He is on the street, the weather is uncertain and it starts to rain, although he does not have an umbrella. Your first reflex is to lean forward and quicken your pace, right? This way he thinks he will get as little wet as possible. You may even accept getting wetter as long as it doesn’t last as long.
Is this behavior justified? Is it possible to build a model to answer this important question? Specifically, does the amount of water received depend on the speed? Is there a speed such that the amount of water received to go from one place to another is minimal?
The effect of tilt and speed
To answer these questions, let’s simplify things, but retain the important elements of the situation. Let us consider a homogeneous rain that falls vertically. Schematically, we can consider that the walker presents vertical surfaces (the front and back of the body) and horizontal surfaces (the head and shoulders) to the rain.
First, let’s look at vertical surfaces. The faster we walk, the more drops will fall on us. From our point of view, the drops fall at an angle, with a velocity component exactly equal to our own walking speed: the faster we go, the more drops we receive. But the faster we walk, the less time it will take us to get from one point to another. So the two effects cancel each other out: more drops per unit of time, but less rain time.
And what happens with horizontal surfaces? When the pedestrian is stopped, rain falls only on these surfaces. When we watch him walk, we see that he receives drops that previously passed in front of him, but he no longer receives drops that now pass behind him: in total, per unit of time, he receives an amount of rain on these horizontal surfaces that is independent of his walking speed. But since the total duration of the trip decreases with increasing speed, the amount of water received on the horizontal surfaces will be less.
Ultimately, it’s a good idea to walk faster.
The problem in mathematical terms
For those who like the mathematical approach to things, here is an explanation that will satisfy you:
Let us denote by ρ the number of drops per unit volume and per to its vertical speed. Let us denote by sh the horizontal surface of the individual and therefore Sv its vertical surface.
If we are standing, the rain will only fall on our head and shoulders, so this is the amount of water that falls on the surface sh.
Although the rain falls vertically, from the point of view of a walker moving at a speed varrives obliquely, in a direction that depends on the speed v.
During an interval of time Ta drop travels a distance a*T. Thus, all the drops that are at a shorter distance will reach this surface: they are the drops from the base cylinder sh and height a*Tthat is: ρ*Sh*a*T
As we have seen, as they move forward, the drops appear to move at an oblique speed, which results from the composition of the speed to and the speed v. The number of drops that reach sh does not change, because the speed v It is horizontal and, therefore, parallel to sh.
On the other hand, the number of droplets that reach the surface Svwhich was zero when the walker was standing, is now equal to the number of drops contained in a (horizontal) cylinder of base Sv and length v*Tsince this length represents the horizontal distance traveled by the droplets during this time interval.
In total, the walker receives a number of drops given by the expression: ρ*(Sh*a + Sv*v)*T
Now we must take into account the time interval during which the walker will get wet. If you have to travel a distance d at a constant speed vthe time interval is given by the quotient d/v (which obviously means that v is not zero). Transferring this to the previous expression, we obtain the final result: ρ(Sh*a + Sv*v)*d/v = ρ(Sh*a/v + Sv)*d
Therefore, we obtain the following double result:
- On the one hand, the amount of water received on the head and shoulders is less the higher the speed.
- On the other hand, the amount of water received in the vertical part of the body is independent of speed, since the reduction in travel time is exactly compensated by the increase in the number of drops received.
Moral: It’s a good idea to lean over and run. But be careful: bending over increases the horizontal surface at the mercy of the rain; so this increase must be compensated by the increase in speed.
Jacques Treiner He is a theoretical physicist at the Université Paris Cité.
This article was originally published in The Conversation.
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