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Running in the Rain simplified equations
By: RTSplayer , 3:01 AM GMT on December 30, 2012
Basically, if it is raining outside and I am standing outside of a store looking to get to my car I would run. I would get to my car faster thus spending less time in the rain.
Spending less time in the rain doesn't guarantee you get less wet.
That's what some are failing to understand.
The terminal velocity of a 6-millimeter raindrop was found to be approximately 10 m/s. This value has been found to vary between 9 m/s and 13 m/s when measurements were taken on different days. The variance has been contributed to different air temperatures and pressures. In comparison, a human being falling to the surface of the Earth experiences a drastically larger terminal velocity of approximately 56 m/s.
Ok, so something as simple as changes in temperature and pressure changed the rate of a rain drop's terminal velocity by a factor of about plus/minus 2m/s.
If you consider a human sprint, which is not a world class sprinter, and is a straight shot, not a shuttle run with turns, would also be about 9 or 10m/s.
If the front of your body is thrice as big as the top of your body, then exposures are:
T = 1 (standard metric for area of top of body)
F = 3T (area of front of body)*
r = rate
x = time
E = x *(rT + rF) total exposure
* I picked a round number that seems close. I doubt it's exactly this ratio.
For terminal velocity: 9m/s
rT = 0.8746*T (sin 61)
rF = 0.4856*3*T = 1.4568T (3*sin 29)
x = 10s
E = 23.314 arbitrary units water.
rT = 0.7894*T (Sin 52.12)
rF = 0.6139*3*T (Sin 37.88)
x = 7.15285s
E = 18.8199 units water
rT = 0.6691*T (sin 42)
rF = 0.7431*3*T (3*sin 48)
x = 5s
E = 14.492 units water*
13m/s terminal velocity:
rT = 0.9333*T
rF = 0.3590*3*T
x = 10s
E = 20.103 units water
rT = 0.7926*T
rF = 0.6097*3*T
x = 5s
E = 13.1085 units water
Ok, so I guess I've proven myself wrong, not sure though, because I don't know if the 3 to 1 ratio of front body surface area to top of the body surface area that I used was high enough. The higher this ratio the more it favors walking.
This also still doesn't account for dynamic motion of the body, and still only incorrectly treats the body as a prism...but what the heck, he wanted to see some math...
The 9m/s terminal velocity calculation actually has exposure rate ratio of just 1.24, which means the Build team runners actually got much wetter (compared to walking,) than THIS attempt to solve the problem suggests they should have, because the actual ratio was 1.6.
The 13m/s terminal velocity ironically causes the person to get less wet, whether walking or running. I guess a faster falling drop is more likely to fall out of the way before you run into it. Additionally, the ratio of exposure rates is higher.
So what happened on Jamie and Adam's experiment, and why did it work for both of them, with and without wind?
Below terminal velocity 5m/s:
rT = 0.7071*T
rF = 0.7071*3*T
x = 10s
E = 28.284 units water
rT = 0.4472*T
rF = 0.8944*3*T
x = 5s
E = 15.652 units water
Simplified explanation still doesn't replicate Jamie and Adam's results, and it under estimates how wet the runner got. It under estimated the ratio of rate of exposure for the runner in the build team's experiment by 40%.
I actually think the scales in at least one of the experiments was screwed up. Remember, in one part of the build team's tests, they got the outrageous conclusion that Torre ended up having lighter clothes after running in the rain than before he started, which they threw that out because it was obviously wrong, but what about the not-so-obviously wrong results?! If this is the case, then the margin of error on the scales is too large to even do the experiment properly.
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