### A Math/Physics Word Problem

If you are walking from point A to point B in the rain, do you get more or less wet depending on how fast you walk?

Sounds stupidly simple, doesn’t it. Not so (for me, at least). Here are some things we can assume for the sake of the problem:

- let’s assume you are rectangular - let’s say, 1 meter, 0.5 meters wide, and 0.5 meters deep
- forget about dripping rain - any drop that hits you counts as one drop
- the rain is evenly distributed and falls at a constant and consistent speed

The qestion is, over a given distance, does the rate at which you move (in a straight line, you can assume) affect how many drops of rain you come in contact with?

If you run fast, you’ll “run through” more drops, right? However, you’ll also be in the rain for less overall time (remember, we’re going a set distance).

It might help to think through the problem in two dimensions.

## Comments

## Sean Egan - October 3, 2004 8:58 PM

http://www.straightdope.com/classics/a3_395.html

## Zach Hale - October 3, 2004 9:32 PM

Okay, I think that if you are walking you will get more wet than if you just stand there. But, I'm not sure about the running thing. I would think that the faster you go, the more you will hit though.

## Zach Hale - October 3, 2004 9:35 PM

Let me clarify that.

If the comparison was standing for 1 minute or walking for 1 minute, the person walking would get more wet.

If the comparison was between someone running for 1 minute compared to walking/standing for 1 minute, he would also get more wet.

You would get less wet if it were a matter of running from Point A to Point B compared to walking from Point A to Point B though.

(I hope I'm right)

## Scott - October 3, 2004 9:39 PM

I remember reading somewhere years ago that it doesn't matter how fast or slow you drive you get the same number of raindrops on your windshield. So you don't need to increase the speed of your wipes if you speed up.

I guess this doesn't exactly relate to your relate, but it is close.

## Andrew Davis - October 4, 2004 12:05 AM

The show Mythbusters on Discovery Channel solved this little question in its first episode last season. They set up a rain corridor inside, for a constant drizzle. Industrial sized fan for wind. Then, donning wetsuits, and pre-weighed coveralls, they ran and walked through the rain a set distance. They then weighed the coveralls again, and found that running through the rain gathered more water, and made them wetter.

I believe they tried it with a few different settings for downpour, and windspeed and came up with the same result. You get wetter running through the rain than walking. (Though it was a while ago, so it may have just been the same settings, repeated to be sure).

It answered the question well enough for me.

## Jonathan Barrett - October 4, 2004 12:23 AM

So running gets you wetter than walking?

What if you just stand still? Surely that gets you wettest of all? So there must be some "sweet spot" of walking speed that gets you the least wet - Wetness vs. speed must be like:

<pre>

WET | _____

\ _____- - - -

\___- - - - -

0 ^ RUN

</pre>

So what is that magic sweet spot? Anyone?

I need to know, I live in Scotland!

## IXG - October 4, 2004 12:26 AM

Walking gathers more rain. Imagine walking at an infinitly slow pace - you would of course accumulate an infinite amount of drops. On the other hand, if you could move at lightspeed, you would only accumulate the sum of all the drops of water in a rectangle multiplied by the course length ....

## jonah - October 4, 2004 12:31 AM

Think of the extreme case: A steady rain is falling and two people are going 10 feet. One person runs the 10 feet at full speed. The other person walks .0001 miles per hour. Who do you think is going to be soaked at the end of the experiment?

## ~bc - October 4, 2004 12:33 AM

Heck with math, you need a "u" after the "q" in the word "question." (3rd to last paragraph) Other than that... interesting brain teaser.

## kteela - October 4, 2004 12:43 AM

i think there's a way to do this. you would need a sponge, 2 rates of speed (walking and running) and several different rates of rain(constant w/ both the walking and running speeds. weighing the sponge before and after each experiment.

i think that the results would show that there's a threshold in the rate of rainfall where running is better and then walking is better.

the wind factor, might take more tests too. and i don't know about the splashing in puddles part of running in the rain. that might make it more complicated.

well, i've thought about this one.

-physics theory geek signing off

## Samual Icky - October 4, 2004 1:07 AM

I think everyone is forgetting something... the splash back back effect. Think about it when you run and hit a standing pool of water. What happens?

## Karl Dubost - October 4, 2004 1:13 AM

Missing factor:

The speed of the rain

You have to consider

d = distance A to B (meters)

t = Time for the person to go from A to B. (seconds)

Speed of the person vp = d/t

Falling Speed of the water is important too.

Imagine one drop of water, only one and imagine you are a flat person. The probability to reach you depends on your speed.

Basically you cant encounter the drop of rain when it's falling from 1 meter to 0.

so vr = 1 / tr .... tr being time in seconds to fall from 1 m to 0.

if you are parcouring your distance in the same time (t) than tr, you will hit the drop of water as a limit, if you do it in less time, you will hit always. If you are longer than one drop of rain to fall. You might avoid it, except if you are not lucky :)

so t > tr to have a chance to not be wet.

tr = 1/vr and t = d/v

so d/v > 1/vr

So your speed to have a probability to not be wet is

d*vr > v

if v is more or equal to vr*d You will be wet for sure.

Hope it helps to solve your problems.

For sure by simplication we take a rain fallin vertically only, the density of the rain doesn't matter much, It's why I have taken one drop of rain.

So basically at a certain point when you run faster, the *probability* of being wet is more important. :)

Probability is the important word here.

## jez - October 4, 2004 2:40 AM

why not just use an umbrella? then you don't get we at all ;)

## Dan Phiffer - October 4, 2004 2:45 AM

We talked about this problem in a college physics class. If I remember correctly it doesn't matter how fast you move (as long as you don't stop). Of course the television experiment seems to contradict that, so maybe I've remembered wrong...

## Brian - October 4, 2004 3:31 AM

Assuming constant distance, rather than time, the walker gets wetter (maybe). There are two parts to the rain that hits the subject:

1. The water in front of you that you walk through (basically a rectangle which is the same height as the person and has a length which is the distance the subject is walking).

2. The water falling on the head of the walker/runner. For the runner, this is a thin flat parallelogram but for the walker it is a tall fat one (Its easier if you draw it).

Therefore the walker gets hit by more rain. Of course this doesn't take into account:

a) If you run the rain may soak through more.

b) If you run you splash more.

## Brian - October 4, 2004 3:40 AM

Correction:

The parallelogram is the same width, it just gets taller when you walk.

## charlie kunkel - October 4, 2004 4:02 AM

The only real way to measure this would be to take different sized sponges and move them at different speeds through a room with rain falling at different speeds. Measure how much water the sponge retained (by weighing it before and after (assuming it never gets near it's saturation point)). Plot results, then profit! ;)

## Ken Wollins - October 4, 2004 5:43 AM

I'll answer your question with a question: If you are walking in the woods, are you more or less likely to be hit by a falling tree than if you are running?

## Hypgnostic - October 4, 2004 6:11 AM

ummmm...uhhhh...

ALL YOUR RAINCOAT ARE BELONG TO ME

and stuff...

## Josh marshall - October 4, 2004 6:11 AM

Logically I would think walking would get more rain on you, if you think of it in absolutes, if you were traveling from point A to point B at a speed where by no rain was falling on you the only rain you would contact would be from the you shaped corridor inbetween the two points. now if you slow it down you are getting hit by rain from above as well as from the front, but just as much rain as the walker (just for a shorter period of time)so the variable left is rain from the front. now the best way to picture this is to freeze time, and imagine the space between the runners-walkers as a nearly infinete series of walls each the thickness of a raindrop and it is made of random spacings of raindrops, now due to the random nature of the raindrops we can assume that at anypoint in this frozen time if the walker and the runner are the same distance from point b there is the same number of raindrops between them and their target now irregardless if the person is running or walking through these walls they have just as many walls to go through, even if theyère going slower any distance they have covered would have contained the same amount of raindrops as the person running. basically leaving you with the fact that both are getting hit with the same amount of raindrops during any given span of time, but the runner is just doing so for less time and so over an equal distance would get hit by less drops. from above anyway.

## Dan - October 4, 2004 6:14 AM

well assuming that everythig is evenly distributed and rectangles you can take the prblem into 2 parts. First if the rain hitting the top of you. The answer for this is the amount of water for your given area per second times the seconds you are in the rain. The walker loses here. Second for the rain hitting you from your direction of travel the equation can be made in much the same way. Take the average amount of water for your given area and multiply by distance traveled. For this both come out equal. Add the two results togeather and the runner wins as fewer drops of rain hit the head of the runner. This of course assumes perfect water absorbtion no wind and an evenly distributed rain. QED

## Chris Currivan - October 4, 2004 6:16 AM

You get wetter walking. There are two components, drops you run into and drops that land on you. Instead of thinking of the drops as moving, think of them as a fixed 2-d grid of points. Your from (which hits the drops) is a line moving through the drops with a horizontal component determined by your speed, and a vertical component determined by the velocity of the falling drops. You pass through a parellogram of drops, and the area of a parellogram is determined by height, not the "skew" angle, which in this case would correspond to the ratio of your velocity to the drops' velocity. So you hit a constant number of drops regardless of your speed. The number that land on you is proportional to the time you spend in the rain, so you get hit by more total drops the slower you go.

