Re: buoyant bullets.

["Richard W. Tarara" <rtarara@saintmarys.edu>]

The following doesn't really answer the question about the difference
between the dropped versus fired bullet, but it is a starting point for
modeling the more complex behavior in air.

I modeled the bullet as a lead ball of 1 cm diameter then wrote a
spreadsheet where I could adjust things like the power of 'v' in the air
resistance term -bv^n.  For 'b' I used .5*(density of air)*C*A where C is a
geometric factor (.5 for spheres) and A the cross sectional area.

Running this with an initial x velocity of 500 m/s and a height of 1.5 m,
calculating the time to hit the ground and the horizontal range,  I get:

Without air resistance    t = .553 s,   x = 276.6 m

With n = 2                      t = .65 s      x = 206 m

With n = 1                      t = .586 s     x = 294 m

This model does not include any real aerodynamic effects, but is interesting
that the velocity dependence is critical.  Any air resistance increases the
time in the air, but the effect on the range is substantially different
between n = 1 and n = 2.  In an experiment we do here, we show that the
effect is n = 2 on a 6 cm, 3 gram foam ball.  I have real doubts, however,
that the modeled bullet would show a v^2 dependence, but the experiment is
beyond our equipment capabilities (since we used stop watches and 1-15 meter
drops).

OF COURSE, because of the lack of aerodynamic modeling, the dropped bullet
would take the same time to hit the ground as the fired bullet in both these
cases since the vertical motion is modeled as independent of the horizontal.

Rick


-----Original Message-----
From: Robert Mathieson <rmathieson@culver.edu>

> I am interested in learning the difference in flight times for high speed
> projectiles-1%, 50%?  Maybe someone will have some quantitative data.
>
> Cheers,
>
> Robert Mathieson
> Culver-Stockton College
> Canton, MO 63435
> (217)231-6000
> rmathieson@culver.edu