tcliftonr@cc.curtin.edu.au (06/11/91)
RE: IMPACT WITH ATMOSPHERE: High speed exits The high altitude record jump from 120 000 ft will also achieve a record airspeed of 750 mph and then impact the atmosphere with 1.6 gees according to the simulation, nearby on the bulletin board. Once these jumpers have achieved the record airspeed (it will be a record regardless), this number will be something of a wow in itself, but a bigger achievement among skydivers will be the handling of sustained excess wind force. The violence of the forces affecting a jumper goes up as the square. Have you ever exited above terminal? You know all about it! There is a realisation of imminent hazard, particularly of the increased tendency to oscillate or rock. Beyond oscillation is the tendency to tumble. In the sustained wind force of the high altitude jump, the jumpers will have to ensure they do not tumble to the point of red- out. If anyone wants to train for excess wind force, they can get the required wind by exiting a plane at an airspeed v where- v2/Vt2 = no of gees so that v = Vt * sqrt( no of gees ) eg: if you are training for 1.21 gees, you can exit at v = Vt * sqrt(1.21) = 120 * 1.1 mph = 132 mph. (ie 120 knots). Similarly 1.44 gees requires another 10% of airspeed. That's if you exit at modest altitudes where Vt is 120 mph or so. Mind you, it wont last long, just a taste of the impact. Here is airspeed for an exit from a plane at 160 mph (142 knots) for an exit impact of 1.8 gees. The simulation assumes a normal posture of the jumper can be maintained against the wind and that the exit is made with the body inclined at 45 degrees to the wind. Since this means that the impact will lift this jumper above the exit point, there is a risk of hitting the tail of the plane. An assumption is made too, that the jumper is normally well-damped, that is, (s)he is a quite stable freefaller. Yet the simulation clearly shows oscillation: 0 v AIRSPEED, mph v 160 -90 a ANGLE OF TILT ON WIND a 90 0 f WIND FORCE, N/kg f 40 ................................................................. . . f . a v . jiggle in . f . a v. . wind force . f . a . v . . due to posture f . a . v . . changes . f a . . v . . . f a . oscillation . v . . . f a . . v . . . f a. . v . . . f a . v . . . f .a . v . .1 . f a . v . . . f a. . v . . .f a. flat . v . . .f a. on . v . . f one a. wind .v . . f gee a. v . . f. a. v 120 mph . . f. a. v. . . f. a. v . . . f . a. v . . .2 f . a. v . . If you want a copy of the simulator, just ask. Roger Clifton Kalgoorlie West Australia.
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/13/91)
Yes, I am interested by the simulator code. Also, one thing does not seem obvious to me: I understand why the velocity will be *high* but why are you expecting stability problems? First of all, gravity will be the only driving force pulling down (no energy will come from the skydiver, he/she will be "passive"). 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 thru the atmosphere. Intuitively to me, there should not be any difference with the feel of "regular" terminal velocity. 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? Just curious. -- 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
mspurgeo@oucsace.cs.OHIOU.EDU (Mike Spurgeon) (06/14/91)
In article <950@lhdsy1.chevron.com>, yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: > Yes, I am interested by the simulator code. > Also, one thing does not seem obvious to me: I understand why the > velocity will be *high* but why are you expecting stability problems? (stuff deleted) > 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? > Just curious. At a guess, the 1.3 G's is because the atmosphere is acting to 'decelerate' the jumper from his previous high velocity. Mass and inertia still apply. Regarding stability, Col. Kittenger felt he could have done without the 6 foot drogue, and that stability was _not_ a problem. Keep in mind that a premature deployment would be fatal. Mike Spurgeon Internet: mspurgeo@oucsace.cs.ohiou.edu
kevin@nodecg.ncc.telecomwa.oz.au (Kevin Spencer) (06/14/91)
In article <3518@oucsace.cs.OHIOU.EDU> mspurgeo@oucsace.cs.OHIOU.EDU (Mike Spurgeon) writes: >In article <950@lhdsy1.chevron.com>, yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: >> Yes, I am interested by the simulator code. Getting briefly back to the original article... I get the impression it was implying that someone had jumped or is going to jump from 120 000 ft. My knowledge of things like the thickness of the atmosphere is a little hazy, but isn't that getting almost out of the atmosphere? Just wondering if someone could fill me in on the details of this jump. If it's able to be done, I could think of little to beat it in terms of a high :-) -- Kevin Spencer _.-_|\ I know it's stolen, but 17 Winchelsea Rd NOLLAMARA / \ the picture's brilliant PERTH, WESTERN AUSTRALIA ----> \_.--._/ Phone: +61 9 345 1025 v kevin@ncc.telecomwa.oz.au
greeny@wotan.top.cis.syr.edu (Jonathan Greenfield) (06/14/91)
In article <950@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: >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? >Just curious. 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). It shouldn't be too surprising that the drag exceeds 1 g. You've experienced a noticeable >1 g drag before, right--when you pull your main? As for energy, sure the diver is inputting energy. The plane gives the diver a lot of potential energy. Whey the diver jumps out, some of that energy is converted to kinetic energy. When the drag kicks in, some of the energy gets converted to heat, sound, etc. One hopes that by the time the diver reaches the ground, almost all of that energy has been expended!! greeny greeny@top.cis.syr.edu "What's the difference between an orange?"
