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Old September 3, 2013, 08:57 PM   #34
johnwilliamson062
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Join Date: May 16, 2008
Posts: 6,809
Quote:
That means that during the entire bullet flight time, the total amount of water in the air is 5 inches x 2 divided by 3,600 seconds
That is the calculation for a unit of space, not the entire flight path. When I read your calculations it seems to me you calculated the entire length of the "tube" would get 5" of rain, not every square inch. At the end of an hour we have a .308 wide 5" high mess of rain, not a tube with 5 cubic inches of rain poured into it. 5 inch per hour is more or less a hurricane BTW.

Quote:
In that 2 second flight time, we're only filling that entire cylinder with 1/360th of an inch of rain
I think in two seconds we are filling every point along the tube with a vertical 1/360th of rain. Along the complete length of the tube for two seconds something like .308X1000X36X2/360 would fall. 61.6 in^3 of rain. Over the course of an hour .308X1000X5, or 55,440 in^3. In two seconds that would be .308*1000*36*2/360/.0019877 or almost 31,000 rain drops along that path. I think in such a torrential down pour the diameter of the rain drop would be at the high end near .13. That results in a rain drop volume of .009in^3 and a total raindrop count for two seconds at about 6,844 along our bullets path. ~3,5drops/second in an are .308X1 yard. About 12 square inches. 18 drops per second on a human standing upright. Having been in several hurricanes I am confident this is a reasonable number. Maybe with all the estimates it is double the correct value. With your calculation one rain drop falls on 77 square(the area of our 1000 yard X .308 flight path) feet every two seconds during a 5in/hour rainfall. That doesn't pass the smell test for me.

I'm not 100% sure, but that is what I get from reading your calculations and developing my own method(I admit based off your method where the heavy lifting was accomplished). There was a time when this sort of problem would have been extremely simple for me, but all I deal with now is simple statistics and the word problem analytic training has faded. I might be off base in my thinking.

Where the bullet is located or whether it is moving is actually irrelevant. In theory the chance that an object with the area of the bullet will get hit by a raindrop is the same whether the bullet is moving or still, at least in our simplified algebraic model. I'd have to consider what adding the bullet flying into raindrops instead of them just landing on top would do, but I am quite certain that would increase the likelihood of contact, not decrease.
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