0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $v^{2}=v^{T}v$ v is a vector $\endgroup$ – Eli Mar 12 at 7:43
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.