# Getting wet from the rain

1. Dec 15, 2007

### mordechai9

Let's say it's raining outside and you're traveling by foot from point A to point B. If you run, will you get more wet or less wet, or will it be the same?

2. Dec 15, 2007

### Parlyne

As long as the rain is coming down at a constant rate, it pretty much has to be less if you run than if you walk.

3. Dec 15, 2007

### pixel01

It is not the same. If you walk under the rain, you get the rain from two direction: from ahead W1 and from front of you W2.
W1 is proportional to the the time you travel : the longer you go, the bigger W1
W1 = S/V*A1*J :
where S : the distance from A to B
V : your speed
J : the flux of the rain (drop/m2/sec)
A1 : cross section area of your body according to vertical direction (perpendicular to)

W2 is calculated as follow:
W2 = A2*J *sin(actan(V/Vo))*S/V
where A2 the cross section area of your body according to the direction you go
Vo : speed of the rain drop.

I think (W1+W2) will depend on V and may be we can find its max/min value.

4. Dec 15, 2007

### pixel01

I tested it in excel, the faster you run, the less wet you get.

5. Dec 15, 2007

### dst

Does it only work for spherical chickens in a vacuum? :D

6. Dec 15, 2007

### mordechai9

I think that's an interesting post Pixel, however, I have a few objections:

1.) You assume the rain falls straight down. This is not necessarily accurate.
2.) Where do you get the sin term for your equation for W2? I don't follow you there.
3.) There are two different fluxes of rain; the flux of rain which hits you from above, and the flux of rain which hits you on your front. Clearly, the second flux is a function of your speed.

7. Dec 15, 2007

### pixel01

@ dst : it works for all A1 and A2, so your spherical chickens are covered.
@ mordechhai: please try to draw the sketch. My model requires the rain drop straight down. If necessary, you can change it for some alpha. You can have a second look at W2 expression: it depends on the speed V

8. Dec 15, 2007

### pixel01

Here is the plot in matlab with different flux J

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9. Dec 15, 2007

### cesiumfrog

(But consider your limit where Vo=0. Or in general, transform to the frame where this is so.)

The amount of water soaking a surface parallel to the constant rain fall is a constant of the path taken, and the amount of water soaking a perpendicular surface is proportional to the time spent, so running is always optimal.

Last edited: Dec 15, 2007
10. Dec 15, 2007

### malty

Um, I may be wrong here, but I think that the location of where a rain drop falls is randomly scattered. So if you were to stand still the odds on more rain hitting you would be significantly higher than if you were to say move quickly. Even though you will get soaked either way, I believe that changing your own position frequently will effectively reduce the chances of getting wetter than if you were to stand perfectly still or move slower.
I suppose an analogy to my understanding would be waiting in the carpark of a shopping mall for an empty space to become available, odds are if you keep moving you'll be less like to find a car space (avoiding the rain), whereas if you stay in the one location in the car park you're practically guranteed to come across a lot (soaked by rain) eventually.

Last edited: Dec 15, 2007
11. Dec 15, 2007

### pixel01

Imagine the rain is not straight down, but inclines with alpha angle. so the water you get is S*J*A2*sin (alpha). Let say you have a hat, so alpha = 0, you are dry and alpha=90 dgees, you get the max wet.
Now, the rain is straght down with Vo, you run at V, so in your frame, the rain is inclined and you can calculate alpha via V and Vo and that's the key.

12. Dec 15, 2007

### pixel01

In my model, if you stand still, or V= 0, you get infinite amount of water in your own ! And if you move at speed of light, you are dry.

13. Dec 15, 2007

### Mk

We now have proof that light can't get wet.

14. Dec 15, 2007

### malty

Does that mean it is possible for one to make a a light rain shield or something??

15. Dec 15, 2007

### pixel01

Oh no, you should run at an infinite speed to be dry. V= c and you still get some. lol

16. Dec 15, 2007

### cesiumfrog

I thought it would have been already evident to you that this is also wrong. If you ran at infinite speed, you would "run into" the drops that had already fallen to face-height, and so your front would still get wet to some non-zero extent. Now that we agree your model is flawed, if you wish to vainly insist that it is still applicable in some range of lower speeds then you will need to provide a clearly argued explanation of your derivation.

17. Dec 15, 2007

### pixel01

I think I have explained the expression for W2 quite clearly already. And after reanalyze the formulae, I haven't find a flaw in it. Imagine, if you could go at infinite speed, you also exposed to the rain at no time at all (t=S/v), so you were not under the rain.
If you lash a rod very fast out in the rain, the rod may still be dry.

Last edited: Dec 15, 2007
18. Dec 15, 2007

### cesiumfrog

19. Dec 15, 2007

### Staff: Mentor

Particle man, particle man ...

20. Dec 15, 2007

### marcusl

We assume that the density of water in the air is constant (so many grams per liter, say), given whatever velocity the drops have. The frontal surface sweeps out the same volume, hence same mass of water, from A to B regardless of how fast you walk or run, so the only difference is how wet you get on top. As stated above, the faster you go the less wet on top. I'm with Cesiumfrog.