> Like Don, I'm confused about what happens at the border case of rain falling straight down. It doesn't make logical sense to me that if the rain is falling at you at 90.00000001 degrees, that you should run infinitely fast, while if the rain is falling behind you at 89.99999999 degrees, you should only walk at what would be a snail's pace, because there would be VERY little horizontal rain speed.

Yeah, you're right, that doesn't make sense. The error I made was in choosing values of the rain angle which were too large, and not even checking small angles. What I did this time was to do a 3-D plot of the total rainfall as a function f of both V and A, in the case where the rain is angled in your direction. I graphed the ENTIRE range of A, from 0 to pi/2 (if A is bigger than that, the rain is moving upward, which isn't a realistic condition). I could send you a .gif of the plot, if you'd like.

It turns out that for A less than pi/4, f(V,A) looks just like the case for the rain angled backward, as V increases, the rainfall decreases, so you want to go as fast as possible. But when A hits pi/4, the graph flattens out, and for A between pi/4 and pi/2, f(V,A) has a global minimum at V = R Sin(A). The border case, A = pi/4 is interesting in that it is *perfectly* flat for V bigger than R Sin(A). That means that if the rain speed is 10 m/s at exactly a 45 degree angle and hitting your back, you get less and less total rainfall (for a fixed travelling distance) as you increase your speed up to 5 sqrt(2) m/s, and then for any speed above 5 sqrt(2) m/s, you get exactly the same total rainfall, no matter how fast you go.

So if the angle of the rainfall from the vertical is 44.999 degrees, you run as fast as possible. If the angle of the rainfall from the vertical is 45.001 degrees, you go at the horizontal velocity of the rain. If the rainfall is exactly 45 degrees from the vertical, you go at LEAST as fast as the horizontal velocity of the rain, and it doesn't matter how far past that you go. [As velocity increases, the tradeoffs between getting hit with more rain per second and having a shorter travel time exactly balance.]

In terms of asymptotes, if you leave off all the constants and units, f approaches 1 as v approaches infinity, for all angles A. For A less than 45 degrees, f approaches the asymptote from above, and has no global or local minima. For A equal to 45 degrees, f decreases until it hits 1 at V = R Sin(45 degrees) = R/sqrt(2), and then lies on the asymptote for larger V. For A bigger than 45 degrees, f decreases, dips slightly below 1 at V = R Sin(A), and then for larger V increases toward the asymptote at 1.

In a check of the limit case A = pi/2 (horizontally moving rain), the local minimum is at V = R Sin(pi/2) = R, with a value of f = 0. You move at the rain's speed, and don't get wet at all.

Corrected final analysis (I hope):

If the rain is coming in from ahead, or coming in from behind at less than a 45 degree angle from the vertical, go as fast as you can.

If the rain is coming in from behind at more than a 45 degree angle, go at the horizontal speed of the rain.