6.1. - John Wiley & Sons · 1 Solutions to sample problems for Chapter 6 Aircraft Design: A Systems Engineering Approach, Wiley, 2012 6.1. Using the Reference [5] or other reliable

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Solutions to sample problems for

Chapter 6

Aircraft Design: A Systems Engineering Approach, Wiley, 2012

6.1. Using the Reference [5] or other reliable sources, identify the tail configurations of the

following aircraft:

Stemme S10 (Germany): T-tail

Dassault Falcon 2000 (France): Conventional/Cruciform

Embraer EMB 145 (Brazil): T-tail

Canadair CL-415: Non-conventional/Cruciform

ATR 42: T-tail/Cruciform

Aeromacchi MB-339C (Italy): Conventional + twin ventral fin

Eagle X-TS (Malaysia):

PZL Mielec M-18 Dromader (Poland): Conventional

Beriev A-50 (Russia): T-tail

Sukhoi Su-32FN (Russia): Twin vertical tail

Sukhoi S-80: Boom-mounted inverted U

Saab 340B (Sweden): Conventional

Pilatus PC-12 (Switzerland): T-tail

An-225 (Ukraine): H-tail

Jetstream 41 (UK): Cruciform

FLS Optica OA7-300 (UK):

Bell/Boeing V-22 Osprey: H-tail

Boeing E-767 AWACS: Conventional

Cessna 750 Citation X: T-tail

Learjet 45: T-tail

Lockheed F-16 Fighting Falcon: Conventional

Lockheed F-117A Nighthawk: V-tail

McDonnell Douglas MD-95: T-tail

Northrop Grumman B-2 Spirit: No tail

Bede BD-10: Twin vertical tail

Hawker 1000: Cruciform

2

6.2. Using the Reference [5] or other reliable sources, identify an aircraft for each of the

following tail configurations:

Conventional aft tail: Piper PA-46-350P Malibu Mirage, Northrop Grumman E-8C North STARS,

Xian MA-600

Xian MA-600 (Courtesy of Antony Osborne)

Let L-13 Blanik (Courtesy of Miloslav Storoska)

3

V-tail: Beech Bonanza, V-Tail version, GA light single engine, Lockheed F-117A Nighthawk

Lockheed F-117A Nighthawk (Courtesy of Antony Osborne)

Canard: Eurofighter Typhoon

Eurofighter Typhoon (Courtesy of Antony Osborne)

4

T-tail: Eclipse 500, Learjet 40, Bell-Agusta BA-609, Fokker 100, Tupolev Tu-154, Orlican VSO-10b

Gradient, PZL-Bielsko SZD-48 Jantar Standard 3

Bell-Agusta BA-609 (Courtesy of Antony Osborne)

Fokker 100 (F-28-0100) (Courtesy of Antony Osborne)

5

Tupolev Tu-154M (Courtesy of Miloslav Storoska)

Orlican VSO-10b Gradient (Courtesy of Miloslav Storoska)

6

PZL-Bielsko SZD-48 Jantar Standard (Courtesy of Miloslav Storoska)

H-tail: Bell Boeing MV-22B Osprey, PZL-Mielec M-28B1R Bryza

Bell Boeing MV-22B Osprey (Courtesy of Antony Osborne)

7

PZL-Mielec M-28B1R Bryza 1R-2 (Courtesy of Jenny Coffey)

Non-conventional: RMT 03Bateleur (Germany), Aeronix Airelle (France), North American

Rockwell OV-10B Bronco

North American Rockwell OV-10B Bronco (Courtesy of Miloslav Storoska)

8

Cruciform: Hawker 800XP, Beriev Be-103 Bekas

British Aerospace HS-125 CC3 (HS-125-700B) (Courtesy of Jenny Coffey)

Tri-plane: P-180 Avanti

(Courtesy of Tibboh)

9

Boom-mounted: Adam A500, MiG-110, De Havilland Vampire T11

De Havilland Vampire T11 (DH-115) (Courtesy of Jenny Coffey)

Twin vertical tail: F-22A Raptor

Lockheed Martin F-22A Raptor (Courtesy of Antony Osborne)

10

Sukhoi Su-30MKI-3 (Courtesy of Antony Osborne)

Inverted V-tail: General Atomics MQ-1 Predator

(Courtesy of US Air Force)

11

6.5. An unmanned aircraft has the following features:

S = 55 m2, AR = 25, Sh = 9.6 m2, lm

Determine the horizontal tail volume coefficient.

6.6. The airfoil section of a horizontal tail in a fighter aircraft is NACA 64-006. The tail aspect

ratio is 2.3. Using the Reference [8], calculate the tail lift curve slope in 1/rad.

