Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!wuarchive!zaphod.mps.ohio-state.edu!rpi!uupsi!rodan.acs.syr.edu!wotan.top.cis.syr.edu!greeny
From: greeny@wotan.top.cis.syr.edu (Jonathan Greenfield)
Newsgroups: rec.skydiving
Subject: Re: Graph of osc'n due to high windforce
Message-ID: <1991Jun24.155154.14380@rodan.acs.syr.edu>
Date: 24 Jun 91 19:51:54 GMT
References: <1991Jun18.150541.5220@rodan.acs.syr.edu> <983@lhdsy1.chevron.com> <1991Jun21.145113.6028@Stardent.COM> <994@lhdsy1.chevron.com>
Reply-To: greeny@top.cis.syr.edu (Jonathan Greenfield)
Organization: CIS Dept., Syracuse University
Lines: 27

In article <994@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes:

>>Well, this all depends on how you look at it.  I did not check your math,
>>so assuming your numbers are correct, the skydiver would feel 1.02265 g.
>>This is because if terminal were constant the whole way down, he would
>>feel exactly 1.0 g (assuming "g" is not changing with altitude).  While
>>we are on the ground we feel 1.0 g because we have a constant (zero)
>>change in velocity.  True, he would feel an added acceleration of
>>0.02265 g upward (more than normal), but the total "g's" he would feel
>>would be more than 1.0.
>
>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.

When you stand up, don't you feel a 1 g force at your feet?  We can not
*feel* gravitational forces, but we can *feel* the forces that offset
gravitation.


greeny                                           greeny@top.cis.syr.edu

"What's the difference between an orange?"