Re: drag coef. and falling bullets

[brian whatcott <inet@INTELLISYS.NET>]

At 23:26 1/19/00 -0800, Leon Leonardo wrote:
> /// The
> objective of the project is to determine the speed
> with which a bullet shot straight up, on a windless
> day, will return to the earth. The bullets we chose to
> study are the 22 lr, the 9 mm and 9 mm hollow point.
> ///
> As for Reynolds number, it seems that so long as the
> flow is turbulent, and well away from Mach one, the
> v^2 dependence is valid.
>
> Oh yes,  anyone have any idea what a 9 mm bullet will
> freely fall at?
> ///
> Leon

Using the usual suspect for a first estimate, we recall
the parachutist falling at 120 mph (54 m/s)
We say he weighs 180 lb (803N),
and we say weight = kAv^2 where air drag balances weight
so 803 = k * 1.11 * 54 * 54 and k = 0.25 in this system of units
for a jumper whose cross section A is roughly 2ft x 6 ft (1.11 m^2)

Blithely assuming that Re is unimportant, we scale to a 9mm
bullet, taking the weight as 115 grains (0.073N) and the side area
as 9 x 27 mm^2 = 2.43E-4 m^2

 We jump immediately to a terminal velocity given by
   0.073 = 0.25 * 2.43E-4 * v^2
V^2 = 1.2E3
V = 35 m/s (78 mph)

This is probably rather too fast for a 9 mm bullet
 dropped on its side.



brian whatcott <inet@intellisys.net>
Altus OK