Using the Reference [8], the lift curve slope for NACA airfoil 64-004 is determined as follows:

(Equ 5.19)

(Equ 5.18)

(Equ 6.24)

(Equ 6.57)

S1 55 m2

Sh 9.6m2

l 6.8m AR 25

b1 AR S1 37.081m

CbarS1

b11.483m

Vbarl Sh

Cbar S10.8

CL1 0.6 1 5.5 deg CL2 0.6 2 5.8deg

CL CL2 CL1 1.2 2 1 11.3deg

Clh

C L

6.085 ARh 2.3

CLh

Clh

1Clh

ARh

3.3031

rad

12

6.8. The airfoil section of a horizontal tail in a GA aircraft is NACA 0012. The tail aspect ratio

is 4.8. Using the Reference [8], calculate the tail lift curve slope in 1/rad.

Using the Reference [8], the lift curve slope for NACA airfoil 0012 is determined as follows:

6.9. The wing reference area of an agricultural aircraft is 14.5 m2 and wing mean aerodynamic

chord is 1.8 m. The longitudinal stability requirements dictate the tail volume coefficient to be

0.9. If the maximum fuselage diameter is 1.6 m, determine the optimum tail arm and then

calculate the horizontal tail area. Assume that the aft portion of the fuselage is conical.

(Equ 6.57)

(Equ 6.47)

(Equ 6.24)

CL1 1.2 1 9.5 deg CL2 0.8 2 6.7deg

CL CL2 CL1 2 2 1 16.2deg

Clh

C L

7.074 ARh 4.8

CLh

Clh

1Clh

ARh

4.8151

rad

S1 14.5m2

Cbar 1.8m VH 0.9 Df 1.6m Kc 1

lopt Kc

4 Cbar S1 VH

Df 4.324m

Sh

Cbar S1 VH

lopt

5.433m2

13

6.10. Consider a single-seat GA aircraft whose wing reference area is 12 m2 and wing mean

aerodynamic chord is 1.3 m. The longitudinal stability requirements dictate the tail volume

coefficient to be 0.8. If the maximum fuselage diameter is 1.3 m, determine the optimum tail arm

and then calculate the horizontal tail area. Assume that the aft portion of the fuselage is conical.

(Equ 6.47)

(Equ 6.24)

S1 12 m2

Cbar 1.3m VH 0.8 Df 1.3m Kc 1

lopt Kc

4 Cbar S1 VH

Df 3.496m

Sh

Cbar S1 VH

lopt

3.57m2

14

6.11. A 19-seat business aircraft with a mass 6,400 kg is cruising with a speed of 240 knot at

26,000 ft. Assume that the aircraft lift coefficient is equal to the wing lift coefficient. The aircraft

has the following characteristics:

S = 32 m2, ARw = 8.7, Wing airfoil: NACA 651-412

Determine the downwash angle (in degrees) at the horizontal tail.

Using the Reference [8], the lift curve slope for NACA airfoil 651-412 is determined as follows:

The wing list curve slope is:

The downwash angle at the horizontal tail is determined as follows:

(Equ 5.20)

(Equ 5.10)

(Equ 6.55)

(Equ 6.56)

CL1 0.1 1 4 deg CL2 1.3 2 10deg

CL CL2 CL1 1.4 2 1 14deg

Clw

C L

5.73

ARw 8.7 CLw

Clw

1Clw

ARw

4.7371

rad

m1 6400kg S1 32.m2

VC 240knot hC 26000ft C 0.00103slug

ft3

CL2 m1 g

C VC2

S1

0.485

CLw CL 0.485

o

2 CLw

ARw2.032deg

dd2 CLw

ARw0.347

deg

deg

15

The Wing airfoil NACA 651-412 has an ideal lift coefficient of 0.4. The corresponding wing setting

angle for this coefficient is 1 deg. If the fuselage has zero degrees of angle of attack,

6.12. Suppose that the angle of attack of the fuselage for the aircraft in problem 11 is 2.3

degrees and the horizontal tail has an incidence of -1.5 degrees. How much is the horizontal tail

angle of attack at this flight condition?

(Equ 6.54)

(Equ 6.53)

iw 1 deg f 0

w iw f 1deg

o dd w 2.379deg

2.379deg ih 1.5 deg f 2.3deg

h f ih 1.579 deg

16

6.13. The horizontal tail of a transport aircraft has the following features:

ARh = 5.4, h = 0.7, Sh = 14 m2, h_LE = 30 degrees

Determine span, root chord, tip chord and the mean aerodynamic of the horizontal tail. Then sketch the

top-view of the tail with dimensions.

