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You have just finished work and it’s raining. This morning the sun was shining so you haven’t taken an umbrella with you and now you have to get to the bus stop while trying to avoid getting really wet. Should you run or walk?

Your instinct tells you that running is the best choice, but is it so? Let’s analyze the problem from a mathematical point of view.

Running or not running?

To simplify the situation, let’s imagine that your body is a rectangular cuboid: a solid that has six rectangular sides.

An example of a rectangular cuboid.

Imagine that there is no wind, so the rain falls vertically. You can thus define the following parameters of the problem:

$d$ = distance to be covered
$v$ = speed of the person
$V$ = speed of the rain
$q$ = rain density in liters per cubic meter

The last two parameters describe rain intensity, in particular $q$ measures how much water is present in a volume of one cubic meter of air if we imagine time has frozen.

Suppose you have to go from point A to point B, covering a distance $d$. Let’s try to understand how much water hits the vertical face and how much water hits the horizontal face of the rectangular parallelepiped that represents ”you”.

The rain hitting the vertical face during the movement is that found at the initial time inside the volume of a prism which has the vertical area as the base and the distance to be traveled as the height.

In the following image you can see two different prisms depending on whether you move slower (upper prism) or faster (lower prism).

The amount of water hitting the vertical face is that found inside a prism. The lower prism represents a faster pace, the upper prism represents a slower pace. In both cases, the height of the prism is the distance to be covered.

But pay attention! For the Cavalieri principle the volumes of these prisms are the same and so the amount of water hitting the vertical face does not change if we go slower or faster!

Let’s call the area of the vertical face $A_{vert}$. The volume of the prism is equal to $A_{vert} \cdot d$ so the amount of water hitting the vertical face $Q_{vert}$ is given by this volume multiplied by $q$:

$$Q_{vert} = A_{vert} \cdot d \cdot q $$

What happens if we consider the horizontal face? The situation is similar, in the sense that the amount of rain hitting the face is always that found inside a certain prism, but now the volume of the prism depends on the speed.

(To make the following image look clearer the distance AB has been represented smaller than before.)

The amount of water hitting the horizontal face is that included inside a prism. The lower prism represents a faster pace, the upper prism represents a slower pace. The height of the prism changes with the speed.

The height of the prism is no longer fixed but corresponds to the distance traveled by the rain in the time spent in going from A to B. The time taken to travel the route is given by $d / v$. To find the distance traveled by the rain we multiply this term by the speed of the rain, $V$, so the prism height is:

$$\displaystyle h = V \cdot \frac{d}{v}$$

The prism volume is given by $A_{horiz} \cdot h = A_{horiz} \cdot V \cdot d / v$ and the amount of rain hitting the horizontal face is obtained by multiplying this volume by water density $q$:

$$\displaystyle Q_{horiz} = q \cdot A_{horiz} \cdot h = q \cdot A_{horiz} \cdot V \cdot \frac{d}{v}$$

We see that for this component, as speed $v$ increases, the quantity of water is reduced more and more (going infinitely fast will prevent even a single drop hitting the horizontal face).

Now adding the two contributions $Q_{vert}$ and $Q_{horiz}$ we obtain the total quantity of water:

$$\displaystyle \begin{aligned} Q_{tot} & = Q_{vert} + Q_{horiz}\\ & = A_{vert} \cdot d \cdot q + q \cdot A_{horiz} \cdot V \cdot \frac{d}{v} \end{aligned}$$

And collecting common terms, we obtain the final formula:

$$ \displaystyle Q_{tot} = d \cdot q \cdot \left[A_{vert} + A_{horiz} \cdot \frac{V}{v}\right]$$

For very low speeds $v$ the amount of water hitting you is very large. For large speeds the second term inside the brackets is very small but you can’t go below the quantity $Q_{vert}$ of water.

Within the brackets you find a sort of effective area you can change by modifying your speed.

Here is the graph of the effective area at different speeds.

Graph of the effective area at different speeds. The parameters are $A_{vert} = 0.6\ m^2$, $A_{horiz} = 0.07\ m^2$, $V = 10\ m / s$.

Above a certain speed, going faster hardly reduces the effective area. Whether you are a normal person running at 5 m/s or a word class sprinter running at more than 10 m/s the amount of water hitting you changes very little.

Summing up: running is better but not much better, and running won’t change the amount of water hitting your torso and legs, but will reduce only the amount of water hitting your head and shoulders.

Real Life is More Complicated

Real life is often much more complicated than the models we use to analyze it. This problem is a good example of how even a seemingly simple question can become complex to deal with it rigorously.

For example, we could add some wind, which would provide the rain with a horizontal speed component. If the wind blows against you then increasing the speed will always lead to a decrease in the water hitting you, just like before.

If the wind blows in your favor, then, depending on the ratio between the horizontal and vertical area of your body, it may be better to run as fast as possible or to proceed at a speed equal to the horizontal speed of the rain (in this way you don’t get wet either on the front nor back of you. For details see the following article by David E. Bell: Walk or Run in the Rain?).

The wind is the easiest variable to add to the model, other variables are more complicated, for example:

  • The shape of our body is not that of a rectangular parallelepiped and moreover it changes according to the speed at which we proceed. When we run the torso is tilted forward and the legs will assume a different position.
  • The intensity of the rain may change over time. If the intensity is about to increase, then it is usually better to run faster reach your destination sooner.
  • There is a maximum level of soaking, beyond which you just can’t get any wetter. For this reason, if the journey is a long one and it’s going to rain pretty heavily all the way, you can relax and go at your preferred speed as one thing is already certain, you will get completely soaked!

It’s interesting to notice that when we find ourselves in these situations, unconsciously and without making precise calculations, our instinct can take into account so many variables (distance to be covered, possible increase in intensity of the rain, type of clothes, danger of slipping) to finally choose the strategy which seems the most sensible.

As a last remark remember that you often have the option to find a safe shelter in a café and wait for the rain to stop. That would give you time to catch up on your favorite blog posts too 😊.

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EnricoDeg

I live in Verona, Italy and I teach mathematics and physics at high school trying to make students understand the beauty and usefulness of STEM subjects. Before being a teacher, I worked for 12 years in the financial industry dealing with risk management, stochastic models for derivative pricing, and IT banking applications. My interests include chess, go, guitar, volleyball, trekking, snowboarding and of course reading and writing about mathematics and physics!

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