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From: SKYDIVE@f15.n233.z1.FIDONET.ORG (SKYDIVE)
Newsgroups: rec.skydiving
Subject: decceleration from 700 mph
Message-ID: <3084.2867C4B5@ehsnet.fidonet.org>
Date: 22 Jun 91 06:31:00 GMT
Sender: daemon@ucbvax.BERKELEY.EDU
Lines: 80

Reply-to: Dave.Appel@p30.f30.n231.z1.fidonet.org (Dave Appel)
Fido-To: philip

In a message to Skydive <17 Jun 91 13:39> uiucuxc!lhdsy1.chevron.co wrote:

 ui> The net freefall deceleration of 750 mph to 90mph (at 2000') is very
 ui> slow and I doubt that it can be felt. After all, in a jump from 12000',
 ui> skydivers do not feel their terminal velocities decreasing (until they
 ui> open their chutes!), yet this is the region where the increase of
 ui> density per altitude change is the greatest (logarithmic profile of
 ui> atmospheric density.)

 My subject is deccelerating from 700 mph to (let's say) 120 mph.

 The jump is to take place from 120,000 feet.  Disregarding
 air friction, it would take 32 seconds to accelerate from 0 mph
 in the vertical plane to 700 mph (at 32 ft/sec/sec).  Disregarding
 air friction, 700 mph would be reached at 103,590 feet.  Since
 there WILL be air friction, the 700 mph will be reached somewhere
 BELOW 100,000 feet. (Assuming the 700 mph figure is anywhere
 correct at all).  It will also be below 100,000 feet because
 acceleration due to gravity is going to be less than 32 f/s/s
 at that altitude.

 I am going to guess and say that it will take 70,000 feet for
 the jumper to deccelerate from 700mph to 120 mph.  Assuming
 constant decceleration, it will take 116.5 seconds to deccelerate
 from 700 mph (1026 ft/sec) to 120 mph (176 ft/sec), if the
 decceleration is to occur during a 70,000 foot segment of freefall.
 This is not 100% correct because as air density increases, 
 decceleration will be greater at the end of the segment.
 We are computing AVG decceleration, therefore the maximum 
 instantaneous decceleration will be greater than what we calculate.
 In other words our calculations for g-force will be minimums,
 actual g-force will be greater.

 I get the 116.5 seconds by: time = distance/(avg speed)
           116.5 sec = 70,000 ft / [ (1026 f/s + 176 f/s )/2 ]

 Acceleration is:   (velocity2 - velocity1) / time
              ( 1026 - 176 ) / 116.5  = 7.3 ft/sec/sec

   since 1g = 32 ft/sec/sec, then  7.3 / 32 = .23 g

 So the skydiver will experience at least .23 g in the vertical
 direction, while deccelerating from 700 mph to 120 mph, assuming
 such decceleration occurs within 70,000 feet of freefall.

 If, as you say, that air density increases logarithmically, I
 can easily imagine .3 g being an instantaneous maximum.
 After thinking about it, I do believe it should be "added"
 to 1.  So the jumper would be experiencing a maximum of 1.3 g
 of wind resistance deccelerating him.  In constant velocity,
 wind resistance is applying 1g of force on the jumpers body,
 since he is deccelerating, there must be greater than 1g force
 on him.

 Please excuse my rudimentary math.  I only took 1 year of college 
 level physics and 1 year of calculus.  But I've forgotten the math
 necessary to account for varying air densities.  The main
 question mark in these calculations is the assumption of 70,000
 feet.  Other than that, the .23 g is a minimum. Actual will be
 greater.

 Hope this helps.

 (Note to our jumper friend in ows-trail-ya: [sorry I forgot yer
 name] Could you send me the source code to
 yer simulation program in a message in this newsgroup?  Or 
 perhaps to me at "dappel@ehsnet.fidonet.org".  It has to be in
 a mail message, not in a file.  I don't have access to files, only
 newsgroups and mail messages. Thx.)

--- XRS! 4.50
--- eecp 1.45 LM2 

 * Origin: The Drop Zone, Dave Appel (1:231/30.30) (Quick 1:231/30.30)
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