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EDG
2010-Jun-04, 07:47 PM
I'm having a bit of a brain-fart here, so this may be a silly question. Obviously in lower gravity, it would take longer for an object to drop vertically from a given height and hit the ground.

How does the lower gravity affect the height that a person can jump though? If the local gravity is half that of Earth's (let's assume that Earth's gravity is 10 m/sē, so half gravity is 5 m/sē), would a person doing the exact same vertical standing jump be able to jump twice as high as on Earth? Presuambly they'd take twice as long to fall from their maximum vertical height, but would they also take twice as long to reach that height from standing on the ground too?

And would jumping horizontally (or hitting a golf ball) go twice as far in lower gravity as it would on Earth as well? (assuming there's no air resistance in all these examples).

I'm just wondering if it all scales linearly, or if there are other factors involved to make it non-linear.

alwaysnumb
2010-Jun-04, 08:12 PM
Well your legs can supply the same amount of energy to the jump. so
U=potential energy from height m=mass g=gravity and h=height
U = mgh
h=U/mg

mass stays the same so call it 1 enegy stays same so call it 1

so h=1/g

so height you can jump is inversly proportional to gravity. Other factors like atmosphere density and your muscles working time will make a small difference.

Yes you can jump 2*higher in 5m/s^2, yes it takes as long to reach the height as it does to drop from that height

With the some horizontal movement I would say you could jump 4 times further as you/golf ball would spend 4 times longer in the air.

korjik
2010-Jun-04, 08:24 PM
I'm having a bit of a brain-fart here, so this may be a silly question. Obviously in lower gravity, it would take longer for an object to drop vertically from a given height and hit the ground.

How does the lower gravity affect the height that a person can jump though? If the local gravity is half that of Earth's (let's assume that Earth's gravity is 10 m/sē, so half gravity is 5 m/sē), would a person doing the exact same vertical standing jump be able to jump twice as high as on Earth? Presuambly they'd take twice as long to fall from their maximum vertical height, but would they also take twice as long to reach that height from standing on the ground too?

And would jumping horizontally (or hitting a golf ball) go twice as far in lower gravity as it would on Earth as well? (assuming there's no air resistance in all these examples).

I'm just wondering if it all scales linearly, or if there are other factors involved to make it non-linear.

Assuming the same work, jump height is

W/mg=h

So at constant work, the height is linear with the change in gravity
(assuming that the gravitating bodies are large enough to not have to worry about curvature)

Time to fall from height h is

sqrt(2h/g)=t

So the time it takes to fall is only sqrt2 longer. Also means that the horizontal displacement is only sqrt2 longer also. I will leave it to the industrious student to show that

:)

IsaacKuo
2010-Jun-04, 08:37 PM
First off, we need to figure out whether the starting vertical velocity is significantly affected. Roughly, the starting velocity will be about the same because the acceleration distance depends only on the degree to which the legs can extend and the amount of acceleration should be mostly the same (slightly higher acceleration due to less gravity to fight).

Second, we note the amount of time it takes to reach the apex of the jump is inversely proportional to the gravity level. This time is equal to the starting vertical velocity divided by the gravitational acceleration.

Third, we note that the average upward speed is roughly constant. It's half of the initial upward speed.

The peak height is equal to the average upward speed times the time taken to reach the apex--therefore, the peak height is inversely proportional to the gravity level.

Not everything scales linearly. You have to apply physics and math to figure out how things scale.

EDG
2010-Jun-04, 09:34 PM
So at constant work, the height is linear with the change in gravity

So the time it takes to fall is only sqrt2 longer.

Thanks, that's gven me something to chew on.

So... on Earth, the work done for a 100kg person to jump up by one metre is 1m * 10m/sē * 100kg = 1000 J.
On our half gravity world, he'd jump 1000/(100*5) = 2 metres
On a world with 1/6th gravity, he'd jump 1000/(100*1.67)= 6 metres.
All good and linear here.

On Earth, he'd take 0.45 seconds to fall from a height of 1 metre at 10 m/sē acceleration.
On our half gravity world, he has to fall a distance of 2 metres for the same jump. That means he actually takes 0.89 seconds to fall... twice the time it takes to fall given the same jump in 1g.
On our 1/6th gravity world, he has to fall a distance of 6 metres for the same jump. Thus, he takes 2.68 seconds to fall, which is six times the time it takes to fall given the same jump in 1g.
So this is actually a linear relationship too!

(Now, if he had to fall a distance of 1 metre on the half gravity world, then it'd take SQRT(2) times longer (0.63 seconds), and it'd take 1.10 seconds, which is SQRT(6) times the time it takes to fall a distance of 1 metre in 1g. But he's got to fall from a greater height on the lower gravity world, so we have to account for that).

So - for the same energy put into the jump - it looks like the max height and the duration of the fall is actually linear in both cases?


This all stemmed from a volunteer education thing I did with some younger kids last week where I was showing them around a model of the solar system - I was trying to figure out an activity for them (rather than just telling them about the planets) and figured that I'd get them to jump up, and then I could say it takes X times longer to fall and they'd jump Y times higher on a given planet. But then I realised that I wasn't entirely sure how that worked, hence the question here :).

neilzero
2010-Jun-04, 10:35 PM
Jumping at two g is likely to produce sprains and/or broken foot, ankle or leg bones when you land on a hard surface

grant hutchison
2010-Jun-04, 11:23 PM
Jumping at two g is likely to produce sprains and/or broken foot, ankle or leg bones when you land on a hard surfaceOn a level surface, the biomechanical forces during landing are the same as those during the initial jump. Since you can't exert enough force to break your ankle when jumping upwards, you won't break your ankle when you land, either.