## Chris - October 4, 2004 6:18 AM

from = front

## timmay - October 4, 2004 6:28 AM

Surely there must be a more critical way to think of this involving thinking of the amount of rain "flux" the person travels through. So when standing still, a constant flux of rain is hitting the person, this same flux is hitting the person the entire time they are moving horizontally, no matter what speed. However, a moving person has the added flux of hitting any drops that are in front of them at head-level, but the person runs into them before they hit the ground. So, compare the added flux due to movement to the lesser flux due to the lesser amount of time spent in the vertically hitting rain. If the former is bigger, then running is more wet, otherwise walking is. Somehow I have a sneaking suspicion walking gets you wetter.... (I think them drops are falling FAST).

## Plasma - October 4, 2004 6:30 AM

I thought the answer would be pretty much common sense. You get wet in two ways when running/walking in the rain: at a constant rate from water falling on the top of your head, and at a rate directly proportional to your running speed from water that you run/walk into.

However, once you think about it you will run into exactly the same amount of water during your trip from A to B no matter what. If you're travelling 100m, you will always pass through 100m x (person's cross-sectional size) of space with rain in it, so therefore this amount of rain will always be the same regardless of speed. However the amount of rain falling down on top of you is purely dependent on the amount of time spent in the rain, it is completely unlinked to speed. So therefore whatever solution gets you out of the rain fastest should be the dryest solution, shouldn't it?

## Plasma - October 4, 2004 6:32 AM

hmm... okay, looks like Chris posted first while I was writing mine, but I'm trying to say the same thing

## Joe Casey - October 4, 2004 6:41 AM

I saw that Mythbusters show, and I think they missed an important point that also is a factor in real life. When running, the experimenter was making splashes when his feet hit the puddles that formed on the floor. Some of the splashed water hit the legs of the suit he was wearing. This didn't happen when walking. A better experiment would use some kind of matting to eliminate puddles, or perhaps weigh the top and bottom of the suits they wore separately.

## Chris Johnson - October 4, 2004 7:01 AM

I think Chris Currivan has the right way of looking at the problem, but has missed one aspect - we actually need to trace a person shape through the grid of points rather than just a line. This means that the width of the parallelogram traced out is dependent on angle, because the width is the diameter of a person measured at that angle. This will have the effect (since humans are taller than they are wide) of making walking better per-unit-time, BUT the effect is _small_, so when you look at rain-per-unit-distance, getting there quicker by running makes a much bigger difference and so is the best option.

## dennis sherman - October 4, 2004 7:04 AM

there is a plane .5 X .5 meter left outside in the

rain. if its outside for two minutes, it will be

struck by fewer drops of water than if it were outside

for ten minutes.

regardless of its velocity, it still occupies .25 of a

square meter. so if plane has to travel over a given

distance at 3 miles per hour, it will be struck by

more drops than if it were traveling at 15 miles per

hour because it would be exposed to the rain for a

longer time. remember, this is for a plane.

if we're talking about a three dimensional rectangle,

then we have to account for the drops struck by the

leading vertical surface of the object resulting from

its forward movement. to do this, the exact forward

velocity of said rectangle must be known. also, the

downward velocity of the drops must be known so as to

know long the drops will hang in front of the

rectangle.

then there is wind. wind works like velocity in

regard to how many drops the rectangle will be struck

by. with a head wind, the object will be struck by

more drops. with a tail wind, it will be struck by

fewer drops. with a side wind, the rectangle will be

struck with the same amount of drops that it would

with no wind at all, plus those blown in from the

side.

so, a horizontal plane traveling a given distance at a

greater speed will be struck by fewer drops than one

traveling more slowly. a vertical surface traveling

more quickly will be struck my more drops than one

traveling at a lesser rate.

this all means that if one were to discount air

movement & splashing water from below, & variances of

absorption & penetration rates resulting from

different strike forces caused by varying speeds in

forward movement, it comes down to how fast the person

is running & the downward velocity of the drops.

## Golightly - October 4, 2004 7:10 AM

1. It depends on how far the distance is from A to B, because after a certain amount of time spent in the rain, you just can't get any wetter, regardless of whether you walk or run.

2. It depends on the volume of rain, because after a certain amount of rain has landed on you, you just can't get any wetter, regardless of whether you walk or run.

3. If "the rain is evenly distributed and falls at a constant and consistent speed" and "you are a rectangle - let’s say, 1 meter, 0.5 meters wide, and 0.5 meters deep" then the question is merely reduced to the amount of time you spend in the rain because the amount of rain that hits you is a constant.

## simple - October 4, 2004 7:16 AM

1. If you move through twice as much space per second, you will gather less than twice as much rain during that second (because you double the amount or airborne water which the front of your body moves through, but the amount hitting the top of your head and shoulders remains a constant.)

2. By doubling your speed, you will reach your destination in half as many seconds.

Therefore, you will be slightly more dry moving at double your walking speed than if you walk.

## Chris Currivan - October 4, 2004 7:22 AM

Re: Chris Johnson

I think it's pretty accurate to think of a person moving through the grid as traversing 2 parallelograms, one for the front and one for the top of the person. If the horizontal of the grid is distance on the ground, and the vertical is time, then the parallelogram for the front of the person has the skew direction (which doesn't affect area) dependent on speed, while the one for the top has the height (which does affect area) determined by time spent in transit. It wold be easier to explain if I could draw it.

## chris grzegorczyk - October 4, 2004 7:23 AM

Lets assume that a rain drop is a point, in the mathematical sense, i.e. it is dimensionless. This means that for some raindrop, call it r, the statement r (belongs to) U, where U denotes the set of points in R^3 which are contained in the closed set defined by your cube. Further, lets assign the rain the falling properties:

Rain falls in sheets of dimensions 1xinfinity, that is, they are 1 unit in width, and infinitely tall.

These sheets of rain contain k rain drops per 1x1. And more so, the entire path of travel contains k^2 rain drops in any 1x1x1 space.

Finally, all rain drops travel @ v units/s with a phase which is normal to the surface of the path. i.e. they fall straight down.

Consider this point of view: The rain is stationary, but you are walking up a slope whose angle is a function of your velocity. So if you were to move with an infinite speed in the original scenario, the rain would not manage to fall at all, and all the rain you come into contact with would be on the front of your cube. Under this new point of view, the slope of your path would be 0. If you had 0 velocity, the slope of your line would be infinite. What we are actually talking about is your position function, which will be linear and depend upon the specific distances involved. So now we can find the amount of rain you come into contact with.

A little geometric observation leads to the realization that the volume occupied by rain through which you travel will be a parallelogram, one for your front, and one for your top. Lets say that we know the function modelling your position, call it f(t), then f(t_0)=0, and f(t_n)=L ( where t_n is your final time, and L the length of the path ). Then the volume of rain you pass through is given by the sum of the rain on your head and the rain on your front:

Let d_t denote the surface area of your top, and d_f denote the surface area of your front, and L is the length of the path, and t_n be the length of the path divided by your velocity ( assumed to be constant ). Denote the position function as f(t), then f(t_0)=0, and f(t_n)=L, which will be linear. Let R be the rain function,

R = sqr( L^2 + t_n^2 ) * [ ( d_t * sin( arctan( L / t_n ) ) )

+ ( d_f * sin( pi/2 - arctan( L / t_n ) ) ) ]

Note that L, d_t, and d_f are all predetermined, so only t_n is variable, and depends on your velocity. At this point an approach can be made at minimizing R by taking the derivative and looking for roots. enjoy...

-c

god forbid i make an arbitrary guess here, but according to this setup, the faster you go, the dryer you stay. It follows simply from the observation that R is minimized as t_n approaches 0, where the limit for R is d_f*(L*t_n).

## Mr. Govorov - October 4, 2004 7:43 AM

It's been said a few times already but lets see if I canmake it simpler than Chris' explanation. Assuming the rain falls straight down you are only going to get wet on two surfaces, the front and the top. No matter how fast you move, you move forward through the same volume of air and therefor the same amount of rain drops. So on the front you get just as wet no matter how fast you move. As for the top side, well obviously the longer you spend in the rain themore falls on you. So you're dryer if you run.

## Phoenix - October 4, 2004 7:48 AM

hey, look @ it this way: you'll move through a constant volume of rain if you move at the same speed over the same distance. aight? so the rate of rainfall is constant. now, the actual volume of rain you go through will then depend on the speed you go through the rain. at higher speed, you'll move through a smaller amount of rain. so you'll actually move through less rain at a higher speed. ciao!

## Nicholas Bretangna - October 4, 2004 7:48 AM

A More important question:

WHY has *this* meme suddenly taken off:

ALL YOUR RAINCOAT ARE BELONG TO ME

-- I had never heard this base phrase until about 5 months ago, and now it seems I bump into it everywhere -- at least 5-6 different utterly unrelated contexts in the last couple months. The game it is based on is like, *many* years old, is it not? Why --NOW--?

