65)2 - 0.014 x 5 x 0.65 1.24 - 0.24 - .05 = 0.95 TO= 0.95 x 5 = 4?75 Lateral force = 2.58 x 4.75 _{x 90.25 x 51.9} 57.3 = 616 lb. Yawing moment = 26.5 x 616 = 16320 ft. lb. v = yawing moment dv NV x T L =n N _v n.. = d N TL C _{d v} dN _

C

x nvZ dv _TL

C

= c m b2 0.115 x 9920 x (56.5)2 4 _{32.2 4} 28300 Checking by T, N, 375 C = 0.018 _{x 9920 (43.02+} 32.2 27950

-24-dN _{- 28300 x 1.26} = 520 dv 2.42 x 28.3 v dN = _{16320 (yawing moment)} dv v = 16320 _{31.4 ft. per second} 520 Lv x _{T L lv} L lv - dL S TL- A dv dL

_A

_{lV= 14750 x} 1.68 _{= 362} dv g TL _{2.42 x 28.3} dL p dL dv dp 31.4 x 362 = p x 111000

. p = 31.4 x 362 = _{0.102 radians per second} 111000

The rudder hinge moment (rudder balanced at 20% of the chord) necessary to give the above 50 rudder angle is = 0.0135 x 90.25 x 31.2 x 4.2

= 161.2 ft. lb.

where 0.0135 = hinge moment coefficient 90.25 = q

31.- = rudder area

4.2 = rudder mean chord

Design of elevator (3 - point landing)

It is desired that the plane stall with 180 up

elevator, 270 being the maximum deflection of the elevator. The following determination of the t ratio is made with

the flaps up. The airplane moments are more readily

obtained in this condition and a plane which lands factorily with the flaps up, will be found to land

satis-factorily with them down.

Stalling angle = 180, CL = -1.50 CD = -0.170 C-Al= -0.06 = 0.141 S k = 0.25 d c aI 35 - 0.310 c 113 = 15 - 0.133 113 d = + 15" c = 113 = 0.355 L 318 dCL)" =- 3.92 (radians) = -0.0685 (degrees) Ct - 0.170 (0.951) - (-1.50 x 0.309) - 0.162 + 0.465 + 0.303 CL C: = - CL 2 ( -k) + Ct

L

d L L c L c S (L is + here) C -0.06 (0.355)-(-1.50)(0.355)(0.06)+(0.303)(0.355)(O.133) 0.141 -0.0213 + 0.0319 + 0.0108 = 0.141 0.0214_0.141 CL = - 0.152 -0.152 -0.0685 = + 2?22 0.25 _{x 1.50 x 57.3} 6 f 2.22 = 18 -3.58 + oc e =358 K = 0) - 12? 2 - 12.2 18 1.90 - 0.57(t) 2 -0.014 x 18 x t 11 -c c oC

-26-"0.678 = - 0.57 (t )2 c + 1.73 t (t)2 -. 0 4 + 1.19 = 0 c c t - 3.04 \9.25 - 4.76 - 3.04 - 2.12 = 0.46 c 2 2

Trimming of tabs on elevator in a vertical dive at 10000 feet. As in the case of the elevator at the stall, this calculation is made assuming the flaps arp up.

CL = 0 (=- 4'6 CD = 0.0100 (profile drag only)

= - 0.06 _{Ct = CD ( oc} is small) -0.06(0.355) - 0.0100 (0.355)(0.133) 0.141 -0.0213 - 0.0005 0.141 oC = 0.155 -0.0685 0.0218 _{= + 0.155} 0.141 = - 2.26 - 2.26 - 4.6 + C(e (CQ = 0) 'Ce = + 2.034

T=

1.90 x 0.46 - 0.57 (0.46)2 - 0.014 (3 x (0.46) = 0.875 - 0.121 - 0.006440 = 0.754 - 0.00644 P 2.34 = 0.754 P - 0.006440 2 02 - 117 P + 363 = 0

Q=

117 e 13700 - 1452 2 3 ?0 down = 117 - 111 2

Tab angle (Journal of the Aeronautical Sciences, Oct.19341 2.00 _{to trim at this speed.}

CM 0

CL

Rudder Tab Angle necessary at best climb at sea level with

one engine cut out.

Thrust = Thp x 375 = 388 x 375= 1470 lb.

Vmph 99.0

Yawing moment produced = 7.4 x 1470 = 10880 ft. lb. Lateral force necessary on rudder = 26.5 = 410 lb.

2.58 x 7V x 25.4 x 31.9 = 410 57.3 410 x 57.3 2.58 x 25.4 x 31.9 = 11.2 11.2 = 1.24 3 - 0.24(3 (2 - 110.0 0 + 1230 (3= 110.0 -1 2 (3= 110.0 -84.7 2 - 0.00910 2 =0 12100 - 4920 = 25.3 2 = 12?7

-28-DISCUSSIOIT AND (ONCLUSIONS

In the performance calculations the two engines were assumed to be supercharged to an altitude of 20000 feet.

The propellers are of the "constant speed" type. As previous-ly stated, the selection of a suitable propeller is a major problem. It is necessary to use very high blade angles. Data on propellers operating on such angles is singularly lacking. The values assumed for the parasite drag of the plane are low, but,it is believed, are a good estimate of what could actually be attained. Wind tunnel tests indicate

that a CL of 2.0 can be obtained with the tapered Clark Y wing section used and with the flaps deflected to 45'.

As regards longitudinal stability, the period is high but it is to be doubted if this can be appreciably decreased. "T" is high, due, of course, to the small value

of

"e"

at an altitude of 20000 feet. A good criterion to

investigate, time permitting, would be the damping at the

speed of best climb.

The lateral stability calculations show a very short period in both level flight and in the vertical dive but the damping is exeellent. In each case the time required

to damp to J is less than one period. The short period should therefore give no trouble. "lv" was chosen at level flight

-29

-so as to make the coefficient E=0. The root of the factor representing spiral motion is then here equal to zero. The assumed value of yv = -0.4 is considered reasonable. It is to be noted that the nv produced is here due to the vertical tail surface only. The fuselage will alter this somewhat.

The velocity in the vertical dive, calculated for a mean altitude of 10000 feet, was obtained by equating the weight to the drag. The plane will dive with the flaps down. The velocity calculations and the lateral stability calculations were made assuming the flight path to be vertical. This condi-tion may be a bit extreme but the results can be taken to apply, without serious error, to any flight path within a few degrees of the vertical, which condition it was desired to investigate. Actually, the direction of the flight path is somewhat uncertain here since we are operating on a very steep part of the L/D curve.

The angular velocity of roll due to the ailerons is

about what might have been expected but that due to the rudder seems low. A more accurate determination of the moment of in-ertia "C" is probably desirable.

The tabs on the rudder and elevator are for trimming purposes and not fcr balancing. These surfaces are already

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