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 walks or runs?  This
is a very commonly asked question.  To develop a quantative 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 you're waiting for the person in his car, and 
you intend to figure out how fast he ran based on how soggy he is 
when he reaches the car.  You know the distance from the building to 
the car is D, and you assume the time he spends in the rain will be 
proportional to his wettness when he arrives.  Thus, you have W = cT, 
where T is his (presumed) travel time, w is his accumulated wettness 
upon arrival, and c is a constant.  Therefore, you 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), you 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?