02-2 - wing contribution

7/23/2019 02-2 - Wing Contribution

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Flight Dynamics

AAE 6710

Lesson 02-2

WingContribution


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Static Longitudinal Stability

We assume that the net aerodynamic force or moment coefficient is equal to

the sum of the individual contributions from fuselage, wing, powerplant and

tail surfaces.


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Wing Contribution


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= = +

+ +

Non-dimensionalise by dividing by

2

&

use cos = 1 and si = =

+

+

+


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Contribution

If we assume >> , and the vertical contribution isnegligible, then

= +

= + +

For applying the conditions of static stability:

= +

=


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Remarks

To make < 0, the aerodynamic center must lie aft of center ofgravity To have > 0, a negative cambered airfoil or an airfoil section

with reflexed trailing edge be used

Normally, the center of gravity is located slightly aft of the

aerodynamic center

The airfoil sections of the wing have positive camber

Therefore, wing contribution to static longitudinal stability is usually

destabilising

Negative camber Reflexed trailing edge
http://en.wikipedia.org/wiki/File:Reflex_camber_line.jpg


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Example 1

For a given wing bodycombination the aerodynamic center lies 0.03

chord length aheadof the center of gravity. The moment coefficient

about the center of gravity is 0.0050, and the lift coefficient is 0.50.

Calculate the moment coefficient about the aerodynamic center.

= 0.005 0.5 0.03 = 0.01

= +

0.0050= + 0.50 0.03


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Example 2

Consider a model of a wing-body shape mounted in a wind-tunnel.

The flow conditions in the test section are standard sea-level

properties with a velocity of 100 m/s. The wing area and chord are

1.5 m2and 0.45 m, respectively.

Using the wind tunnel force and moment-measuring balance, the

moment about the center of gravity when the lift is zero is found to

be -12.4 Nm.

When the model is pitched to another angle of attack, the lift and

moment about the center of gravity are measured to be 3675 N

and 20.67 Nm, respectively. Calculate the value of the moment coefficient about the

aerodynamic center and the location of the aerodynamic center.


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Solution

= 12 = 12 1.225100

= 6125 /

= = 12.461251.50.45 =0.003

= = 0.003


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=

=3675

61251.5 =0.4

= =

20.6761251.50.45 =0.005

= +

=

= 0.005 0.0030.4 =0.02


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Example 3

An airplane is equipped with a wing of aspect ratio 6 ( = 0.095 )and span efficiency factor of 0.78, with an airfoil section giving =0.02.Calculate, for between 0 and 1.2, the pitching moment coefficient of thewing about the c.g. (i.e. ) which is located 0.05 ahead of a.c. and 0.06 under a.c. Repeat the calculations when chord wise force components are

neglected. Assume =0.008, = 1, = 5.


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Solution

=0.02 =0.05 =0.06

= = + 1 = . + 1

= +

=0.008+

3.1460.76

=0.008+0.068

=

1 + = 0.095

1 + 0.095 0 . 7 6 6 57.3= 0.0688 = 3.94


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=

+

+ +

= 0.05 + 0.008 + 0.068 0.05

0.0688

+ 1 5 /57.3

+ 0.06 0.0688 + 1 5 /57.3 0.008 + 0.068 0.06

+0.02= +

+ + +

= +

= 0.02 0.05


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Figure Example 3

0 0.2 0.4 0.6 0.8 1 1.2-0.06

-0.05

-0.04

-0.03

-0.02

-0.01

0

0.01

0.02

0.03

CL

Cmcg

Cmcg

Cmcg,approx

Cmac

Cm(Lift Vert)

Cm(Drag Ver)

Cm(Lift Hor)

Cm(Drag Hor)

CG: 0.05 ahead of ac0.06 under ac


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Figure Example 3

CG: 0.05 behind ac0.06 under ac

0 0.2 0.4 0.6 0.8 1 1.2

-0.02

-0.01

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

CL

Cmcg

Cmcg

Cmcg,approxCmac

Cm(Lift Vert)

Cm(Drag Ver)

Cm(Lift Hor)

Cm(Drag Hor)


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Figure Example 3

CG: 0.05 behind ac0.06 above ac

0 0.2 0.4 0.6 0.8 1 1.2

-0.02

0

0.02

0.04

0.06

0.08

0.1

CL

Cmcg

Cmcg

Cmcg,approxCmac

Cm(Lift Vert)

Cm(Drag Ver)

Cm(Lift Hor)

Cm(Drag Hor)

02-2 - wing contribution

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