Re: buoyant bullets.

[John Mallinckrodt <ajmallinckro@CSUPomona.Edu>]

On Mon, 3 Nov 1997, Richard W. Tarara wrote:

> 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.

Rick,

This value of b will be pretty reasonable for n = 2 (for which it 
is dimensionally correct as well), but it is simply wrong (if for 
no other reason than that it is dimensionally *incorrect*) for any 
other value of n.

> 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

The result for n = 1 doesn't make sense to me.  With drag, one 
*could* imagine having enough extra time in the air that the 
bullet would actually travel farther, but ...  I suspect this is 
at least partially an artifact of using an inappropriate model 
(i.e., n=1) as well as an improper value for b.

> 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).

I won't argue strongly for a strict n=2 dependence for the bullet, 
but I'd be *very* surprised to see an n=1 dependence since that 
usually implies that viscosity is the primary effect--clearly not 
the case for a fired bullet.

> 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.

I don't follow this.  If you have modeled a projectile subject to 
nonlinear drag as having independent vertical and horizontal 
motions, you have simply made a mistake in writing out the force 
terms.

John
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