Re: Does angle of rain affect whether to run/walk through the rain?

Start with the answer to 873909296.Ph

If v is not vertical but at some angle q _r then what we called "w" in the previous answer has a larger x component. This addition to the x component is v sin(q _r). So the components of w are then
w_x = v_x + s
where v_x = v sin(q _r) and
w_y = v_y
where v_y = v cos(q _r)
The angle that w makes with the vertical is then tan^-1(w _x / w_y) = q_w

Using the areas of the person's body as in the previous answer we now get, for the total rain collected,
RC = w cos(q_w) D T t + w sin(q_w) D 5 T t
where, as before, t = x / s

It can be shown that cos(q_w) = v cos(q_r) / w
and
sin(q_w) = (v sin(q_r) + s ) / w

It can also be shown, although it's not necessary for the answer to this exercise, that
w = (v² + 2vs sin(q_r) + s²) ^0.5

So the rain collected is now
RC = D x [ (v T cos(q_r) / s ) + (5 v T sin(q_r) / s ) + 5 T ]

As before, if s = 0 we obtain an infinite amount of rain collected by the body, so don't stand in the rain even if the rain is not falling straight down. This makes sense.

If s approaches infinity then the first two terms in the brackets become zero and we are left, as before, with the term 5 T so that the total rain collected is D x 5 T, which is the same as before where the rain was falling straight down. If you think about going infinitely fast and sweeping up all the rain drops that are in a body-shaped row that is x units long, then this answer makes sense, as before.

John Link, MadSci Physicist