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Re: Buoyant Bullets



At 18:06 10/30/97 -0400, Ludwik wrote:

John, did you try to invent a retarding force formula (nonlinear in any way
you wish) which would produce a lift equal to the weight? That would be
even more amazing (even if only one value of v satisfied --> Lift=m*g).


Ludwik seems readily amazed today - so let me do my little to help.
How about a retarding force formula *greater* than the bullet's weight.

Gee whizz! That means that a bullet fired straight down at mach 2
would DECELERATE!

Brian

brian whatcott <inet@intellisys.net>
Altus OK

Am I missing something here, or is this some kind of a trick question? It
seems obvious to me that a bullet fired straight down at mach 2 would have
a retarding force on it that was greater than g and so it would be slowing
down. Anything that, for whatever reason, finds itself traveling at a speed
greater than its terminal velocity for the local conditions, will be slowed
down regardless of what direction it is traveling in. My distant recall of
the nature of aerodynamics, is that the terminal velocity of any object
will always be less than the local speed of sound.

If this is an inside joke that my absence for the past few days has caused
me to miss, never mind.

Hugh

************************************************************
Hugh Haskell
<mailto://hhaskell@mindspring.com>

The box said "Requires Windows 95 or better." So I bought a Macintosh.
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