Grant Hutchison

EDG
2010-Jun-04, 11:43 PM
In lower gravity, you would feel obviously lighter... so would that mean that it's actually going to be very hard to put the same amount of work into jumping as you would on a 1g planet? Is it likely that a person would think they're trying to jump the same height, but are actually putting more energy into the jump because of the lower gravity?

alwaysnumb
2010-Jun-04, 11:51 PM
For humans to live on a planet with higher gravity you gotta be real carefull. If you trip you will hit the deck much sooner and at a higher velocity than on earth. In 2gs your gonna break something like your wrists if you manage to put your arms out in time or more likely your skull as it smacks down hard. So jumping would be something only a complete idiot would attempt. Makes me think of the apollo astronaut who did it on the moon and came down onto his pack.

NEOWatcher
2010-Jun-07, 12:30 PM
In lower gravity, you would feel obviously lighter... so would that mean that it's actually going to be very hard to put the same amount of work into jumping as you would on a 1g planet? Is it likely that a person would think they're trying to jump the same height, but are actually putting more energy into the jump because of the lower gravity?
I would think that would only depend on muscle control. It might be an interesting experiment for the Mythbusters.
Put them in a high speed elevator (Like the one to the observation deck of the Willis [aka Sears] tower) and check the height of thier jump when still, then going up, and going down.


For humans to live on a planet with higher gravity you gotta be real carefull. If you trip you will hit the deck much sooner and at a higher velocity than on earth. In 2gs your gonna break something like your wrists if you manage to put your arms out in time or more likely your skull as it smacks down hard. So jumping would be something only a complete idiot would attempt. Makes me think of the apollo astronaut who did it on the moon and came down onto his pack.
Yes and no;
A trip and fall will be from the same height no matter what g you are in because of your height. So; in this case a higher G is more dangerous.
But; (as Grant indicated above) the G is going to also change the height of your jump. So; the act of jumping is not going to change the fall, although once your feet land back on the ground, then the rest of the fall will be effected.
So; landing from a jump is not the issue, the issue is if you can control the landing.

joema
2010-Jun-07, 01:44 PM
...How does the lower gravity affect the height that a person can jump though?...
Early popular level books on space exploration depicted space-suited astronauts jumping 20 feet high. The assumption was 1/6th gravity = 6 times the jumping height.

However that neglects several factors: inertia, spacesuit mass, constriction of movement, etc.

A massive object -- even if weightless -- is hard to get moving. Note the difficulty Apollo astronauts have in moving a large rock, even in 1/6th gravity:

http://www.youtube.com/watch?v=TnbM6h_2g-I&feature=related

During Apollo 15, astronaut John Young jumped vigorously upward -- he reached about .437 meters or 1.43 feet:

http://www.youtube.com/watch?v=icayHrkoBvc&feature=related

An unencumbered athlete on the lunar surface (say in a pressure dome) could definitely jump higher, but it wouldn't be close to 6 times his jumping height on earth.

NEOWatcher
2010-Jun-07, 02:07 PM
...An unencumbered athlete on the lunar surface (say in a pressure dome) could definitely jump higher, but it wouldn't be close to 6 times his jumping height on earth.
I'm having a hard time understanding why.
The power of the jump is coming from the muscle strength and force against the ground. As long as that athlete has reached the proper squat, I can't see the difference.

joema
2010-Jun-07, 11:32 PM
You may be correct. I'll have to study this further... I was thinking that while gravity (hence weight) decreases to 1/6th, inertial effects do not. A heavy object still feels heavy and resists changes in movement, whether 1/6th g or zero g.

I was also thinking that you can't multiply by 6 the high jump record, as that is partially achieved by body contortions around the center of gravity. E.g, the high jump record is 2.45 meters (8.04 feet). The same athlete could not jump 48 feet high on the moon, even if unencumbered.

But if you multiply the *vertical* jump achievable by a terrestrial athlete (say 40 inches) by 6, that equals about 20 feet, which is still pretty high.

Jens
2010-Jun-08, 06:02 AM
On a level surface, the biomechanical forces during landing are the same as those during the initial jump. Since you can't exert enough force to break your ankle when jumping upwards, you won't break your ankle when you land, either.


That's certainly true. However, I would think that in 2G, a person simply standing up would be putting more load on the ankle than one in 1G. So in fact, if your bones were weak, I think you'd have a better chance of breaking them in 2G without even jumping.

NEOWatcher
2010-Jun-08, 12:07 PM
...I think you'd have a better chance of breaking them in 2G without even jumping.
True, but I don't think the jumping would add anything to it because the max impact would still be the same as the initial jump.

Murphy
2010-Jun-10, 03:29 PM
Interesting thread, I've been trying to imagine how life on a low gravity world would work. It would be an interesting environment in many aspects, for instance the way you could move much more easily and jump great distances, but also the fact that it would be quite slow movement.

Low gravity would have a huge effect on the evolution of animals that lived on such a planet. Imagine what a chase between the alien equivalents of a Cheetah and Gazelle would look like in low grav. They could do huge leaps, but only is slow motion, so there would be plenty of time to think about the next move, etc.

swampyankee
2010-Jun-10, 11:48 PM
I've been thinking a bit about this*.

First, anybody who is capable of climbing stairs "normally" can lift a minimum of their body weight with either leg. So, the minimum amount of force that would be available for a vertical jump would be equal to the person's body weight. For me, that would be about 825 Newtons**. This would permit an acceleration of about 9.8 m s-2 for about 0.4m, and (if my arithmetic is vaguely correct ;)) a vertical velocity of about 2.8 m s-1 on Earth. Since the force available for vertical acceleration would increase as the local gravity decreases, the achievable height would vary roughly by g-2.


