Derivation of Aerodynamic Forces

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SUMMARY

The derivation of Lift (L) and Drag (D) forces in aerodynamics is accurately represented by the equations L = N cosα - A sinα and D = N sinα + A cosα, as stated in "Fundamentals of Aerodynamics" by John Anderson. The derivation utilizes the cofunction identities sin(90 - x) = cos x and cos(90 - x) = sin x to relate the forces to the normal (N) and axial (A) components. The discussion provides a clear step-by-step breakdown of how these equations are formulated from the geometric relationships in the diagram.

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VyRianS
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Good day all:

I'm having trouble deriving the forces for Lift and Drag in the attached diagram.

The text I have (Fundamentals of Aerodynamics - John Anderson) states the solution to be:
L = N cosα - A sinα
D = N sinα + A cosα

Can anyone guide me through this? I think the cofunction identity is used somewhere:
sin(90 - x) = cos x
cos(90 - x) = sin x

But that's as far as I've gotten.

Help is appreciated, thanks!
 

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VyRianS said:
Good day all:

I'm having trouble deriving the forces for Lift and Drag in the attached diagram.

The text I have (Fundamentals of Aerodynamics - John Anderson) states the solution to be:
L = N cosα - A sinα
D = N sinα + A cosα

Can anyone guide me through this? I think the cofunction identity is used somewhere:
sin(90 - x) = cos x
cos(90 - x) = sin x

But that's as far as I've gotten.

Help is appreciated, thanks!
angles.jpg


For ##L##:
##\cos (\alpha + x)=\frac{L}{R}##
So ##L=R\cos(\alpha + x)=R\cos x \cos \alpha -R\sin x \sin \alpha \ \ \ \ (1)##
Now ##\cos x=\frac{N}{R}## and ##\sin x=\frac{A}{R}\ \ \ \ (2)##.
Then (1) becomes: ##L=N \cos \alpha -A\sin \alpha##.

Similarly for ##D##:
##D=R \cos y= R \cos(\frac{\pi}{2}-\alpha -x)=R\sin(\alpha + x)=R\cos x \sin \alpha + R\sin x \cos \alpha##.
Using (2) again, you get ##D=N\sin \alpha + A\cos \alpha##.
 

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