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bushindo
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I've always wonder whether it is better to run very fast through a constant rain, or simply to walk through it. Intuitively, running leaves you less soaked, but how much of an improvement are we talking about here. This question will settle it once and for all.

Assume that all rain drops are identical in size, and that they are falling at 9m/s straight down vertically. Their density is 200 drops/meter2. You'll be walking or running through the rain. Let's make it simple and assume that you are a rectangle, and that all water that hit you will either hit in the head portion, or the chest portion. Let's say that your head portion has a surface area .06 meter2, while your chest potion has a surface area of .5 meter2. Let's say that water drops from head hits and chest hits count the same.

Now, you are supposed to get across a 200 meter field,

1) Let's say you take your time and walk at 1.5 m/s, how many drop of water would hit you during your walk?

2) suppose you ran across at a speed of 8 m/s, how many drops of water would hit you during the run?

Edited by bushindo
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I propose that the faster we go the drier we will be. I contend below that the chest will be struck by the same amount of water independent of speed but that the head will be drier the faster we go.

Consider the amount of water which will strike the chest. At the time you set out running / walking the rain which will hit you is contained in a "tube of space" this tube has a regular cross section in the vertical plane (that of your chest) and is sloping upwards towards a plane at your destination 200m away. How "steep" this tube is depends on how fast you are travelling (the slower you travel the further up the plane the point of interception is and vice versa). However the tube will have a constant volume independent of the slope (think of a stack of playing cards stacked on a slight angle which has the same volume as a stack of cards aligned on on top of the other - then consider the stack to consist of an infinite number of cards of "tending to zero" thickness). Therefore the volume of water which strikes your chest is constant and, if we consider the special case of someone travelling at infinite speed, the amount of water will be the surface area of the chest (0.5 sqm) times the distance (200m) times the density (200 drops / metre cubed - not squared as stated in the puzzle I take it) = 20,000 drops

The head is different because it lies in a plane at right angles to the plane at our destination. In this case the cross section of the tube which will strike the head is a function of how fast you are travelling and, consequently will vary how much water strikes you. If we consider the head to be a rectangle of width W and depth D. If we consider our walker / runner setting out then the water which strikes the backmost part of the head will be at a distance H above the front most point of the head at the start. We can therefore say that the tube of water which will strike the head will have a cross section of W x H and the same calculation can be done as above for the chest. H will be smaller the faster our traveller moves. This follows because raindrops are falling at a constant speed so the plane containing the rain which will strike the head at any point (including the point of departure) will be shorter if the time for the depth of the head to pass that point is less.

Therefore the faster we move the less rain will strike our head but it has no impact on the volume which hits our chests.

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Not gonna make any math, but I'd say some of these calculations seem to consider time has frozen while you walk this field, since some of the arguments state that you'll walk into the same amount of rain as you cross, either walking or running. But of course rain continues to pour down, so if you take one step into the field and just stand there looking around pondering whether or not you should walk or run you'll get soaked over time, cause rain is not going to stop while you decide.

As some of you have pointed out it's a simple problem of ratios. Regardless of the speed we use (and taking into account that we have perfect rain that falls fully vertically down) we'll hit roughly the same amount of drops with our chest, that is, we'll intercept approx. the same amount of droplets as we cross the field running or walking. So speed matters not on how wet our chest gets. That leaves the other part of the equation, that is how many drops hit us from above. And there it's quite obvious that speed matters, since the more we spend under the rain the wetter we'll get. So the fastest the speed, the less the time, the drier we'll get across...

See? No math at all. The mythbusters show was great but as some of you pointed out it neglected to consider the amount of water hitting the head, so it's useless as proof for this question...

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i think if you think about what actually happens, a person hunches over when they are going through the rain so only their head, shoulders, and a little bit of their back gets wet. (the shoulders shield your front) when running you run into drops which can no longer be shielded by your shoulders because of your speed so your front as well as your head and shoulders now get wet. just kind of what i think happens could be way off

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For those who say it does not matter running or walking will be the same.

Would that hold true no matter how hard it is raining?

My thoughts, mostly from expirience, is the lighter the rain, the better it would be to run, as it gets heavier running or walking start to not really matter.

Maybe someone would like to compare the two extremes...Run a formula for a very slow rain, and one for a very hard rain, and compare and see if it really does'nt matter what you do, walk/run.

Just curious. ;)

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What ever happened to the use of spoilers, btw? :blush:

First, note if you stand still you get infinitely wet.

Second, if your speed is infinity you encounter 20,000 drops

[200 times the volume you sweep out in your trip = 200x.5]

So speed helps, and the answers lie in decreasing order between these limits.

The general formula is Drops = A x D x V x T, where

A = your cross sectional area - as seen by the rain

D = raindrop density

V = velocity of the raindrops in your coordinate system

T = time spent in the rain.

Define the apparent angle of the rain as THETA where tan [THETA] = your speed / 9 m/s.

Then

A = .06 cos[THETA] + .5 sin[THETA]

D = 200 drops /m3 -- assuming 200 drops/m2 is a typo

V = sqrt{(your speed)2 + (9)2} - since the velocities are at right angles.

T = 200/your speed.

At 1.5 m/s you are hit with 34,400 drops

At 8 m/s you are hit with 22,700 drops - only 13% higher than the minimum.

The in-between values are shown in the graph.

post-1048-1244094596.gif

Wow, that's some crazy math, let me geuss you're an English Professor? :lol::lol: j/k

I did'nt see this before I put this Post #30. Maybe you'd want to try and see? I would'nt even know how to attempt it. AxVxGxPxH(sin*45(2y)+100)p^2-3xy(cos)[TEATH]sin)...well you get the point...But I agree on the speed having something to do with it somehow.

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For those who say it does not matter running or walking will be the same.

Would that hold true no matter how hard it is raining?

My thoughts, mostly from expirience, is the lighter the rain, the better it would be to run, as it gets heavier running or walking start to not really matter.

Maybe someone would like to compare the two extremes...Run a formula for a very slow rain, and one for a very hard rain, and compare and see if it really does'nt matter what you do, walk/run.

Just curious. ;)

I'm sure in reality running does matter. But the problem given in the OP isn't a very accurate model, for example it assumes the rain drops vertically. Also we (or at least I wasn't) aren't considering the rain from above since you are 2D. If we consider this then running would be better I'm sure.

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I'm sure in reality running does matter. But the problem given in the OP isn't a very accurate model, for example it assumes the rain drops vertically. Also we (or at least I wasn't) aren't considering the rain from above since you are 2D. If we consider this then running would be better I'm sure.

  1. I think the OP meant to describe a rectangular parallelepiped.
    Otherwise his head and chest would themselves not be rectangles.
  2. Vertical rainfall is a reasonable assumption, just means there is no wind.
  3. Also, the density was probably meant to be 200 drops/m3
With those changes, the OP gives a solvable puzzle.
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