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/16/91)
In article <3518@oucsace.cs.OHIOU.EDU> mspurgeo@oucsace.cs.OHIOU.EDU (Mike Spurgeon) writes: >In article <950@lhdsy1.chevron.com>, yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: >> Yes, I am interested by the simulator code. >> Also, one thing does not seem obvious to me: I understand why the >> velocity will be *high* but why are you expecting stability problems? >(stuff deleted) >> 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? >> Just curious. > >At a guess, the 1.3 G's is because the atmosphere is acting to >'decelerate' the jumper from his previous high velocity. Mass >and inertia still apply. But this happens all the time and at every jump, even from 3 grands! The atmosphere is denser at lower altitudes, so a skydiver who has terminal velocity constantly decelerates (ie: the terminal velocity is not really terminal: with today's jumpsuit, terminal velocity at 10,000' is about 120mph, at 2000' about 90mph) Now the density has a bigger change at low altitudes (like where "normal" skydives happen) than at high altitudes (logarithmic profile in density, or is it pressure, or both?) so a negative 1.3G does not make sense to me at all. > >Regarding stability, Col. Kittenger felt he could have done >without the 6 foot drogue, and that stability was _not_ a >problem. I would agree. Once the skydiver achieves a "fast enough" velocity (that is, when a change of body attitude creates an instantaneous response) then any skydiver with a little experience should be able to stay or regain stability! -- 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
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/16/91)
In article <1991Jun14.115812.28436@rodan.acs.syr.edu> greeny@top.cis.syr.edu (Jonathan Greenfield) writes: >In article <950@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: > >>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? >>Just curious. > >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). It shouldn't be too surprising >that the drag exceeds 1 g. You've experienced a noticeable >1 g drag before, >right--when you pull your main? > Yes, but there are no sharp increase of density. The only sharp increases/decreases are found in the temperature profile of the atmosphere and these start happening at the end of the troposphere (10,000 meters, 30,000'.) By then, the atmosphere is so tenuous that I doubt very much that it can have influence on the freefall speed of a *freefalling* object. >As for energy, sure the diver is inputting energy. The plane gives the diver >a lot of potential energy. Whey the diver jumps out, some of that energy is >converted to kinetic energy. When the drag kicks in, some of the energy gets >converted to heat, sound, etc. One hopes that by the time the diver reaches >the ground, almost all of that energy has been expended!! Yes, the plane gives potential energy to the skydiver (vertical because of the altitude, horizontal due to its motion) but these "external" influences end (by that I mean the skydiver becomes "passive") when terminal velocity is reached. The only way for a skydiver to input energy then is to use muscles to turn, track, or whatever else. The net freefall deceleration of 750 mph to 90mph (at 2000') is very slow and I doubt that it can be felt. After all, in a jump from 12000', skydivers do not feel their terminal velocities decreasing (until they open their chutes!), yet this is the region where the increase of density per altitude change is the greatest (logarithmic profile of atmospheric density.) -- 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
greeny@wotan.top.cis.syr.edu (Jonathan Greenfield) (06/19/91)
In article <970@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: >Yes, the plane gives potential energy to the skydiver (vertical because >of the altitude, horizontal due to its motion) but these "external" >influences end (by that I mean the skydiver becomes "passive") when >terminal velocity is reached. The only way for a skydiver to input >energy then is to use muscles to turn, track, or whatever else. Actually, the "horizontal" energy is kinetic. In any case, even after 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"). >The net freefall deceleration of 750 mph to 90mph (at 2000') is very >slow and I doubt that it can be felt. After all, in a jump from 12000', >skydivers do not feel their terminal velocities decreasing (until they >open their chutes!), yet this is the region where the increase of >density per altitude change is the greatest (logarithmic profile of >atmospheric density.) You are undoubtedly correct. I was simply trying to establish that an upward acceleration of >1 g is reasonable. I don't mean to make any claims about exactly how large that acceleration will be. (I don't know enough about the atmosphere.) greeny greeny@top.cis.syr.edu "What's the difference between an orange?"
