Does Running in the Rain Make You More or Less Wet?

[mordechai9]

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

 

[Parlyne]

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

 

[pixel01]

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

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

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

 

[pixel01]

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

 

[dst]

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

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

 

[mordechai9]

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

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

Please continue...

 

[pixel01]

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

 

[pixel01]

Here is the plot in MATLAB with different flux J

 

[cesiumfrog]

W2 is calculated as follow:W2 = A2*J *sin(actan(V/Vo))*S/V

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

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

 

[malty]

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

 

[pixel01]

I think W2=A2*J*S. Particularly, consider the limit where Vo=0.

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

 

[pixel01]

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

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

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

 

[Mk]

And if you move at speed of light, you are dry.

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

 

[malty]

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

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

 

[pixel01]

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

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

 

[cesiumfrog]

Oh no, you should run at an infinite speed to be dry.

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

 

[pixel01]

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

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

 

[cesiumfrog]

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

 

[Dale]

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

Particle man, particle man ...

 

[marcusl]

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

 

[Integral]

I see that no one has factored in the puddles that I am able to skirt when walking but hit full on at a run. Thus splashing up more from the standing water on the ground.

Oh yeah, this was for a spherical chicken in a vacuum.

 

[cesiumfrog]

And if the rain is coming from behind then running faster isn't necessarily better: I think the optimum is v_0\ cosec\ \theta..

 

[noagname]

also the mythbusters did this test and they said no more rain hits you if you walk or run

 

[jachyra]

i think maybe the misunderstanding between pixel and cesium is that pixel starts running JUST before rain actually starts to fall and cesium starts to run while rain is already falling.

without doing any math i like to thing of this problem as I am Hiro Nakamura of the show Heroes. If rain is already falling at an angle and I stop time (move at infinite speed...ok its not possible but just bear with me), then I can walk from point a to point b, but I'll get wet AT LEAST a certain amount in the front of my body because i am walking through all the droplets frozen in time. But if I chose to walk so slow that it takes me the length of time that the rainstorm lasts to go from point a to point b, then I'll get REALLY really wet... both on the head and in the front of my body.

so basically my conclusion is that when it starts raining... run!

 

[Shooting Star]

Do we really need computer programs to solve this? I would have thought a simple continuity argument from v=0 to v=infinity would have sufficed. (There’s really no need to invoke relativity.) Of course, I may be wrong and anyway, the programs will give a good idea about the graph of wetness and speed.

But nobody has asked the really interesting question: Why did the spherical chicken cross the road in the rain?

 

[mordechai9]

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

You didn't explain your expressions whatsoever! You simply wrote two equations and then said "these correspond to these quantities". Cesiumfrog is just asking for a derivation, which is an extremely simple request, and if you want your answer to be really accepted you have to give a derivation, not just punch out two equations and say "tada!".

 

[stewartcs]

Why did the spherical chicken cross the road in the rain?

To see if he would get wet or not.

 

[stewartcs]

It seems like a read somewhere the answer to this was that you would get the same amount of wetness either way. Don't remember where I read it though.

Intuitively I would think this to be true. Since there is a certain amount of rain that the rather large cross-sectional area of your front side would be running into (as compared to that of your head and shoulders), any benefits of running faster to avoid the amount of time your head (plus any other horizontal areas) is exposed to the rate of rain fall would cancel out (if not cause you to get wetter).

I'll see if I can find that paper on this subject.

 

[Shooting Star]

Just think that you are moving with speed very close to zero, that is, infinitesimally slowly. If the rain is eternal, then the amount of water falling on you from the top would tend to infinity.

Now mentally compare it with the situation when you make a high speed dash. Surely, this time the amount of water or the number of drops you sweep up in the front and the top must be finite, even though the rain will be slanted wrt you.

Please think.

 

[Shooting Star]

To see if he would get wet or not.

Good!

Anyway, it'll taste the same in both cases.

 

[pixel01]

Sorry, I have rechecked the model and the formula for W2 had a mistake. When V increases, the flux of rain according to the moving direction also increases, so the new flux J' in that direction must be multiplied by the speed V. Then the function of wetness according to speed V is calculated as :

W = W1+W2 = S*J*A1/V + S*J*A2*sin(atan(V/Vo)).

So if the speed goes to infinity, W1 leads to zero, W2 will approach S*J*A2 which is constant.
If you run, you can get more wet or less wet depending on the ratio of A1/A2.
Here I draw some curves according to different A1/A2 ratio, r = 0.1:0.7

 

[cesiumfrog]

pixel, care to explain how you obtained the formula? For example, what physical reasoning led to your sin of arctan term?

