> I'll let gremlinn take it from here.

Ok, time for some simplifications. We'll use Newtonian mechanics (no relativistic effects), point-sized raindrops, constant rainfall velocity and raindrop density in the air, a perfectly flat ground, and a person in the shape of a cube 1 meter to each side (it makes the calculations nicer). Also if the person walks in the north/south direction, the rain doesn't have any velocity component to the west/east. I could try adding that in later, maybe.

Let's say the rain's velocity is R m/s, falling at an angle A from the vertical, in the opposite direction the person walks. The person walks with velocity V m/s. So, a quick calculation shows that in the reference frame of the person, the rain falls with velocity V' = [R^2 + V^2 + 2RV Sin(A)]^0.5 at an angle of A' = Arctan[(R Sin(A) + V)/(R Cos(A)].

Let's say the density of water drops in the air is D drops/m^3. The amount of water that hits the person per second can be calculated from the volume of the cylinder projected over the visible surface area with height V' (the apparent velocity).

The visible surface area of the cube can be shown to be Cos(A') + Sin(A') m^2, and the height is V' meters. So the number of waterdrops that hit the person per second is DV'[Cos(A') + Sin(A')].

Now, if the person has to walk a length of L meters, the total time spent walking will be L/V seconds, so the total amount of water that hits him will be DLV'[Cos(A') + Sin(A')]/V. Plugging everything back in, here's how to solve the problem:

Minimize, as a function of V (V>0), the value of

DL[R^2 + V^2 + 2RV Sin(A)]^0.5 * [Cos(Arctan[(R Sin(A) + V)/(R Cos(A))]) + Sin(Arctan[(R Sin(A) + V)/(R Cos(A))])]/V

D and L are just multiplicative constants here so they don't affect what value for V is optimal.

So, I picked a couple values (R = 8 m/s, A = 30 degrees) and plugged it into Mathematica.

I plotted the rainfall as a function of V. As expected, as V approaches 0, the amount of water that hits you goes up to infinity. But there is NO minimum in the curve. As V increases, the value decreases to an asymptote. The value of this asymptote turns out to be the total volume of water directly in front of you, and is independent of the rain speed (if you move really, really fast, you essentially just scoop out everything in a horizontal line without regard to how fast the rain is falling).

Ok, so the answer is: run as fast as you can. Just don't fall down.