G Vertical Hgt
0.05 15.96
0.1 7.77
0.2 3.68
0.5 1.23
1 0.41
1.5 0.14
2 0


Since I'm positing that a person can just lift his or her body Earth-based body weight with one leg, my estimate is that somebody can exert enough force to provide 9.8 m s-2 vertical acceleration on Earth; this means that there's nothing left to jump at 2 g. The height estimate for Earth -- 0.41 m -- is about what I can jump. When I was younger (and fitter), I could generate considerably more vertical acceleration, although I could never jump terribly well. Muggsy Bogues, about 13cm shorter than me, could dunk (http://smith.mn/nbashort/), which means he could generate enough height to get his hand about 30cm above the rim. I'll leave the creation of his g vs height table as an exercise for the student. ;)

------------

* As opposed to doing a complete biomechanical analysis.
** I'm quite far from being a star athlete.

EDG
2010-Jun-11, 01:47 AM
Low gravity would have a huge effect on the evolution of animals that lived on such a planet. Imagine what a chase between the alien equivalents of a Cheetah and Gazelle would look like in low grav. They could do huge leaps, but only is slow motion, so there would be plenty of time to think about the next move, etc.

I don't think it's really going to be slow motion... wouldn't it take about the same time to reach max height but then just take longer for them to get back on the ground due to the lower gravity? The jumps themselves would be fast, but the 'flight' would be longer.

WayneFrancis
2010-Jun-11, 02:20 AM
For humans to live on a planet with higher gravity you gotta be real carefull. If you trip you will hit the deck much sooner and at a higher velocity than on earth. In 2gs your gonna break something like your wrists if you manage to put your arms out in time or more likely your skull as it smacks down hard. So jumping would be something only a complete idiot would attempt. Makes me think of the apollo astronaut who did it on the moon and came down onto his pack.

For someone living on a planet with higher gravity our bones would naturally be stronger. For bone and muscle I think you'd find that physically we'd be pretty much ok as far as muscle and bone density. evolution wise I think our circulatory system might have a bit more of a problem long term.

Its kind of like people that do martial arts. They do lots of things that for those that don't train like they do would cause bones to break. It isn't just the fact that they are hitting the blocks of wood/cement/etc correctly that stops them from breaking their wrists. Their bone density is much higher from constantly training and hitting hard objects.

Barabino
2014-Sep-13, 07:16 AM
Jumping at two g is likely to produce sprains and/or broken foot, ankle or leg bones when you land on a hard surface

On the morning of which I am speaking there lay full length on the brink of a little cliff, and gazing into the pool beneath her, a woman of my world. To me she is lovely, exquisite, the very embodiment of beauty; to you she would seem a strange half-human monster. To me, as she lay there with her breasts against the rock and one arm reaching down into the water, her whole form expressed the lightness and suppleness of a panther. To you she would have seemed unwieldy, elephantine, and grotesque in every feature. Yet if you were to see her moving in her own world, you would know, I think, why her name in our speech is the equivalent of Panther in yours.

If you or any of your kind were to visit our world, and if by miracle you were to survive for a few moments in our alien atmosphere, gravity would make it almost impossible for you to support yourselves at all. But we, since our bones, like our buildings, are formed largely of artificial atoms, and are far more rigid than steel, since moreover our muscle cells have been most cunningly designed, can run and jump with ease. It is true, however, that in spite of our splendid tissues we have to be more solidly built than the Terrestrials, whose limbs remind us unpleasantly of insects.

The woman on the rock would certainly have surprised you, for she is a member of one of our most recent generations, whose skin and flesh are darkly translucent. Seeing her there, with the sunlight drenching her limbs, you might have taken her for a statue, cut from some wine-dark alabaster, or from carbuncle; save that, with every movement of her arm, sunken gleams of crimson, topaz, and gold-brown rippled the inner night of her shoulder and flank. Her whole substance, within its lovely curves and planes, looked scarcely solid, but rather a volume of obscure flame and smoke poised on the rock. On her head a mass of hair, flame-like, smoke-like, was a reversion to the primitive in respect of which she could never decide whether it was a thing for shame or complacency. It was this pre-historic decoration which first drew me toward her. In a closer view you would have noticed that on her back and the outer sides of her limbs the skin's translucency was complicated by a very faint leopard-like mottling. I also bear that mottling; but I am of the sort whose flesh is opaque, and my bronze-green skin is of a texture somewhat harsher than I should choose. In her, how well I know it, the skin is soft and rich to the exploring hand.

Olaf Stapledon, Last men in London (1932)

grapes
2014-Sep-13, 10:05 AM
I was also thinking that you can't multiply by 6 the high jump record, as that is partially achieved by body contortions around the center of gravity. E.g, the high jump record is 2.45 meters (8.04 feet). The same athlete could not jump 48 feet high on the moon, even if unencumbered.

Yes, the energy calculations are based upon the the change in height of the center of mass. In a high jump, the center of mass starts high (4-5 feet), and doesn't even need to get higher than the bar in order for the body to clear it. The difference achieved is still around that 40 inch value you mention.


But if you multiply the *vertical* jump achievable by a terrestrial athlete (say 40 inches) by 6, that equals about 20 feet, which is still pretty high.
Should be able to get the high jump record to 24 feet then, right? :)

Nowhere Man
2014-Sep-13, 12:10 PM
Thread necromancy alert, four years, for an apparent non-sequitur.