Given:

(Equ 6.63)

(Equ 6.66)

(Equ 6.65)

(Equ 6.64)

Sh 14 m2

ARh 5.4 h 0.7 h 30deg

ARh

bh

Ch

Sh bh Ch

bh Sh ARh 8.695m

Ch

bh

ARh

1.61m

Ch2

3Ch_r

1 h h2

1 h

Ch_r

Ch3

2

1 h h2

1 h

1.875m

h

Ch_t

Ch_r

Ch_t h Ch_r 1.312m

17

6.15. The vertical tail of a transport aircraft has the following features:

ARV = 1.6, V = 0.4, SV = 35 m2, V_LE = 45 degrees

Determine span, root chord, tip chord and the mean aerodynamic of the vertical tail. Then sketch the side-

view of the tail with dimensions.

Given:

(Equ 6.75)

(Equ 6.77)

(Equ 6.80)

(Equ 6.79)

(Equ 6.78)

Sv 35 m2

ARv 1.6 v 0.4 v 45 deg

ARv

bv

Cv

bv Sv ARv 7.483m

Sv bv CvCv

bv

ARv

4.677m

Cv_r

Cv3

2

1 v v2

1 v

6.296mCv

2

3Cv_r

1 v v2

1 v

v

Cv_t

Cv_r

Cv_t v Cv_r 2.518m

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6.16. The aircraft in problem 11 has other features as follows:

h = 0.18, ho = 0.23, h = 0.97, l = 12 m, Sh = 8.7 m2

Determine the aircraft static longitudinal stability derivative (Cm) and discuss whether the

horizontal tail is longitudinally stabilizing or destabilizing.

d

dh

C

l

S

SChhCC h

hLoLmhwf

1 (6.67)

From Problem 6.11, we have:

Given:

The wing-fuselage lift curve slope and tail lift curve slope are not given. We assume that the

wing-fuselage lift curve slope and tail lift curve slope are equal to the wing lift curve slope.

So:

Since Cm is negative, the horizontal tail is longitudinally stabilizing and aircraft is statically

longitudinally stable.

So

(Equ 5.19)

(Equ 5.18)

(Equ 6.67)

CLw 4.731

rad S1 32 m

2 d_d 0.347

deg

deg ARw 8.7

b1 ARw S1 16.685m

CbarS1

b11.918m

h 0.18 ho 0.23 h 0.97 l1 12m Sh 8.7 m2

CLh CLw 4.731

rad CLwf CLw 4.73

1

rad

Cm CLwf h ho CLh hSh

S1 1 d_d( )

l1

Cbarh

5.1861

rad

19

6.17. This is an open-ended design problem, which has no single distinct solution, and can have several

acceptable designs. See the solution for Example 6-2 for an example of the design process.

6.18. This is an open-ended design problem, which has no single distinct solution, and can have several

acceptable designs. See the solution for Example 6-2 for an example of the design process.

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6.19. Figure 6.29 shows the original design for the empennage of a transport aircraft with a

horizontal tail area of 12.3 m2. The wing reference area is 42 m

2, and wing aspect ratio is 10.5.

Figure P6.19a. Side-view of the aircraft in problem 19

The aircraft is spinnable and the designer found out that the vertical tail is not effective for spin

recovery. Move the horizontal tail horizontally such that the vertical tail becomes effective in

recovering from spin. Then determine the horizontal tail area such that the horizontal tail volume

coefficient remains unchanged. Assume that the sketch in figure 6.29 is scaled.

Solution:

An experimental rule for the vertical tail effectiveness to achieve a recoverable spin is as

follows: At least 50 percent of the vertical tail planform area must be out of the horizontal tail

wake region to be effective in the case of a spin. The horizontal tail wake region is considered to

lie between two lines. The first line is drawn at the horizontal tail trailing edge by the orientation

of 30 degrees. The second line is drawn at the horizontal tail leading edge by the orientation of

60 degrees.

Figure P6.19b. Side-view of the aircraft when 30o-60

o rule is applied.

30o 60o

ach acwf

6 m

ach acwf

6 m

21

The Figure P16.19b indicates that the vertical tail is graphically located to be inside the horizontal tail

wake region. Thus, the horizontal tail moment arm needs to be adjusted. The arm may be gets shorter or

longer. A longer arm is recommended, since it does not require the horizontal tail area to be increased.

Figure P6.19c. Side-view of the aircraft when horizontal tail is moved aft.

The new arm illustrates that the more 50 percent of the vertical tail planform area is out of the

horizontal tail wake region. Since the sketch is scaled, the new arm is measured to be 7.3 m.

