## Wednesday, August 20, 2008

### Running in the rain

I was sent a letter yesterday which asked me to look and comment on the last claim on this page. Since that page might not remain forever, I am copying the paragraph in question:

You get less wet by running in the rain. Actual mathematical equations devoted to this popular question have suggested it is true, though not for the simple reasons you might think. Complexities include factoring in the number of rain drops hitting the walker’s head versus smacking the runner’s chest.

Well, lets do the calculations. While this sound like a complex problem it is actually a simple case of Galilean relativity, so it is easy to calculate.
The first step in dealing with a problem like this is to formulate it correctly. As I understand this problem, we need to prove or disprove the statement that if you stand still under the rain for some time T, you will be more wet than if you were running for the same time T.
The second step is to simplify the problem. In this case the simplification is to assume that the rain is uniform. That is, the amount of rain drops per square cm is the same in all the are we are dealing with, and there are no sudden changes in the wind.
Now, what we want to calculate is how wet will a person become. Clearly this depends on many factors - the size of the drops, the body area of the person, the speed of the rain drops etc. Because of the uniformity assumption we made, we can say that all the factors except for the speed of the rain are fixed - they don't change with time, and they don't change if the person is moving. We can now define the "wetness" (B) as a simple multiplication between the matrix A and a vector v. "A" will be the constant which we get from the problem condition. Since we are in 3d space A is a diagonal matrix with A1,A2,A3 (the corresponding constant for each one of the three directions) on the diagonal. "v" is the speed of the rain ( it is a vector v=(v1,v2,v3)). The multiplication result, B, is also a vector. The size of B multiplied by the time T is then the answer.

So if the person stands still under the rain we will get:

B=Av

T|B|=T$\sqrt{(A_{1}v_{1})^2+(A_{2}v_{2})^2+(A_{3}v_{3})^2}$

Now lets look what happens if a person starts to run. Since we can choose the coordinate system in any way we want we can assume the the person runs on the x axis with some speed w (for simplicity, we will assume that the speed is constant). Now it is time to use the Galilean transformation. From the coordinate system of the person the speed of the rain is no longer v=(v1,v2,v3). Instead it now becomes v'=(v1-w,v2,v3). Since nothing else changed, we will get:

B'=Av'

T|B'|=T$\sqrt{(A_{1}v_{1}-A_{1}w)^2+(A_{2}v_{2})^2+(A_{3}v_{3})^2$

It is perfectly possible that this is less than the previous result, but it can also be large. It all depends on the direction of w. If for example w=v1, you will get less wet by running. But if w=-v1 you will get more wet if you will run.

This solution was done under the assumption that the rain is uniform. However, while this assumption is not realistic it is very close to reality and on a short time interval it should be extremely close to reality. It is possible to solve the problem without this assumption, but it will only introduce extra steps without changing the final result.

#### 1 comment:

- A - C - said...

Very nice. you have an interesting blog. thanks for posting.

A.