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## Question

A while ago that asked whether

how wet you get moving a fixed distance in the rain

depends on whether you walk or run.

Let's assume there is a speed that keeps you the driest,

but you don't have the stamina for it. My question is what

speed should you run to get only twice as wet as optimal?

Assumptions you can make:

1. You are [a rather squarish] 6 feet tall, 2 feet wide and 6 inches thick.
2. You always move in a standing position, i.e you never lean.
3. Shelter is 1000 feet away, on level ground.
4. The rain is falling at a speed of 10 feet per second.
5. At any given place or time there are 1000 raindrops per cubic foot.

Enjoy! ## Recommended Posts

• 0

A while ago that asked whether

how wet you get moving a fixed distance in the rain

depends on whether you walk or run.

Let's assume there is a speed that keeps you the driest,

but you don't have the stamina for it. My question is what

speed should you run to get only twice as wet as optimal?

Assumptions you can make:

1. You are [a rather squarish] 6 feet tall, 2 feet wide and 6 inches thick.
2. You always move in a standing position, i.e you never lean.
3. Shelter is 1000 feet away, on level ground.
4. The rain is falling at a speed of 10 feet per second.
5. At any given place or time there are 1000 raindrops per cubic foot.

Enjoy! My take on this

Assuming infinite speed, then the lowest number of rain hits is 12,000,000.

We want to get twice as wet, so we want the head drop counts to be 12,000,000. The rain is falling at the rate of 10,000 drops/sec/ft2. That means we need to get across the 1000 ft field in 12,000,000/10,000 = 1,200 seconds.

That works out to be 5/6 ft/sec, which is probably should be called a strolling pace rather than a running pace =)

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Since you are (essentially) a box that's always in a standing position and moving through rain (assumed to be) falling straight down, you will get wet on top and on the side pointing the direction you are moving (assumed to be the largest side... no running sideways).

Like Bushindo started with, if you run infinitely fast, only the side of the box will get wet... your head will remain dry. If you imagine this face of the box as carving a space out through time, you'll see the volume will be 6 feet * 2 feet * 1000 feet = 12000 ft3 regardless of how long it takes to get there. This is because if you skew a rectangular prism parallel to two sides, the volume doesn't change.

To get twice as wet, we need the face of the box representing your head to carve out an equal area in the falling rain. I'll simplify this by ignoring the last 6 inches of the journey (your thickness), and the result is simply another skewed rectanglular prism (but one for which the distance separating the two parallel lines you are skewing along is dependant on time). So the volume/time will be 2 feet * 6 inches * 10 ft/sec = 10 ft3/sec. Since we want this volume to equal 12000 ft3, we divide them. This is 12000/10 = 1200 seconds. The needed speed is 1000 ft / 1200 seconds = 5/6 feet per second.

So it looks like Bushindo was right.

If we want to include the time it takes to get into the shelter (your thickness further will be needed, and the rain falling on your head will carve out a rectanglular prism cut diagonally in half). This will result in a total volume of 1000 ft / speed * 10 ft3/sec + .5 ft / speed * .5 * 10ft3/sec.

12000 ft3 = 1000 ft / speed * 10 ft3/sec + .5 ft / speed * .5 * 10ft3/sec.

12000 ft3 * speed = 1000 ft * 10 ft3/sec + .5 ft * .5 * 10ft3/sec.

12000 ft3 * speed = 10000 ft4/sec + 2.5 ft4/sec.

12000 ft3 * speed = 10002.5 ft4/sec.

speed = 10002.5 / 12000 ft/sec = 0.8335416666666 ft/sec

So you need to go unnoticably faster than 5/6 feet/sec.

What speed would I need to run sideways to get just as wet as running infinitely fast running normally?

We already found the area carved out running infinitely fast is 12000 ft3/sec.

The area carved out by the side of the box would simply be a quarter of the 12000 ft3/sec, so 3000ft3/sec. This leaves 9000 ft3/sec for the rain falling on your head.

