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From: yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille)
Newsgroups: rec.skydiving
Subject: Re: Skydiving at high wind force
Message-ID: <995@lhdsy1.chevron.com>
Date: 24 Jun 91 16:36:39 GMT
References: <1991Jun24.134604.8779@cc.curtin.edu.au>
Organization: Chevron Oil Field Research, La Habra, CA.
Lines: 187

In article <1991Jun24.134604.8779@cc.curtin.edu.au> tcliftonr@cc.curtin.edu.au writes:
>RE: GRAPH OF OSC'N DUE TO HIGH WINDFORCE
>
>yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes-
>
>PY> Also, one thing does not seem obvious to me: I understand
>why the velocity will be *high* but why are you expecting
>stability problems?
>
>- Increased wind forces means increased torques and increased
>likelihood of rocking and tumbling.  See the graph of the high
>speed exit.  Do the net-people want me to simulate higher
>speed exits - which can red-out?
Stability is obtained *only* because of speed. The main reason why
students (and other experienced skydivers who mess up the exit) get unstable
right at exit is because they actually *loose* velocity as they jump out of
the airplane, the reason being that a skydiver will loose horizontal
velocity (jumprun velocity) faster than the vertical gain. After 2-3 seconds,
the velocity is back up again and it is easier to control one self.
>
>
>PY> First of all, gravity will be the only driving force
>pulling down (no energy will come from the skydiver, he/she
>will be "passive").
>
>RC:  Sure, there is indeed one gee of weight pulling down.
>But the amount of wind force pushing up is equal to v2/vt2,
>and this ratio of squares is easily greater than one.  This
>can occur if v increases above vt or if vt decreases below v.
Please describe your notation! What is v?, what is vt?
('cause v2/vt2 = v/t2 to me!)
>
>
>PY> Second, the atmosphere density will be *small* up there.
>The terminal velocity will be reached when the force due to
>the gravity pull * mass of skydiver will equal the force of
>drag the skydiver is creating due to his/her passage through
>the atmosphere.
>
>RC:  Sure, terminal velocity is reached momentarily when
>weight equals drag, so there is no net force.  That is when
>v2/vt2 = 1.  But vt changes when we are falling into thicker
>air.
>
Once we reach terminal velocity, weight equals drag, always. The reason
why terminal velocity decreases is because drag increases. So I do not
understand why you say "terminal velocity is reached MOMENTARILY".
>
>PY> Also, why the 1.3 g? Should it not be 1? How could it be
>above 1 since the skydiver is not inputting energy? He/she
>won't have a rocket up their butt, will they?
>
>RC:  By "gee" I meant a normalised unit of force rather than
>an acceleration.  Thus weight is 9.8 or 10 N/kg and the wind
>force can indeed be 1.3 gee = 13 N/kg.  The simulator uses
>9.81 N/kg as force due to gravity.

I guess I have to see your equations. But, for an accurate jump
description, you should use the fact that g is dependent on altitude.
In other words, g = g[y]. It makes a difference, try it.
>
>greeny@wotan.top.cis.syr.edu (Jonathan Greenfield) writes:
>Date: 14 Jun 91 15:58:12 GMT
>
>JG>  If one goes from a velocity of ~750 mph to normal, low
>altitude terminal (around 600 mph difference), then obviously,
>there must be a net upward acceleration.  In other words, the
>deceleration due to atmospheric drag must be greater than the
>acceleration due to gravitational pull, in order for the diver
>to slow down.  How large the upward acceleration is depends
>upon how quickly the atmospheric density is changing.  If
>there is a fairly sharp increase in density at some point, one
>would expect to experience a significant upward acceleration
>(a shock).
>
>
>RC:  Yes, dead right.  But it is an increase of density over a
>short period (rather than a "point") that gives rise to
>impact.