## Steve - October 4, 2004 8:02 AM

Graphed out with speed across the bottom (standing still -> walking -> running) and rate of wetness up & down on the left: I think you'll get an upside down bell curve. Like others have said - standing still will get you very wet, but running at a very high speed will also accumulate all of the rain drops in the time/space in question. There is a sweetspot of speed for every rate of rain and distance to cover.

The answer is decided by the rate of rain: If it is raining very hard then "time outside in the rain" is the problem. Running in a very hard rain actually helps because you are spending less time in the rain. IE: Walking might get you 20 drops per second, and running might get you 25 drops per second. So walking 30 seconds gets 600 drops, and running for 15 seconds will equal 375 accumulated drops.

If it is raining lighlty then the problem is avoiding the rain drops, ie: walk slowly and avoid accumulating more drops in the time/sapce in question.

Speed of walker, distance to be covered and rate of rain all dictate which walker get wetter.

## Nicholas Bretangna - October 4, 2004 8:34 AM

A More important question:

WHY has *this* meme suddenly taken off:

ALL YOUR RAINCOAT ARE BELONG TO ME

-- I had never heard this base phrase until about 5 months ago, and now it seems I bump into it everywhere -- at least 5-6 different utterly unrelated contexts in the last couple months. The game it is based on is like, *many* years old, is it not? Why --NOW--?

## Branden from Maui No Ka Oi - October 4, 2004 8:41 AM

I think many people here are correct, but looking at the problem from different viewpoints.

1. If you measure getting wet by distance, walking gets you wetter.

2. If you measure getting wet by time, running gets you wetter.

Since the rainfall is constant, when moving horizontally (and using the wall/plane idea from previous posts) it is safe to assume that there is a consistent number of drops in front of you at each step. Let's say that in these walls/planes of drops in front of you, there are 10 drops per foot. If the distance you must travel is 10 feet, there are 10 drops per foot, so 100 drops that you must walk through to cover that distance. If we measure by distance, we see that by running 10 feet, you must go through 100 drops. We see that by walking 10 feet, you must go through 100 drops. If we measure by time, we see that by running for 2 seconds, we cover 20 feet and must go through 200 drops. Still measuring by time, we see that by walking for 2 seconds, we cover 5 feet and must go through 50 drops. Distance traveled is important in important in determining horizontal wetness, not time.

Let's say your surface area from above is 2 square feet. Your area from above will never change from 2 square feet, no matter how fast or slow you move. If you freeze time, your chance of getting hit from above would be the same as if you were standing still. If you move forward in the smallest possible increment of time, your chance of getting hit from above would be the same as if you were standing still. If you multiply this out to cover any distance, you would find that moving across that distance, your chance of getting hit from above would be the same as if you were standing still. So in getting hit from above, the important variable is time. The longer you are in the rain, the more you will get hit from above, regardless of how much distance you cover. So let's say that standing still, you get hit from above by 2 drops per second. If you run for 10 seconds, you get hit by 20 drops from above, if you walk for 10 seconds, you get hit by 20 drops from above. If you stand for 10 seconds, you get hit by 20 drops from above. So if you run 10 feet in 2 seconds, you will get hit by 4 drops from above. If you walk 10 feet in 5 seconds, you will get hit by 10 drops from above. Time spent in the rain is important for getting hit from above, not speed or distance traveled.

1. Sooo, if you measure getting wet by distance, let's say 10 feet. In measuring by distance, horizontal "wetness" is constant, as you must cover 10 feet regardless of your speed. So you will walk through 100 drops no matter your speed. Let's also say you get hit by 2 drops per second from above, and that running 10 feet takes you 1 second, and walking 10 feet takes you 2 seconds. So in "getting hit from above", we see that by running, we get hit by 2 drops, and by walking we get hit by 4 drops. So by walking 10 feet, you get hit by 104 drops. By running 10 feet, you get hit by 102 drops. So you get wetter by walking 10 feet, than you do by running 10 feet.

2. If you measure getting wet by time, then getting hit from above is a constant. If you walk for 10 seconds, and get hit by 2 drops per second from above, you will be hit by 20 drops from above. If you run for 10 seconds and get hit by 2 drops per second from above, you will be hit by 20 drops from above. Let's say that by walking for 10 seconds, you cover 5 feet. If you walk through 10 drops per foot, you will go through 50 drops horizontally. If you run for 10 seconds, you cover 20 feet. At 10 drops per foot, you will go through 200 drops horizontally. So by walking 10 seconds, you get hit by 70 drops. By running 10 seconds, you get hit by 220 drops. So you get wetter running 10 seconds than you do by walking 10 seconds.

I apologize if somebody else already posted ideas similar to this (it's late where I am and I need sleep). Hopefully what I posted is somewhat close to being correct. Peace.

## Branden from Maui No Ka Oi - October 4, 2004 8:51 AM

I would like to clarify my ideas a bit further.

1. If you measure getting wet where distance is constant, walking gets you wetter.

2. If you measure getting wet where time is constant, running gets you wetter.

This makes sense in a common sense way too. Let's say that you are standing under a roof, and you need to get under the roof across the street. Rainfall is constant, and the distance is constant. We are in state #1, where walking gets you wetter. If you were trying to move from point a to point b, and it is raining, what is your natural instinct? To run there.

Forgive me, I can't think of a good example for #2. Perhaps someone else can come up with one. Good night!

## Jochem - October 4, 2004 9:04 AM

The way Chris Currivan describes it seems to be very good. I would like

to modify it a little.

Let's first assume that the raindrops hang (discreetly) still in a

3D-grid. Secondly we assume our 'person' is also in 3D.

In this way, it seems to me, we can reduce the problem to vector

calculus. The horizontal component of the vector is the speed of

the person going a given distance from a to b. The vertical component

of the vector is the speed with which the rain is falling.

In the same way, it seems, we can control for windspeeds. As the

wind is also relative to the person we can add or subtract the vector

of the wind to or from the vector of the person.

Furthermore I would like to assume we have a perfectly clean 'world'

in which there are no puddles from which water can splash or other

unwanted effects can occur.

Now we can for ourselves picture a 'person' going diagonally through

the 'rain'.

As far as I can see (and I'm no mathematician) we now have a

mathematical model that covers the point in the original question by

Steven. Can someone do the math? (I think that any highschool student

that did reasonably well at math should be able to calculate this...)

## Andrew Mudie - October 4, 2004 9:07 AM

What if you were running on a treadmill in the rain, and someone was standing next to you, who would get wetter?

## Dutch VandeStaten - October 4, 2004 9:11 AM

If rain is falling at a constant rate of say X cubic feet per second, then a constant size body would get hit by a volume of water Y that is linearly proportional to the amount of time T spent in the rain.

Y = X*T . Thus running through rain would cause a lesser volume (Y) to hit you. Consider someone going the speed of light, as compared to close to zero movement. Which would you think would get more wet. Besides, this is a linear problem, thus there is no optimal speed betwwen zero and infinity. Hope this doesn't seem too easy. VandeStaten

## k000k - October 4, 2004 9:15 AM

We also assume the rain is falling straight down and at a constant rate right?

If you ran instantaneously from A -> B, then the amount of raindrops he crashes into is proportional to his height. Think of it like a net, a tall person will scoop more rain out of the air than a short person. This amount of rain is constant so long as the amount of rain falling is. If you could freeze time at that instant, there would be a dry tunnel in the shape of the person's cross section in the otherwise rain saturated air (ala bugs bunny running thru a snowbank)

Going faster or slower can't alter this 'net' effect. The only thing speed can alter is how much rain falls on your top surfaces (head/shoulders) while you're scooping this rain out of the air.

Running will ALWAYS keep you drier than walking, no argument..

## Wang Tango - October 4, 2004 9:18 AM

There is no rain.

## Kevin - October 4, 2004 9:27 AM

couple of points:

running and walking don't always present the same profile. very few people run standing straight up. Runners may present a wider top profile if they lean forward (i.e. they now get water on their back too).

for those wanting to view the rain as a static grid of rain drops -- don't. The rain is falling. A drop that a walker may have gotten hit on the head with would hit the runner in the front. a drop the walker would miss might hit the runner in the shin. a drop you're assuming the runner would miss (i.e. off the back of the head) might hit the runner in the back due to leaning forward (see above).

Don't build strawman arguments by building bogus examples:

why are some insisting that standing still gets you wettest? if you assume the person would stand in the rain until they finish the distance you've created an example that is bound to fail. For the standing in rain problem you would have to compare on time. The person standing still would probably be drier after the same amount of time.

Additionally use real numbers. .0001 miles per hour is .5 feet per hour. Who the heck walks that slow? Even garden snails move at .03 miles per hour.

I think mythbusters had it right -- running the same distance gets you wetter. However, they measured "wetness" in how much weight gain their clothes gained. Is it not possible that running drives more of the water into the cloth while walking allows more of it to run off? So is wetness the number of drops you hit (which they didn't measure) or the amount your clothes absorb? I think most people judge it on how much wetness is in their clothes.