Fred

Jeff Root
2014-Sep-13, 01:28 PM
It sequiturs for me just fine.

I have *got* to read 'Last and First Men'. I had it out of the
library once, but didn't get to it before I returned the book,
reading only 'Sirius' in that collection. Almost 40 years ago!

-- Jeff, in Minneapolis

Noclevername
2014-Sep-13, 03:21 PM
Thread necromancy alert, four years, for an apparent non-sequitur.

Fred

If you read the full post, it's a "yes-sequitur"; IE, it relates to the topic of the thread.

Barabino
2014-Sep-13, 03:31 PM
I have *got* to read 'Last and First Men'. I had it out of the
library once, but didn't get to it before I returned the book,
reading only 'Sirius' in that collection.



I'm happy I've sparked some interest in Olaf Stapledon... My favorite is the "Other men" civilization in "Star maker"... a parody of racial and class conflicts (based not on skin color but on flavour)... that's so 30's :-D

eburacum45
2014-Sep-13, 03:39 PM
It sequiturs for me just fine.

I have *got* to read 'Last and First Men'. I had it out of the
library once, but didn't get to it before I returned the book,
reading only 'Sirius' in that collection. Almost 40 years ago!

-- Jeff, in Minneapolis

Yes, I recommend it wholeheartedly. Starmaker is another book by Stapledon that is even more expansive in time and space, and is (more-or-less) connected to the Last Men saga. A few details in cosmology have changed since these books were written, but that seems to make very little difference to the whole.

Noclevername
2014-Sep-13, 03:42 PM
A quick search finds that both books are available free online at the Australian Project Gutenberg site:

http://gutenberg.net.au/ebooks06/0601101h.html (Last And First Men)
http://gutenberg.net.au/ebooks06/0601841.txt (Star Maker)

eburacum45
2014-Sep-13, 03:48 PM
As far as jumping on the Moon is concerned, I would be more concerned about somersaulting by accident. On Earth we can keep quite good control of our balance while jumping, but a leap on the Moon - especially inside an air-conditioned gym in a hab sopmewhere, with no bulky spacesuit to weigh you down - would last much longer. If your feet and head are moving at different speeds, this could easily develop into a forward roll in mid air or worse. Presumably you could compensate for this with practice, but you might end up landing on your back (or your head) the first few times you tried it. Landing awkwardly on your neck is not recommended.

DaveC426913
2014-Sep-13, 05:27 PM
True, but I don't think the jumping would add anything to it because the max impact would still be the same as the initial jump.

The acceleration upon landing is much greater in magnitude (despite being opposite in direction) because it will not be linear, meaning while some portions are lower deceleration, other portions will be higher deceleration.

When taking off, you have fine control to spread the acceleration evenly over the entire lift off. It is much more difficult to spread the deceleration evenly when coming down.

Bones break, not because of an average acceleration, but because of a peak.

Nowhere Man
2014-Sep-13, 11:23 PM
If you read the full post, it's a "yes-sequitur"; IE, it relates to the topic of the thread.
All right, all right, but I'll stand by the "Thread necromancy alert, four years" part.

Fred

Jeff Root
2014-Sep-14, 02:45 AM
When taking off, you have fine control to spread the
acceleration evenly over the entire lift off. It is much
more difficult to spread the deceleration evenly when
coming down.
That's it! I just *knew* there was something wrong with
the assertion that it is the same on takeoff and landing.

The forces are limited to what your muscles can do when
taking off, but not when landing. You don't always have
time to adapt to sudden, unexpected changes in force,
especially if you can't see the ground under you.

-- Jeff, in Minneapolis

Jeff Root
2014-Sep-14, 02:50 AM
All right, all right, but I'll stand by the "Thread necromancy
alert, four years" part.
Did you take into account the time dilation due to the
lower/higher gravity, as well as all the jumping about?

-- Jeff, in Minneapolis

Barabino
2014-Sep-14, 01:36 PM
http://www.newscientist.com/article/dn26192-low-gravity-makes-astronauts-prone-to-falling-over.html

swampyankee
2014-Sep-14, 02:16 PM
Did you take into account the time dilation due to the
lower/higher gravity, as well as all the jumping about?

-- Jeff, in Minneapolis

If time dilation due to gravity is an issue, people would be puddles; they'd not be jumping.

NEOWatcher
2014-Sep-15, 12:14 PM
When taking off, you have fine control to spread the acceleration evenly over the entire lift off. It is much more difficult to spread the deceleration evenly when coming down.
Please explain that.
Once your feet leave the ground, you are traveling upward at a certain speed. How can the speed be any different when your feet hit the ground.
In between is just non-linear ballistic trajectory.


Bones break, not because of an average acceleration, but because of a peak.
I never said anything about "average". You even quoted me as say "max".

Hornblower
2014-Sep-15, 12:46 PM
Please explain that.
Once your feet leave the ground, you are traveling upward at a certain speed. How can the speed be any different when your feet hit the ground.
In between is just non-linear ballistic trajectory.


I never said anything about "average". You even quoted me as say "max".In principle, the leg action is exactly reversible for achieving a soft landing. The challenge is in getting the repetitions needed to condition the reflexes without injuring yourself on the initial attempts. If you mistakenly come down on your heels with your knees locked, it will hurt. I was just now doing some practice jumps to a height of about 6 inches and I could control the action of my ankles and knees to make a soft landing. Gymnasts and paratroopers with proper training learn to do it from much greater heights.

