[bushindo]

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

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

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

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

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

Edited June 3, 2009 by bushindo

[Guest]

Haven't worked anything out but... it should be the same, running or walking.

[Guest]

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

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

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

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

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

Who can run 18 miles an hours??????????????????????????????? And at that speed wind turbulence would be a factor I bet

[bushindo]

Who can run 18 miles an hours??????????????????????????????? And at that speed wind turbulence would be a factor I bet

Usain Bolt

[Guest]

lol yeah this is one fast running rectangle

[Guest]

Haven't worked anything out but... it should be the same, running or walking.

check that... still no math but... the chest portion should be same running or walking ( faster ur run the faster the rain hits u). But the head part... the shorter time in the rain the better so running is better.

[Prof. Templeton]

It shouldn't matter how fast you go. The amount of drops in the area you have to pass through is constant, rain falls out of the area as fast as it falls in. You might think of it like freezing time, suspending the drops in midair, then walking through them. It won't matter how fast you go, you're going to hit them all. That said, since the distance is the same, it should then only matter how much time you spend out in the rain. So run!

Edited June 3, 2009 by Prof. Templeton

[Guest]

your going to hit the same amount of horizantal rain (rain you walk into not falls on you) but the time difference changes the vertical rain so it still is time out ratio that gives you rain i think

so with almost no math

((walking rain-horizontal rain )*1.5/8)+horizontal rain= running rain

so at distance as the limit goes to zero just the ratio of 1.5/8ths as much

or if the distance is suffiecient and the speed increase ration is sufficient it would tend towards horizontal rain

but the horizantal rain is a constant per distance no matter what speed you travel

anyway i might feel like doing the math later but im pretty sure my logic works pretty well.

Edited June 3, 2009 by final

[Guest]

Exact numbers cannot be solved without knowing the frequency of raining falling from the sky. In other words, how many drops hit the ground per second.

You could only solve for a relationship.

Because it is assumed to be constant rain, both head and chest will be hit with rain, constantly. Therefore the answer is just a ratio of times spent in the rain.

Walking across the field takes 200/1.5 = 133.33s.

Running takes 200/8 = 25s.

You will get 133.33/25 = 5.32 times as wet walking as you would running.

[Guest]

It shouldn't matter how fast you go. The amount of drops in the area you have to pass through is constant, rain falls out of the area as fast as it falls in. You might think of it like freezing time, suspending the drops in midair, then walking through them. It won't matter how fast you go, you're going to hit them all. That said, since the distance is the same, it should then only matter how much time you spend out in the rain. So run!

I want to add one thing to your conclusion, you are right, but the faster you go the less rain will hit your back and your head and the more it will it the front of you.

For those of you that don't know the Mythbusters did do this one and had clothing on that they weighed before and after. They came to the conclusion that it is worse to run, but they didn't take one thing into their calculations, that is they had nothing over your head. Think of it this way if you stand still and the rain is falling straight down the majority will hit your head and if you are a rectangle only your head will be hit, but the faster you go the more you run into, so the more your front would get wet, hence the reason they concluded it was worse to run.

[Guest]

The guys on MythBusters (tv show) challenged this myth with an elaborate experiment. it was actually pretty funny to watch.

http://kwc.org/blog/archives/2003/2003-11-...ythbusters.html

The first time they did it they found that walking in the rain resulted in being less wet than if you ran in the rain. However, I think they re did the experiment on a later show and found the opposite.

Like some people have said before, I really don't see how it makes that much of a difference. Personally I like to enjoy the rain while i'm in it, so I walk lol

[Guest]

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

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

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

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

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

We need to know the frequency of raindrops. The 200 drops/meter² falls in a sheet (let's assume). How many sheets fall in a time frame?

[Guest]

Maybe intuitively it seems like running through the rain is better, but what about when I'm driving and watching the rain on my windshield? I know the rainfall is not exactly constant, because that's not how nature is, but when I'm stopped at a red light, I have practically nothing hitting the glass, but when I start going, there's always more. Is this my imagination, or something else?

[Guest]

what angle am I at when I am running?

