tcliftonr@cc.curtin.edu.au (06/24/91)
RE: GRAPH OF OSC'N DUE TO HIGH WINDFORCE yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes- PY> Yes, I am interested by the simulator code. Roger Clifton (back from a spell in the Bush) replies: RC: Anyone who wants it, please email your request direct to me so I can REPLY a file straight back to you. NEWS has been relayed many times to get to this corner of the world, which makes your signature less than a perfect address. 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? RC: There is always some incipient oscillation in all skydiving postures, damped out by sheer practice. We all rocked as students, sometime. But our confidence that we wont rock and tumble is based only on wind-forces near one gee, the zone familiar to us. - 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? 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. 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. PY> Intuitively to me, there should not be any difference with the feel of "regular" terminal velocity. RC: I tend to agree. Certainly we can assume skydiving at terminal is the same regardless of the speed at which terminal occurs. - There is no need to assume otherwise while we are confident that the nature of the drag (the wake, wash, etc) is the same. It changes at supersonic speeds and should feel different there. This difference has got to be the important discovery waiting for high altitude jumpers. But there may be subtle differences in the sub sonic range too. 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. 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. - 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. 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. 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. - 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. 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. 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. PY> A more physical answer is: the reason why the velocity is terminal is due to friction with the atmosphere. The potential energy is being transformed in kinetic energy which, in turns, is transfered as heat due to the skydiver "frictionning" with the atmosphere! RC: Yes. I think the transfer to the atmosphere is as kinetic energy of the air eddying in the burble. JG>. I don't mean to make any claims about exactly how large that acceleration will be. (I don't know enough > about the atmosphere.) RC: Please try! Anyone willing to have a bash, please ask me to post the necessary equations. 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. Mike Spurgeon> Somebody check on the Colonel's jump. All good libraries should have _something_ about it. I believe it's also in the book, 2000 Unforgettable Jumps. 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!
<ICRJP@ASUACAD.BITNET> (06/24/91)
A friend and & were wondering. How does the skydiver plan on reaching
120,000 ? I told him that I thought that it was by balloon. How were
the previous jumpers taken to altitude. Also, I thought the record was
held by a Russian jumper? ( who did not use a drougue)
Randy Palmer
rpalmer@dz.inre.asu.eduyzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/24/91)
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