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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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