[Guest]

as i understood the question drops are assumed to be of negligible width and height (if its center is off your body the drop isnt wide enough to possibley hit you no matter how incredibley small the distance from you is) and the 200 drops is per cubic meter (itd be a really bad rain if it was every square meter)

but u might want to wait for bushindo if your gonna do alot of math

[Guest]

From personal experience in the rain, I have reached the following conclusions:

1. Drizzle or light rain: walking is better, as you get a lot more wet if you run. The rain drops seem to "miss" while walking

2. Heavy rain: Either case you're soaked!! So run baby run!

[bushindo]

as i understood the question drops are assumed to be of negligible width and height (if its center is off your body the drop isnt wide enough to possibley hit you no matter how incredibley small the distance from you is) and the 200 drops is per cubic meter (itd be a really bad rain if it was every square meter)

but u might want to wait for bushindo if your gonna do alot of math

yeah, sorry about that. Let's say that the rain density is 200 drops/meter³, or 200 drops per cubic meter. Thanks final.

Edited June 3, 2009 by bushindo

[Guest]

As others have mentioned, as long as the rain is constant, it does not matter whether you run or walk. You will have to pass through the same amount of water either way.

However, if the rainstorm is just beginning, then it will most likely get worse later on. Thus, it is better to run, because you can get out of the rain before the frequency picks up.

On the other hand, if the rainstorm is ending, it is better to walk, because then you give the rainstorm the chance to die down, as opposed to hurrying through and getting as wet as you can.

Maybe intuitively it seems like running through the rain is better, but what about when I'm driving and watching the rain on my windshield? I know the rainfall is not exactly constant, because that's not how nature is, but when I'm stopped at a red light, I have practically nothing hitting the glass, but when I start going, there's always more. Is this my imagination, or something else?

Just as in the case when you are walking, the speed of your car makes no difference--it has to pass through the same amount of water either way to get from one point to another (assuming the rain is constant). However, the frequency is different. If you drive slow, you accumulate the same amount of water, but slowly, over a longer period of time. If you drive fast, naturally the rain drops will hit more frequently, but you will get where you are going in less time, leading to the same amount of accumulated water either way. And as others have said, your speed does affect which parts get wetter--the faster you go, the more is absorbed by the front.

[Guest]

as i understand it if your completely vertical this is true your running into the water (same amount to walk through as only movement forward collects water)

but if you have a horizontal surface think of it this way by your(some of your) logic it is the same if you stand still for 5 hours and then walk a block or just walk the block. It just doesnt make sense. Id call that mindfreak idiot if this happens hed want to learn that trick.

people say it doesnt matter, because people are on the other side that running cures everything, but people that say it doesnt matter are more wrong then the people that think running cures all.

This is my logic it might have some errors so please point them out but im almost positive running does nothing for a vertical surface but reduces water on a horizontal surface by the ratio of the speeds.

Edited June 4, 2009 by final

[Guest]

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

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

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

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

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

Well, if walking, then you'd be going 1.5 meters per second and the rain is faling at 9 meters per second, and at 200 drops per cubic meter u'd have 300 in the 1.5 meter and 300*9 would be 2,700 drops. if the field is 200m,and you go 1.5m per sec, then it will take 133.33 secs to cross the field. 2,700*133.33=359,991 and-oh shoot,I forgot to take into acount the size of the body. Try again tommorow after my math final when my gears are in full motion and I have a better chance of an 'aha' moment. Plus, SCHOOL WILL BE OVER! This site will keep my jucies flowing. better stop typing and get rest for my exams, huh? "are you of the computer yet? BED! NOW!"

"OKAY! ten more minutes mom?"

But seriously, maybe with sleep the answer will come to me.

[Guest]

Amarnath_Caterpillar

The number of drops hitting is directly related to the time taken to travel 200m . Assuming no two/more successive drops at a location hit the area during the trace.

