From: yzarn@lhdsy1.lahabra.chevron.com (Philip Yzarn de Louraille)
Newsgroups: rec.skydiving
Subject: Re: 7th floor drop--Acceleration of G, minus wind resistance
Date: 15 Sep 92 16:54:34 GMT
Organization: Chevron Oil Field Research, La Habra, CA.

In article <689@zds-ux.UUCP> ryoder@zds-ux.UUCP (Robert Yoder) writes:
>In article <190l08INNdsa@bigboote.WPI.EDU>, megazone@penny.WPI.EDU (MegaZone23) writes:
>[text deleted]
>
>> Velocity(t)=velocity(initial)+acceleration(9.8m/sec^2)*time(in seconds)-
>>   (Cd(coeff. of drag)*surface area*density air*vel.^2)/(2*grav. const. 9.8))
>
>This equation is incorrect.  Converting each term to units yields:

... deleted stuff ...

>Therefore:
>Vt = sqrt(2 * g * m / (Cd * p * A))

The following involves more math/physics so those of you who are allergic
can pass...
Others have to press the space bar or the return key.


Yes, this is the correct equation for the terminal velocity. It also
illustrate a point, that is, that the terminal velocity is not really
terminal but will decrease as a skydiver goes down due to the fact that
the air density is increasing as one gets closer to the ground.
Actually, g, also, varies with altitude, but much slower than the density
and that variation is usually ignored.

The above equation can be rewritten in a way which will make more sense
to a skydiver, after all, how many how us think in terms of coefficient of
drag, surface area and mass?

Assumptions: skydiver keeps same body position (Cd, m, and A constant for the
jump)

Rearranging the Vt eq. gives:

Vt^2 p / g = 2 m / Cd / A = constant!

Let f[y] be my notation for a function which depends on y (altitude)

Which means, that, for a jump,

Vt^2[y] p[y] / g[y] = Vt^2[sl] p[sl] / g[sl], where sl is for sea level.

Hence, Vt^2[y] = Vt^2[sl] p[sl] / p[y] g[sl] / g[y] ~= Vt^2[sl] p[sl] / p[y]

since g[sl] / g[y] ~= 1 for types of jumps made by skydivers.

Hence the Vt equation becomes:

Vt[y] = sqrt(Vt^2[sl] p[sl] / p[y])

So, instead of thinking in terms of coefficient of drags and surface
areas, we can think in terms of terminal velocities, which make sense to
me. Once someone knows his/her terminal velocity at sea level, by using
standard charts for the air density as it varies with altitude, one can
determine their terminal velocity for any altitude.

I used sea level for a reference point rather than, say, my drop zone ground
altitude, because like this, people can compare their terminal velocities.

For relative work, we all have basically an "average" terminal velocity at
sea level, which reflects the fact that people do move their arms, legs,
change the shape of their torso, etc...

I have another formula which I will post if people are interested, that
calculates how much weight should people in a team put in order to make
everyone in the team have the same terminal velocity without having to
strain or watch for it.

I think the above is correct but if someone finds a flaw, please let me
know, thanks.
-- 
  Philip Yzarn de Louraille                 Internet: yzarn@chevron.com
  Research Support Division                 Unix & Open Systems
  Chevron Information & Technology Co.      Tel: (310) 694-9232
  P.O. Box 446, La Habra, CA 90633-0446     Fax: (310) 694-7709
Newsgroups: rec.skydiving