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Assume you have a sphere travelling at a velocity of $30$m/s, at angle $30^\circ$ relative to the local horizontal plane. For the purposes of the below equations please ignore more complicated influences on drag like turbulence.

It has a drag coefficient of $0.9$, and a frontal surface area of $0.2$m$^2$. I'm using a sphere because I assume the drag coefficient and frontal area is constant no matter the direction it travels. Therefore, using a basic formula for drag, this can be expressed by

$$D=\frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot V^2$$

My question is can you resolve this Drag force into its components by

$$D_x = D \cdot \cos\left(\frac{\pi}{6}\right)$$ $$D_y = D \cdot \sin\left(\frac{\pi}{6}\right)$$

Would this adequately resolve the drag force into its two components or do you need to resolve the velocity into its two components? e.g.

$$v_x = V \cdot \sin\left(\frac{\pi}{6}\right)$$ $$v_y = V \cdot \cos\left(\frac{\pi}{6}\right)$$

Then use those velocity components to calculate the drag acting in those directions? i.e.

$$D_x = \frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot (v_x)^2$$ $$D_y = \frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot (v_y)^2$$

I also assume the drag would be acting opposite the direction of the sphere, therefore become negative.

Are all of these interpretations correct? If they are incorrect please indicate which ones.

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  • $\begingroup$ which technique gives a drag-force-magnitude that does not depend on the coordinate system chosen? $\endgroup$ – JEB Aug 10 at 2:40
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Drag acts in the opposite direction of velocity. Therefore, if $$v_x = V \cdot \cos\left(\frac{\pi}{6}\right)$$ $$v_y= V \cdot \sin\left(\frac{\pi}{6}\right)$$

Then $$D_x = -D \cdot \cos\left(\frac{\pi}{6}\right)$$ $$D_y =-D \cdot \sin\left(\frac{\pi}{6}\right)$$

This is because, in general, if for vectors $\mathbf a$ and $\mathbf b$ the following is true $$\mathbf b\propto-\mathbf a$$ then it's also true that $$b_x\propto-a_x$$ $$b_y\propto-a_y$$

However, it is not true in your case that $$D_x = \frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot (v_x)^2$$ $$D_y = \frac{1}{2} \cdot \rho \cdot A \cdot C_d \cdot (v_y)^2$$

You just need to take the drag force and break it into components using the correct trigonometric functions as done above. You don't replace all instances of $a$ with $a_i$ for each general vector $\mathbf a$.

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In the general case, we can write the drag force in vector form $$\vec F_D=\frac {1}{2}\rho A C_D(Re)|\vec v_0-\vec v|(\vec v_0-\vec v)$$ here $\vec v_0$ is wind speed, $\vec v$ is body speed, $Re$ is Reynolds number.
In the particular case of $\vec v_0=0$, we have $$\vec F_D=-\frac {1}{2}\rho A C_D(Re)|\vec v|\vec v$$ The projections of the force on the coordinate axis have the form $$(F_D)_{x}=-\frac {1}{2}\rho A C_D(Re)|\vec v|v_x, (F_D)_{y}=-\frac {1}{2}\rho A C_D(Re)|\vec v|v_y$$ In this case $|\vec v|=\sqrt {v_x^2+v_y^2}$. For a spherical particle, I can recommend the empirical formula for the drag coefficient $$C_D=\frac {21.1}{Re}+\frac {6.3}{Re^{0.5}}+0.25$$ For $Re>>1$, we have $C_D=0.25$ (this is not 0.9).

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  • $\begingroup$ Sorry, I was just using that as an example drag coefficient, not a realistic one. I was more interested in the difference between decomposing the drag force versus the velocity. $\endgroup$ –  Bodmas12 Aug 10 at 7:01
  • $\begingroup$ @Bodmas12 Did you get an answer? $\endgroup$ – Alex Trounev Aug 10 at 7:34

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