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?

  • $\begingroup$ Apoligies for my tagging, I'm a bit lost here :( Also, apologies if my question doesn't belong on this site. $\endgroup$
    – going
    Jun 5 '12 at 10:50
  • 7
    $\begingroup$ mathoverflow.net/questions/5954/… $\endgroup$
    – t.b.
    Jun 5 '12 at 10:58
  • 4
    $\begingroup$ There are also other solutions to this problem! $\endgroup$
    – AD.
    Jun 5 '12 at 11:10
  • $\begingroup$ @t.b. - I see that is closed as off topic. SOrry about that. I will flag for moderators. Thanks. $\endgroup$
    – going
    Jun 5 '12 at 11:20
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    $\begingroup$ I wouldn't do that. MO has this "research level" standard that isn't in place here. This site is open for mathematical questions on all levels and there's no close vote here, yet, and why not wait a little? The worst that can happen is that it's going to be closed, but I wouldn't bet on it happening. There is mathematics in your question and there are lots of recreational maths papers on this topic, so I'd just wait if I were you. $\endgroup$
    – t.b.
    Jun 5 '12 at 11:23

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.

  • $\begingroup$ @DonAntonio Thanks, I am not alone with this weird humor. +1 $\endgroup$
    – AD.
    Jun 5 '12 at 16:34
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    $\begingroup$ Any time. Apparently some around here have the sense of humour of two pounds of garlic, so I'm forced to erase my answer in order to stop these bricks from downvoting it. $\endgroup$
    – DonAntonio
    Jun 6 '12 at 12:00
  • $\begingroup$ The article cited in the answer accepted at MO, which is available online, treats both the box-shaped human and the spherical cow, with the result that a box-shaped human should run as fast as possible at low tailwind speeds (relative to the crosswind, taking into account the box dimensions) but at the speed of the tailwind at high tailwind speeds, whereas a spherical cow has an optimal finite speed for any non-zero tailwind speed. $\endgroup$
    – joriki
    Aug 7 '12 at 14:28
  • $\begingroup$ @joriki: Interesting. Intuitively I'd have thought that a spherical cow was roughly equivalent to a cubical box. On further thought, however, in this case it is easier to view things from the rest frame of the falling rain, where the target describes a path through the rain. The cow wants to carve out the shortest possible cylindrical tunnel through the rain to the target. The ideal is then to move perpendicularly to the target's path. If there's a headwind this would require moving backwards in time (or into the ground), but with a tailwind this ideal is achievable exactly. $\endgroup$ Aug 7 '12 at 16:02
  • $\begingroup$ In contrast, a tall box-shaped human can (in some situations) (locally) minimize the cross section of his tunnel by moving in the direction of his vertical axis, even though the distance he covers through the rain is nor minimal. Conversely, a short fat box-shaped human can minimize the cross section of his tunnel by choosing its direction to be as close to horizontal as he can make it. (I.e., run as fast as possible). $\endgroup$ Aug 7 '12 at 16:05

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.

  • 1
    $\begingroup$ The conclusion does not sound right intutively. Suppose that the rain is falling with an extremely small vertical velocity, and you have a tailwind of, say 1 m/s. Then, if you walk at exactly 1 m/s together with the rain, neither your front nor your back is going to get wet, and you get only the negligible amount of moisture due to $v_v$. However if you then increase your speed to 2 m/s, now you're going to plow into a lot of rain-filled air along the way, too much to offset the decrease in the negligible top-area rain you get from getting there faster. $\endgroup$ Jun 5 '12 at 14:12
  • $\begingroup$ Mathematically, at least part of the problem is that the rain can hit either our front or our back, so the $(v+v_h)$ factor should be $|v+v_h|$, and then you need two different cases in the analysis when $v_h$ is negative. $\endgroup$ Jun 5 '12 at 14:28

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