Derivation of Drag Equation

[start: speculation]

I've always wondered about the how the Drag equation came about. It was presented in a very arbitrary fashion, like a equation that was found through experiment. 'Derivation' consisted of the use of dimension analysis of variables.

But while idly scribbling around yesterday night, I had a little idea how this drag equation $$F_d = \frac{1}{2}C_d\rho Av^2$$ came about.$$\LaTeX $$

Imagine a ball moving through the air. As it moves through the air, the little molecules of air collide with it. This is actually just a simple exercise of using the conservation of momentum, since its a big ball colliding with a really tiny ball.

$$Mv_i = Mv_f + mu_f$$
M is mass of ball, $$v_i$$ is the initial velocity of the ball. m is the mass of the small molecule. $$u_f$$ and $$v_f$$ is the respective final velocities of small molecule and ball.

We need a relation between final velocity of ball and initial velocity of ball, dependent of the ratio of masses of the 2 colliding things. So we use the equation that the speed of approach is equal to the speed of recede. $$v_i = u_f - v_f$$

Substituting $$u_f$$ away, and rearranging, we get $$ v_f = \frac{M-m}{M+m} v_i$$

But the reality is that the ball is ploughing through this sea of molecules, and even as it bounces one away, another comes in to collide with it. The ball slowly loses its kinetic energy this way. The amount of 'mass' of little molecules that the ball meets is equal to $$\rho Avt$$. How to incorporate this into our previous relation from applying conservation of momentum?

First of all, we could find the rate of change of $$v_f$$, this could give us acceleration involved.
$$\frac{dv_f}{dt} = [ -\frac{dm}{dt} (\frac{1}{M+m} ) -2(\frac{M-m}{(M+m)^2}) \frac{dm}{dt} ] v_i$$
Here, we implicitly assume the $$v_i$$ is constant, that is we pretend we are finding the deceleration felt by a ball kept steady while it is being bombarded by a stream of air moving with speed $$v_i$$. This is over simplification for the moment, but lets continue. m is changing, and its change rate is represented by $$\frac{dm}{dt} = \rho Av_i$$.

Cleaning thus the equation,
$$\frac{dv_f}{dt} = -\rho Av_i ^2 [\frac{3M-m}{M^2 + 2Mm}]$$

The drag force is then simply F=ma, to see that it does not depend on the mass at all, notice that a term M could be factorised from the denominator from the previous equation, and that this cancels out with the M is the force equation.

Next is the question, where does $$C_d$$ come in? The answer is that model describes what happens due to air molecules colliding with the object. There are other components of drag force. There air rubbing lengthwise with the object, generates friction and contributes to drag. Air that speeds up and slows down while moving around an object generate pressure differences, and that also contribute to drag. These components of drag vary in strength with the shape, size and velocity of the object. Therefore the coefficient of drag is an measure of the strength of a particular component.

[speculation: end]