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/20/91)
In article <1991Jun18.150541.5220@rodan.acs.syr.edu> greeny@top.cis.syr.edu (Jonathan Greenfield) writes: >In article <970@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: > >>Yes, the plane gives potential energy to the skydiver (vertical because >>of the altitude, horizontal due to its motion) but these "external" >>influences end (by that I mean the skydiver becomes "passive") when >>terminal velocity is reached. The only way for a skydiver to input >>energy then is to use muscles to turn, track, or whatever else. > >Actually, the "horizontal" energy is kinetic. In any case, even after 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"). Yes, you are right, the horizontal energy is kinetic. No, you are wrong about the potential energy *not* being transformed into kinetic energy. A simple answer to that is "The Law of Conservation of Energy". 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! What happens is that the skydiver gives up the *kinetic* energy, not the potential energy, because it is first transfered to kinetic, then heat. The skydiver is not inputing energy, he is transfering it (or being used for transfer, like a catalyst, because this happens transparently to the person in question!) >>The net freefall deceleration of 750 mph to 90mph (at 2000') is very >>slow and I doubt that it can be felt. After all, in a jump from 12000', >>skydivers do not feel their terminal velocities decreasing (until they >>open their chutes!), yet this is the region where the increase of >>density per altitude change is the greatest (logarithmic profile of >>atmospheric density.) > >You are undoubtedly correct. I was simply trying to establish that an >upward acceleration of >1 g is reasonable. I don't mean to make any claims >about exactly how large that acceleration will be. (I don't know enough about >the atmosphere.) No, the upward acceleration you come up with is wrong, it is not greater than 1 g, that would be quite noticeable! Let us go through it (with apologies to my fellow netters who are probably bored to f*** death with this technical discussion!) First trust me: around 12,000 feet, terminal velocity is around 120 mph, around 2000 feet, it is about 90 mph. (plus or minus 5 mph, maybe 10 max for face down, stable position, small jumpsuit.) (120 - 90) = 30 mph; from 12,000 to 2000 feet say 60 seconds freefall time (it is a bit too fast, probably 65 to 70, but what the heck!) 30 mph * 1.6 km per mile = 48 km per h = 48000 meter per h = 13.3 meter per s. 13.3 meter per s / 60 s (freefall time) = 0.222 meter per s per s (whao,these are acceleration units!) so the deceleration is 0.222 m/(s^2). In units of g, g being 9.81 m/(s^2), we get: 0.222 m/(s^2) / 9.81 g/m/(s^2) = 0.02265 g. Notice. not 1.02265 but 0.02265 g. (I would not pay too much attention to the last 3 digits). At all times, the skydiver (and all objects on earth) are subjected to 1 g, even in freefall (that's why we keep on falling). So assuming that we say that g, the gravitational acceleration due to mother earth, does not change with altitude (which is wrong, but let's simplify) then the net acceleration a skydiver would feel at 2000 feet would be (1-0.02265)g. One down, 0.02265 up. Of course, this is about as big as the *freefall* upward acceleration will get (assuming an opening at 2 grands) because, the atmosphere gets thinner as altitude goes up, so the deceleration is also smaller. Ah, by the way, in this discussion, negative (upward acceleration) = deceleration. I hope I did not mess up with my units transformation. But who knows? -- 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
mspurgeo@oucsace.cs.OHIOU.EDU (Mike Spurgeon) (06/20/91)
In article <983@lhdsy1.chevron.com>, yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: > In article <1991Jun18.150541.5220@rodan.acs.syr.edu> greeny@top.cis.syr.edu (Jonathan Greenfield) writes: > >In article <970@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: > > > 1 g, even in freefall (that's why we keep on falling). So assuming that > we say that g, the gravitational acceleration due to mother earth, does > not change with altitude (which is wrong, but let's simplify) then the > net acceleration a skydiver would feel at 2000 feet would be > (1-0.02265)g. One down, 0.02265 up. > Of course, this is about as big as the *freefall* upward acceleration will get > (assuming an opening at 2 grands) because, the atmosphere gets thinner > as altitude goes up, so the deceleration is also smaller. Try thinking about when the shuttle hits the atmosphere... In the event we're discussing, the jumper will be above _most_ of the atmosphere. That allows him to attain a speed of approx. 750 mph in the first 20 seconds or so (that's the time line I've read on Colonel Kittinger's jump from 102,000'). _THEN_, just like the shuttle, he slams into the atmosphere. Prior to this _sudden_ decelleration (sp?) he _was_ experiencing 1 G. The friction of the atmosphere, for our purposes, didn't exist. We're talking slowing from 750 mph to 90 (or 120) in a fairly short period of time. We're also looking at 120,000', not 12,000' to opening altitude. I'm fairly certain the jumper _will_ feel more than 1 G. I'd bet it is on the order of the original post (1.3). 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. Mike Spurgeon Internet: mspurgeo@oucsace.cs.ohiou.edu
joep@Stardent.COM (Joe Peterson) (06/21/91)
> Notice. not 1.02265 but 0.02265 g. (I would not pay too much attention
Well, this all depends on how you look at it. I did not check your math,
so assuming your numbers are correct, the skydiver would feel 1.02265 g.
This is because if terminal were constant the whole way down, he would
feel exactly 1.0 g (assuming "g" is not changing with altitude). While
we are on the ground we feel 1.0 g because we have a constant (zero)
change in velocity. True, he would feel an added acceleration of
0.02265 g upward (more than normal), but the total "g's" he would feel
would be more than 1.0.