NEOWatcher
2014-Sep-15, 01:21 PM
In principle, the leg action is exactly reversible for achieving a soft landing. The challenge is in getting the repetitions needed to condition the reflexes without injuring yourself on the initial attempts. If you mistakenly come down on your heels with your knees locked, it will hurt. I was just now doing some practice jumps to a height of about 6 inches and I could control the action of my ankles and knees to make a soft landing. Gymnasts and paratroopers with proper training learn to do it from much greater heights.
Yes; but that would be true at any G force.

The only difference would be any horizontal movement. It wouldn't be in scale with what we normally experience, so I can see problems in misjudging the landing. But; that doesn't have to do with the issues of vertical speed.

DaveC426913
2014-Sep-15, 02:20 PM
Please explain that.
Once your feet leave the ground, you are traveling upward at a certain speed. How can the speed be any different when your feet hit the ground.
In between is just non-linear ballistic trajectory.

I am talking about the pressure your legs feel as you accelerate while still in contact with the ground.

During your initial jump, you will consciously or unconsciously make the push as uniform as possible.
But during the landing, it is extremely difficult to keep it anywhere near as uniform. For one, it takes a moment to acknowledge that your feet have made contact; in that moment, you're not applying your muscles to decelerate. That results in a correspondingly higher deceleration for the remainder of your landing.

From here:
https://physics242.wordpress.com/tag/momentum/

https://physics242.files.wordpress.com/2013/10/weight-graph.jpg

Lift off is smooth and maxes out at ~450N, whereas landing maxes out at ~950N+.




I never said anything about "average".
I didn't say you did.

DaveC426913
2014-Sep-15, 02:25 PM
In principle, the leg action is exactly reversible for achieving a soft landing. The challenge is in getting the repetitions needed to condition the reflexes without injuring yourself on the initial attempts. If you mistakenly come down on your heels with your knees locked, it will hurt. I was just now doing some practice jumps to a height of about 6 inches and I could control the action of my ankles and knees to make a soft landing. Gymnasts and paratroopers with proper training learn to do it from much greater heights.
As you say: in principle.

The initial question was:
Would landings in higher g's result in more injuries?
This was followed up with the claim that
Take-offs exert the same gs as landings, so no.
We are refuting this second claim.

True, In a controlled experiment (using a spherical chicken of uniform density in a vacuum), g's should be the same. But in reality, would there be more injuries upon landing? Yes.

profloater
2014-Sep-15, 02:26 PM
Yes; but that would be true at any G force.

The only difference would be any horizontal movement. It wouldn't be in scale with what we normally experience, so I can see problems in misjudging the landing. But; that doesn't have to do with the issues of vertical speed.

When landing both the muscles and the tendons will stretch elastically to absorb energy, this is different from muscle action to jump. This property is exploited in minimum energy gait walking which is very efficient at 1 g. The training to land is mostly about keeping the joints flexed to avoid bone lock up which can break bones for even a small fall.

It is clear one can fall succesfully from a far greater height than one can jump up, and the elastic properties are inherent to make this possible. Once exceeded tendons snap and this can take a long time to heal because of low blood supply.

NEOWatcher
2014-Sep-15, 02:31 PM
I am talking about the pressure your legs feel as you accelerate while still in contact with the ground.
Ok; that made it clearer. Although; I'm not sure how much of a difference that would make.[/QUOTE]


I didn't say you did.
Because I didn't understand what you were saying... It came across to me as something different. Sorry.

NEOWatcher
2014-Sep-15, 02:33 PM
When landing both the muscles and the tendons will stretch elastically to absorb energy, this is different from muscle action to jump. This property is exploited in minimum energy gait walking which is very efficient at 1 g. The training to land is mostly about keeping the joints flexed to avoid bone lock up which can break bones for even a small fall.

It is clear one can fall succesfully from a far greater height than one can jump up, and the elastic properties are inherent to make this possible. Once exceeded tendons snap and this can take a long time to heal because of low blood supply.
But; what does that have to do with my comment?

DaveC426913
2014-Sep-15, 02:41 PM
Ok; that made it clearer. Although; I'm not sure how much of a difference that would make.
Well, we're talking about the likelihood of injury. That's going to occur as a direct result of max acceleration.

For a random sampling of jumps on our massive planets, whose initital accelerations will span a range up to the point of breaking a bone, a bunch of those will exceed the limit of gs on landing. So, at sick bay many more injuries from landing than from take off.




Because I didn't understand what you were saying... It came across to me as something different. Sorry.
No, I can see why it came across that way. It looked like it was a direct response-to/refutation-of your words, but it wasn't intended to be.

NEOWatcher
2014-Sep-15, 02:45 PM
For a random sampling of jumps on our massive planets, whose initital accelerations will span a range up to the point of breaking a bone, a bunch of those will exceed the limit of gs on landing. So, at sick bay many more injuries from landing than from take off.
We don't have that random sampling yet, so there's no way to quantify it yet. That's why I leave it as an unknown.

DaveC426913
2014-Sep-15, 03:06 PM
We don't have that random sampling yet, so there's no way to quantify it yet. That's why I leave it as an unknown.
The sampling will show differences only in scale. The curve will be qualitatively the same.

Graph in post 38 shows a "typical" jump. It is not unreasonable to assume that almost all jumps would look similar in the one relevant property: initial acceleration is going to be smoother than final deceleration, making final peak acceleration higher than initial peak acceleration.

I grant that there could, in principle, be a few expert jumpers who could have a final peak acceleration only slightly above initial peak acceleration. But I do not grant that even a single jumper would have a final peak acceleration lower than initial peak acceleration.


