Re: A Parent's Day gem

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Regarding John Mallinckrodt's observation and simulation:
> ....
> You can't, however, ignore the effects of air resistance even on a
> symmetrical bullet and, because its dependence on the velocity is greater
> than linear, the fired bullet will receive more "lift" than the dropped
> bullet.  I just ran a couple of simulations and find that, for realistic
> parameters, this effect can pretty easily cause a 10% larger flight time
> for the bullet.

and
> ....
> No.  The significance of the air resistance is not important; the 
> form of the dependence on v is. When the air drag is not linearly 
> proportional to v, horizontal motion contributes to force in the 
> vertical direction as long as there is some vertical motion.  For 
> example, if F = kv^2 then
>
>   F_vert = -k*sqrt(v_vert^2 + v_horiz^2)*v_vert.
>
> In the case of the horizontally fired bullet, v_vert << v_horiz at 
> all times so, to a good approximation,
>
>   F_vert = -k*|v_horiz|*v_vert.
>
> This is a far greater upward drag force than is experienced by the 
> bullet that is simply dropped.
> ....

I thought I would point out that the horizontally fired bullet problem is
exactly solvable in the limit that John describes above, i.e.
|v_y| << |v_x| for a quadratic velocity damped model for the air drag,
F_vec = - k*v_vec*|v|.  In addition the vertically dropped bullet problem
is also exactly solvable with the same quadratic (in velocity) drag.  The
|v_y| << |v_x| inequality will be automatically satisfied throughout the
trajectory for the horizontally fired bullet if
v_0x >> max{sqrt(h*g),h*sqrt(g*k/m)}.  Here h = height above the ground the
bullet is fired from.  In the vertically dropped bullet case the time T it
takes to land is:

T = sqrt(m/(k*g))*arccosh(exp(k*h/m)).

OTOH, for the horizontally fired bullet case the falling time T is given
implicitly as a solution to:

A + A*A/2 - ln(1 + A) = 2*(h/g)*(v_0x*k/m)^2 , where A = v_0x*k*T/m .

In the horizontally fired case there is a closed form formula for the
distance fallen as a function of the elapsed time.  Unfortunately, this
function cannot be explicitly inverted to find the time elapsed in terms of
the distance fallen.

Consider the model that the air resistance drag is proportional to the
velocity of the bullet to the power p, i.e. F_drag = k*|v|^p.  If p = 1 the
vertical falling time is independent of the initial horizontal velocity
v_0x.  If the power p has a value p > 1 (such as the p = 2 quadratic case 
above) then the falling time increases the faster that the value v_0x
becomes.  OTOH, in the unphysical case that 0 < p < 1 then the falling time
*decreases* the faster that v_0x becomes.

David Bowman
dbowman@gtc.georgetown.ky.us