Joe Peterson
C-20351
joep@stardent.comyzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/24/91)
In article <1991Jun21.145113.6028@Stardent.COM> joep@Stardent.COM (Joe Peterson) writes: >> Notice. not 1.02265 but 0.02265 g. (I would not pay too much attention > >Well, this all depends on how you look at it. I did not check your math, >so assuming your numbers are correct, the skydiver would feel 1.02265 g. >This is because if terminal were constant the whole way down, he would >feel exactly 1.0 g (assuming "g" is not changing with altitude). While >we are on the ground we feel 1.0 g because we have a constant (zero) >change in velocity. True, he would feel an added acceleration of >0.02265 g upward (more than normal), but the total "g's" he would feel >would be more than 1.0. Nooooo. I said one down (assuming g not changing with altitude, and remember, Earth is attracting us, so g is toward the center of the Earth) and 0.023 (rounded) up, which would mean his acceleration is less than one. Look at it this way, since we agree that the skydiver's terminal velocity is slowing down as he/she gets lower in the atmosphere, how could his/her acceleration be greater than g? I don' get it. -- 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
joep@Stardent.COM (Joe Peterson) (06/24/91)
> Nooooo. I said one down (assuming g not changing with altitude, and > remember, Earth is attracting us, so g is toward the center of the > Earth) and 0.023 (rounded) up, which would mean his acceleration is less > than one. > Look at it this way, since we agree that the skydiver's terminal > velocity is slowing down as he/she gets lower in the atmosphere, how > could his/her acceleration be greater than g? I don' get it. Well, what I was saying was that the skydiver would FEEL a force larger than 1.0 g. Just like when you open your parachute, you feel a force much larger than 1.0. If a person is falling at a constant velocity in a constant gravitational field, he/she will feel the same force as when lying on the ground. On the ground, there is the force due to gravity acting downward and the "normal" force acting upward (the support of a solid surface). These are equal and opposing forces. In freefall, the force of gravity downward would be equally opposed by the force of friction upward (once terminal velocity is reached) if we assume no slowing down as the atmosphere gets more dense. Now, if the person is slowing down, the force due to friction is constantly higher than that due to gravity. This is an additional force felt by the skydiver pushing upward, making him feel heavier. Therefore, he is "pulling more than 1.0 g's." It is the same as being in an elevator and starting to go up. The net force upward is greater than 1.0 g's for a second or so. Does this make sense? Joe Peterson C-20351 joep@stardent.com
greeny@wotan.top.cis.syr.edu (Jonathan Greenfield) (06/25/91)
In article <983@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: >>Actually, the "horizontal" energy is kinetic. In any case, even after 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"). > >Yes, you are right, the horizontal energy is kinetic. >No, you are wrong about the potential energy *not* being transformed >into kinetic energy. A simple answer to that is "The Law of >Conservation of Energy". 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! What happens is that the skydiver gives up the *kinetic* >energy, not the potential energy, because it is first transfered to >kinetic, then heat. I disagree. Of course energy is conserved--but in the system (diver + atmosphere), not just in the diver. As I stated in a previous post, the potential energy is lost via drag to heat, sound, etc. However, it cannot be that the energy is first transformed into the skydiver's own kinetic energy. As long as the diver's velocity is fixed, he cannot be gaining any kinetic energy (KE = 0.5*mass*velocity^2). Molecules in the atmosphere, of course, can gain kinetic energy. >>>The net freefall deceleration of 750 mph to 90mph (at 2000') is very >>>slow and I doubt that it can be felt. After all, in a jump from 12000', >>>skydivers do not feel their terminal velocities decreasing (until they >>>open their chutes!), yet this is the region where the increase of >>>density per altitude change is the greatest (logarithmic profile of >>>atmospheric density.) >> >>You are undoubtedly correct. I was simply trying to establish that an >>upward acceleration of >1 g is reasonable. I don't mean to make any claims >>about exactly how large that acceleration will be. (I don't know enough about >>the atmosphere.) > >No, the upward acceleration you come up with is wrong, it is not greater >than 1 g, that would be quite noticeable! This is just a misunderstanding...sorry that I was not clearer. I did not mean that there was a *net* upward acceleration >1 g--simply an upward acceleration (working against gravitational acceleratIon). My qualitative argument only supported that it is possible to have a *net* upward acceleration. Since I have no quantitative data regarding the atmosphere, I couldn't say anything more than that. Apparently, I misunderstood your original post. I assumed that the figures being quoted had not yet been adjusted to reflect the downward acceleration of gravity (and, therefore, a net acceleration). Looks like that was a poor assumption. Maybe I'm too used to feeling that 1 g upward acceleration under my feet every day. :) greeny greeny@top.cis.syr.edu "What's the difference between an orange?"
greeny@wotan.top.cis.syr.edu (Jonathan Greenfield) (06/25/91)
In article <994@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: >>Well, this all depends on how you look at it. I did not check your math, >>so assuming your numbers are correct, the skydiver would feel 1.02265 g. >>This is because if terminal were constant the whole way down, he would >>feel exactly 1.0 g (assuming "g" is not changing with altitude). While >>we are on the ground we feel 1.0 g because we have a constant (zero) >>change in velocity. True, he would feel an added acceleration of >>0.02265 g upward (more than normal), but the total "g's" he would feel >>would be more than 1.0. > >Nooooo. I said one down (assuming g not changing with altitude, and >remember, Earth is attracting us, so g is toward the center of the >Earth) and 0.023 (rounded) up, which would mean his acceleration is less >than one. >Look at it this way, since we agree that the skydiver's terminal >velocity is slowing down as he/she gets lower in the atmosphere, how >could his/her acceleration be greater than g? I don' get it. When you stand up, don't you feel a 1 g force at your feet? We can not *feel* gravitational forces, but we can *feel* the forces that offset gravitation. greeny greeny@top.cis.syr.edu "What's the difference between an orange?"
