# Aerospace Derivation of Aerodynamic Forces

1. Mar 8, 2016

### VyRianS

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!

File size:
12.8 KB
Views:
45
2. Mar 8, 2016

### Samy_A

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 $D=N\sin \alpha + A\cos \alpha$.