Relative Rain

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 sideways
velocity (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?