By the way, with your new formula, if the rain stops falling (Vo=0) then the wetness will still depend on the speed of travel. Doesn't that seem wrong to you?

 

[pixel01]

pixel, care to explain how you obtained the formula? For example, what physical reasoning led to your sin of arctan term?

By the way, with your new formula, if the rain stops falling (Vo=0) then the wetness will still depend on the speed of travel. Doesn't that seem wrong to you?

In the formula, the wetness W = W1 + W2. If the rain drops stop falling, we have sin(atan(V/Vo))=1, => W2= constant, then W depends only on W1 which in turn depends on V.
But when Vo=0, then J must be zero as well (according to the vertical direction) son W1 is equal zero ==> W must be constant.
I have assumed Vo = constant in my model (have not thought of Vo=0).
If necessary, I can make it into the model so it will cover the case when the rain drops stay still.

 

[Mephisto]

i once solved this problem. You had to have a function
E(v), which is Extra area exposed to the rain, which was a function of how fast you run, because if you run really fast, your pants will get exposed too, plus your arms etc.
I modeled the extra area as E(v) = c*sqrt(v), for some constant c that i can't remember, and arrived at conclusion that you get least wet at 16 m/s... now... you can't run that fast, so just run as fast as u can was my conclusion

my basic equation was

Amount of rain = Rain accumulated on your head + Rain accumulated on your Extra area + Constant rain that you have to "carve out" with your body in front of you.
Then i took derivative wrt v, and found the minimum

 

[PRyckman]

On discovery channel the myth busters answered this question. Using rain coming straight down they found the faster you went the more wet you became. Even if traveling faster meant spending less time in the rain. You must also consider that when traveling at a speed you are receiving rain on a larger surface area (your chest legs face etc) than you would just standing still.

 

[cesiumfrog]

you so did not cite mythbusters as an authority...

 

[PRyckman]

you so did not cite mythbusters as an authority...

well they follow the scientific method, I suppose you would have to watch the episode to see if you agree that they covered all the bases with their experiment

 

[cesiumfrog]

since when do they follow the scientific method?? Watch the chicken-windshield saga for a rare example of them going back and effectively admitting mistake in their method, conclusion, and high school physics calculations. In real science, you can usually only make weak conclusions, whereas on that "blow stuff up" show everything is conclusively "busted" or "confirmed".

 

[PRyckman]

Well certainly they have to jazz it up to make people watch the show, They definitely make mistakes as I'm sure any science lab would. I have seen a few shows where they went back and admitted to mistakes and re did the experiment.

I'm not trying to say that they have the definitive answer, but all we can do in this thread is talk math, at least they have a real world experiment to draw conclusions from

 

[kanato]

These are my thoughts.

Just defining terms: For a person moving through the rain they have cross sectional areas A1 and A2 above and in front of them. Rain is coming down with velocity vector -(v_x,v_y) and person is moving with velocity v\hat{x}. Density of the rain is R (drops/m^3), so the current density of the rain is J = -R(v_x,v_y). S is the distance traveled in the rain. Assume all variables are positive except possibly v_x.

If we transform into the person's moving reference frame then in that frame the rain is coming down with velocity (-v-v_x, -v_y) which makes an angle \alpha = \tan^{-1}\left(\frac{v+v_x}{v_y}\right) with the y-axis. So the current density is J = -R(v+v_x,v_y)

Then the flux through each surface is the current density dotted with the area, choosing area normals so that the result is positive.
F_1 = R A_1 v_y
F_2 = R A_2 |v+v_x|

So here we integrate with respect to time. Since these are time independent, we just need to multiply by the time traveled in the rain, S/v.
W_1 = S R A_1 (v_y / v)
W_2 = S R A_2 |1 + v_x / v|

For the simplified case where the rain is coming straight down:
W_1 = S R A_1 (v_y / v)
W_2 = S R A_2

So this says that the amount you get is a simple function of the form (a + b/v) so that the faster you run, the less you get wet. (The result is obviously not valid at v = 0, because we assumed that time spent in the rain is proportionate to 1/v). If the rain were suspended so that v_x = v_y = 0 (time stopped) then W_1 = 0 but W_2 isn't. If you move at infinite velocity, same result. It also says that if the rain is coming down at an angle from behind you (so v_x is negative), then if you run at a v = |v_x| then you will get the least wet. In this case, in your frame of reference, the is rain coming straight down on you. If you run faster then your front hits the rain in front of you and slower and the rain coming down hits your back. This seems intuitive to me.

I don't see how the factor of \sin(\alpha) comes in into your derivation, pixel01. Can you explain that? Have I missed something? It's late so there's probably a mistake in here somewhere.

