Published by Adrien - Monday, October 7, 2024 - Other Languages: FR, DE, ES, PT

**Source:** The Conversation under Creative Commons license

You've certainly experienced this situation before, whether in light rain or during a storm. Let's approach the problem from a physics standpoint and try to calculate the amount of water that will fall on you as a function of your speed.

You're outside, under uncertain skies, and it starts to rain while you don't have an umbrella. The instinct is to lean forward and speed up your pace, right? That way, it feels like you'll get less wet. You might even accept getting wetter as long as it doesn't last as long.

But is this behavior justified? Can we create a model that allows us to answer this rather critical question? In particular, does the amount of water you receive depend on your speed? Is there a speed at which the amount of water you receive, while traveling from one place to another, is minimal?

Let's keep this simple while retaining the key elements of the situation. Consider homogeneous rain falling vertically. To simplify, we can assume that the walker presents vertical surfaces (the front and back of the body) and horizontal surfaces (the head and shoulders) to the rain.

Let's start with the vertical surfaces. As you move forward, you encounter the raindrops: the faster you go, the more you'll come across. From your point of view, the raindrops fall at an angle, with a component of speed exactly equal to your walking speed: the faster you go, the more raindrops you'll receive. But, to travel from point A to point B, you'll take less time, and the faster you go, the less time you spend getting rained on! Thus, the two effects exactly balance out: more raindrops per unit of time, but less time spent in the rain.

What about the horizontal surfaces? When the walker is standing still, they only receive rain on these surfaces. When you watch someone walk, you'll see that they receive raindrops that used to pass in front of them, but they no longer receive drops that now pass behind them: overall, per unit of time, the amount of rain they receive on these horizontal surfaces is independent of their walking speed. However, since the total walking time decreases as speed increases, the amount of water received on horizontal surfaces will be lower.

In summary, it indeed makes sense to pick up the pace.

For those who love a mathematical treatment of things, here's something to satisfy you:

Let ρ represent the number of raindrops per unit volume, and a represent their vertical velocity. Let Sh be the horizontal surface area of the individual, and Sv the vertical surface area.

If you're standing still, you'll only receive rain on your head and shoulders, which is essentially the amount of water that falls on the Sh surface area.

Even if the rain falls vertically, from the perspective of a walker moving at speed v, the rain arrives at an angle, in a direction that depends on the speed v.

During a time interval T, a raindrop covers a distance a·T. So every raindrop within that distance will hit this surface: these are the raindrops found in the cylinder with a base of Sh and a height of a·T, which gives:

ρ.Sh.a.T.

As we've seen, once you start moving, the raindrops appear to be moving at an angle due to the combination of speed a and speed v. The number of raindrops hitting Sh remains unchanged because speed v is horizontal and, therefore, parallel to Sh. However, the number of raindrops hitting the Sv surface, which was zero when the walker was motionless, is now equal to the number of raindrops contained in a (horizontal) cylinder with a base of Sv and a length of v·T, since this length represents the horizontal distance traveled by the raindrops during this time interval.

In total, the walker receives a number of raindrops given by the expression:

ρ.(Sh.a + Sv.v). T.

We now need to account for the time interval during which the walker will get wet. If the walker needs to cover a distance d at a constant speed v, the time interval is given by the ratio d/v (this obviously assumes v is non-zero!). Substituting into the above expression, we obtain the final result:

ρ.(Sh.a + Sv.v). d/v = ρ.(Sh.a/v + Sv). d.

Thus, we get two key results:

- On the one hand, the amount of water received on the head and shoulders decreases as speed increases.

- On the other hand, the amount of water received on the vertical part of the body is independent of speed, as the shorter travel time is exactly offset by the number of raindrops hitting the walker.

The takeaway: you are indeed right to lean forward and pick up the pace! But be careful, leaning forward increases Sh, so the increase in speed must compensate for that!

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