# Flight Design Questions Solved

The value of the mean aerodynamic chord of the wing (cw ) and the location of

the aerodynamic centre would be needed in chapter 6. These are worked out

below.

For a wing which is symmetric about the root chord, w c is given by :

b/2

2

w

W 0

2 c = c dy S ∫

In the present case :

c = 2.636 m for 0 y 4.636 ≤ ≤

= ( )

( ) 2.636-1.318 2.636- y-4.636 for 4.636 y 13.245 8.609

≤ ≤

= 3.3458-0.1531 y for 4.636 y 13.245 ≤ ≤ (5.11)

Consequently,

( )

4.636 13.245

2 2

W

0 4.636

2

c = 2.636 dy + 3.3458 – 0.1531y dy 58.48

 ∫ ∫

{ ( )

13.245

2 2

4.636

2 = 2.636 ×4.636 + 11.194 + 0.02344 y – 1.0245 y dy 58.48 ∫

( ) ( ) ( ) 2 0.02344 3 3 2 2 = 32.211+11.194 13.245-4.636 + 13.245 -4.636 -1.0245 13.245 -4.636

58.48 3

     

= 2.295 m

Location of aerodynamic centre

From Eq.(5.11) the spanwise location (ymac) where a chord of 2.295 m is located

is given by :

2.295 = 3.3458 – 0.1531 ymac

Or ymac = 6.863 m

To obtain the location of a.c. of the wing, the wing chord at ymac is projected on to

the root chord. The quarter chord of this is the a.c. of the wing. (Fig. E 5.1 a).

Dept. of Aerospace Engg., Indian Institute of Technology, Madras 5

Airplane design(Aerodynamic) Prof. E.G. Tulapurkara

Chapter-5

The leading edge sweep of the outboard wing is 6.07o

. Hence, the location of a.c.

from the leading edge of the root chord is given by :

( ) 2.295 6.863-4.636 tan6.07+ = 0.811m

4

V) Twist

Taking guidelines from the value of wing twist in IPTN-N-250-100, a value of

0 ε=3 (wash-out) is chosen at this stage of preliminary design.

VI) Incidence (iw)

This angle is chosen such that during cruise, the fuselage is horizontal and wing

with α=iw produces the desired lift. From item(I) above, CLcruise = 0.476

From Eq.(5.6)

   

     

Lα 2

2 2 1

2

2 2

2πA C =

tan Λ

A β 2 + 4+ 1+

η β

2 A = 12, β = 1-M = 1-0.4312 = 0.902

η = (average slope of lift curve of airfoil on wing) / 2π .

Generally, η is taken as unity.

This quantity 1

2

Λ depends, A, λ and 1

4

Λ . The taper ratio and 1

4

Λ are those of an

equivalent trapezoidal wing (ETW). The steps to obtain the dimensions of ETW

are given below.

Equivalent trapezoidal wing :

It may be pointed out that the procedures to obtain the aerodynamic data and

stability derivatives are generally available for straight tapered wings. Reference

5.6, section 2 suggests that the geometrical parameters of practical wings can be

approximated by an ETW. A certain amount of judgement is involved to obtain

ETW. The procedure is as follows.

(a) The actual wing and the ETW have the same span and area.

Dept. of Aerospace Engg., Indian Institute of Technology, Madras 6

Airplane design(Aerodynamic) Prof. E.G. Tulapurkara

Chapter-5

(b) The length and location of tip chord is the same for the two wings. The swept

back portion of the two wings are nearly the same.

In the present case the actual wing has the following values.

Area of wing = 58.48 m2 ; wing span = 26.49 m; root chord (cr) = 2.636 m;

tip chord(ct) = 1.318 m ; the constant chord (2.636 m) extends upto 4.636 m on

either side of the root chord ; leading edge sweep of the outboard wing is 6.07o ;

quarter chord sweep of the outboard wing is 3.9 o through out the .

The ETW is taken as having area of 58.48 m2

, span of 28.49 m, tip chord of

1.318 m and quarter chord sweep of 3.9o through out the same-span.

Consequently, the root chord (cre) of ETW is given by :

( re ) 26.49 58.48 = c +1.318

2

or re c = 3.1m

Taper ratio of (λe ) of ETW = 1.318 / 3.1 = 0.425

The ETW as superposed over actual wing is shown in Fig.E5.1b.

The quarter chord sweep 1

4 ETW

    Λ

 

is chosen to be 3.9 o

. The value of 1

2 ETW

    Λ

 

can be worked out by geometry. However, from section 1 of Ref.5.6, the value of

1

2

tan Λ is given by :

1 ( )

4

e

1

2 e

4 1- λ tan = tan – 0.5-0.25

A 1+λ

  Λ Λ    

= 0.06817 – 0.0336 = 0.03454

o

1

2

Λ = 1.98

With the above data, the value of CLα of the wing is obtained as :

Lα 22 2

2

2π×12 C =

12 × 0.902 0.03454 2+ 4+ 1+

1 0.902

     

= 5.793 rad-1 = 0.1011 deg-1

From Fig.5.5 a, αol of the airfoil is o -3

Dept. of Aerospace Engg., Indian Institute of Technology, Madras 7

Airplane design(Aerodynamic) Prof. E.G. Tulapurkara

Chapter-5

Since, the wing has a twist of o 3 (wash-out), from Eq.(5.10) the angle of zero lift

of the wing ( αoL ) is :

o α = 0.4 (-3) = -1.8 oL

Hence, wing incidence (iW) is given by :

0.476 = 0.1011 (iW + 1.8) or iW = 2.90

Remark :

IPTN-N-250-100 which also uses NASA MS (1)-0317 airfoil has iw = 20

This value would have been obtained after wind tunnel tests on the model of the

final configuration of the airplane.

VII) Dihedral (Γ)

At this stage of the preliminary design a value of 0 Γ=3 is chosen based on

Table 2.1.

VIII) Vertical location

For airplanes with propellers mounted on wings, the preferred vertical location is

the high wing configuration. This is chosen in the present case.

Flaps and ailerons

As mentioned in example 4.11, a double slotted flap, similar to ATR-72-200 has

been chosen. Based on this airplane, the flap span is chosen as 70% of the

exposed wing span (i.e. wing span minus the fuselage width). The flap chord is

40% of the wing chord in the constant chord section. The ailerons start from the

spanwise location where the flap ends and extend upto the wing tip.

Figure E 5.1a shows the tentative locations of flaps and ailerons.

The wing parameters, at this stage of preliminary design, are :

Wing area (S) = 58.48 m2

, span (b) = 26.49 m

Root chord (cr) = 2.636 m, constant chord central section upto 4.636 m

from root on either side.

Tip chord (ct) = 1.318 m.

Airfoil : NASA MS(1) – 0317 at root, t/c = 17%

Dept. of Aerospace Engg., Indian Institute of Technology, Madras 8

Airplane design(Aerodynamic) Prof. E.G. Tulapurkara

Chapter-5

NASA MS(1) – 0313 at tip, t/c = 13%

Aspect ratio = 12

Quarter chord sweep of outboard panel = 3.9o

Mean aerodynamic chord = 2.295 m

Twist ( ε ) = 3o (wash-out)

Wing incidence (iw) = 2.9o

Dihedral angle (Γ) = 3o

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