I still do not believe in the "impact" theory. I do not think it can be
felt.
>

>- In the ten seconds after reaching local terminal velocity of
>750 mph, the jumper in the simulation travels about 750
>mph/3600sph*10s * 1600 m/mi ~  3 km.  About 10 000 ft, through
>which the stratospheric density increases about 60% and
>accordingly, terminal increases 30%.  As drag is v2/vt2, drag
>increases 60% to 1.6 gees.
>
"terminal increases 30%"??? You must mean decreases! And I still have to
understand your gee definition.
>
>PY> (... with today's jumpsuit, terminal velocity at 10,000'
>is about 120mph, at 2000' about 90mph)
>
>RC:  90 mph seems too low.  Drag is proportional to rho and
>vee squared, so vt is half sensitive to changes in rho.  Over
>10 000 to 2 000 ft, the density only changes 1.6% per grand *
>8 grand, or 13%.  Accordingly terminal decreases 6.5%.  That
>would be 120 to 112 , or 128 to 120 mph.
I think that you are right, 90-95 mph is about the sea level terminal
velocity. At 2000', terminal should be about 105-110+, so 112 is
perfectly reasonnable to me too.
>
>
>PY> Now the density has a bigger change at low altitudes (like
>where "normal" skydives happen) than at high altitudes
>
>RC:  Sure.  Also the terminal decreases slower per grand than
>at low altitudes.  But the critical rate is density change per
>second, not density change per grand.
Agree. I will look at that before I make another (hopefully intelligent)
response to that point.
>
>- When the high altitude jumper can cover a hell of a lot more
>grand per second than a low altitude jumper can, it allows
>density to increase far faster per second than at low
>altitudes.
Hmm. Not obvious to me, but I will look at the equations.
>
>
>greeny@wotan.top.cis.syr.edu (Jonathan Greenfield) writes:
>Date: 18 Jun 91 19:05:41 GMT
>
>JG>  When the diver reaches "terminal" velocity, he is
>inputting energy--he keeps giving up potential energy by
>falling (which is not replaced with kinetic energy, since his
>velocity is "terminal").
>
>RC:  Yes, that is the concept of terminal.  Usually the
>assumption of a constant density atmosphere is adequate.  But
>when density and thus terminal is changing fast (ie per second
>of descent), the speed must decrease as kinetic energy is
>surrendered to drag.
I disagree there. This is not the concept of terminal. The skydiver is
not inputting energy at all, he is transferring it, this is not the same
thing. And that the atmosphere is assumed to have a constant density
represents an adequate model is also wrong. Christ, the density change
is bigger at low altitude (where we usually jump) and *has* to be taken
into account. Otherwise, what is the point of using a computer to do
modelling. Sure the equations are a bit harder to derive but so what?
>
>
>PY>  The net freefall deceleration of 750 mph to 90mph (at
>2000') is very slow and I doubt that it can be felt.
>
>RC:  Deceleration is "per second" and at 750 mph,  the
>deceleration is not "very slow" at all.  See earlier graph.
>The familiar "blast handles" on older gear refers to the
>capacity of a high airspeed (during emergency ejection) to
>drag on rip-cord handles.  They were real cows to pull when
>you wanted to.
Deceleration is "unit of length per second squared" ;-)
Ejecting from an aircraft is *totally* different from what we are
talking about here! The reason why it is *unsafe* (an understatement) to
eject above 300 mph (around 15,000') is because the terminal velocity
there is way below that number, hence, the jumper-pilot is hitting a
wall! I have jumped off a Hercules flying off at 160 knots at 12,000'
Since terminal velocity there is lower (around 125 mph) when I hit the
air, I got slapped in the face quite hard!
>
>PY>  No, the upward acceleration you come up with is wrong, it
>is not greater than 1 g, that would be quite noticeable!
>
>RC:  Well, the net acceleration is weight minus drag.  So that
>is exactly one gee of weight minus approximately one gee of
>drag.  Only when airspeed or terminal speed is changing fast
>is there much imbalance.  And only these high-altutude-record-
>setting jumpers will get to experience SUSTAINED excess drag.
When are these people going to make their jump?
>
>PY> (..apologies to my fellow netters who are probably bored
>to f*** death with this technical discussion!)
>
>RC:  Not at all!  We are surely bored to death with anything
>less!

Excellent!


-- 
  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