## k000k - October 4, 2004 9:31 AM

Golightly [7:10 AM October 4, 2004]

1. It depends on how far the distance is from A to B, because after a certain amount of time spent in the rain, you just can't get any wetter, regardless of whether you walk or run.

That's an interesting point, the measure of wetness should take into account the new rain that touches you AND the drops that fall off of you, which are equal when your as wet as you can get.. However, the problem specifically states 'Forget about dripping rain', so I guess we're assuming that you never get completely saturated.

## k000k - October 4, 2004 9:42 AM

Kevin [9:27 AM October 4, 2004]

couple of points:

running and walking don't always present the same profile. very few people run standing straight up. Runners may present a wider top profile if they lean forward (i.e. they now get water on their back too).

-----------------

Ok, taking that into account, lets say a runner can run 3x their walk rate. If their top cross section area gets more than 3x their standing profile, then he's getting drier. To generalize: If a runner can make his walk/run speed ratio larger than his walk/run top cross section surface area ratio, he gets drier by moving faster.

How well the water absorbs into the clothing depends on the clothes. Leather soaks up a lot less water than a cotton sweatshirt.. We'll need some constants defined for various fabrics and materials.

## Jochem - October 4, 2004 9:43 AM

Kevin,

The idea of using a static grid for the raindrops is workable when

you use a vertical speed for the person in the model. By doing this

we might reduce the computational complexity with several orders

of magnitude.

In the model I proposed you can simply modify the form of the person

going (either straight up or bent and running).

Of course, this still isn't a perfect representation of reality but

at least it allows us to do some theoretical computations. I don't

expect the results to be really viable in the real world.

## Dennis Beasley - October 4, 2004 10:04 AM

Waling is better because if you run you will fall on your a$$ and get hurt as well at getting wet and dirt from the fall.

## Penance - October 4, 2004 10:16 AM

Walking or running, I always end up equally wet, and I'll tell you why. I and the rest of modern man elegantly solved this problem long ago: it's called an umbrella. Take a trip to the local drug store and $5 later you're good to go.

## parvin - October 4, 2004 10:17 AM

Here's the answer, which explains the main source of confusion here...

... drumroll ...

There's an ambiguity in what it means to "get more wet". If by "get more wet" you mean get hit by the most water, then walking gets you far more wet. If, however, "get more wet" means to have more water on your clothes at the end, then under some conditions running gets you more wet (this is especially true with non-absorbant clothes, such as raincoats).

*problem dissolved*

## Jeff - October 4, 2004 10:21 AM

I like the arguments where people took the extremes. If you disagree then we can try it. You go at a pace of 1/(3x10^8)m/s and I'll go 3x10^8 m/s. You'll just be standing there getting wet whereas I'll be engulfed in a ball of fire as I get to my destination at the speed of light - with not a drop on me.

## Gummi - October 4, 2004 10:25 AM

I don't know about the math side, but being an experienced Floridian in running and walking in the rain, I get just as wet either way from the rain coming down (or the wind blowing it). But running in the rain does get me wetter due to hitting un-avoidable puddles. Plus unless you are an athlete in good shape, running for a minute will cause most people to break a sweat, adding to the overall wetness.

On a side note: (trying to) stand, walk or run in tropical depression or hurricane wind and rain will get you drenched every time!

## Allen - October 4, 2004 9:14 PM

what a coincedence , i just got done watching a show called mythbusters , and they actually came to the conclusion that you get wetter running from point a to point b than by walking .

## Bruce Anderson - October 4, 2004 10:28 PM

The posters who have divided the problem into two parts--the rain impinging upon the top of the walker/runner, a constant function of time, and the rain impinging upon the front of the walker/runner, the product of distance and cross section, have neglected to consider that raindrops captured by the first function cannot be considered in the second.

If the runner leaned forward to some extent, the raindrops captured by the head and shoulders could not be captured by the chest. Since sprinting yields a speed of say, 10 m/s, and raindrops fall, maybe, 40 m/s, the required angle of lean would be the arctangent of 1/4, or about 14 degrees. Then raindrops which barely missed the head would barely miss the feet, and the runner would only be subject to rain impinging from above, since raindrops that barely missed the back of the head would also barely miss the heels.

Of course, in a drenching downpour with very large drops with higher terminal velocities, the runner would have to lean more, or run faster, or both. In a mist, where the vertical motion of the raindrops is nearly zero, the solution reduces to the runner staying drier because he intercepts the same volume as the walker on the front but is subjected to the constant bombardment from above for less time.

Regards,

Bruce

## shell - October 4, 2004 11:24 PM

Assuming the position of rain particles is random enough, we can arrive at a spread rate, or areal density, for rain. Let's call that <i>R</i><span style="vertical-align:sub; font-size:80%">s</span>. Let <i>L</i>, <i>W</i>, and <i>H</i> be the Length, Width, and Height of the man-rectangle.

We are then interested in the incidence area the man-box makes with the rain.

Treating rain from the frame of reference of the man-box, we get

<span style="font-size:150%"><i>Wetness = W <i>R</i><span style="vertical-align:sub; font-size:80%">s</span> t A</i></span><br>

where <i>A</i> is the incidence area.

Since we are dealing with a simple man-box, no integration is needed, so <br>

<img src="http://shell.slact.net/stuff/wetness.gif"><br>

Remembering that we are covering a given distance,

<img src="http://shell.slact.net/stuff/wetness-d.gif">

## shell - October 4, 2004 11:31 PM

I should add that the velocities are <u>absolute values</u>, and the rain is falling strictly vertically.

the equation can be expanded to work with windy-rain.

Also, keep in mind that this solution works for the extremes, too.

Taking the version where the distance does not come into play, if the man-box did not move, only W*H would be rained upon.

Similarly, if the man-box's speed approaches infinity (huh? speed of light, you say? we're in Newtonian-physics land, where all chickens are spherical, and the density of a cow is perfectly uniform), the limit of the area approaches L, which makes sense.

Note that if we took a man-sphere, or someone morbidly obese, his velocity would not matter as the area would remain constant.

## PDaveBall - October 5, 2004 12:20 AM

If you can sprint 10 meters per second then you need to be in a different career.

## Scott the redneck Engineer - October 5, 2004 12:36 AM

Hey all, pretty spiffy theories. And great explanations too! But if you've ever driven a pick up truck you know the truth. It don't matter what you do. You'll get just as wet either way. The water gets distributed in other directions, depending on what you are doing.

Here goes my rant: When you're at a stop light in the rain, the vertical component of atmospheric rain and its random distribution sorta assures there is the same amount of rain hittin' your windshield as well as the stuff behind the cab in the bed of your truck. Barring horizontal hurricane winds (and I'm tired of them!) when you accelerate on the green light, your fwd velocity increases the amount of rain incident on your windshield, while reducing it on the stuff in the bed, kinda like a vertical plane moving. Again, velocity determines if the front gets wetter or the back less wet. (And veee-hicle aerodynamics figger in this too, no doubt, but running people are sorta the same. Wetter in front, drier on back.) Because I've done that.

I've also found that driving at highway speeds in torrential downpours cause me to have to speed up the wiper blades, but my stuff in the bed stays dry -until I have to stop. Then it gets wet, unless I jump out and put it in the cab. (If I got room for it, natch!)

So walk if you wanna get wet evenly on the front and back, run if you wanna get the front wetter. It's your pick! But wear a hat. Ain't nothin' worse than a wet-ass head. Dang wet hair gets cold in the office...'till it dries out.

Maybe that's an engineering observation versus pure science? And real practice versus scientific work? Maybe. But it's what I do for a living. And I do love turning boffin's theo's into working stuff.

Scott [Engineer]

## Brendan - October 5, 2004 1:18 AM

i don't "know the answer" but I do think about it almost every time it rains and I'm driving. I think the way to solve the question is this:

consider a volume of air that you are passing through and the amount of water inside that volume. If the rain is coming down at a constant rate, then the volume in a large enough space would stay constant for all practical purposes. So then you could determine the probability of each raindrop hitting you based on what percent of the total volume is made up of water. This volume of rainwater would increase the longer the length of time you spend in the volume. So if the space you were travelling through (say our hallway) contains a volume of 10 billion raindrops, then consider two speeds of travel. Say if I walk down the hallway it takes me 30 seconds. If the hallway receives raindrops at a rate of 10 thousand drops per second, then the volume of raindrops contained in the hal;lway during my 30 second walk is 300 thousand raindrops. Then say I can run the hallway in 8 seconds, then the volume of the raindrops during the time I spend in the hallway is 180 thousand raindrops. The calculation to determine how many drops would hit you is complex, but your volume remains constant, as does your surface area and (presumably) your orientation.

It would seem to me that the faster you travel, the less raindrops are in the volume, so the less percentage of the volume is made up of raindrops, so the probability of being hit by any one of them is less. This combined with there being less raindrops would indicate you would get less wet by running.

