Early summer math

22.05.2008 20:02

Getting caught in an early summer shower a while ago got me thinking. If you have to run a certain distance through the rain and want to remain as dry as possible, is it better to slowly walk or run as fast as possible? It's a popular question and I remember seeing Mythbusters and Brainiac episodes on a similar topic. But since I didn't feel like experimenting I tried to come up with a theoretical solution.

First there are some things that needed to be defined:

I defined rain as a homogeneous mixture of air and water, moving downwards with a constant velocity (since raindrops reach their terminal velocity well before hitting the ground).

The measure of wetness is the amount of water accumulated on you during the exercise, and that amount is proportional to the volume of air/water mixture you displace during movement. The rationale here is that the air will move around you as you move through the rain while water droplets will stick to you since they cannot follow the air flow due to their inertia.

To make calculation simpler I also presumed that the person is of a rectangular shape (the following calculation is done in two dimensions, but accounting for the depth is trivial). You can think of the rectangle sides a and b as your projections to the vertical and horizontal plane.

Now with these things defined, it's pretty simple to get to a result. You basically have to calculate the volume of the hole you bore through the rain. Here is the situation with the ground as the frame of reference, where va is your velocity, vb is the velocity of rain droplets and d is the distance you have to cross in rain.

It may be simpler to think with the water droplets as the frame of reference. In that case the rain is stationary and you are moving up and right through it with the velocity va - vb.

The displaced volume is then the sum of volumes of three parallelograms:

\mathcal{P} = \mathcal{P}_1 + \mathcal{P}_2 + \mathcal{P}_3
\mathcal{P} = a \cdot b + a \cdot h + b \cdot d
\mathcal{P} = a \cdot b + a \cdot \frac{v_b \cdot d}{v_a} + b \cdot d

Now if you look at these three terms: the first and the third one are constant. Only the second one depends on your velocity va and it's an inverse relationship.

So the conclusion of this purely theoretical endeavor is that the faster you run, the dryer you'll be and even if your speed goes to infinity, you're still going to get wet. Also note that the amount of water accumulated on your front side is the same regardless of your speed, it's only the amount that falls on the top of your head that varies.

Posted by Tomaž | Categories: Ideas

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