Re: Falling pellets

[brian whatcott <inet@INTELLISYS.NET>]

At 21:33 12/19/98 -0800, John M wrote:
> Jim,
>
> The viscosity of air has very little to do with the terminal velocity of a
> lead pellet as large as 1 mm in diameter beyond determining that the
> Reynold's number is very large for any appreciable speed.  I calculate a
> terminal velocity of 15 to 20 m/s based on equating the dynamic drag (1/2
> C_drag rho A v^2) to the weight of the pellet.  At this velocity the
> Reynold's number is on the order of a thousand and the drag
> coefficient is approximately 0.5.
>
> John

I estimate a similar value from modeling considerations, like this:

let us suppose that we can scale cross section area and
terminal speed as between a parachute jumper and a shot.

 We will not expect great accuracy in estimating the
shot's terminal speed, computed in this way.
 We take the jumper's area as 0.6 x 1.63 = 1m^2
and the shot cross section as pi x 0.001^2/4 = 0.8E-6 m^2
we recall 54 m/s as the jumpers terminal speed
so we estimate
drag force: A.Cd.rho.v^2/2
equals
weight: 170 lbs = 77.3  x 9.8 newtons = 757N
at 54 m/s
i.e. 757 = 1.Cd.rho.54^2/2 = 1458 Cd.rho
so Cd.rho = 0.52


For the shot then
Weight: 4/3 pi r^3  x  11370 kg/m^3 = 5.8 micronewton
and 5.8E-6 = A . 0.52 v^2/2
v^2 = 0.00022/A = 275 (m/s)^2
v = 17 m/s
(You will notice I completely ignored the buoyancy of air on
the shot,the ratio of densities being high)

Brian
brian whatcott <inet@intellisys.net>
Altus OK