Several people have pointed out my skydiver results are wrong.
A look at the sum involved shows I divided by 2.6 rather than 0.26 so the
answer was low by a factor of root 10 = 3.16 so my result should have been
300 kph or 190 mph which is almost exactly the figure Tom Watson gives from
his practical experience. Adjust Cd and frontal area as desired.
I used Cd=1 for the rain drop and it has been pointed out that a figure of
0.5 is appropriate for a sphere, which the water drop approximates. This
brings the result up by root 2 to about 21 kph which compares well with the
25 kph reported.
Experiments in the shower this morning :-) indicate that a drop size of
under the 0.1cc that I used may be more appropriate - this would produce a
lower terminal velocity as the drop mass/area scales as cube/square
respectively of diameter. Haven't tried catching rain drops yet.
Russell McMahon
_____________________________
{Quote hidden}> That Cd is pretty high; I'd guess that the real Cd is somewhere
> closer to 0.7.
>
> Regardless, your math appears to be in error:
> sqrt (2000/0.26) = sqrt (7692)
> = 88 m/s
> = 289 f/s
> = 196 mph
> = 316 kph
>
> That number's pretty high, probably because your guesstimate of
> A is small by a factor of 2 or 3 (remember, the skydiver's
> carrying a big parachute on his back).
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