Total surface area=0.56m^2

After making several other assumptions like number of drops hitting a area= 200drops/m^2/sec,etc i figured out the answer would be close to the solution below

Case1: Time to travel 200m=133.33sec

Number of droplets hitting = Number of drops hitting the total rectangle area standing at a location for 133.33sec

Ans=200*0.56*133.33=14932 drops (Units=drops/m^2/sec*m^2*sec=drops)

Case2: Time to travel 200m=25sec

Number of droplets hitting = Number of drops hitting the total rectangle area standing at a location for 25 sec

Ans=200*0.56*133.33=2800 drops

Ratio= 133.33/25= 5.3332 times more wet than running

[Guest]

ok i was trying to get out of this without doing the math but i guess its hard to convince someone that im right without math (for reference always assume im right and your wrong it will save you alot of time and me alot of work)

I reread the problem as stated and if you are a 2d person that exists entirely in the vertical plane then no drop can hit you if you stand still. So technically the only water that hits you, you have to walk into

so

add all area todethor .56*200=112cubic meters that you walk through

each cubic meter has 200 drops so 112 *200=22400 drops.

you can solve it without ever using speed (or anything dependent on it) so speed doesnt enter into it.

What i think the problem might be is a 3d figure made of two rectangles (one in each plane) I think this because otherwise way too much info is given (even if you ignore the speed because that would be just to trick you)

so...

I have two basic assumptions

these assumptions are the head is in the horizontal plane only. The surface area can be 3d but i think for this problem how much of it lies in each plane needs to be known so i assumed it was the top of the head as any part of it in the vertical plane could have just been added to the body

second the body area is only in the vertical plane so math ho

this can be treated as two different objects and then added togethor so as in the previous problem but only for the .5 of the body

.5*200*200=20000 drops hit body

for the head you have a lil more math

200 drops/cubic meter and the drops move at 9 m/s so 200 drops/cubic meter* 9m/s=1800 drops/(square meter*second)

or 1800 drops per square meter per second (remember this is falling straight down so that is y this doesnt apply to the body)

so 1800 drops/(square meter*second)*.06square meter/head=108 drops per second/head (1 head so per head is stupid but if you think of it as a unit like inch then it might help you follow it more)

anyway for walking 200meter/(1.5meters/second)=133.3333seconds

so drops on head for walking = 108 drops/second *133.3333 seconds= 14400 drops on head

14400+20000=34400

for running body is the same as explained above but the head drops becomes

200/8=25 seconds

25 seconds * 108=2700

2700+20000=22700 drops

so as i explained in my (first?) post

running=(walking-body)ratio of speeds +body

running=(34400-20000)1.5/8+body=22700

the problems and arguements people are having i think come from the two different interpretations but I would offer the thought that either way it is not a direct ratio of the speeds (unless poeple are thinking the rectangle exists entirely in the horizantal plane)

so bushindo will have to explain which way he meant

p.s. i think its hilarious that what bushindo thinks is easy gets this many arguments and different answers

(the only question that has more then one answer on a math test is Name)

Edited June 4, 2009 by final

[bonanova]

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

Assume that all rain drops ... are falling at 9m/s straight down vertically.

Their density is 200 drops/meter².

Let's say that your head ... has a surface area .06 meter², your chest ... .5 meter².

you are supposed to get across a 200 meter field,

1) Let's say you walk at 1.5 m/s

2) suppose you ran across at a speed of 8 m/s.

, how many drops of water would hit you during the run?

What ever happened to the use of spoilers, btw?

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

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

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

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

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

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

D = raindrop density

V = velocity of the raindrops in your coordinate system

T = time spent in the rain.

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

Then

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

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

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

T = 200/your speed.

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

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

The in-between values are shown in the graph.

[Guest]

wow i my mind redefining values by a new ( whats it called, plane isnt right, i mean it is a new plane but thats not what its called but i guess its close enough) plane is a lil overcomplicated for this problem but you got it right if thats the way the problem was suppose to be interpreted. I only use recoordinating if it eliminates cos and sin or an extra variable like polar or cylindrical or if the plane of perspective (is that what im thinking of...) is accelerating

[Guest]

(a bit off-topic)

We should have more riddles like this one, as it helps in real life to know these kinds of things...

[Guest]

I don't think it's clear which plane this rectangle is in. But I agree with final that the OP makes it sound like you can't get hit from above and you have to walk/run into rain drops.

It won't make a difference if you walk or run. I calculate 22400 drops for both.

There are 200 drops per meter cubed. Which averages out at 1 drop every 0.005 meters per square meter.

Your area is 0.06 + 0.5 = 0.56m^2

So you will encounter a rain drop every 0.005 / 0.56 = 0.0089m

So over 200 meters you will encounter 200 / 0.0089 = 22400

I dont think the speed comes into it at all, given the assumptions we are making.