Re: buoyant bullets.

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

I'm probably being very dense here, but why isn't the x and y motion totally
separable.  That is

Fx = -kvx^n

Fy = -kvy^n

Isn't it only the horizontal motion through the media that adds a horizontal
retarding force and the vertical motion that adds a vertical retarding
force?  This is how I set up my calculation..with the x and y forces,
accelerations, velocities, and displacements calculated separately--good old
introductory, algebra-based physics.

Even with your formulae, since vy << vx, there should be little difference
in the results.

Rick


-----Original Message-----
From: John Mallinckrodt <ajmallinckro@CSUPomona.Edu>
To: phys-l@atlantis.uwf.edu <phys-l@atlantis.uwf.edu>
Cc: AJMALLINCKRO@CSUPomona.Edu <AJMALLINCKRO@CSUPomona.Edu>
Date: Monday, November 03, 1997 7:13 PM
Subject: Re: buoyant bullets.


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

> Well for n = 2 with b correct, then I don't see why there would be any
> difference in my model between between dropping the bullet and firing the
> bullet.  What couples the motion such that the vertical falls are not
> identical?

I showed explicitly how the motions are coupled for n = 2 in a message
last week.  It is, however, simple to write the force components for
arbitrary n as follows:

If
      |F| = k * v^n
then
      F_x = -k * [(v_x)^2 + (v_y)^2]^[(n-1)/2] * v_x
and
      F_y = -k * [(v_x)^2 + (v_y)^2]^[(n-1)/2] * v_y

Thus, unless n = 1, the motions are coupled.

John
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