The horizontal tail volume ratio is:

The new arm is 7.3 m, so the new horizontal tail area will need to be:

(Equ 5.19)

(Equ 5.18)

(Equ 6.24)

(Equ 6.24)

Sh 12.3 m2

Sw 42 m2

Lh 6 m AR 10.5

b AR Sw 32.404m

CbarSw

b1.296m

VH

Sh Lh

Sw Cbar1.356

Lh 7.3m

Sh

Cbar Sw VH

Lh

10.11m2

acwf

7.3 m

30o 60o

ach

22

6.20. A fighter aircraft has the following features:

S = 57 m2, AR = 3, Sh = 10.3 m2, Sv = 8.4 m2, lm, lv = 6.2 m

Determine the horizontal and vertical tails volume coefficients.

(Equ 5.19)

(Equ 5.18)

(Equ 6.24)

(Equ 6.72)

Sw 57 m2

ARw 3 Sh 10.3 m2

Sv 8.4 m2

Lh 6.8m Lv 6.2m

b ARw Sw 13.077m

CbarSw

b4.359m

VH

Sh Lh

Sw Cbar0.282

Vv

Sv Lv

Sw b0.07

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6.22. The airfoil section of the vertical tail for a twin-jet engine aircraft is NACA 66-009. Other

features of the aircraft are as follows:

S = 32 m2, AR = 10.3, SV = 8.1 m2, ARV = 1.6, l m, 32.0

d

d, V = 0.95

Determine the aircraft static directional stability derivative (Cn). Then analyze the static

directional stability of the aircraft.

bS

Sl

d

dCKCC VVt

VLfnnVV

1 (6.73)

NACA 66-009:

(Equ 5.20 revised)

Selected based on page 322

(Equ 6.73)

Sw 32 m2

ARw 10.3 Sv 8.1 m2

Lv 9.2m d_d 0.32 v 0.95 ARv 1.6

CL1 0.6 1 6 deg CL2 0.6 2 6 deg

CL CL2 CL1 1.2 2 1 12deg Clv

C L

5.73

CLv

Clv

1Clv

ARv

2.6781

rad

Kf1 0.75

Cn Kf1 CLv 1 d_d( ) vLv Sv

b Sw 0.231

24

6.23. The angle of attack of a horizontal tail for a cargo aircraft is -1.6 degrees. Other tail

features are as follows:

Sh = 12 m2, ARh = 5.3, h = 0.7, airfoil section: NACA 64-208, h = 0.96

If the aircraft is flying at an altitude of 15,000 ft with a speed of 245 knot, determine how much

lift is generated by the tail. Assume that the tail has no twist.

Solution:

The following MATLAB m-file is utilized to calculate the tail lift coefficient:

clc clear N = 9; % (number of segments-1) S = 12; % m^2 AR = 5.3; % Aspect ratio lambda = 0.7; % Taper ratio alpha_twist = 0.00001; % Twist angle (deg) a_h = -1.6; % tail angle of attack (deg) a_2d = 4.521; % lift curve slope (1/rad) alpha_0 = -1.3; % zero-lift angle of attack (deg) b = sqrt(AR*S); % tail span MAC = S/b; % Mean Aerodynamic Chord Croot = (1.5*(1+lambda)*MAC)/(1+lambda+lambda^2); % root chord theta = pi/(2*N):pi/(2*N):pi/2; alpha=a_h+alpha_twist:-alpha_twist/(N-1):a_h; % segment's angle of attack z = (b/2)*cos(theta); c = Croot * (1 - (1-lambda)*cos(theta)); % Mean Aerodynamics chord at each

segment mu = c * a_2d / (4 * b); LHS = mu .* (alpha-alpha_0)/57.3; % Left Hand Side % Solving N equations to find coefficients A(i): for i=1:N for j=1:N B(i,j) = sin((2*j-1) * theta(i)) * (1 + (mu(i) * (2*j-1)) /

sin(theta(i))); end end A=B\transpose(LHS); for i = 1:N sum1(i) = 0; sum2(i) = 0; for j = 1 : N sum1(i) = sum1(i) + (2*j-1) * A(j)*sin((2*j-1)*theta(i)); sum2(i) = sum2(i) + A(j)*sin((2*j-1)*theta(i)); end end CL_tail = pi * AR * A(1)

25

where the zero lift angle of attack of the airfoil section (NACA 64-208) is -1.3 degrees and it has

the following lift curve slope.