There are a couple other factors to consider. First, running doesn't necessarily expose you to more raindrops because as your volume moves forward you still occupy the same amount of space and you leave your old space just as quickly as you enter your new space. The only way runnign would make you more likely to be hit by a raindrop is if you flail your arms. This would increase your probability of being hit by a raindrop somewhat, but not enough to offset the lesser volume of raindrops. So, running still keeps you drier.

The main reason running keeps you drier is that the volume of raindrops is dependent ont he time you spend in the rain, so the faster you travel, the less raindrops there are to hit you. If, however, you are going to drive for an hour regardless of distance and the rain will be constant for that entire hour, your speed wouldn't make any difference.

## Another Scott, none of the ones from above - October 5, 2004 1:25 AM

Living in Portland, OR I have had the chance to run/walk in the rain almost everyday of my life.

Regardless of whether or not you run or walk in the rain, you will get equally as wet, running only gets it over with faster, which will lead to getting dry sooner.

/wishes it didn't rain so much here

## HyperX - October 5, 2004 2:30 AM

Running will get you more wet. When you walk you can use an umbrella ;)

## eCow - October 5, 2004 2:40 AM

Sounds like more people come from places affected by drought. Those of us in wetter climates have learned that it is wiser to get out of the rain than stand there counting drops. We normaly chose to run because after enought years you learn from experence that unless you are wearing rain gear the less time out in the rain the dryer you stay. That and you look less stupid than those just standing out there getting wet.

## Dave - October 5, 2004 3:19 AM

We did this problem in a college calc class many years ago. IIRC, it comes down to how much volume in space you're sweeping out per unit time. Running sweeps out more space per unit time, therefore you get wetter.

## Hoot Al - October 5, 2004 4:32 AM

Looks like everyone covered this question quite well.

My thought: If I am going from point A to point B, I think you will get wetter by going faster. How did I reach such a profound conclusion at 2:20 am? You are driving your automobile at 30 mph and it begins to rain. Set your windshield wipers to accomodate the rain. If you decrease your speed and the rain remains constant, your wipers are moving too fast for the volume of rain. This tells me the faster (walker versus runner) you go, you accumulate more water thus the wetter you would become.

## TAVO - October 5, 2004 9:44 AM

If you are going from point A to point B then walking would get you wetter. Why? Because you are get rained on longer walking than if you were running. Longer you are in the rain then the wetter you become. That is all.

## Mike Greene - October 5, 2004 11:08 AM

Maybe think of this. You have rain wetting you from two dirrections (assuming verticle rain). Rain hitting you from above, and rain that you hit from the side as you move, the front. THeoretically, as you move faster, you have more drops hitting you from the front, and slower you have more drops hitting you from above.

The exposed surface area of the front is something like 6 feet by 1.75 feet, the exposed surface area of your top is much less, something like 1 foot by 1.75 feet. (Assuming person is a rectangular prism. :) )

That means it could make sense for a dryer walker.

## Tom Kleingard - October 5, 2004 11:36 AM

if you were a really fast midget running would be drier...or would it because you would have less rain hitting you from the front, but the splashes would reach up to more of your body. Most likely even into your face.

## Andrej Milas - October 5, 2004 12:37 PM

I believe this also comes down to exposed surface area, which will vary as per run/walk speed as well as lol, dps drops per second.

Assuming that the rain is coming straight down the surface area most like effected by walking is your head and shoulders.

however...

Still assuming the rain is coming straight down and your running you will have somewhat of a bend to your head and back which will expose more of your back during your time in the rain.

therefore..

The amount of "wetness" should be derived from X (Exposed Surface Area) x DPS drops per second x seconds in the rain. Running or walking is irrelevant unless it would decrease or increase the aforementioned variables

## M Claf - October 5, 2004 1:49 PM

This is just a simple calculus problem. You must first create an expression that relates the amount of water that one encounters as a function of velocity. We will assume that the person is walking perfectly vertical and therefore do not have to take into acount body lean, which could make for another whole problem. You then will be able to determine the optimal velocity. We know that the ammount of water that hits your head and shoulders will be a constant per unit time no matter what speed you are traveling, therefore you can use the distance/velocity relationship substituted for time in that part of the expression. Good luck!

## It's not really that difficult. - October 5, 2004 2:40 PM

When you are outside and it starts raining, and you really, really don't want to get wet, you don't walk inside. You run inside. Why? Because you get less wet the less time you spend outside.

## Sillyness - October 5, 2004 3:21 PM

You guys don't get it. That test was useless! The rain from the beginning of the run dripped off the guy who walked. The runner still had the rain from back there cause it hadn't. That's all. It has nothng to do with who got wetter. But in my honest opinion, the guy who walks is gonna be friggin cold and wet for longer... therefore, wetter all together.

## Tim - October 5, 2004 3:32 PM

The faster you run the wtter you get until you run faster than the rain is falling, then the opposite is true.

## Pastor M - October 5, 2004 3:58 PM

More surface area to be hit with drops while running.

More time in the rain while walking.

Running gets you a little more wet IMO

## epistemology - October 5, 2004 4:03 PM

The question is about how wet I will get at different speeds going a given distance in a steady rain; let's use a mile for convenience. If I cover this distance in 1 second, there can be little doubt that I will get less wet than if I walk the distance in an hour.

If I made a 2-D cutout of the front of my body, and swept it through the mile in an instant, it would hit all the drops that were in the cutout's area, for the whole mile. But this would be true no matter what the speed. The cutout is going to sweep through that exact same volume (area X distance) no matter how fast, hitting all the same number of drops along the way, since the rain is steady. And fewer hit my head the quicker I go.

The seeming paradox is that the quicker I go, the more drops I run into. But this is <b>NOT</b> the case. In a given <b>time</b>, I will hit more drops if I run, but the front of my body will sweep through the exact same number of drops after traveling a mile, no matter how fast I run.

## Wen-Sheng - October 5, 2004 5:16 PM

I would assume that if I was walking fast or running, it would be because I had forgotten to bring an umbrella.

## Carpe Noctem - October 5, 2004 5:51 PM

For all practical purposes, as in moving from point A to point B and not simply running around in the rain, your resulting degree of dampness should correlate inversely with your speed.

## Edna - October 5, 2004 7:00 PM

My answers:

1) I don't know

2) If you're a rectangle, how are you going to walk?

3) I think you will get rained on at exactly the same rate regardless of your speed, because the raindrops fall at the same rate and there is an equal probability of a drop being in any given point at any given time. As you move (assuming you are on a conveyer belt because rectangles don't walk, or run) you will run into the same number of drops over any given time period, regardless of your speed. The only variable is time spent in the rain. I'm guessing that if the raining stops in three times the length of time as it takes to walk to point B (your top suface area is .25 square meters which is 1/3 of the total surface area [top of .25 + front of .5 = .25/.75= 1/3]) with potential of being hit), you are at break-even. If the length of commuting time is less than 3 times, you are better off standing still, and then walking. Unless you're late and don't care about getting wet. Then you probably ought to get on your conveyer belt.

## Ryan - October 5, 2004 8:33 PM

I think this one might be a bit easier than it looks. Correct me if I'm wrong. Let's take a stretch of 100 ft. of sidewalk in the rain with no wind. Let's put a barrel with an open top on a skateboard. Now, put another barrel atop that one, with open top AND open front. The bottom barrel is to catch the water for the experiment, where the fullest barrel=wettest person.

Barrels would look something like this:

(

(_)

.-. <<--skateboard ;OP

Now, let's 'walk' the barrels. While walking, the barrels will be in the rain longer. Rain will be pouring into the top of the barrel the entire time the barrel is exposed as it moves the 100 feet. The barrels will also catch 100 ft. worth of horizontal rain, which should be a constant given the 0 wind factor.

Now, let's 'run' the barrels. While running, the barrels will be in the rain for a shorter time, and thus gathering less vertical rain than the walker. However, the 'runner' will be catching the same number of horizontal drops as the walker, as they are travelling through the same 100 ft. of water.

The walker's barrel should finish with more water in it than the runner. However, if you were to stretch this experiment over a 10-mile distance, you might imagine that BOTH barrels would fill and overflow. Same would occur with runners and walkers. There is a point at which their clothes would saturate and they could not hold anymore water. At this point, they would be equally wet.

It's 4:31. I get off at 4:30. I'm outta here.

## Rick - October 5, 2004 10:26 PM

Assume rainfall is 1) constant and uniform (same volume per time at each location in a horizontal plane and drop size distribution uniform over time & area), 2) drops are falling vertically (straight down, no wind), 3) travelers (runner / walker) are identical in size and shape, and 4) traveler completely absorbs all drops contacted.