tcliftonr@cc.curtin.edu.au (06/25/91)
In article <994@lhdsy1.chevron.com>, yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: > Since we agree that the [extreme high altitude] skydiver's terminal > velocity is slowing down as he/she gets lower in the atmosphere, how > could his/her acceleration be greater than g? I don' get it. The wind force is 16 N/kg upwards, his/her weight is 9.8 N/kg downwards. So the net force is 6 N/kg upwards, giving a net acc'n of 6 m/s2 upwards. Just as I feel my weight reaction push back up on my feet at 9.8 N/kg, or one gee, this re-entering jumper feels 16 N/kg or 1.6 gee pressing back up on torso and limbs. And it is just that force which will be used for skydiving with - for the deliberate accelerations. We are familiar with about one gee of wind force to work with. But these high-altitude-challengers will get much more than one gee, up to 16 N/kg, for about thirty seconds. Pioneers!
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/26/91)
In article <1991Jun24.154627.285@Stardent.COM> joep@Stardent.COM (Joe Peterson) writes: > >> Nooooo. I said one down (assuming g not changing with altitude, and >> remember, Earth is attracting us, so g is toward the center of the >> Earth) and 0.023 (rounded) up, which would mean his acceleration is less >> than one. >> Look at it this way, since we agree that the skydiver's terminal >> velocity is slowing down as he/she gets lower in the atmosphere, how >> could his/her acceleration be greater than g? I don' get it. > >Well, what I was saying was that the skydiver would FEEL a force larger >than 1.0 g. Just like when you open your parachute, you feel a force >much larger than 1.0. If a person is falling at a constant velocity in >a constant gravitational field, he/she will feel the same force as when >lying on the ground. On the ground, there is the force due to gravity >acting downward and the "normal" force acting upward (the support of >a solid surface). These are equal and opposing forces. In freefall, >the force of gravity downward would be equally opposed by the force >of friction upward (once terminal velocity is reached) if we assume >no slowing down as the atmosphere gets more dense. Now, if the person >is slowing down, the force due to friction is constantly higher than >that due to gravity. This is an additional force felt by the skydiver >pushing upward, making him feel heavier. Therefore, he is "pulling >more than 1.0 g's." It is the same as being in an elevator and starting >to go up. The net force upward is greater than 1.0 g's for a second or >so. Does this make sense? I knew that at some point the elevator analogy would be brought up! The reason why when a skydiver opens his/her parachute, he/she feels it and *it is greater* than 1 g only because the person goes from 100mph (say) to 10-15' per second (3 mph???) in a *short* time and since acceleration is measured in unit of length/second/second then you obtain a fast deceleration say (100 - 3)*1.6*1000/3600 [mph*km/miles*meters/km*seconds/hour]= 43 meters/second = velocity loss. Say the parachute takes 2 seconds to open 43/2 = 21.5 m/s/s Since g = 9.81 m/s/s, the deceleration is 21.5/9.81 [m/s/s*g/(m/s/s)] = 2.2 g deceleration. Yes, it is above one, but only because the amount of time it took to slow down was short. And this is the net difference. Now, I agree that the net upward force on a skydiver in freefall in an atmosphere (Earth!) will be greater than 1 g, in the case discussed previously (a few exchanges ago), it would be around 1.025 g? *But* 1 g would be cancelled because of the net downward gravitational attraction. So the skydiver would *feel* 0.025 g which is unfeelable by human standards! Using your analogy, when a person is in an elevator, he/she does not feel 1+ g but the acceleration of the elevator. When I am standing up, Earth is pulling down 1 g, the surface I am standing on compensates (so I don't sink into the floor) and I will argue that I don't feel the floor pushing me up (to avoid sinking into it!) I can also say that I do not feel my weight either because my muscles are used to and can compensate (for now!) for the gravitational pull. When you do analyze physical problems, you analyze all forces that apply but the resultant (or summation of all the forces) is what will determine what happens to the system under study. A person at rest on the ground does not pull one g! Yet this is what you argue a skydiver feels in freefall. I disagree. I look forward to your next entry, this is stimulating. E.F.S. -- 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
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/26/91)
In article <1991Jun24.155154.14380@rodan.acs.syr.edu> greeny@top.cis.syr.edu (Jonathan Greenfield) writes: >In article <994@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: > >>>Well, this all depends on how you look at it. I did not check your math, >>>so assuming your numbers are correct, the skydiver would feel 1.02265 g. >>>This is because if terminal were constant the whole way down, he would >>>feel exactly 1.0 g (assuming "g" is not changing with altitude). While >>>we are on the ground we feel 1.0 g because we have a constant (zero) >>>change in velocity. True, he would feel an added acceleration of >>>0.02265 g upward (more than normal), but the total "g's" he would feel >>>would be more than 1.0. >> >>Nooooo. I said one down (assuming g not changing with altitude, and >>remember, Earth is attracting us, so g is toward the center of the >>Earth) and 0.023 (rounded) up, which would mean his acceleration is less >>than one. >>Look at it this way, since we agree that the skydiver's terminal >>velocity is slowing down as he/she gets lower in the atmosphere, how >>could his/her acceleration be greater than g? I don' get it. > >When you stand up, don't you feel a 1 g force at your feet? We can not >*feel* gravitational forces, but we can *feel* the forces that offset >gravitation. YES! Just my point. -- 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