The output of this m-file is:

CL_tail = -0.0181

Thus, the horizontal tail is producing a lift with the amount of

NACA 64-208:

(Equ 6.57)

(Equ 6.28 and 6.12)

CL1 0.5 1 6 deg CL2 0.8 2 6 deg

CL CL2 CL1 1.3 2 1 12deg Clh

C L

6.207

1

rad

CLh

Clh

1Clh

ARh

4.5211

rad

hC 15000ft C 0.0015slug

ft3

CLh 0.0181 Sh 12 m2

h 0.96 VC 245knot

Lh1

2C VC

2 Sh CLh h 1280.3 N

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6.24. The sideslip angle of a vertical tail for a maneuverable aircraft during a turn is 4 degrees.

Other vertical tail features are as follows:

Sv = 7.5 m2, ARV = 1.4, V = 0.4, airfoil section: NACA 0012, V = 0.92

If the aircraft is flying at an altitude of 15,000 ft with a speed of 245 knot, determine how much

lift (i.e. side force) is generated by the vertical tail. Assume that the tail has no twist.

The following MATLAB m-file is utilized to calculate the vertical tail lift coefficient:

clc clear N = 9; % (number of segments-1) S = 7.5; % m^2 AR = 1.4; % Aspect ratio lambda = 0.4; % Taper ratio alpha_twist = 0.00001; % Twist angle (deg) a_h = 4; % tail angle of attack (deg) a_2d = 2.488; % lift curve slope (1/rad) alpha_0 = 0; % zero-lift angle of attack (deg) b = sqrt(AR*S); % tail span MAC = S/b; % Mean Aerodynamic Chord Croot = (1.5*(1+lambda)*MAC)/(1+lambda+lambda^2); % root chord theta = pi/(2*N):pi/(2*N):pi/2; alpha=a_h+alpha_twist:-alpha_twist/(N-1):a_h; % segment's angle of attack z = (b/2)*cos(theta); c = Croot * (1 - (1-lambda)*cos(theta)); % Mean Aerodynamics chord at each

segment mu = c * a_2d / (4 * b); LHS = mu .* (alpha-alpha_0)/57.3; % Left Hand Side % Solving N equations to find coefficients A(i): for i=1:N for j=1:N B(i,j) = sin((2*j-1) * theta(i)) * (1 + (mu(i) * (2*j-1)) /

sin(theta(i))); end end A=B\transpose(LHS); for i = 1:N sum1(i) = 0; sum2(i) = 0; for j = 1 : N sum1(i) = sum1(i) + (2*j-1) * A(j)*sin((2*j-1)*theta(i)); sum2(i) = sum2(i) + A(j)*sin((2*j-1)*theta(i)); end end CL_V_tail = pi * AR * A(1)

where the zero lift angle of attack of the airfoil section (NACA 64-208) is 0 and it has the

following lift curve slope.

27

The vertical tail angle of attack is equal to the sideslip angle (i.e. 4 deg).

The output of this m-file is:

CL_V_tail = 0.1059

Thus, the vertical tail is producing a lift with the amount of

NACA 0012:

(Equ 5.57 revised)

(Equ 6.70 revised)

CL1 0.8 1 8 deg CL2 0.8 2 8 deg

CL CL2 CL1 1.6 2 1 16deg Clv

C L

5.73

1

rad

CLv

Clv

1Clv

ARv

2.4881

rad

hC 15000ft C 0.0015slug

ft3

VC 245knot

CLv 0.106 Sv 7.5 m2

v 0.92

Lv1

2C VC

2 Sv CLv v 7118N

28

6.25. An aft horizontal tail is supposed to be designed for a single piston engine aircraft. The

aircraft with a mass of 1,800 kg is cruising with a speed of 160 knot an altitude of 22,000 ft. The

aircraft center of gravity is at 19% MAC and the wing-fuselage aerodynamic center is located at

24% MAC.

S = 12 m2, AR = 6.4, Sh = 2.8 m2, lm, 06.0owfmC

Determine the horizontal tail lift coefficient that must be produced in order to maintain the

longitudinal trim.

The aircraft longitudinal trim equation is:

0howf LHhoLm CVhhCC (6.29)

In this problem, the tail efficiency is not given, so it is assumed to be 1. The solution is as

follows:

(Equ 5.19)

(Equ 5.18)

(Equ 6.24)

(Equ 5.1)

(Equ 6.29)

Cmowf 0.06 Sw 12 m2

AR 6.4 Sh 2.8 m2

Lh 3.7m

m1 1800kg h 0.19 ho 0.24 h 1

hC 22000ft C 0.00118slug

ft3

VC 160knot

b AR Sw 8.764m

CbarSw

b1.369m

VH

Sh Lh

Sw Cbar0.63

CL2 m1 g

C VC2

Sw

0.714

CLh

Cmowf CL h ho

h VH0.152