Front component of wetness: First consider the traveler as a vertical planar region moving horizontally from A to B. Water will be absorbed as the vertical planar region “runs into” raindrops on the path from A to B. The “vertical area” of the traveler can be defined as the traveler’s average frontal area as projected on a vertical plane. Although traveler’s shape may vary some based on speed of travel due to different motions in running vs walking, traveler’s projected area is approximately constant. Total water volume of drops in any vertical plane (let’s call it “drop density”) is constant due to 1) above. Water absorbed frontally depends on traveler’s average frontal area (a constant), drop density (a constant) and distance of travel from A to B (a constant), and is therefore, approximately constant for runners & walkers. Note that “still-standers” would avoid this component of wetness, but would also never get to B.

Top component of wetness: Next consider the traveler as a horizontal planar region moving horizontally from A to B. Clearly, faster travel from A to B exposes the traveler to less wetness due to less time spent on the path, given the same horizontal area. However, the horizontal projected area of the traveler is highly dependent on speed of travel. Arms and legs extend significantly while running, presenting a larger horizontal surface area, while walking creates a lesser area (especially if the walker were trying to minimize the horizontal area). I estimate running horizontal area at 2.5 to 4 times walking horizontal area. Therefore, a runner would have to travel 2.5 to 4 times faster than a walker to get the same amount of wet.

So in real situations, I think this can make the runner and walker fairly close in wetness after a traverse from A to B. The foot splash factor increases the projected wetness for the runner, however. Also, the rate of drop absorption may depend on impact angle, reducing the absorption rate for faster runners.

I come out at a wash. The guys who did the study are right for the conditions studied. In the end who cares anyway?

Assume rainfall is 1) constant and uniform (same volume per time at each location in a horizontal plane and drop size distribution uniform over time & area), 2) drops are falling vertically (straight down, no wind), 3) travelers (runner / walker) are identical in size and shape, and 4) traveler completely absorbs all drops contacted.

Front component of wetness: First consider the traveler as a vertical planar region moving horizontally from A to B. Water will be absorbed as the vertical planar region “runs into” raindrops on the path from A to B. The “vertical area” of the traveler can be defined as the traveler’s average frontal area as projected on a vertical plane. Although traveler’s shape may vary some based on speed of travel due to different motions in running vs walking, traveler’s projected area is approximately constant. Total water volume of drops in any vertical plane (let’s call it “drop density”) is constant due to 1) above. Water absorbed frontally depends on traveler’s average frontal area (a constant), drop density (a constant) and distance of travel from A to B (a constant), and is therefore, approximately constant for runners & walkers. Note that “still-standers” would avoid this component of wetness, but would also never get to B.

Top component of wetness: Next consider the traveler as a horizontal planar region moving horizontally from A to B. Clearly, faster travel from A to B exposes the traveler to less wetness due to less time spent on the path, given the same horizontal area. However, the horizontal projected area of the traveler is highly dependent on speed of travel. Arms and legs extend significantly while running, presenting a larger horizontal surface area, while walking creates a lesser area (especially if the walker were trying to minimize the horizontal area). I estimate running horizontal area at 2.5 to 4 times walking horizontal area. Therefore, a runner would have to travel 2.5 to 4 times faster than a walker to get the same amount of wet.

So in real situations, I think this can make the runner and walker fairly close in wetness after a traverse from A to B. The foot splash factor increases the projected wetness for the runner, however. Also, the rate of drop absorption may depend on impact angle, reducing the absorption rate for faster runners.

I come out at a wash. The guys who did the study are right for the conditions studied. In the end who cares anyway?

I am a bigger NERD than you.

## Olyama - October 6, 2004 2:55 AM

Why not accept the fact that your going to get wet and decide how quickly you want to get there rationally!!

## Karisma - October 6, 2004 5:58 AM

Why do you think people run, not walk to get out of the rain ?

## Fraser - October 6, 2004 10:07 AM

Without any mathematical or anecdotal evidence to back me up, but I imagine that the difference in wetness between walking and running is probably a few percent at best (except for very short distances covered by a sprint) so the eventual answer isn't going to matter much in the real world

Having said that, go ahead and run if you can, cause it's usually nicer out of the rain than in it, so better to get it over with.

## me - October 6, 2004 10:34 AM

The question and the answer are both

GAB

General Abstact Bullshit

## PhysicsNewbie - October 6, 2004 2:28 PM

'''''''''''

(A)'''''(B)

The same reasoning for why walking gets you wetter has been provided, but I think pictures and the concept of density will help some people better understand why. Looking at the picture above if run infinitely fast (freeze time) you hit all the rain drops in front of you from a to B. If you walk slowly you hit the raindrops in front of you at a slower rate but still hit the same amount since the density is the same. The differnce is that you now are hit by rain drops above you and thus you are more wet when walking. Constant wind still implies constant density and thus no effect.

## PhysicsNewbie - October 6, 2004 2:29 PM

'''''''''''

(A)'''''(B)

The same reasoning for why walking gets you wetter has been provided, but I think pictures and the concept of density will help some people better understand why. Looking at the picture above if run infinitely fast (freeze time) you hit all the rain drops in front of you from a to B. If you walk slowly you hit the raindrops in front of you at a slower rate but still hit the same amount since the density is the same. The differnce is that you now are hit by rain drops above you and thus you are more wet when walking. Constant wind still implies constant density and thus no effect.

## Joe Bob - October 6, 2004 3:41 PM

There are so many variables that it would probably take longer to define them than to actually solve the problem.

IF all things remain consistent, IF the speed of the standee, walker, and runner, the speed of the rain in relation to the person and the area covered, and the surface area of the person are all constants AND have absolute relationships, then it is relatively easy to reach a (very simplified) answer.

Basically, if these are all constants and the relationships are a straight line, then the only difference between running and walking is that your front would get wetter than your back (and shoulders).

If you're caught in the rain, you have to figure out all those - and more - variables to make the "right" decision. Or just carry an umbrella in your damn bag.

## Rick - October 6, 2004 9:29 PM

This matter is of little import, practically immaterial, and easily resolvable by both runner (rainsuit) and walker (umbrella). No great good will be done by having the answer. Still...

Further thinking: Consider the motion of the arms and legs as factor in runners & walkers wetness after traveling from A to B:

Assume rainfall is 1) constant and uniform (same volume per time at each location in a horizontal plane and drop size distribution uniform over time & area), 2) drops are falling vertically (straight down, no wind), 3) travelers (runner / walker) are identical physically (i.e., in size, shape, clothing, hair, etc.)

Front component of wetness: First consider the traveler as a vertical planar region moving horizontally from A to B. Water will be absorbed as the vertical planar region “runs into” raindrops on the path from A to B. The “vertical area” of the traveler can be defined as the traveler’s average frontal area as projected on a vertical plane. Although traveler’s shape may vary some based on speed of travel due to different motions in running vs. walking, traveler’s projected area is approximately the same for both. Total water volume of drops in any vertical plane (let’s call it “drop density”) is constant due to 1) above. Water absorbed frontally depends on traveler’s average frontal area (a constant), drop density (a constant) and distance of travel from A to B (a constant), and is therefore, approximately constant for runners & walkers.

Second, consider water absorbed frontally due to motion of the traveler’s extremities while running/walking. Due to back and forth motion of arms and legs, projected vertical areas of the traveler’s extremities effectively “run into” more raindrops than traveler’s static vertical projected area, as described above. Also, they absorb water on both front and back surfaces since they are moving in both directions. Water absorbed on these regions will depend on drop density (constant), horizontal extension of extremities (depends on speed of travel) and number of back and forth motions over the path A to B (depends on number of strides and therefore also on speed of travel). Since stride length varies by probably no more than a factor of 2 in normal running vs normal walking, the number of back and forth motions over a given distance varies similarly—half as many for running. By contrast, extension of the extremities, especially the arms, varies by more, perhaps 3 or 4 times. Average projected vertical area of the extremities would seem about the same for running vs. walking. Given all this, water absorbed frontally due to motion of the extremities would be more for runners than for walkers.

Note that “still-standers” would avoid both these components of wetness, but would also never get to B.

Top component of wetness: Next consider the traveler as a horizontal planar region moving horizontally from A to B. Clearly, faster travel from A to B exposes the traveler to less wetness due to less time spent on the path, given the same horizontal area. However, the horizontal projected area of the traveler is highly dependent on speed of travel. As noted, arms and legs extend significantly while running, and present a larger horizontal surface area, while walking presents a lesser area. Also, in running, the arms are often held horizontally extended, as opposed to swinging at the sides, as in normal walking. It would be reasonable to estimate average horizontal area while running at 4 times that while walking; i.e., the runner would have traverse A to B in one quarter the time to get the same amount of wet. For normal runners (<10 mph or 6 min/mile) & walkers (>3 mph), runners would get wetter. Additionally, runners are more exposed to the splash component of wetness.

It would be expected then, that the runner will get wetter than the walker during a traverse from A to B. I think this explains the experimental result. I would be interested in hearing others’ criticisms of these arguments.

## Rick - October 6, 2004 10:20 PM

Fact is... it probably matters less how wet you got than that you got wet. Runner & walker may suffer the same.

## ryan h - October 7, 2004 2:03 AM

it comes down to this:

if you run faster than the rain falls, you get wetter. if the rain falls faster than you run, you get less wet. this all assumes that you are a running cube, with equal surface area facing up and facing down.

