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If someone goes from his car to his front door in a rainstorm, will he get more wet, less wet, or equally wet if he runs (rather than walks)? This is a very commonly asked question. To develop a quantitative answer, let's first consider a spherical man, and assume he moves to his car in a straight horizontal path with velocity u. The raindrops are falling at an angle such that its velocity is v_z in the downward direction, v_x in the horizontal direction (straight into the man's face), and v_y in the sideways horizontal direction (the man's left to right). The intensity of the rain is such that each cubic foot of air contains G grams of water.

Relative to the rain's frame of reference the raindrops are stationary and the man and his car both have an upward velocity v_z and a sidewaysvelocity (right to left) of v_y. In addition, the man has a forward horizontal velocity of u + v_x. Clearly the amount of rain encountered by the man is equal to G times the volume of space he sweeps out as he moves relative to this stationary mist of raindrops. Since he is spherical with radius R, this swept volume is essentially equal to pi*R^2*L, where L is the distance travelled (relative to the rain's frame of reference).

If D is the horizontal distance to the car, and the man moves straight to his car with velocity u, the time it takes him is D/u. His total velocity relative to the falling rain is

________________________________

V_t = / (u + v_x)^2 + (v_y)^2 + (v_x)^2 (1)

so the distance he moves relative to the rain is (D/u)*V_t. Therefore, the amount of rain he encounters in the general case for arbitrary direction of rainfall is

_____________________________________

| / v_x\2 /v_y\2 /v_z\2

W = G*D*pi*R^2 | ( 1 + --- ) + ( --- ) + ( --- ) (2)

\| \ u / \ u / \ u /

We can easily incorporate other assumptions, such as the man having some non-spherical shape. It's just a matter of geometry to compute how much volume he sweeps out relative to the rain's frame of reference.

Notice that if v_x = v_y = 0 then the rain is falling vertically with a total velocity v = v_z. In this case equation (2) reduces to

____________

| / v \2

W = G*D*pi*R^2 | 1 + ( --- ) (3)

\| \ u /

which shows that the key parameter is the ratio of the rain's vertical speed to the man's horizontal speed. Of course, if v was zero (which would mean the rain was motionless relative to the ground), then L would always equal D, regardless of how fast the man runs. On the other hand, for any v greater than zero, the amount of rain he encounters will go down as his horizontal velocity u increases.

Incidentally, suppose we're waiting for the person in his car, and we intend to figure out how fast he ran based on how soggy he is when he reaches the car. We know the distance from the building to the car is D, and we assume the time he spends in the rain will be proportional to his wettness when he arrives. Thus we have W = cT, where T is his (presumed) travel time, w is his accumulated wettness upon arrival, and c is a constant. Therefore, we perceive his velocity to be

v = Dc/W (4)

If the rain is falling vertically with a constant velocity C, then for an appropriate choice of units we have

_____________

| / C \2

W = | 1 + ( --- ) (5)

\| \ V /

where V is the person's true speed (with time measured according to his wrist-watch, rather than according to his wettness). Substituting this into equation (4) gives the relation between the man's "proper velocity" and his "wettness velocity"

_______________

v/c | / v \2

----- = | 1 - ( --- ) (6)

V/C \| \ c /

This shows that, although the man can reduce his travel time to an arbitrarily short duration (according to his wrist-watch), we will never perceive him to have moved with a velocity greater than c.

Needless to say, equation (6) is the Lorentz transformation factor that plays a central role in the theory of special relativity, raising the question: Was Einstein all wet?

Well, if you want to go from A to B, running is better, because you will be exposed to the rain for less long. If you have a speed of zero, you will take an infinite amount of rain. If your speed is infinite, on the other hand, the rain will stand still in comparison. Hence you will take all the rain that is inside your path and no rain on the top of your head and your shoulders. The interesting thing to note is that you will take all the rain inside your path anyway, as long as you move, but the amount of rain that falls on top of you changes. So, running is better.

However, if the time you're exposed to the rain is fixed, walking is better. To see this, take the two extremes: Standing still and running. If you stand still, theoretically only the top of you (e.g. your head and your shoulders) will get wet (assuming the rain falls down straight). If you run, you will pick up the additional raindrops that are in front of you. The faster you run, the more space you will travel during a given amount of time, and hence the more raindrops you will pick up.

with help ... http://www.mathpages.com