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bonanova

Staying dry in the rain

Question

For the purposes of this puzzle, consider our old friend Albert to have the shape of a rectangular paralellepiped (when he was born the doctor remarked to his mother, I don't explain them, ma'am I just deliver them) just meaning a solid having (six) rectangular sides. At a recent physical exam, Albert was found to be 2 meters tall, 1 meter wide and .20 meters thick (front to back.) He maintains his geometric rectitude by never leaning forward when he walks or runs. 

So anyway, Albert, alas, has found himself caught in a rainstorm that has 1000 raindrops / cubic meter that are falling at a constant speed of 10 meters / second, and he is 100 meters from his house.

Just how fast should Albert run to his house so as to encounter as few raindrops as possible?

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On 3/20/2018 at 10:05 PM, bonanova said:

For the purposes of this puzzle, consider our old friend Albert to have the shape of a rectangular paralellepiped (when he was born the doctor remarked to his mother, I don't explain them, ma'am I just deliver them) just meaning a solid having (six) rectangular sides. At a recent physical exam, Albert was found to be 2 meters tall, 1 meter wide and .20 meters thick (front to back.) He maintains his geometric rectitude by never leaning forward when he walks or runs. 

So anyway, Albert, alas, has found himself caught in a rainstorm that has 1000 raindrops / cubic meter that are falling at a constant speed of 10 meters / second, and he is 100 meters from his house.

Just how fast should Albert run to his house so as to encounter as few raindrops as possible?

 

This is a thought exercise I've considered in the past.  The answer that I've always come up with is that it doesn't matter when disregarding thickness.  If the raindrops are constant and fill any cubic meter equally, he'll encounter the same number of raindrops on his "face" regardless of speed. 

The .20m thickness, though, needs to be considered.  We know that when he moves at 0m/s, he will encounter some amount of rain along that thickness and he will have made no progress.  So to minimize the number of raindrops that lands on top of him, we just increase his speed infinitely.  We also know that this can cause severe issues, however.

If you'd like some math to support this argument:

 

...you should probably ask someone else.

Edited by Molly Mae

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First thought is that no matter how fast Albert runs his front always sees 2000 rain drops.  i.e. his front is 2 square meters and he is seeing 1000 drops/cu meter .                       On the other hand the top of Albert  0.20  square meters.  if Albert was 1 sq meter moving at 1 m/sec ,then his top sees  (  1,000 / cu meter falling at 10 m/sec = 10,000 drops ) however his profile is only 0,20 sq meters and therefore he sees  ( 0.20 x 10,000 drops  = 2,000 drops ); If he moves at 2 m/sec ; then he sees 1,000 drops at 4 m/sec he would see 500 drops,etc So Albert should run as fast as possible.

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On 3/22/2018 at 10:27 AM, Molly Mae said:
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This is a thought exercise I've considered in the past.  The answer that I've always come up with is that it doesn't matter when disregarding thickness.  If the raindrops are constant and fill any cubic meter equally, he'll encounter the same number of raindrops on his "face" regardless of speed. 

The .20m thickness, though, needs to be considered.  We know that when he moves at 0m/s, he will encounter some amount of rain along that thickness and he will have made no progress.  So to minimize the number of raindrops that lands on top of him, we just increase his speed infinitely.  We also know that this can cause severe issues, however.

If you'd like some math to support this argument:

  Reveal hidden contents

You're right. And it's not much of a puzzle after all.

You have to allow Albert to lean forward as he runs to make this puzzle at all interesting. But even then, the obvious solution is to have Albert lie horizontally and crawl at infinite speed. Only the top of his head gets wet then.

I ran into this puzzle a few years back and "solved" it, more interestingly but also more incorrectly, by multiplying his front and top areas respectively by sin theta and cos theta where theta was determined by his speed compared to the speed of the rain, and some other stuff. It was nonsense. :blush:

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First thought is that no matter how fast Albert runs his front always sees 2000 rain drops.  i.e. his front is 2 square meters and he is seeing 1000 drops/cu meter .                       On the other hand the top of Albert  0.20  square meters.  if Albert was 1 sq meter, moving at 1 m/sec ,then his top sees  (  1,000 / cu meter falling at 10 m/sec = 10,000 drops) however ,Albert has only a 0.20 sq meter profile therefore his top would see ( 10,000 x 0.20  = 2,000 drops ) if he moves at 2 meters / sec  then he is hit on top by 1,000 drops. if then he is moving at 4 m/sec 2,000 divided by 4 = 500 drops.  There fore Albert should run as fast as possible

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