Originally Posted by
gregf83
The power to overcome aerodynamic drag is proportional to V^3. Drag (force) is proportional to V^2.
This man got it right. The reason for the additional factor is that the higher speed means higher force (v * v)
in less time (1/t where t = k/v).
P = f * d / t = k1* v * v / (k2 / v) = k3 * v * v * v.
How well the f = k * v * v as the sole drag component assumption holds is subject to debate.