 

[Shooting Star]

For the simplified case where the rain is coming straight down:
W1 = SRA1(vy/v)
...
(The result is obviously not valid at v = 0, because we assumed that time spent in the rain is proportionate to 1/v).

I assume W1 is the amount of rain (wetness) on the top of your head. This relation indicates that your result may possibly be very correct, because as v tends to zero W1 should tend to infinity, as I have pointed out earlier. This is one good example of Physics calculation.

(When I say "possibly correct", no offence is intended. I just mean that I haven't gone into the math thoroughly, which I will do.)

 

[pixel01]

It may take some time for me to rewrite it in detail. It was more complicated as I expected at first. It is a nice question isn't it

 

[mgb_phys]

But two years later in 1997, two meteorologists, Thomas C. Peterson and Trevor W. R. Wallace, from the National Climatic Data Center in North Carolina in the USA, published another paper called "Running In The Rain". They had read the "Raindrops Keep Falling On My Head" paper, and found a few mistakes, corrected them, and added in stuff like "rain driven by the wind".

Their equations showed that in a light rain with no wind at all, running will give you only a 16% reduction in wetness, as compared to walking. But if you're running rapidly and leaning forward in a heavy rain that is driven by the wind, you will end up 44% less wet than if you had walked. At this stage, Petersen and Wallace showed that they were fair-dinkum scientists and decided to do the experiment. Luckily they were roughly the same build, so they bought two identical sets of sweat shirts, pants and hats. They also bought two very large plastic bags to wear underneath these clothes, so that any rain which ended up on their clothes would not get soaked into their underclothes. They then measured out a 100 metre track behind their United States National Climatic Data Center office and waited for some rain. Soon, some heavy rain came along - falling at around 18 mm (or 3/4 of an inch) per hour. They made sure that they weighed the clothes both before and after the rain.

Dr. Wallace ran the hundred metres at around 14.4 kph, and his clothes picked up 130 grams of water. Dr. Petersen walked his hundred metres at a much more leisurely 5 kph, but his clothes soaked up 217 grams of water. Running, instead of walking meant that you got less wet by 40%, which was pretty darn close to their predicted 44%.

Their results can be summed up as:

"When caught in the rain without a mac,
walk as fast as the wind at your back,
but when the wind's in your face,
the optimal pace
is fast as your legs can make track".

 

[stewartcs]

This is kind of neat...

http://www.dctech.com/physics/features/0600.php

 

[Mephisto]

thanks for sharing that mgb, that's pretty interesting.
I personally think that this problem is too complicated to be resolved using algebra, and is much easier to just do the experiment. The problem is that you cannot just treat a person like a rectangle here, and when we run, our shape changes, and even worse, if you also want to consider the direction of the rain, you have to consider the angle of the surfaces that make us up as well - which is a nontrivial function of velocity; and then different people have different proportions and different ways of running...
it's a dissaster.
But I do believe that the faster you run the less wet you get - it makes more intuitive sense.

 

[chroot]

All of the mathematical models offered here and elsewhere are just that: models, and nothing more. Experiment is the arbiter in science. Experiment alone determines which models may be correct, and which cannot be correct. In this case, I'd listen to the MythBusters more than to anyone else, even if their show can be a bit sloppy at times.

- Warren

 

[disregardthat]

Does anyone have some data on amounts of litres of water per cubic metre in a given speed that is usual?

 

[Shooting Star]

I personally think that this problem is too complicated to be resolved using algebra, and is much easier to just do the experiment.
…
But I do believe that the faster you run the less wet you get - it makes more intuitive sense.

By algebra, did you mean mathematics or just simply algebra? I think you meant mathematics. Do you really believe what you said? I’m not joking. I would sincerely like to know. In my opinion, it is after all an extremely simple problem. We just have to sit down and concentrate on the problem, and do the math, which some people are doing.

And you have a fine intuition.

 

[Huckleberry]

I don't think the faster you run the less wet you would be. If you run too fast you will run into more falling rain. It would effectively change the angle at which the rain is acting against you, exposing a wider surface area for the drops to hit your body. How fast you should run depends on how the rain is falling in comparison to your velocity. Ideally you would want the wind at your back running at exactly the wind speed. Then the rain would be coming at the smallest surface area of your body, straight down.

This is assuming you are looking for the least amount of rain over an unspecified [distance]. If you have a specific destination in mind for shelter, of course, you will be better off getting out of the rain as fast as possible.

 

[Shooting Star]

We are considering the simplest case here. Afterward, we can refine it.

The rain is falling vertically, and the man is moving at a uniform speed in a st line betteen two shelters. I'd be grateful if you could give the basic mathematical formulae.