Flaneur in the rain - i

A common misconception about this puzzle is that running quickly through the rain causes the runner to collect more rain on the front of their body. In fact this is not the case.

For both the vertical (head), and horizontal (body) rain collection, the runner simply collects the amount of water that exists in a rectangular prism whose dimensions are given by the details of their motion.

Head

The dimensions of the rectangular prism that is incident on the head of the runner is simply given by the cross sectional area of the head, times the velocity of the falling water, times the amount of time the runner is exposed to the rain. I.e. if the runner stands still in the rain for $t_0$ seconds, and the rain falls at $v_\text{rain}$ , a rectangular column of height $v_\text{rain}t_0$ will fall on their head. In crossing the road, the runner spends $d/v_0$ seconds in the rain, in which time they collect an amount of water $\rho A_\text{head}v_\text{rain}d/v_0$ on their head. Clearly, this quantity can be reduced by increasing the velocity with which one crosses the road.

Body

Following the argument above, we can see that the runner moves through a rectangular prism of dimensions $A_\text{body}d$ . Notice that this has no dependence on the velocity of the runner, the speed of the falling rain, or anything else that can be changed by walking at differing speeds. The only parameter that can be modified is the cross sectional area that the body presents to the direction of motion, $A_\text{body}$ . By shifting the body by $\pi/2$ , the runner presumably slims his profile relative to the rain and therefore collects a smaller amount of water.

All together

Putting these together, we see that in crossing the road, a runner will collect a mass of water given by

$m_w = \rho d\left(A_\text{body} + A_\text{head}\frac{v_\text{rain}}{v_0}\right)$

Which can be reduced by minimizing $A_\text{body}$ and maximizing $v_0$ .