Running in the rain, good or bad idea? I originally wanted to ask this on the Skeptiks site, but while having a look around I came across this site which seems to have some complicated maths, but not a reliable answer.
The conclusion it attempted to arrive at:

So here we have it - more mathematical advice to avoid getting wet.
  Because we divide by VP in this equation, maximising our velocity now
  emerges as a good idea, assuming there is a shelter available.

However, this mentions that the longer you stay in the rain the wetter you get. So, my question is: Is there a point where if the nearest shelter is at a certain distance it is better to walk to that shelter rather than run? or Is it always better to run no matter the distance of the shelter?
 A: Let's assume that the cow is spherical .. sorry, assume that the walker is a box with height $h$, width $w$ and length $l$. The box needs to cover a horizontal distance $D$ while being hit by a minimal amount of water.
Further assume that the air is uniformly filled with $\rho$ kilograms of water per cubic meter, moving straight downwards at a uniform speed $v_r$ (since we imagine that the raindrops have reached terminal velocity).
If the box is walking at speed $v_h$, how much water hits him? In each infinitesimal moment $\Delta t$, the water in the air falls a vertical distance of $\Delta t v_r$, and $\Delta t v_r w l\rho$ amount of water hits the top of the box. The total amount of water to hit the top during the trip is $\frac{D}{v_h} v_r w l \rho$.
Similarly, during each infinitesimal moment $\Delta t$ the front of the box pushes into a volume $\Delta t v_h h w$ of rain-filled air. The rain is moving downwards, but hits the box nevertheless, so the total amount of water to hit the front is $\frac{D}{v_h} v_h h w \rho = D h w \rho$ -- in other words this amount is independent of $v_h$. The only influence $v_h$ has is that the larger $v_h$ is, the less water hits the top of the box. So under the above simplifying assumptions one should indeed attempt to maximize speed.
Things get rather more interesting if the rain is falling at an angle -- then the amount of rail to hit the front/back of the box does depend on $v_h$ and matters are not so simple anymore. If I remember correctly, then if there is a tailwind, there's an optimal walking/running speed above which the amount of water to hit the front increases with speed faster than the amount of water to hit the top decreases. (But I don't have time to derive that right now).
More specifically, if there's so much tailwind that, standing still, you would receive more water on your back than on your top, then you should run exactly at the wind speed in order to minimize the amount of water hitting you.
A: Intuitive Idea :
Assume that rain is falling at some constant velocity at some angle and our body's front area and top area are constant.Also there is shelter at some distance.
So,You're going to pass through a 
volume of watery air, and you're going to absorb all the water that's 
in the air that you pass through. The amount of water per unit volume 
doesn't change over time - the number of drops falling out of that 
volume are the same as the number falling in - so it doesn't matter 
when you get to a particular piece of that volume or how fast you're 
going when you get there. Imagine if we could stop gravity for a few 
minutes so the raindrops would just stay where they are. Would it 
matter how fast you went?Hence there will always be some constant amount of water you will face inspite of your velocity.
So where our velocity comes into picture?
Our velocity also has a role to play because rain has vertical velocity and horizontal velocity.So,we will get wet in top area due to the vertical velocity of rain as well as in front area due to horizontal velocity.
Recapitulating:Our wetness is due to three reasons of which one is constant and other two are due to rain's velocity. 

Mathematical Approach:
I am measuring the amount of wetness by the mass of water that falls on our body.
Let the density of water in surrounding be $\rho$.
Front area of our body $A$ and top area $a$.
Suppose we are walking at velocity v = $v$ i.
And Rain's velocity is $v_h$ i + $v_v$ j.
{Ofcourse if angle from ground $\theta$ is given then $v_h$ = $v_r\cos\theta$ and $v_v$ = $v_r\sin\theta$.}

Now, the Wetness
$$\begin{align*}W &= \text{mass of the water that fell on the body}\\  
                  &= \text{density} \times \text{volume covered during the run}\\
                  &=\rho\ *\ [\underbrace{A(v+v_h)t}_\text{horizontal contribution} + \underbrace{av_vt}_\text{vertical contribution} ] \\
                  &=\rho\ *[\ Avt + av_vt + Av_ht]\\
                  &=\rho\ *[\ Ad + av_vt + Av_ht]\\
                  &=\rho\ *[\ Ad + av_v\frac{d}{v} + Av_h\frac{d}{v}]\\
\end{align*}$$
here $d(=vt)$ is the nearest distance from the shelter.
Note Here I have taken the horizontal velocity of rain and our velocity in the opposite direction that's why the relative velocity is $v+v_h$.

So, to minimize the wetness we have to increase $v$ as it is in the denominator.
Also you can compare this relation with the Intuitive Idea that I have given above about the constant term and other terms arriving due to the horizontal and vertical velocity of rain.

If anybody finds any mistakes the please notify.
