# Aerodynamic Force Decomposition

We know that aerodynamic drag is proportional to the square of the velocity of the incoming flow: $$D=kV^2.$$

If I decompose velocity into arbitrary orthogonal $$x$$-$$y$$ directions:$$V_x=V\cos\theta\\V_y=V\sin\theta$$ we have $$D_x=kV_x^2\\D_y=kV_y^2.$$

Now my total drag is $$D=\sqrt{D_x^2+D_y^2}=k\sqrt{V_x^4+V_y^4}\le k(V_x^2+V_y^2)=kV^2.$$

So what's wrong with my calculation, or my idea? Are there any physical concept that I misused?

• $v^{2}=v^{T}v$ v is a vector – Eli Mar 12 at 7:43

## 2 Answers

Your decomposition is wrong. D is a vector of magnitude $$1/2 k V^2$$. It’s x component is $$1/2 k V^2 cos(\Theta)$$; the $$V^2$$ in that remains the same.

So you're saying

$$\vec D = k||\vec v||^2$$

which is a vector equal to a scalar, and then breaking down the vector side into a coordinate system and discovering it doesn't work (and it shouldn't, vectors aren't equal to scalars). That is a very general physical concept that is unrelated to fluid mechanics.

In coordinate free geometric formulation of physical laws, you would salvage it by introducing a rank-3 tensors so that:

$$D_i = k_{ijk}v_jv_k$$

but I don't see how that would work just yet. I suppose you need the r.h.s to be:

$$k||\vec v||\vec v$$

If you can write that as a tensor relationship, you can rotate it to any coordinate system and it is guaranteed to work.