## ryan h - October 7, 2004 2:04 AM

i meant "facing up and facing in the direction you are running."

## Mike Iafeta - October 7, 2004 8:58 AM

OK, I'm officially 'stumped'!

Stephen, if you have an answer, could you email it to me please?

## kevin - October 7, 2004 3:36 PM

Let's look at a somewhat special case.

First you have to assume that the distance to be travelled is relatively short. If it is not, you reach saturation with either method and neither has any benifit as far as "wetness" is concerned.

This is a variation of the "two surfaces" argument which has been proposed many times previously. The special case is this, while walking through the rain, you hunch your head and shoulders foward to shield the front of your body from the rain, and walk at a speed that keeps your hunched-foward back mostly dry. When doing this, your head, neck, and shoulders get quite wet bacause you spend longer with the rain falling on you, but you basically eliminate the rain you would encounter while "walking into it". This of course breaks down in high wind or if the distance is so great that the water dripping off your head and shoulders onto the rest of your body becomes significant.

This is more an engineering approach than a scientific one, and this is the way I handle rain if I don't have an umbrella and the distance is greater than about 30' or so. For distances shorter than that I find I can maintain a sprint of sufficient speed to minimise the wetness (though leaning foward to minimise the amount of rain encountered by the front of my body is still an issue)

## Stephen A. Meigs - October 8, 2004 12:17 AM

We may assume that rain is distributed continuously in space with constant density. Relative to the reference frame of the box, the equation of continuity for the motion of continuous matter (the rate of change of mass density is minus the divergence of the linear momentum density) together with the divergence (Gauss) theorem show that we may calculate the rate at which water mass is entering the box by calculating (relative to a normal pointing inwards since we are interested at the rate at which water enters the box rather than leaves it) the surface integral of the linear momentum density (relative to the reference frame of the box) on the upper and front surfaces. On the upper surface, the surface integral is the product of the density of the water, the downward component of the velocity of the water, and the area of the top of the box. On the front surface, the surface integral is the density of the water times the forward velocity of the box times the area of the front of the box. Thus water enters the top of the box at a rate not depending on forward velocity, while water enters the front of the box at a rate directly proportional to the velocity of the box. Hence, the total amount of water mass that enters the top of the box during the trip is proportional to the time of the trip, which is inversely proportional to the speed of the box, while the total amount of water mass that enters the front of the box does not depend on the speed of the box. Therefore, the slower box gets wetter.

## Charlie Finn - October 10, 2004 2:57 AM

Walking sideways presents even less surface area to the rain and therefore you will as wet! Though in practicallity who wants to be seen walking sideways in the rain. Run through it all and get over it.

## argie - October 10, 2004 7:24 AM

Returning from the Pacific Northwest, where we have engaged in an humble experiment to judge the veracity of at least two of the obviously aleph-null number of possible answers to this question, our small team was recently able to ascertain -- using a single unemployed Maxwellian Daemon (nicknamed "Worbert Niener" and fortuitously available due to the USA's enthusiastic and patriotic Dark Ages Retro Movement) to move the holes about within one of the team member's pet Peruvian Ambulatory Sponge (Spongatus Perambulatus), which happens to be precisely of the size 1m x 0.5m x 0.5m -- that it makes no difference whatsoever at what velocity the sponge moves about or even whether it ever arrives, except in the degenerate case wherein v approaches c, at which point said mobile holes are insufficiently round to avoid overlap with the enumerable transiting droplets. At this juncture, however, with mass approaching infinity, the number of contacted droplets per sponge become an infinitesimal fraction and, per Cantor, it is easily demonstrated that there is no wetting whatsoever.

Therefore, in both the wetted and unwetted situation, no wetting occurs.

(disclaimer: For the sake of simplicity, reciprocity effects have been ignored, and droplet size has been assumed larger than 1.616 × 10^^-35 m. or so.)

## blueboy - October 10, 2004 1:18 PM

This is just utterly stupid to spend time on. Just go out and try it for cryin out loud. If you walk very slowly and rain is falling at a contant, say 1"/hr, if you walk for 1 hr you will receive 1" of rain. If you walk the same distance in 1/2 hr you will only receive 1/2" of rain. The only factor here that presents an unknown would be the individuals body metabolism in regard to how much heat would be expended to cause a drying effect. One would need to develope a constant for this also.

## James - October 10, 2004 3:48 PM

But if you just run though, you will *think* you are less wet then the person who slowly walks though. Plus, you will get where you are going faster and be dry again sooner, and isn't that what counts? ;-)

## james martin - October 10, 2004 5:02 PM

The extra rain your front collected as you moved into it would be cancelled by the fact your back would get less, as you moved out the road of it, idiots!

As for your top; now listen carefully: The less time you spend in the rain the less water will hit you, period.

Senario: a volcano erupts spitting dropplets of molten toxic fire into the atmoshhere and onto a nearby town where a game of football, soccer, rugby, whatever you like is taking place.

One of the sides is made up of sensible people that know why they have been blessed with legs that have the ability to vary the rate of velocity (in this example the change would be to 'as fast as posible') that they can travel over the ground to the relative safety of the changing rooms, and, just as important as this ability; when not to let the brain interfere with the natural increase in velocity that certain situations require. i.e, a volcano erupts spitting dropplets of molten toxic fire onto you.

The other side however; is made up of mentaly constipated morons that for the same reason that they would have lost the game anyway; namely: trying to invent over complicated tactics and strategies thats real purpose is to give them a sense of purpose while they are trying to learn to trust there real instincs; decide to try and outdo hell by walking to the fitting rooms.

Ok, from all of all the players in the two teams which side is left with the most able bodied members that are able to pass there gentic makeup onto the the next generation after the storm?

p.s, sorry about any bad punctuation.

## D Bacon - October 11, 2004 10:03 AM

This has now been given a mathematical formula by Imperial College of Science, Technology and Medicine. Other variables which need to be considered are:

angle of rain.

ratio between profile areas,i.e. area of top of head against area of front of body.

and probably a few more no doubt.

Anyway the conclusion from Imperials work is RUN! But it does depend!!! Running into the rain you'll probably get more wet than runnign with it!!!

## Stephen A. Meigs - October 11, 2004 5:56 PM

If there's a tailwind that is sufficiently strong, you are best off walking (or running?) at the exact speed of the tailwind. That way, your front (and back) won't get wet at all. The condition for when you should move at tailwind speed is v_f > ((A_t)(v_d) + (A_s)(v_s))/A_f, where

v_f is the forward component of the rain velocity

A_t is the top area

v_d is the downward component of the rain velocity

A_s is the lateral (side) area

v_s is the lateral (side) component of the rain velocity

A_f is the frontal area

Here, v_w and v_s are to be calculated with respect to a

reference frame fixed relative to the ground (with axes pointing in the obvious directions).

If the inequality is not satisfied, run as fast as possible.

## Jesse - October 12, 2004 1:40 PM

The rain has a flux F (drops/m^2/s)

The box has dimensions L x W x H

The distance to travel is D

The box velocity is vb

The rain velocity is vr

If you imagine the box sitting still for 1 second, the # drops N would be N = F*L*W

But if the box is moving it will not only capture all the drops hitting its top, but also all the drops that are moving in front of it. This effectively increases the box's length by the ratio between the box speed and the drop speed times the height L' = L + H*vb/vr.

So the total drops N would be

N = F*L'*W* (D/vb) = F*W*D*( L/vb + H/vr )

So if you're going a fixed distance, it's always better to run. If you're going for a fixed time, it's better to stay still. Relativistic effects are ignored for simplicity.

## Gabi Geva - October 12, 2004 5:52 PM

We may think of a pendulum and assume a man is attached to the moving cord of the pendulum. taking into account the following assumptions:

1. the rain velocity is fixed.

2. the size of the rain drop is fixed.

3. no wind

A. if the pendulum moves i,e, v>0 than for the same time the pendulum might do different distances (different speed...) i.e. different exposure surface to the rain and thus if the man is running he enlarges his exposure surface and thus get more drops per time unit.

## Stephen A. Meigs - October 15, 2004 11:25 AM

Kevin's idea of leaning forward is the most practical solution. Moving forward makes the water come at you at an angle. Point yourself in that direction and theoretically your front and back won't get wet, because (relative to a reference frame travelling along with you) the rain will be travelling parallel to those surfaces.

## chere vhermelle - October 21, 2004 12:25 AM

well actually considering the velocity of the rain and the distance travelled by it we can tell if you will get more wet,and concidering your vellocity too.

well eventually the faster you are the less you'll get wet...that is if the distance travell of the rain is grater.but even if you are fast if the rains velocity is grater and its distance travelled from the clouds to you is lesser there's no sence of having a grater speed....rigth?????

therefore wetness is directlly proportional to the velocity times the distance of the rain and inverselly proportional to your speed and the distance you travelled...

## jerB - October 21, 2004 12:29 AM

well as for me I think chere was a little bit rigth there..."may point siya don!"so I'll consider my self one of the pro for him...thats all!

