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From: joep@Stardent.COM (Joe Peterson)
Newsgroups: rec.skydiving
Subject: Re: Graph of osc'n due to high windforce
Message-ID: <1991Jun24.154627.285@Stardent.COM>
Date: 24 Jun 91 15:46:27 GMT
References: <1991Jun18.150541.5220@rodan.acs.syr.edu> <983@lhdsy1.chevron.com> <1991Jun21.145113.6028@Stardent.COM> <994@lhdsy1.chevron.com>
Organization: Stardent Computer, Concord MA
Lines: 28
In-reply-to: yzarn@lhdsy1.chevron.com's message of 23 Jun 91 18:20:17 GMT


> Nooooo. I said one down (assuming g not changing with altitude, and
> remember, Earth is attracting us, so g is toward the center of the
> Earth) and 0.023 (rounded) up, which would mean his acceleration is less
> than one.
> Look at it this way, since we agree that the skydiver's terminal
> velocity is slowing down as he/she gets lower in the atmosphere, how
> could his/her acceleration be greater than g? I don' get it.

Well, what I was saying was that the skydiver would FEEL a force larger
than 1.0 g.  Just like when you open your parachute, you feel a force
much larger than 1.0.  If a person is falling at a constant velocity in
a constant gravitational field, he/she will feel the same force as when
lying on the ground.  On the ground, there is the force due to gravity
acting downward and the "normal" force acting upward (the support of
a solid surface).  These are equal and opposing forces.  In freefall,
the force of gravity downward would be equally opposed by the force
of friction upward (once terminal velocity is reached) if we assume
no slowing down as the atmosphere gets more dense.  Now, if the person
is slowing down, the force due to friction is constantly higher than
that due to gravity.  This is an additional force felt by the skydiver
pushing upward, making him feel heavier.  Therefore, he is "pulling
more than 1.0 g's."  It is the same as being in an elevator and starting
to go up.  The net force upward is greater than 1.0 g's for a second or
so.  Does this make sense?
					Joe Peterson
					C-20351
					joep@stardent.com