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

(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 = KN/m we get

where it should be clear that the terminal velocity
Vmax is obtained when the acceleration is zero, that is, when
Vmax = , which is . Integration of this equation will be much simpler if we rewrite it by
using aN = g/C. This puts the problem in the
form

(4)

where we note that a = Vmax.

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

so the solution is

(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.

Jim Carr
2000-02-17