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From: bleck@ai.mit.edu (Olaf Bleck)
Newsgroups: rec.skydiving
Subject: Re: High altitude landings
Message-ID: <15939@life.ai.mit.edu>
Date: 16 May 91 05:11:37 GMT
References: <13377@exodus.Eng.Sun.COM>
Sender: news@ai.mit.edu
Organization: MIT Artificial Intelligence Laboratory
Lines: 38

In article <13377@exodus.Eng.Sun.COM>, brent@terra.Eng.Sun.COM (Brent Callaghan) writes:

|> The only uncertainty is the landing.  The DZ will
|> be in the Rockies at 9,500 MSL.  How much harder
|> will the landing be than at sea level ?
|> 
|> What's the increase in rate
|> of descent and ground speed ?

First order calculation:

Assume steady state fall rate under canopy, so weight equal to
force on canopy:

W=Pressure*Area=0.5*density*velocity^2(squared)*area of canopy

Weight is constant at both sea level and 9,500ft, as is area, so

density1*velocity1^2 = density2*velocity2^2

=> velocity2 = velocity1 * Square Root (density1/density2)

Using standard atmosphere,

density of air at sea level = 1.19 kg/m^3
density of air at 3000meters = 0.885kg/m^3

so, velocity(3000m) = 1.16*velocity(sea level)

Not too bad.  This is assuming I didn't miss some key point in my
calculation by sleeping through class as an undergrad!  Of course this
is a much hairier aerodynamic problem than this calcualtion, but I
suspect this will get you kind of close, and I'm not current enough to figure it out.

Believe it or don't,

-Olaf
A-12241