Walking in between raindrops...

[Kyrisch]

On the contrary of what the title suggests, this is an actual serious question born of curiosity that came to mind while spending a miserably dreary morning outside in the pouring rain waiting for the bus.

 

Is there a greater chance of a mass getting hit by a raindrop if said mass is moving or stationary?

 

I can't work out the answer. Is there some advanced math formula for stuff like this?

[cosine]

If you assume that the rain is randomly distributed, then I would expect that your chances of being rained on don't change whether you are moving or still. I mean this is assumming random distribution though, otherwise if some places have less rain on then than others, then the trivial solution would be to go and stay where the least rain is. If the place of least rain is changing, but the position of the rain is moving to discrete points and not moving along a continuous line, I think that would be a pretty interesting problem.

[MattC]

Absolutely - you will get wet faster if you move fast (there are exceptions). To understand why, take two extremes:

 

1) You are standing still. The number of drops hitting you is basically the average number of drops falling through an area equal to your profile (from above, looking down) in a given amount of time.

 

2) You are running very, very fast. In addition to always having a 'profile' of a certain area, which is hit by the aforementioned number of drops, you are also walking into drops that are already below the level of your head/shoulders. These drops hit your legs, your stomach, even the fronts of your feet (the parts perpendicular to the surface of the ground.

 

This is assuming the rain is falling straight down, which happens sometimes. Other times, the rain is falling at an angle - then there is a certain speed at which you can walk in the same direction as the failling rain drops (say, they are falling down and to the right - you walk to the right) which will minimize these drops that you walk into. You can walk in the opposite direction and have the opposite effect.

 

Ultimately, if you want to minimize the number of rain drops hitting you, stand very, very still ... under a tree.

[ecoli]

2) You are running very' date=' very fast. In addition to always having a 'profile' of a certain area, which is hit by the aforementioned number of drops, you are also [i']walking into[/i] drops that are already below the level of your head/shoulders. These drops hit your legs, your stomach, even the fronts of your feet (the parts perpendicular to the surface of the ground.

 

Not only that, when I run I tend to lean foward. This would expose me to more drops if the rain was coming strait down.

[starbug1]

Well, Kyrisch, Matt C has the basic answer for the "rain" problem. But because rain, and weather in general, are so unpredictable, there would be no mathematical equation (not that I know of.) to show the rate of the "wetness factor."

 

Now, standing miserable in the rain is another problem altogether. A tree, like Matt C suggested, is not always ideal because the raindrops coming from the trees are usually bigger (if the tree has leaves. If the tree does not have leaves, well, then it wouldn't be very helpful, would it?)

Of course there are also the "variables of wind, temperature (of the air and rain), and, let's call it "types of rain" There is also the fact that you are standing in an area most likely to get your feet wet, if not by the ground, then because the rain tends to "splatter" so to speak.

 

It has been proven that you pick up water (rain) faster when walking or running, but put to practical use I've found that myself and others tend to become "less miserable" when walking running, or doing some sort of active body movement. Although you may be getting more wet, you are activating body heat, which usually makes a cold wet day more bearable. That's if, and only if, you don't have to wait long!

 

Basically, there is no set equation for "normal conditions" because there are no "normal conditions" in rain patterns. I wouldn't be surprised, however, if there were a mathematical equation for a "controlled experiment of rain" using equal rain (water) distribution and wind conditions that are controlled or nonexitent.

...so pack your raincoats and galoshes because weather can be unpredictable?

[cosine]

Not only that, when I run I tend to lean foward. This would expose me to more drops if the rain was coming strait down.

 

Haha, you mean the mathematical formula shouldn't just assume that you're a point?

[YT2095]

stationary or moving slowly is best.

 

 

I saw it done on MythBusters )

[insane_alien]

well in scotland the rain usually falls almost horizontally so your soaked whatever happens

[JonM]

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

 

 

you can figure it out for youself... enter your speed, height and other factors in order to figure out how to minimize the rain that falls on you

[Sisyphus]

Wouldn't you always stay the driest by moving the fastest? Sure, you would obviously get more wet per unit of time, but you would also get to where you're going faster, and presumably be out of the rain. This, of course, is assuming that the limiting factor is not length of rainstorm but distance you have to travel. Again, take the extremes. Moving infinitely fast, you get all of the water that is in the air between you and your destination on you. Standing still, however, you get infinitely wet, since you'll never reach your destination.

 

I guess it depends on the context. If you're stuck outside and have to wait out the rain, stand still or move as slowly as possible. If you're going from one dry place to another, move as quickly as possible.

 

EDIT: That website seems to think I'm wrong about moving as quickly as possible (at least when the rain is not vertical), but it's still definitely a bad idea to move as slowly as possible!

[cosine]

Wouldn't you always stay the driest by moving the fastest? Sure' date=' you would obviously get more wet per unit of time, but you would also get to where you're going faster, and presumably be out of the rain. This, of course, is assuming that the limiting factor is not length of rainstorm but distance you have to travel. Again, take the extremes. Moving infinitely fast, you get all of the water that is in the air between you and your destination on you. Standing still, however, you get infinitely wet, since you'll never reach your destination.

 

I guess it depends on the context. If you're stuck outside and have to wait out the rain, stand still or move as slowly as possible. If you're going from one dry place to another, move as quickly as possible.[/quote']

 

This is an interesting point, though the original post notes he was waiting for a bus... so it would not be from one dry spot to another. Although physically speaking, I guess he could run infinitely fast to where the bus would have taken him...

[Ophiolite]

If you ran really fast - supersonic that is - the shock wave should blast those raindrops out of the way. Even if you do get hit by any, your elevated surface temperature from frictional heating will immediately evaporate them. You will arrive at your destination dry, but exhausted.

[starbug1]

How do we come about running supersonic?

[ecoli]

it's called the suspension of disbelief.

[Kermit]

Didn't they make a Mythbusters episode out of this?

[aj47]

Ah i've been thinking about this all week cos its been raining so much here.

 

I came to the conclusion that instead judging how fast I should walk to mimimise how wet I get, I should just be prepared for English weather and always bring a coat with me. Its seem to have worked so far.

[ydoaPs]

Didn't they make a Mythbusters episode out of this?

yes

[s pepperchin]

If we assume that the average density of raindrops is constant than we can calculate the how much strikes the object.

 

We start by looking at a unit time of 1 second, In the picture the height h is the distance the rain falls in 1 second and the distance d is how far the object travels in the same second. The wire frame shape represents the volume of rain which strikes the object.

 

The volume that strike the object is:

 

[math]Vfront + Vtop =(Afront + Atop){\sqrt{h^2 + d^2}}[/math]

 

then we can substitute for h and d

 

[math]h=Vrain*t[/math]

 

[math]d=Vmass*t[/math]

 

 

[math]Vfront + Vtop =(Afront + Atop){\sqrt{(Vrain*t)^2 + (Vmass*t)^2}}[/math]

 

We could use the ratio of the two velocities to determine an angle for the volume of rain striking the object. Then we could use the equation of the volume as a function of that angle to figure out when the Volume is maximized. If the rain is at terminal velocity than it would only depend on the velocity of the mass.