Re: walk to enjoy the rain

[David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>]

Regarding Joseph J. Scherrer's solution:

> > One question - does one get wetter walking in the rain or running. We are
> > thinking of doing an expt with two students, two lab coats, and measure
> > mass of coat before and after. Saving it for a rainy day.
>
> By running, theoretically, as may be inferred from the limiting case as
> v--> infinity.

The solution to this question depends on the conditions under which the
question is asked.  If the *time* spent in the rain is held fixed and the
running/walking speed is varied, then under these circumstances Joseph's
solution is correct.  Also the minimal amount of wetting in this
circumstance occurs for a person standing upright along the direction of
the rainfall so as to present a mimimal cross sectional area to the
falling rain.

*But*, if, instead of the time, the *(horizontal) distance traveled in
the rain* is held fixed, then the solution is different for this (more
practical case).  This problem is a simple application of the ideas that:

1. The current density j for a for a fluid of density [rho] moving with
   velocity v is j = [rho]*v.
2. The vector velocity v of the rain moving with respect to the person is
   the vector difference v = v_p - v_r where v_p is the vector velocity
   of the running/walking person w.r.t. the ground and v_r is the
   (fixed for an assumed monodisperse distribution of rain drops)
   terminal velocity of the rain drops w.r.t. the ground.
3. The current I of rain encountered by the person is I = |j|*A where A
   is the cross sectional area of the person presented perpendicular to
   the direction of the vector v.
4. The mass M of water deposited on the person is M = I*t where t is the
   time spent in the rain.
5. The time t spent in the rain is t = L/|v_p| where L is the fixed
   (horizontal) distance tha must be covered by the person in the rain
   (say the distance between the person's parked car and the door to the
   person's office building).
6. The projected cross section A is assumed composed of essentially two
   parts, a 'horizontal' part A_h composed of the area of the tops of the
   person's shoulders and head, and a 'vertical' part A_v composed of the
   person's face and fronts of the neck, torso, arms, and legs.  If we
   assume the rain is falling vertically and the person is moving
   horizontally so that the surface of A_h is perpendicular to the
   direction of v_r and A_v is perpendicular to the direction of v_p,
   then the projected cross section A is given by:
   A = (|v_p|*A_v + |v_r|*A_h)/sqrt(|v_p|^2 + |v_r|^2)
8. Using the assumptions in 6. for the directions of v_p and v_r we note
   that 2. implies that |v| = sqrt(|v_p|^2 + |v_r|^2) and this along with
   1., 3. and 6. imply that I = [rho]*(|v_p|*A_v + |v_r|*A_h).
9. Combining this last result of 8. with 4. and 5. gives us the final
   result that:

   M = [rho]*L*(A_v + A_h*|v_r|/|v_p|).

The first term here represents the water deposited on the person's front
vertical surfaces and the second term is the water deposited on person'
upper horizontal surfaces.  The first term is *independent* of both the
speed of the person's walking speed and the speed of the falling rain.
This means that the front of a person will tend to get just as wet no
matter how fast they go from their car to their office building.  But the
second term is proportional to the ratio |v_r|/|v_p| (because it is
proportional to the total time spent in the rain) so we see that the
amount of water deposited on the head and shoulders is inversely
proportional to the person's walking/running speed.  We conclude that
running will keep the person's head and shoulders drier than walking, but
it will have no effect on the person's front side.

It should be noted that running also has the added benefit that it
minimised the time spent in the rain and makes the drying out process
start as soon as possible.

Btw, in case anyone is inclined to misunderstand the meaning of [rho] in
the above derivation, it is the average mass density of rain water in the
air.  This means it the mass density of liquid water (i.e. 1000 kg/m^3)
*times* the volume fraction of the air occupied by the rain drops.

David Bowman
dbowman@georgetowncollege.edu