So using the volume/time found previously, 10 ft3/sec, we just need to divide the volume by that.

9000/10 = 900 seconds. This leaves the speed at 1000/900 ft/sec = 1.111111 ft/sec. So going a little bit faster (.277777777 ft/sec, which means 33% faster) and moving sideways towards your destination will have you get half as wet.

Conclusion: Always run sideways in the rain. Hmm... or maybe crawl on your side!

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Wait a minute... before I even try to tackle this question...

There is actually a difference in the speed you walk vs how much wetter you will get?!

What is the best pace to get wet the least??????? SO CURIOUS!

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Think of this in terms of extremes.

If you go super fast, you will only get 1 block of rain hitting you.

If you stand still, you will get every block of rain hitting you (in that location).

Theoretically the faster you move, the less rain should hit you.

Now, you should get the same water contact on you whether standing still for one second, or running for one second. Standing still will be more water on top of you (head), while running will be making contact with your chest.

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I was too slow to edit, but here's why faster is better.

No matter how fast or slow you move, the front of your rectangle will ALWAYS hit a rectangle of water with the size of 6x2x1000.

It is impossible for you to not hit every water droplet in front of you (assuming equal and even distribution of water droplets).

Therefore the only discrepancy is how much water hits the top of your rectangle. The slower you go, the more hits your top.

So faster is better.

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Good analysis all, but it's actually faster than the answers given so far. An object presenting a frontal area of F and a top area of T

traveling a distance x at a speed s

in a rain of density d and speed sr

will intercept a number of raindrops D given by

D = d x [F + T sr/s] which is minimal when s tends to infinity:

Dmin = d x F = 1,000,000 F [for the values in the OP]

D = 2 Dmin when F = T sr/s, or when s = sr T/F

The OP tells us the man does not lean, so from the given dimensions we know that T = 1.

The OP does not dictate whether the man faces forward, sideways or somewhere in between.

His forward orientation is not in the list of things that can be assumed.

F when facing forward is 12.

F has extreme values of 6 x sqrt[4.25] = 12.37 and 3.

The OP asks: what speed should you run to get only twice as wet as optimal?

One stays driest by minimizing F and running infinitely fast: Dmin = 3,000,000 drops.

The speed that doubles the optimal drop count is 10 x 1/3 = 3.33... feet/second.

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Good analysis all, but it's actually faster than the answers given so far. An object presenting a frontal area of F and a top area of T

traveling a distance x at a speed s

in a rain of density d and speed sr

will intercept a number of raindrops D given by

D = d x [F + T sr/s] which is minimal when s tends to infinity:

Dmin = d x F = 1,000,000 F [for the values in the OP]

D = 2 Dmin when F = T sr/s, or when s = sr T/F

The OP tells us the man does not lean, so from the given dimensions we know that T = 1.

The OP does not dictate whether the man faces forward, sideways or somewhere in between.

His forward orientation is not in the list of things that can be assumed.

F when facing forward is 12.

F has extreme values of 6 x sqrt[4.25] = 12.37 and 3.

The OP asks: what speed should you run to get only twice as wet as optimal?

One stays driest by minimizing F and running infinitely fast: Dmin = 3,000,000 drops.

The speed that doubles the optimal drop count is 10 x 1/3 = 3.33... feet/second.

Well, your answer is based upon the assumption that the man can turn sideways while running.

If he is facing 'forward' the minimum raindrops is 12,000,000

If he can run while sideways, then the minimum would be 1,000,000

How did you get an optimal*2 of 3,000,000?

Should optimal *2 be either 2,000,000 or 24,000,000

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Well, your answer is based upon the assumption that the man can turn sideways while running.

I just tried it; it works.

Do you find that this is inconsistent with the OP?

If he is facing 'forward' the minimum raindrops is 12,000,000

If he can run while sideways, then the minimum would be 1,000,000

Fmin is 3, not 1. [6 feet x 6 inches]

How did you get an optimal*2 of 3,000,000?

