SKYDIVE@f15.n233.z1.FIDONET.ORG (SKYDIVE) (06/22/91)
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)
--
SKYDIVE - via FidoNet node 1:233/13 (ehsnet.fidonet.org)yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/27/91)
In article <3084.2867C4B5@ehsnet.fidonet.org> SKYDIVE@f15.n233.z1.FIDONET.ORG (SKYDIVE) writes: >Reply-to: Dave.Appel@p30.f30.n231.z1.fidonet.org (Dave Appel) >Fido-To: philip > > 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. So far my numbers agree with yours. > > I am going to guess and say that it will take 70,000 feet for > the jumper to deccelerate from 700mph to 120 mph. Let's see. A terminal velocity of 120 mph is reached around 12000'. Say that, using your above figures, for 700 mph we use 100,000' for that mark, we get: So 100,000 -12,000 = 88,000'. So it will take 88,000' to go from 700 to 120 mph. Since we know it would be below 100,000 to reach 700 mph (if it is possible) then I also accept your 70,000' number. Just double checking. > 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 agree again: this is not 100% correct at all. And I am going to stop checking the arithmetic because we are dealing with a non-linear case here and with all the above assumptions, it just does not make sense to get numbers anymore without a more accurate model. For instance, below you calculate an average velocity for a velocity which changes non-linearly with time. That just does not make sense to me, even as a first order approximation. > > 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. No, you cannot add the one g to your above number, *unless* you are one of these person who can pull one "g" while at rest with your feet on the ground! ;-} Yes, the air density behaves as I said it before (you don't seem to believe it, though). Actually, the air density decreases exponentially with altitude from the ground up. And you can say that it increases logarithmically while going down. > > 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. No excuses are necessary. Your arithmetic is OK, but the assumptions taken make its application not realistic, even as a first order evaluation, because 1) there are too many assumptions, 2) the atmosphere is a non-linear body. This said, I have a computer model for freefall but it is *only* valid in the troposphere (ground to the altitude where the temperature stops decreasing linearly [Standard Atmosphere model]). I can modify the program (and I will) but it will take some times to make sure of its validity. Once done, (several weeks from now?) I will post my numbers and we will be able to start this all over again! Yahoo. -- 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