I am trying to make a simple 2d drag simulator for an object in freefall with an x and y component of wind. If I knew the coefficient of drag on the bottom and side of the object could I use theta to get the approximate CD at an angle?

Probably and oversimplified function but something like:

$CD_{\theta} = CD_{bottom} + (CD_{bottom} - CD_{side})*\frac{\theta}{\pi/2}$

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  • $\begingroup$ Don't forget the object will accelerate to constant velocity of any cross wind and then have no angle to the relative wind. Also, in spite of the rectangular shape, when released, there may be a "lift" component to the left, as it has an Angle of Attack (which grows less and less in the free fall, becoming drag only at the base). $\endgroup$ Commented Jan 8, 2020 at 12:13
  • $\begingroup$ You should transform the title in a question as this is standard on this Q&A website. $\endgroup$
    – Manu H
    Commented Jan 8, 2020 at 12:31

1 Answer 1


Let's try to derive something. Assume the x-axis points to the right and the y-axis points up. I'm gonna use the subscripts b and s for bottom and side. The incoming velocity vector is decomposed in two components along the axes, so $v_s=v\sin\theta$.

Writing the total drag force as a vector:

$\vec D=\begin{bmatrix}-D_s\\D_b\end{bmatrix}=\begin{bmatrix}-C_{D_s}\frac{1}{2}\rho S v_s^2\\C_{D_b}\frac{1}{2}\rho S v_b^2\end{bmatrix}$

Taking the magnitude of the vector:

$D=C_D\frac{1}{2}\rho S v^2 = \sqrt{{C_{D_s}}^2\left(\frac{1}{2}\rho S\right)^2 v_s^4+ {C_{D_b}}^2\left(\frac{1}{2}\rho S\right)^2 v_b^4}$

This can be further simplified to: $C_D = \sqrt{{C_{D_s}}^2\sin^4\theta+{C_{D_b}}^2\cos^4\theta}$

I don't think this can be simplified further.

Edit: S in this case is of course just the area of the rectangle, I got confused before

  • 1
    $\begingroup$ Ok, that makes sense to me. Do you have any suggestions on how to add varying surface area? You mention it is more complicated, but I am trying to set up the functions in sympy and am curious where that complexity comes from (and if it's worth looking in to). Looking a what you have written here the surface area for the drag magnitude seems to be where everything blows up in complexity. $\endgroup$
    – Boto
    Commented Jan 8, 2020 at 15:59
  • 1
    $\begingroup$ Sorry, I got confused, of course S is just the surface area of the rectangle. You could use this, take "long cylinder" for the bottom Cd and "short cylinder" for the side Cd. But tbh the results from this won't be very exact at all. To get a better estimate for the Cd you could for instance use a potential flow method. $\endgroup$
    – Daniel
    Commented Jan 8, 2020 at 19:33

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