This means that the likelihood of injury on landing will always be significantly more than injury upon take-off.

profloater
2014-Sep-15, 03:17 PM
But; what does that have to do with my comment?OK not just your comment but previous ones too, the vertical jump was answered at the beginning, the issue of landing is a little different because it is not symmetrical with take off.

DaveC426913
2014-Sep-15, 03:29 PM
It is clear one can fall succesfully from a far greater height than one can jump up, and the elastic properties are inherent to make this possible. Once exceeded tendons snap and this can take a long time to heal because of low blood supply.
While it's true that one can fall successfully from a great height than one can jump, I'm not sure if that applies here.

A brick can fall successfully from a greater height than it can jump too.

I'm not poking fun here. The reason anything can fall successfully is only partly to do with its elastic properties in absorbing a landing, and greatly to do with its inert tensile strength. i.e. the bones in your body, while the impact will softened by some degree due to muscles, still take a beating, even if they don't break.

NEOWatcher
2014-Sep-15, 05:06 PM
Graph in post 38 shows a "typical" jump....
That wasn't there when I responded. Judging from the times, you must have modified it after I started to compose my response.



It is not unreasonable to assume that almost all jumps would look similar in the one relevant property: initial acceleration is going to be smoother than final deceleration, making final peak acceleration higher than initial peak acceleration.
But; you keep comparing acceleration against the deceleration. I can understand that part. I am talking about deceleration at 1G vs others.

I don't see how the deceleration speed can be appreciably different at 1G, 2G or whatever if the takeoff speed is going to be essentially the same at 1G, 2g, etc.

WayneFrancis
2014-Sep-16, 04:01 AM
I'll point out that in 2G jumping should be just as hazardous as jumping at 1G and here is why. Given the same person the jumping in both environments the jump and landing will pretty much be the same. The 2G environment will result in a much shorter jump. There will be differences but not that much. I'd find it VERY difficult, read that as impossible, to jump from a standing position and land as to injure myself purely because I don't have the power or technique to get enough force where even landing straight legged any real damage could be done. Increasing the gravity to 2G just means my jump is going to be shorter. The total energy wouldn't change and I bet the peak would also be very similar. If we start to introduce lateral movement the injury on landing will go up but again it would be very similar in both environments. So the change in Gs isn't the biggest factor. The klutz effect would be the deciding factor in if I got injured.
Funny story, when I was active duty in the USMC I was in an exercise where my platoon was assigned to ambush the battalion as they patrolled a wooded area. I was a M60E3 A gunner at the time. Our 3 M60 teams had "intel" where the patrol was going to be so we set up on a ridge overlooking the patrol route. At the bottom of the ridge there was a river giving us an additional barrier from assault. Essentially the battalion was in for a rude awakening since our cover and concealment was optimal and their ability to get to us was very limited. The rest of our platoon was set up to contain the battalion in the kill zone or protect the 60 teams. All was good until we get a radio message from our scout fire team that the battalion was actually patrolling one ridge line over. So we immediately picked up and started the about 1km run to the next ridge line. We bolted down the hill and everyone just went to leap the river gully. With the speed obtained from the run down the hill the leap looked doable even for me and surprisingly I did make that long lateral jump. The landing was a bit hard but the only adverse effect was with my 60. I was running and carrying it, with the assistance of the sling over my neck and shoulder, holding the stock and the barrel. The landing had the 60 continue forward until the sling pulled tight and it essentially bounced back straight towards my face. Luckly I tilted my head down and it slammed into my helmet. I'm pretty sure this was before I went to jump school so I wouldn't have been really schooled in soft landings that I know now. I'm sure I could have injured myself there and I was probably lucky. But all 32 of us made that jump. For those curious we got to the next ridge line just minutes before the battalion came through and it ended with minimal losses from our side. We then spent the next few days teaching the battalion some advanced tactics to deal with those types of ambushes and what other things they might want to keep an eye out for. We also did a debriefing with the entire battalion on the improvisation we needed to do when the mission changed including the jump across that gully. We really should have had a better plan where that jump would not have had to be such an option like prepping 3 areas on the far side of the river bank to climb up more easily.

That said introducing the lateral component changes the dynamic a lot. It is like the bad argument about bicycle helmets trying to claim you shouldn't need them because falling from a bike should be the same as falling from a standing position because your head has the same distance to fall all while ignoring the lateral speed and external object you deal with during a bike crash that you probably don't while standing.

Jens
2014-Sep-16, 10:14 AM
Just a clarification, but the rate of injury is higher on landing than takeoff regardless of how strong the gravity is, right? It certainly seems to be true on earth.

And another comment: the chance of breaking a bone increases with gravity even if you are standing still, so wouldn't jumping also be more dangerous?

NEOWatcher
2014-Sep-16, 12:28 PM
Just a clarification, but the rate of injury is higher on landing than takeoff regardless of how strong the gravity is, right? It certainly seems to be true on earth.
Like they say, it's not the fall that hurts, but the landing. ;)


And another comment: the chance of breaking a bone increases with gravity even if you are standing still, so wouldn't jumping also be more dangerous?
Actually, that sparked another thought.
With all the bantering we've been trying to do to hash it out. This one actually makes sense to me. The force at the time of "touch down" will be the same no matter what the gravity, but now you are trying to stop twice the weight in the same amount of time and distance.

DaveC426913
2014-Sep-16, 02:20 PM
That wasn't there when I responded. Judging from the times, you must have modified it after I started to compose my response.
Yes, I found it moments afterward. It's always a debate for me whether to edit the existing post or add yet one more post onto the thread.