I didn't. Dmin is 3,000,000 drops.

Should optimal *2 be either 2,000,000 or 24,000,000

See above.

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You're right.

I messed up with my sizing.

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Good analysis all, but it's actually faster than the answers given so far. An object presenting a frontal area of F and a top area of T

traveling a distance x at a speed s

in a rain of density d and speed sr

will intercept a number of raindrops D given by

D = d x [F + T sr/s] which is minimal when s tends to infinity:

Dmin = d x F = 1,000,000 F [for the values in the OP]

D = 2 Dmin when F = T sr/s, or when s = sr T/F

The OP tells us the man does not lean, so from the given dimensions we know that T = 1.

The OP does not dictate whether the man faces forward, sideways or somewhere in between.

His forward orientation is not in the list of things that can be assumed.

F when facing forward is 12.

F has extreme values of 6 x sqrt[4.25] = 12.37 and 3.

The OP asks: what speed should you run to get only twice as wet as optimal?

One stays driest by minimizing F and running infinitely fast: Dmin = 3,000,000 drops.

The speed that doubles the optimal drop count is 10 x 1/3 = 3.33... feet/second.

If you cannot assume...

If you cannot assume an orientation in determining the optimal case, why assume one in the answer? After all, it wasn't stated that we were to look for the minimum speed needed. It seems the answer is a function of the desired run orientation... which would make the speeds possible to become twice as wet anywhere from 3.33 ft/s to still infinite (to impossible for many orientations).

(Just for fun... it's 10/( 6 - 12*|cos x| - 3*|sin x| ) ft/s, where x is the angle away from straight forward. Impossible for negative values, infinite speed for denomiator of 0.)

"Let's assume there is a speed that keeps you the driest,

but you don't have the stamina for it. My question is what

speed should you run to get only twice as wet as optimal?"

This paragraph seems to me to indicate speed is the only variable intentionally being scrutinized / optimized over. Perhaps one interpretation for the answer is dependant on the orientation by finding the optimal speed (always infinite) for that orientation, and then finding the speed to get twice as wet as that keeping that same orientation.

(Just for fun... it's 10/( 12*|cos x| + 3*|sin x| ) ft/s where x is the angle away from straight forward. x=0 degrees gives the answer Bushindo and I got, and x=90 degrees gives the answer given by bonanova)

If you cannot assume a static orientation (spin infinitely fast a possibility), constant run speed, constant wind speed / angle of rain (is that 10 ft/s its absolute speed or just its downward velocity or both?), you soak up every drop of rain touched (so no limit on "wetness") and don't drip any, or even the path taken (still a fixed distance as with Bushindo's puzzle?) to the shelter... things can get tricky quick.

Did I miss any applicable assumptions? Besides crazy ones like assuming there is not a person who will throw a bucket of water if you take too long, etc.

Could examining any of these (or other such assumptions) provide an interesting puzzle? Perhaps "what is the most wet you can get given a minimum speed, always moving directly towards the shelter, staying a box shape, and the other given assumptions?" Perhaps the runner is like the Wicked Witch of the West or Gumby and will melt according to how wet they get... "oh, what a world!" Maybe Gumby and the Wicked Witch of the West are having a race in the rain... that may make a funny picture. Two green streaks ending in puddles, one with eyes (and perhaps some clay clumps) and the other with black clothes, black hat, and broom...

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All good points.

There are two parts to the puzzle.

OP does not tell you how to achieve optimal dryness.

That is left to the solver. I think everyone got the point

that the frontal drop count does not depend on speed.

It only depends on the area the drops hit. So, frontal drops

are minimized for minimal frontal area. The top (head)

drop count does depend on speed and tends to zero as

Speed increases.

So it's just: minimize the total drop count, then find the

speed that doubles it.

The OP rules out crawling I guess the final distinction to be made is the difference

between restricting and assuming. Regarding orientation,

the OP restricts only that the man does not lean. If you make

further restrictions, by assuming, you limit the search space

unnecessarily.

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