Resolve Drag into Vector Components

Question

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.

Answer 1

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 $\mathbf a$.

Answer 2

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).