Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!zaphod.mps.ohio-state.edu!caen!spool.mu.edu!uunet!lhdsy1!yzarn
From: yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille)
Newsgroups: rec.skydiving
Subject: Re: Graph of osc'n due to high windforce
Message-ID: <994@lhdsy1.chevron.com>
Date: 23 Jun 91 18:20:17 GMT
Article-I.D.: lhdsy1.994
References: <1991Jun18.150541.5220@rodan.acs.syr.edu> <983@lhdsy1.chevron.com> <1991Jun21.145113.6028@Stardent.COM>
Organization: Chevron Oil Field Research, La Habra, CA.
Lines: 24

In article <1991Jun21.145113.6028@Stardent.COM> joep@Stardent.COM (Joe Peterson) writes:
>> Notice. not 1.02265 but 0.02265 g. (I would not pay too much attention
>
>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.
-- 
  Philip Yzarn de Louraille                 Internet: yzarn@chevron.com
  Research Support Division                 Unix & Open Systems
  Chevron Information & Technology Co.      Tel: (213) 694-9232
  P.O. Box 446, La Habra, CA 90633-0446     Fax: (213) 694-7709