Mythbusters asked the question of whether it's better to run or walk in the rain, assuming you're trying to avoid getting soaked. They did it one way, and then again a different way, and I think they finally decided it was better to run. My friend Dennis asked me this question, he lives in Portland, where this is a nontrivial everyday problem. After thinking about it while out for a walk with the baby, I came up with the following basic model.

Suppose that the density of rain in the air is $\rho$ and the rain velocity is $v$ at an angle $\theta$ from the vertical, with positive theta meaning the wind is blowing into your face. In a time $dt$ the amount of rain that falls on your head is $A_v \rho v \cos(\theta) dt$ where $A_v$ is the cross sectional target area you present vertically. The amount of rain that falls on your front is $A_h \rho (v \sin(\theta) + v_h) dt$ where $v_h$ is your horizontal running velocity. Suppose everything is constant during your dash for the front door so that gusting and your acceleration and soforth are ignored, then rather than integrating with time we can simply multiply by the time taken. The time it takes you is $L/v_h$ where $L$ is the length to your front door. Then the total water falling on you during your dash is:

$\rho v \frac{L}{v_h} (A_v \cos(\theta) + A_h(\sin(\theta)+v_h/v))$

This expression has a term that is constant with $v_h$ and independent of $v$, and two terms that go like $1/v_h$. This reflects the fact that we need to cover the distance L regardless of how fast we do it (intercepting at least the rain that is in front of us right now). The term in $\cos(\theta)$ is always positive since $\cos$ is an even function and $\theta$ is between $-\pi/2$ and $\pi/2$, and the term in $\sin(\theta)$ can be positive or negative depending on whether we're running into the rain or with the rain.

Is there an optimum running velocity? Let's take the derivative with respect to $v_h$ and look for where it goes to zero (a local minimum or maximum).

$\frac{dW}{dv_h} = - \rho v \frac{L}{v_h^2}(A_h \sin(\theta) + A_v \cos(\theta))$

This goes to zero at $v_h \rightarrow \infty$ but does it go to zero anywhere else? If we multiply through by $\frac{-v_h^2}{\rho v L}$ we get $(A_h \sin(\theta) + A_v \cos(\theta)) = 0$ which occurs at $A_h \sin(\theta) = - A_v \cos(\theta)$. This is true when $\theta = -\tan^{-1}(A_v/A_h)$. This suggests that at this certain angle, increasing the speed of running reduces the rain falling on your head but increases the rain falling on your front by an equal amount.

So unless you happen to be in this special wind condition, run hard Dennis!

3 Responses leave one →
1. Adam permalink
November 23, 2014

Hi Daniel! I love the idea of modelling this and I think you've described it in a way that's easy to follow. I just wanted to ask about theta: why is theta between positive and negative pi/2?

• November 23, 2014

theta is between +- pi/2 because it's measured relative to the vertical coordinate axis, and the rain is falling down. For brief periods rain can fall up because of wind gusts or whatever, but the average velocity has to have a downward component.

• Adam permalink
November 23, 2014

Right, that makes sense! Thank you!