I don't see how the deceleration speed can be appreciably different at 1G, 2G or whatever if the takeoff speed is going to be essentially the same at 1G, 2g, etc.

An object resting at 2g is already under more strain than at 1g - putting it that much closer to breaking, before even considering movement.

At, say, 5g, you could be right on the threshold of breaking a bone just by pushing off. So, any (typical) jump will result in about twice the force when coming down (as we see in the graph). Any jump in 5g might break your legs on lift-off, but will always break your legs on landing.

NEOWatcher
2014-Sep-16, 03:31 PM
An object resting at 2g is already ...
See my post above. Sometimes it's just the simple thoughts that help. And your response this time would have done it too.

DaveC426913
2014-Sep-16, 05:23 PM
I'll point out that in 2G jumping should be just as hazardous as jumping at 1G and here is why.
just as hazardous? I think we're demonstrating that it will definitely be more hazardous.





Given the same person the jumping in both environments the jump and landing will pretty much be the same.
Do you mean
"...the jump and landing of 1g as compared to 2g will be the same...",
or do you mean
"...the jump as compared to the landing will be the same..."

I'm arguing that the landing will incur a significantly larger peak force. See the chart in post #38. Max force on take off is ~450N; max force on landing is 950N+.

DaveC426913
2014-Sep-16, 05:31 PM
It is like the bad argument about bicycle helmets trying to claim you shouldn't need them because falling from a bike should be the same as falling from a standing position because your head has the same distance to fall
Who argues this?

What a dumb argument. I'd like to meet the person making this argument.

I'd wait till he's standing, unsuspecting, and push him over and watch him crumple and land on his feet/butt.

Then, I'd put him on a (stationary) bike, and push him over. His hands might go out, but his legs will be useless - his head will be the next thing to hit the ground.

Then I'll run, before they arrest me for attempted murder.

Jens
2014-Sep-17, 01:09 AM
Who argues this?

What a dumb argument. I'd like to meet the person making this argument.


Unfortunately, you can't meet him. He had a fall off his bicycle and is still comatose. :)

WayneFrancis
2014-Sep-17, 01:24 AM
Like they say, it's not the fall that hurts, but the landing. ;)


Actually, that sparked another thought.
With all the bantering we've been trying to do to hash it out. This one actually makes sense to me. The force at the time of "touch down" will be the same no matter what the gravity, but now you are trying to stop twice the weight in the same amount of time and distance.

I agree you are trying to stop 2x the weight but the weight isn't the weight isn't the problem it is the force and the force will remain the same.
f=ma
a=\frac{dv}{dt}
dv will be lower. I'd wager dt will be higher resulting in a much lower a
so now we have to think about why dv will be lower
F=G\frac{m_{1}m_{2}}{r^{2}}

Before we can "jump" we have to exceed the F of gravity otherwise we are just doing squats. Landing we do have a greater force over all because of the higher gravity but the extra force from the jump is much less. Or am I just getting this wrong?
I do agree with Jen but would point out the the more dangerous comes just from the extra gravity and less from the jump.

WayneFrancis
2014-Sep-17, 01:50 AM
Yes, I found it moments afterward. It's always a debate for me whether to edit the existing post or add yet one more post onto the thread.




An object resting at 2g is already under more strain than at 1g - putting it that much closer to breaking, before even considering movement.

At, say, 5g, you could be right on the threshold of breaking a bone just by pushing off. So, any (typical) jump will result in about twice the force when coming down (as we see in the graph). Any jump in 5g might break your legs on lift-off, but will always break your legs on landing.

So you are saying that the moment before contact that F>\frac{m_{1}m_{2}}{r^{2}} which I agree with. But I disagree that you have to have the situation you propose. If you are on the threshold then any jump would break your legs because to jump you would have to exceed the threshold that is keeping you on the ground in the first place. I agree that landings are harder then take offs but we can in theory create machines that can land with a peak load the same as the take off. Just like I can land with less injury to myself then my take off if I accept that my landing will be asymmetric to my jump. Ie for a jump I'll bend a bit and straighten my legs rapidly. On landing I can increase the distance and time I land by more fully collapsing and going into a roll. Essentially this is how I'd land with a T-10c parachute.

Is it easy to do? No! Can it be done and done by humans? Yes.

WayneFrancis
2014-Sep-17, 01:52 AM
Who argues this?

What a dumb argument. I'd like to meet the person making this argument.

I'd wait till he's standing, unsuspecting, and push him over and watch him crumple and land on his feet/butt.

Then, I'd put him on a (stationary) bike, and push him over. His hands might go out, but his legs will be useless - his head will be the next thing to hit the ground.

Then I'll run, before they arrest me for attempted murder.

Oh there are lots of people that make the argument, as bad as it is, that bicycle helmets don't help. I call these kinds of people "stupid"

DaveC426913
2014-Sep-17, 01:34 PM
But I disagree that you have to have the situation you propose. If you are on the threshold then any jump would break your legs because to jump you would have to exceed the threshold that is keeping you on the ground in the first place.
I did not say on the threshold before pushing off, I said on the threshold by pushing off. i.e. the peak acceleration is the threshold.


... but we can in theory create machines that can land with a peak load the same as the take off.
Yes we can. That's not what he discussion is about. It's about injuries. Implicitly human injuries.


Is it easy to do? No! Can it be done and done by humans? Yes.
Is it easy to do? No! Can it be done and done by humans? Maybe. With practice, and enough margin for error.

Still, the question was not 'is it possible', the question was 'will there be more injuries'. The answer to that - for any high-g colony not entirely inhabited by mechanized humans - is definitely 'yes'.