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/26/91)
In article <1991Jun24.153922.12763@rodan.acs.syr.edu> greeny@top.cis.syr.edu (Jonathan Greenfield) writes: >In article <983@lhdsy1.chevron.com> yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) writes: >>No, you are wrong about the potential energy *not* being transformed >>into kinetic energy. A simple answer to that is "The Law of >>Conservation of Energy". 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! What happens is that the skydiver gives up the *kinetic* >>energy, not the potential energy, because it is first transfered to >>kinetic, then heat. > >I disagree. Of course energy is conserved--but in the system (diver + >atmosphere), not just in the diver. As I stated in a previous post, the >potential energy is lost via drag to heat, sound, etc. However, it cannot >be that the energy is first transformed into the skydiver's own kinetic energy. >As long as the diver's velocity is fixed, he cannot be gaining any kinetic >energy (KE = 0.5*mass*velocity^2). Molecules in the atmosphere, of course, >can gain kinetic energy. > I strongly disagree and let see if I can make you change your mind. In the absence of atmosphere, the potential energy (of a falling object) would be changed into kinetic energy. Do you agree? If so, then, since the atmosphere does not "know" that a body is there, how could the mere presence of an atmosphere change the laws of physics? How could the potential energy of the falling object suddenly stop being transferred into kinetic energy and instead change to be transferred into heat? (well, according to your model, the potential energy would gradually transfer itself from kinetic to heat, as "terminal" velocity would be reached.) I just don't see a law of physics that would allow that. Also, if the potential energy was not transformed into kinetic energy (or stopped being transformed into kinetic energy and instead transfered itself into heat) then the skydiver would slow down due to the friction with the atmosphere. He/she does not because of the potential energy (he/she is still above ground) that is still there. The diver's velocity is fixed (say) so he/she is not gaining any kinetic energy, but he/she is transfering it. ...deleted text... >assumption. Maybe I'm too used to feeling that 1 g upward acceleration under >my feet every day. :) Really? Are you feeling 1 g all the times? (or are you just kidding?) ;-) -- 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
jnrees@athena.mit.edu (Jim Rees) (06/26/91)
In article <1000@lhdsy1.chevron.com> you write: >I strongly disagree and let see if I can make you change your mind. >In the absence of atmosphere, the potential energy (of a falling object) >would be changed into kinetic energy. Do you agree? >If so, then, since the atmosphere does not "know" that a body is there, >how could the mere presence of an atmosphere change the laws of physics? >How could the potential energy of the falling object suddenly stop being >transferred into kinetic energy and instead change to be transferred >into heat? (well, according to your model, the potential energy would >gradually transfer itself from kinetic to heat, as "terminal" velocity >would be reached.) I just don't see a law of physics that would allow >that. No laws of physics are being broken here. Without atmosphere, there is only one force on the skydiver, so he accelerates without bound until he hits the ground. With an atmosphere, the skydiver's acceleration starts at 9.8 m/s^2 and gradually reduces to zero when he hits terminal velocity. At this point, the skydiver's kinetic energy is no longer increasing, but his potential energy is decreasing as he falls. Where is this energy going? Since the skydiver isn't getting any of it, it must be going into the air in the form of heat, sound, turbulence, etc. Now in between exiting the aircraft and hitting terminal velocity, the skydiver's kinetic energy is increasing, but not as fast as his potential energy is decreasing. The difference is transferred to the air. >Also, if the potential energy was not transformed into kinetic energy >(or stopped being transformed into kinetic energy and instead transfered >itself into heat) then the skydiver would slow down due to the friction >with the atmosphere. No, because nothing is taking away the kinetic energy that the skydiver has once he has reached terminal. The friction with the atmosphere only prevents the skydiver from gaining any *additional* kinetic energy. >>assumption. Maybe I'm too used to feeling that 1 g upward acceleration under >>my feet every day. :) > >Really? Are you feeling 1 g all the times? (or are you just kidding?) ;-) >-- Yes, one *feels* the push of the ground. Have you ever made a balloon jump? You know that weird feeling when you first step off when there is negligible air friction. That is the *feel* of no forces, even though there is a net force of 1g. The reason why gravity cannot be felt is because it acts uniformly on all the particles in our body. Our entire body accellerates uniformly in a gravitational field. On the ground, however, the normal force the ground provides to prevent us from accellerating towards the center of the earth does not act uniformly on our bodies. There is much more pressure in your feet, for example, than your head. When you see pictures of astronauts in the space shuttle, they float around, apparently experiencing a zero-g environment. But in fact, there is still a force of 1g on them (ok, since they're further from the center of the earth by a couple percent, the force is a little less than 1g if 1g is defined as the gravitational force at ground level). They are, at all times, accelerating uniformly towards the center of the earth, but never get any closer since they're in orbit. Other than the force of gravity, there are no other significant forces on the astronauts, and they feel like they're in a zero-g environment (which, by the way, would feel the same way). Jim Rees D-13359