## chere - October 21, 2004 12:43 AM

pare nakuha mo na ba yung cd ng bold na pinahihiram ko sayo????bigay mo na lang sakin bukas...

## Ken - October 21, 2004 2:46 PM

Rain falls at a slant with the wind, so walking or standing you must lean into the rain and create a shadow of dryness, using an umbrella or just your head & shoulders One side of the street will usually be dryer than the other side.

Often the wind is gusty and there is no escape!

Wind & rain defy simple variables, and at any moment some drops may be blowing at you side on, top down and with varying intensity. Often a summer shower will be over quickly, so walking or running the sun may break through before you arrive! Go figure!

No two raindrops fall the same way.

## Tara.. - November 18, 2004 4:44 PM

thanks Ken Wollins you answered my extra credit question =)

## HeroreV - November 28, 2004 10:21 PM

Who cares? It's irrevelant. Even if walking at a certain speed got you less wet than running faster, you would still look like a dumbass if you slowly walked through the rain. Running/jogging may get you wetter, but at least people won't think you're retarded.

## Kris - January 12, 2005 12:19 AM

Wow. You see, I'm doing a Science Project about this same thing and was wondering if this has ever been efficiently tested. Perhaps I'll use this interesting debate of a page for one of my sources, eh? I have yet to do the experiment, but, though, when I'm finished, the results will be faulty anyway. I'm using sprinklers. Thanks for the interesting information and opinions on the subject. Quite useful really. Know any other good sites about this particular subject?

## mark - January 13, 2005 2:48 AM

The key is that you walk "through" the same number of drops no matter how fast you go. So you stay the dryest by going as fast as you can.

my notes:

how wet you get = (rain that falls on top of you) + (rain that hits the front of you)

rain that falls on top of you = (area of the top of you)(rate of rain fall)(time you are in the rain)

rain that hits the front of you = (area of front of you)(rain density)(distance you have to walk)

Since time doesn't even come into play for the "the rain that hits the front of you," you are going to bump "into" the same number of drops whether you are walking slow or sprinting.

And since you don't actually walk through more drops by going faster, but more drops fall on top of you if you're slow, the way to stay the dryest is to go as fast as possible, assuming you are 'rectangular.' In reality, running probably causes the area of the top of you to increase since you are sticking your legs and arms out further, so that makes things more complicated.

## nobody special - January 22, 2005 12:36 AM

Here's what I get (after some kindergarten level geometry) : the inclined walk visualization of chris works just fine for this.

If the dimensions of the man are H (tall), W (wide) and T (thick), the terminal velocity of the raindrops is v1, the man's speed is v2, and the density of drops per unit volume is n, then the number of drops hitting the man over a time t is :

N = nV = nWA = nW[(H + tv1)(T + tv2) - (t^2)v1v2 - HT] = nWt(Hv1 + Tv2)

If the time in the rain, t, is fixed, then N = Cv2, so, it's better to stay still.

If the distance d, is fixed, then t = d/v2, and hence :

N = nWd[H + T(v1/v2)], so this is minimized by maximizing v2. Hence, the faster you run, the drier you stay.

## BArry Brown - February 5, 2005 10:33 PM

This reminds me of the 1/2 the distance paradox. Mathematically, you never reach your goal by moving 1/2 way there and stopping. In real life, you will hit your nose on the wall. There are more variables to be included than have even been thought about if you are going to imitate life:

1. Rain, even steady (artificial) rain is not of a constant density. This is an assumption that allows us to make mathematical models, but it is not any part of real life.

2. Neither is the assumption that the sizes of the vertical and horizontal components of the body remain the same whether walking or running.

3. How much water is absorbed by the hair and clothing probably has a minimum velocity associated with it until some point has been reached. In other words, most cloth will not absorb easily until it is already wet.

4. Dripping water has to be included in real life. Just because it hit your head and then rolls down to your shoulder instead of bouncing off, doesn't mean that it doesn't add to the wetness.

5. Speed of rainfall has no effect - rain cannot fall faster or slower than you run, it is moving vertically at some density in relation to your position.

6. Type of clothing.

7. Type of shoes.

8. Splash pattern while walking vs running

9. Height of splashing water

10. Size and amount of standing water

There are more, but the fact is, we do not have the mathematical models necessary to simulate even walking or running in the rain. We can perform experiments to learn what happens during each, then we can begin to model what happens to the point where we can reach the same conclusions as real life. But even the best simulation is only as good as the past information. If you are walking across a parking lot with only 1/4" of water and no puddles, then haul butt. But if the water is 1" deep, the faster you run, the more you are going to splash and the wetter you will be from the knees down, for example. Real life, even in such as simple context as this is very much more complex than most people want to believe. All of the simplifications given here allow you to show either through math or modeling what happens, but it is only a crude approximation of real life.

B

## Cranky - June 21, 2005 12:00 PM

I have not finished reading all the posts, and it seems the thread has long since died off, however, I am going to offer e tidbit of what I have garnished from the many times I have considered this problem.

A person has height, width, and depth, but we can ignore width to simplify the problem, and like people said before, there becomes to parallelagrams to consider. 1 representing the bodies front, and 1 representing the bodies top. do to height and distance being constant, the first parralelagram will contain the same amount of space, and therefor an equal amount of raindrops for the 2 speeds. depth is the same for the second, but time is the factor, therefore a faster speed will reduce the amount of raindrops taken on the head.

Depending on the proportions of the subject, the difference this second parrallelagram makes can be significant or insignificant. for example and incredible tall and thin person would not be as affected as a shorter larger person.

We have been talking about rain that is at a consistent rate, but often times moving thru rain without an umbrella is caused by a situation of unexpected rain, in which case the rate of rainfall has a high probability of increasing during your travel, and so getting out of the rain before it gets thicker is always better.

Additionally, when wind is involved, more specifically a tail wind, there becomes a sweet spot where you can occupy the same rainspace throughout your travel, where moving faster would cause you to run into more drops, and moving slower would cause more drops to run into you.

## Erin Jello - September 6, 2005 10:26 PM

science fair project being done

how do you measure the rain and what would answer the question has this experiment been done already?

help!

## Erin Jello - September 6, 2005 11:30 PM

forget it

## Joe - November 26, 2005 10:17 PM

I can't believe how LONG this thread is! none of you are ever going to get rained on because you all sit indoors at computer trying to figure out what makes you wetter. It's only water, the stuff we're made of! It might be a serious topic if you lived on Venus, but not here.

## chris - December 5, 2005 3:23 PM

i think the faster u run the dryer u stay. because if u use physics u can find out that the faster u move the less water u get on u. So if u go slower u get wetter. Standing still will get u the most wet. The reason for this is that moving faster gets u to the place faster and hence less water on u. If u stand there u get the most water on u because u dont move and the rain could last for hours. Also if u walk u are moving faster than standing there but if u run u are in the rain a shorter amount of time therefore u get less rain on u and then u are not as wet as just standing there or walking. So run and get less wet and stay dry.

## fuck u all - December 5, 2005 3:25 PM

get a life u computer nerds

## Blair - February 2, 2006 11:34 PM

Wow....i just wanted a "Yeah runnin get you better", or "Yeah walking gets you wetter" answer. now i dno what to belive oh well i guess i'll go with the mith busters on this 1 and asume walking gets you less wet

## Carman - March 15, 2006 4:43 PM

Just a question off topic(don't know how to post my own stuff here).

Does a puddle splash more if you drive fast through it or slow? many people say its bigger if you go fast but my smartass brother thinks its bigger if you go slow?!?!

this is posibly moderate speed compared to speeding up.

I hope you guys answer this!!!

## Andrew - March 21, 2006 12:29 AM

I just analysed this using a model with

- two independently choosable body areas, vertical and horizontal, plus an inclination angle

- two independently choosable wind velocity components, vertical and horizontal

- a fixed value of rain mass density

- a fixed value of the distance to be travelled

I won't bore y'all with the gory high school algebra, except to say that in all of the 3 scenarios:

a) no horizontal wind component

b) no vertical wind component

c) any combination of horizontal and vertical wind components

the choice for minimising wetness (i.e. the total mass of water intercepted) is the same.

RUN FOR YOUR LIFE!

More specifically, the plot of wetness vs. running speed is an asymptote to zero as running speed increases.

The only twiddle left is to determine the optimal inclination angle, so you know exactly how much to lean your bod as you run as fast as you can through the rain (and granted this could present some practical difficulties :).

It turns out that whether you lean forwards or backwards, there is only a WORST angle corresponding to a maximum wetness; there is no "best" angle corresponding to a minimum.

Can't be sodded to post all the equations - sorry.

-Andrew

## Carman - March 31, 2006 3:58 PM

Thank you Andrew, you gave me enough info to tell of my brother off!!!

## fart man - May 21, 2006 3:17 PM

I like rain

## vanessa - June 26, 2008 1:46 PM

this problem was solved in a mythbuster episode. apparently, you get a lot more drenched if you run under the rain.