NEOWatcher
2014-Sep-17, 04:49 PM
I agree you are trying to stop 2x the weight but the weight isn't the weight isn't the problem it is the force and the force will remain the same.
f=ma
a=\frac{dv}{dt}
dv will be lower. I'd wager dt will be higher resulting in a much lower a
so now we have to think about why dv will be lower
F=G\frac{m_{1}m_{2}}{r^{2}}

Before we can "jump" we have to exceed the F of gravity otherwise we are just doing squats. Landing we do have a greater force over all because of the higher gravity but the extra force from the jump is much less. Or am I just getting this wrong?
I do agree with Jen but would point out the the more dangerous comes just from the extra gravity and less from the jump.
I think the confusion might be mixing velocity with acceleration.
The velocity at landing is similar to either environment, and is roughly the same at take-off.

But; the acceleration is twice as high in the 2G environment. So; as you go from feet hitting the ground, down to a complete stop of the legs, time is taking place. The velocity is changing to go to a stop, and the acceleration over that time of the gravitational pull is different.

WayneFrancis
2014-Sep-18, 04:48 AM
...

Still, the question was not 'is it possible', the question was 'will there be more injuries'. The answer to that - for any high-g colony not entirely inhabited by mechanized humans - is definitely 'yes'.

to which I'll be OCD and say not with training. But I fully agree with you. Just like falling down now can cause me to break my wrists or even just skin my knee. Increasing the forces will increase the resulting injuries.

WayneFrancis
2014-Sep-18, 07:43 AM
I think the confusion might be mixing velocity with acceleration.
The velocity at landing is similar to either environment, and is roughly the same at take-off.

But; the acceleration is twice as high in the 2G environment. So; as you go from feet hitting the ground, down to a complete stop of the legs, time is taking place. The velocity is changing to go to a stop, and the acceleration over that time of the gravitational pull is different.

Isn't the acceleration in a 2G environment going to be lower? Because it will take 2x the power to cover the same distance but we are not scaling the power with the gravity. Simply put if I jump in a 2G environment my jump will be a lot shorter. My upward velocity will be less thus my down ward velocity will be less. I'm trying to understand what you are saying but I think I'm missing something.

swampyankee
2014-Sep-18, 10:27 AM
I think most people calculated on the basis the jumpers would jump up to the same energy, essentially weight times height, which would mean the landing velocities would be the same, regardless of the local gravity.

My somewhat different basis, that a given person's jump is from the total force that their legs can produce minus their weight, actually results in lower takeoff and landing velocities at higher "g," which would mean that landing speed would decrease at higher "g". Counter-intuitively, this means a person would be more likely to be hurt jumping from the ground in low gravity. To clarify: say a person can produce three times their body weight in force from both legs (I can do about 2.8 times mine). In 1 "g," that means that person has twice their body weight in force from their legs to produce vertical acceleration on Earth, but the same person can use only 1 times their body weight for a vertical jump in 2g. Less force for vertical acceleration, same distance to apply it (the legs won't be longer), and the vertical velocity will be less.

NEOWatcher
2014-Sep-18, 11:43 AM
Isn't the acceleration in a 2G environment going to be lower? Because it will take 2x the power to cover the same distance but we are not scaling the power with the gravity. Simply put if I jump in a 2G environment my jump will be a lot shorter. My upward velocity will be less thus my down ward velocity will be less. I'm trying to understand what you are saying but I think I'm missing something.
Your jump height will be lower, but your takeoff speed will be the same. Therefore; the acceleration from take-off to peak takes much less time to occur, which means a higher acceleration. Same for the descent.
-or- You are going from a fixed velocity to 0 in a shorter distance, thus a faster acceleration.

WayneFrancis
2014-Sep-22, 12:21 AM
Your jump height will be lower, but your takeoff speed will be the same. Therefore; the acceleration from take-off to peak takes much less time to occur, which means a higher acceleration. Same for the descent.
-or- You are going from a fixed velocity to 0 in a shorter distance, thus a faster acceleration.

That makes zero sense to me. If I hole 110kg of weight and try to jump there is no way my take off speed will be the same. I'd imagine that I'd struggle to even be able to straighten my legs properly.
a=\frac{\Delta v}{t} I can't see v being higher and I'd imaging that t would be longer.

For example if increase the amount of weight I try to bench press it takes me longer for each repetition. So if either v is lower or t is higher a has to be lower.

swampyankee
2014-Sep-22, 01:32 AM
That makes zero sense to me. If I hole 110kg of weight and try to jump there is no way my take off speed will be the same. I'd imagine that I'd struggle to even be able to straighten my legs properly.
a=\frac{\Delta v}{t} I can't see v being higher and I'd imaging that t would be longer.

For example if increase the amount of weight I try to bench press it takes me longer for each repetition. So if either v is lower or t is higher a has to be lower.

This is the basis of my calculation: the force that is available for the vertical jump is the maximum force that can be produced minus the person's weight. Using me as an example:

My mass is 80 kg
My weight is 780 N
I can produce a maximum of about 2400 N with my legs.
Here, on Earth, that means I have 1620 N available to accelerate myself upwards on Earth. On the Moon, where my weight would be about 125 N, I would have 2275 N available for jumping. On a planet with 2 g surface acceleration, my weight would be 1560 N and only 840 N would be available.
Since the distance I can apply that force is the same everywhere, the kinetic energy I can generate decreases as g increases.

WayneFrancis
2014-Sep-22, 06:29 AM
I think I'm on the same page as you swampyankee but it seems opposite to what NEO is saying or I'm just not getting it at all