joep@Stardent.COM (Joe Peterson) (06/27/91)
> Say the parachute takes 2 seconds to open 43/2 = 21.5 m/s/s > Since g = 9.81 m/s/s, the deceleration is 21.5/9.81 [m/s/s*g/(m/s/s)] > = 2.2 g deceleration. Yes, it is above one, but only because the amount > of time it took to slow down was short. And this is the net difference. > Now, I agree that the net upward force on a skydiver in freefall in an > atmosphere (Earth!) will be greater than 1 g, in the case discussed > previously (a few exchanges ago), it would be around 1.025 g? > *But* 1 g would be cancelled because of the net downward gravitational > attraction. So the skydiver would *feel* 0.025 g which is unfeelable by human > standards! Hmmm, you say it is above 1.0 only because the time is short. That is not true. It would be greater than 1.0 even if the time were very very long (anything short of infinite!) but would be very close to 1.0 (in fact, I agree that it would be too small to notice). When you say the 1.0 g would be cancelled out, I think you are normalizing the 1.0 g we feel on the ground. The force is still there, but it feels normal! > When I am standing up, Earth is pulling down 1 g, the surface I am > standing on compensates (so I don't sink into the floor) and I will > argue that I don't feel the floor pushing me up (to avoid sinking into > it!) I can also say that I do not feel my weight either because my > muscles are used to and can compensate (for now!) for the gravitational > pull. I think we are confusing "experiences" here. Since it is natural to continuously feel normal gravity, people normalize this experience and think of forces in relation to it. But I still maintain that you feel the force of your weight (but as you say, you ignore it). When you are standing up, don't you feel the floor "pushing up?" If you have ever used a leg press weight machine, you know the sensation of the machine pushing your feet toward you. In this case, you would be working against the force of the weight you selected instead of your own weight. If you were weightless for a while (zero g's), a return to normal gravity would certainly make you aware of your weight! > A person at rest on the ground does not pull one g! Yet this is what you > argue a skydiver feels in freefall. I disagree. Why not? Is the person not feeling normal gravitational force? Granted, this force is directed vertically, and when people talk about "pulling g's," they usually are referring to force along an arbitrary axis caused by some vehicle. Have you ever done "zero g's" in the jump plane? This is because the plane starts to follow the same parabolic path you do as you truely freefall (since we can really ignore air friction now!). > I look forward to your next entry, this is stimulating. Cool! > E.F.S. Absolutely! Joe Peterson C-20351 joep@stardent.com
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/30/91)
In article <1991Jun26.162046.8601@athena.mit.edu> jnrees@athena.mit.edu (Jim Rees) writes: >No laws of physics are being broken here. Without atmosphere, there is >only one force on the skydiver, so he accelerates without bound until >he hits the ground. With an atmosphere, the skydiver's acceleration >starts at 9.8 m/s^2 and gradually reduces to zero when he hits >terminal velocity. At this point, the skydiver's kinetic energy is no >longer increasing, but his potential energy is decreasing as he falls. >Where is this energy going? Since the skydiver isn't getting any of >it, it must be going into the air in the form of heat, sound, >turbulence, etc. Now in between exiting the aircraft and hitting >terminal velocity, the skydiver's kinetic energy is increasing, but >not as fast as his potential energy is decreasing. The difference is >transferred to the air. > We both agree that the difference is transfered to the atmosphere, we disagree on how. My point is that the potential energy is first transfered into kinetic then lost to the atmosphere. Yours is that the potential energy is transfered to the atmosphere directly. (In both cases, the objects/skydivers have reached terminal velocity, this is written in case new people are reading this and wondering what we are talking about.) I guess we won't be able to convince each others. A final point from me, *potential* energy is just that, a "reserve" of energy belonging to the skydiver (in this case), not the atmosphere. In order to be transfered to the atmosphere, it *has* first to become "real" (kinetic) to the skydiver who can then loose it (or transfer it.) >>>assumption. Maybe I'm too used to feeling that 1 g upward acceleration under >>>my feet every day. :) >> >>Really? Are you feeling 1 g all the times? (or are you just kidding?) ;-) >>-- > >Yes, one *feels* the push of the ground. Have you ever made a balloon >jump? You know that weird feeling when you first step off when there >is negligible air friction. That is the *feel* of no forces, even >though there is a net force of 1g. The reason why gravity cannot be >felt is because it acts uniformly on all the particles in our body. >Our entire body accellerates uniformly in a gravitational field. On >the ground, however, the normal force the ground provides to prevent >us from accellerating towards the center of the earth does not act >uniformly on our bodies. There is much more pressure in your feet, >for example, than your head. > A "g" is an acceleration, not a force. It seems to me that you are using both intermittently. When you are on the ground, standing still (or moving at a constant velocity), you are *not* feeling (or pulling) a g. No way. I have made ballon jumps, I have made helicopter jumps and I have been on roller-coasters and for the first two cases, when I jumped, I could feel the acceleration of *one* g at the very beginning of the jump. In the case of the roller-coaster (or in a swing), when the machine is reaching the top of an arch and starts to go down, people feel less than one g (but more than