Back to the Introduction to Drag Forces.

2. General Solution for Simple Drag Parametrizations

The simplest parameterization of this problem takes the form of a drag force that goes as a constant K times a power of the speed. Assuming vertical free-fall, we can look at this in one dimension where

$$F_{\rm net} \; = \; m \frac{dv}{dt} \; = \; mg - K_N \; v^N$$ (3)

with the assumption that ``down'' is the positive direction. You will need to alter all of the signs if you need a different convention and insert the appropriate unit vectors to do the problem in more than one dimension.

It will prove convenient to rewrite this equation in terms of some different constants. With C = K_N/m we get

$$\frac{dv}{dt} \; = \; g - C v^N$$

where it should be clear that the terminal velocity V_max is obtained when the acceleration is zero, that is, when V_max = ${\root N \of {g/C}}$, which is ${\root N \of {mg/K_N}}$. Integration of this equation will be much simpler if we rewrite it by using a^N = g/C. This puts the problem in the form

$$\frac{dv}{dt} \; = \; C \left[ a^N - v^N \right] .$$ (4)

where we note that a = V_max.

The differential equation in Eq. (4) is solved by separation of variables. Some simple algebra gives

$$\frac{dv}{a^N - v^N} \; = \; C dt$$

so the solution is

$$\int_{v_0}^{v} \frac{dv}{a^N - v^N} \; = \; \int_{t_0}^{t} C dt$$ (5)

where v₀ and t₀ are the initial conditions. We will only be looking at t₀ = 0 and v₀ = 0, that is, free-fall from rest, here. You can insert other initial conditions if desired. My solution of the special cases in the next two sections will include enough detail that it should be clear how to handle those other initial conditions.

Note that the right hand side just gives C (t - t₀) , which reduces to Ct with our initial conditions.

For the cases of interest, N=1 and N=2, the left hand side is easily integrated. (The case for N=3 can be integrated but I don't see how to write v(t) in a simple form. In any case, N=3 is not relevant to our focus on simple models.) Details for the linear and quadratic drag solutions are given in the next two sections.

Next, Linear (N=1 case) Drag.