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Best velocity to cross a rainy space

  1. Apr 15, 2008 #1
    I found this problem in the book Resnick and Halliday many years ago (1992) and I had it in my mind until 2005 or so when I was able to get some advance.

    "If you want to walk through a distance L in a rainy day, in order to get wet the little as possible what would you do, run as fast as possible, walk slowly or some intermediate velocity?"

    I knew that the answer would be run as fast as possible because with an infinity velocity the drops of water would be static and that would be the least amount of water that you would find through your path, but how to prove it, and maybe that was not the right answer.

    So I tried to solve the problem counting the number of drops that would collide with a box of dimensions A, B, C (height, length and width) during its path trough a L distance during a "uniform" rain. By uniform I mean the velocity of the drops is constant and equal for every drop, the drops are separated the same distance from each other, and the rain falls at an angle the same for every stream. Check the figure.

    After much struggle I came to a formula that I don't have right now (sorry :redface:) but I think the problem is very interesting and I was able to prove that the best velocity is as fast as possible, although it depends on the geometry of the object!!! that was something I was not expecting, so if the geometry of the body is appropiate the best possible it could be an intermediate value (something between zero and infinite).

    I am going to check my papers to post the formula and main steps, and again I would like to share my results check if they are correct and get some feedback from the community.

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  3. Apr 15, 2008 #2
    Also I did a search in the Internet and found that Mythbusters tried to solve this problem weighting the clothes during the walk and they found that it is better to walk than to run (to get wet the least). So we came to different results. But, a regular box is very different from a moving human body, also they have a very limited range of velocities to try (something between zero and 60 mi/h maybe?) so that could be the reason of the different results.
  4. Apr 15, 2008 #3


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    The amount of water deposited on the front face will be BCL rho, where rho is the density of rain drops. It is independent of the velocity u at which you walk (or run).
    The amount deposited on the top is ACl rho X v/u, where v is the vertical velocity of the drops. This gets smaller for high u, so I agree with you.
    An experiment by mythbusters would be ruined by the fact that the water will run off the front at an inderterminant rate.
  5. Apr 15, 2008 #4
    I would tend to think (i.e. I have no formal proof to provide at the moment) that your density 'rho' would depend on the walking speed ([tex]\vec{u}[/tex]). For example, in an extreme case, consider that the speed vector of the rain [tex]\vec{v}[/tex] is equal to your speed vector [tex]\vec{u}[/tex] (on other words, raining horizontally... an extreme case!). In that scenario, you would simply travel at the speed of rain droplets, never coliding with them and not getting wet at all!
  6. Apr 15, 2008 #5
    Interesting problem. You might want to consider another vision of the same problem. Suppose that the box remains stationnary and that the velocity vector of the rain is equal to [tex]\vec{v}-\vec{u}[/tex]. Also, instead of having to travel a distance L, suppose that the rain lasts for [tex]L/\|\vec{u}\|[/tex] seconds.

    This problem is analogous to the previous, but expressed from the reference frame of the box.

    Under this description, I think it's easier to get a feeling for the problem. You have a surface of the object which is exposed to rain going at a certain speed for a certain amount of time. I have not taken the time to think it through, but off the top of my head, here is what I think: The total amount of rain 'cought' by the object should be equal to the density of rain, multiplied by rain speed, integrated over the exposed surface and over the exposure time.

    As [tex]\vec{u}[/tex] increases, the exposure time decreases, but the exposed surface may increase or decrease, depending on object shape and rain orientation, and the effective rain velocity ([tex]\vec{v}-\vec{u}[/tex]) may also decrease or increase.

    ...I'll try to reflect on it some more and come back with a clearer answer.
  7. Apr 16, 2008 #6
    Thank all of you guys! I am going to search my papers and bring the formula so we can discuss it. The formula works well for the case of a horizontal rain (extreme case), so I have the feeling that it is correct.
  8. Apr 16, 2008 #7


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    You should lean forward into the rain, and run (preferably with straight legs, just using your ankles) so that the rain only falls on your head! :smile:

    Doesn't everyone? :confused:
  9. Apr 16, 2008 #8

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    Wouldn't you have effectively "collided" with the drops which were occupying the volume your body is taking up?
  10. Apr 16, 2008 #9
    Well, it's a matter of initial conditions: do you reach the speed prior to the first rain drops hitting you (in which case the only rain would be behind you and you would be dry) or do you reach the speed afterwards (in which case you would be surrounded be rain drops although there would be no droplets in front of you, and you would be wet from being rained on during acceleration)? But to have collided with exactly the amount of rain contained in the volume your body is taking up would require very precise timing!

    In any case, you would not get any wetter once your precise crusing speed has been reached!

    And consider this: if you ran at lightning speed for a given amount of time, you would actually get more wet than if you ran at the speed of rain for the same amount of time (because you would catch up on the rain drops and collide with them). Being lightning fast is way overrated, being rain fast is so much sweeter!:rolleyes:
    Last edited: Apr 16, 2008
  11. Apr 16, 2008 #10
    Just my initial thoughts...

    Conceptually, as far the box example goes, and assuming horizontal, uniform rain, the amount of water will be, at a minimum, equal to the rate of rain fall multiplied by the time spent in the rain. If the forward movement is fast enough to collide with the next line of rain before it hits the ground, then the object will take be hit by more rain. So, anywhere between no movement up to the rate at which you will not be hit by the next "line" of rain. This assumes there is enough space between lines of rain and waves of rain that a box can slip in between. Otherwise, assuming there is a destination where there is no rain, speed for a box may be the best route, since it will inevitably be hit by rain from forward motion anyways.

    But for humans, having a much smaller area on top than on front, it's probably some intermediate speed.
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