My brother claims that a person will get less wet walking through rain that is falling vertically than running through it. He says that when running less rain may fall on top of the head and shoulders, but far more rain will come in contact with the front part of the person's body. Therefore, the best way to get the least wet in a rainstorm is to walk slowly through it. Can anyone confirm this?
Cecil Adams at the Straight Dope tested it with a small experiment, and then cited Thomas Peterson and Trevor Wallis of the National Climatic Data Center, who also did an experiment:
"One rainy day the two men donned identical sweat suits and hats, which they'd weighed before the test. For added accuracy, they wore plastic garbage bags underneath the sweat suits to keep their underclothes from wicking away any water. They then set out through the downpour on a 100-meter course. Wallas ran; Peterson walked.
"When they finished, the men weighed their clothes again to find out how much water they'd soaked up. Peterson's had absorbed about seven and half ounces (212 g), while Wallis's sopped up only four and a half (127 g)."
In short, running will keep you drier than walking.
The original paper is:
Peterson, T. C. and Wallis, T. W. R. (1997), Running in the rain. Weather, 52: 93–96. doi:10.1002/j.1477-8696.1997.tb06281.x
MythBusters also tried it... twice, and the second time they confirmed it
TLDR: If the rain is coming down without angle it is better to maximise speed.
Let's assume that the cow is spherical .. sorry, assume that the walker is a box with height h, width w and length l. The box needs to cover a horizontal distance D while being hit by a minimal amount of water.
Further assume that the air is uniformly filled with ρ kilograms of water per cubic meter, moving straight downwards at a uniform speed vr (since we imagine that the raindrops have reached terminal velocity).
If the box is walking at speed vh, how much water hits him? In each infinitesimal moment Δt, the water in the air falls a vertical distance of Δtvr, and Δtvrwlρ amount of water hits the top of the box. The total amount of water to hit the top during the trip is Dvhvrwlρ.
Similarly, during each infinitesimal moment Δt the front of the box pushes into a volume Δtvhhw of rain-filled air. The rain is moving downwards, but hits the box nevertheless, so the total amount of water to hit the front is Dvhvhhwρ=Dhwρ -- in other words this amount is independent of vh. The only influence vh has is that the larger vh is, the less water hits the top of the box. So under the above simplifying assumptions one should indeed attempt to maximize speed.
The central argument of this answer is theoretical in nature. We do not allow answers based uniquely on common sense or pure logic. Answers which are wholly based on a theoretical model are generally downvoted and may be deleted. See FAQ: What are theoretical answers?