zero!) because the machine does not go straight down but at a steep (but not 90 degrees) angle. >When you see pictures of astronauts in the space shuttle, they float >around, apparently experiencing a zero-g environment. But in fact, >there is still a force of 1g on them (ok, since they're further from >the center of the earth by a couple percent, the force is a little >less than 1g if 1g is defined as the gravitational force at ground >level). They are, at all times, accelerating uniformly towards the >center of the earth, but never get any closer since they're in orbit. >Other than the force of gravity, there are no other significant forces >on the astronauts, and they feel like they're in a zero-g environment >(which, by the way, would feel the same way). > Allow me to disagree! astronauts are not "apparently experiencing a zero-g environment"! They are experiencing zero-g! The reason for that is that they are experiencing a centrifical force which exactly compensate for the gravitational attraction of Mother Earth. The source of this centrifical force is the revolution of the spaceship about the Earth. Now even though (in the case of a circular orbit) the ship would have a constant speed, it would not have a constant velocity. It is this change of velocity which creates the acceleration which compensates for the gravitational acceleration. Actually, it is a little bit more complicated than that, because it is not the accelerations which cancel each others but the forces! "They are, at all times, accelerating uniformly towards the center of the earth..." No they are not. The concept is wrong. You cannot accelerate and not move. What is going on is that they are subjected to a gravitational force which is cancelled by the centrifical force. The resulting force (henceforth, the acceleration) on them is zero! > -- 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
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/30/91)
In article <1991Jun25.194227.8803@cc.curtin.edu.au> tcliftonr@cc.curtin.edu.au writes: >The wind force is 16 N/kg upwards, his/her weight is 9.8 N/kg downwards. >So the net force is 6 N/kg upwards, giving a net acc'n of 6 m/s2 upwards. > Allow me to agree with your numbers since I cannot verify them right now. First of all, a force is unit of mass * acceleration which is equal to a Newton or N. Not a N/kg, let us get our units straight, because a N/kg is a unit of acceleration, not force (I am not being picky, they are very different concepts.) Assuming that the net force is 6 N or 6 kg*m/s/s upwards, then the skydiver is going to slow down. We agree on that. >Just as I feel my weight reaction push back up on my feet at 9.8 N/kg, >or one gee, this re-entering jumper feels 16 N/kg or 1.6 gee pressing >back up on torso and limbs. > No, this jumper is experiencing .6 g (again, using your numbers) upwards because it is the *resultant* or *summation* of all the forces (and not the accelerations) which gives the skydiver his/her motion. One cannot separate the two forces, they are, in the condition described above, indessociable (inseparable.) -- 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
yzarn@lhdsy1.chevron.com (Philip Yzarn de Louraille) (06/30/91)
In article <1991Jun26.182628.1006@Stardent.COM> joep@Stardent.COM (Joe Peterson) writes: >> Say the parachute takes 2 seconds to open 43/2 = 21.5 m/s/s >> Since g = 9.81 m/s/s, the deceleration is 21.5/9.81 [m/s/s*g/(m/s/s)] >> = 2.2 g deceleration. Yes, it is above one, but only because the amount >> of time it took to slow down was short. And this is the net difference. >> Now, I agree that the net upward force on a skydiver in freefall in an >> atmosphere (Earth!) will be greater than 1 g, in the case discussed >> previously (a few exchanges ago), it would be around 1.025 g? >> *But* 1 g would be cancelled because of the net downward gravitational >> attraction. So the skydiver would *feel* 0.025 g which is unfeelable by human >> standards! > >Hmmm, you say it is above 1.0 only because the time is short. That is >not true. It would be greater than 1.0 even if the time were very very >long (anything short of infinite!) but would be very close to 1.0 (in fact, >I agree that it would be too small to notice). When you say the 1.0 g >would be cancelled out, I think you are normalizing the 1.0 g we feel >on the ground. The force is still there, but it feels normal! If a person were to experience a total acceleration greater than one g directed upwards, after a certain time (which would be short and depend on the original downward velocity, say 5-10 sec) that person would accelerate towards the sky because the Earth would compensate for only *one g*. But this does not happen because the total acceleration is less than one g and directed upwards because the skydiver slows down (since the atmosphere is getting thicker.) > >> A person at rest on the ground does not pull one g! Yet this is what you >> argue a skydiver feels in freefall. I disagree. > >Why not? Is the person not feeling normal gravitational force? Granted, >this force is directed vertically, and when people talk about "pulling >g's," they usually are referring to force along an arbitrary axis caused >by some vehicle. Have you ever done "zero g's" in the jump plane? This >is because the plane starts to follow the same parabolic path you do >as you truely freefall (since we can really ignore air friction now!). > I think I finally understand why we disagree. I think you are confusing acceleration and force. They are not the same thing. I agree (how could I not) that Mother Earth directs on me a force, but I am not experiencing an acceleration of any kind (including one g) when I am standing still. Once again, what is important when you analyze a system is the resultant or summation of all forces, not just one. The summation of the acceleration is not the concept to use to analyze such systems. This is Newton's first law. -- 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