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?

  • $\begingroup$ $v^{2}=v^{T}v$ v is a vector $\endgroup$
    – Eli
    Mar 12, 2019 at 7:43

2 Answers 2


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.


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