Why a man on the Moon can jump 21 times higher than on the Earth

[A.T.]

Do you observe the guy seemed to be overestimating the height of his jump? It looked like 3 feet at most to me, but he thought it was 4 feet.

Well, that's how you know it was just a TV-studio.

 

[sgphysics]

An experimental way to estimate it could be an inclined rowing machine, such that the person pushes off with legs against 1/6 of its weight (instead of the string).

A good suggestion. Some of those machines have the possibility to add weights that are moved upward when you row. I propose to use an accelerometer, which are in most phones today, to measure the acceleration achieved as function of added mass.

 

[5P@N]

On the moon less energy is expended pushing the body up through the acceleration zone,due to reduced gravity, and this the energy which increases the velocity over the Earth jump.

on the Earth m x g x s ...on moon m x g/6 x s ... so the energy available for the velocity increase = m x g5/6 x s (s = squat length)

I would like to propose the following method so as to answer how much higher a person could jump on the moon in the absence of any conclusive model or experiments...
But before I do so, I would like to initially get a more precise conversion factor for the moon's gravity in relation to the Earth.
So I take the moon's gravity $(1.622_{\frac{m}{s^2}})$, and divide it by that of Earth $(9.807_{\frac{m}{s^2}})$, to get: $.16539206689$. Granted the difference is less than one percent, but it makes me happy.
Now for my glorious method...
I Calculate what the vertical displacement of the subject was from the crouching position. There is the vagary of attempting to determine how tall the subject is when he or she is crouched furthest down...however, I have found this image taken originally from Harvard:

...and I'll use the first figure's height as my basis for calculation. This subject appears to be unable to crouch any further, which is what we want. Unfortunately, we have no way of knowing how tall the model is when standing (which is necessary to serve as a basis of comparison with a subject of a different height). So I visited the http://www.fas.harvard.edu/~loebinfo/loebinfo/Proportions/humanfigure.html wherein this figure originated, and found the height of it while standing to be 5 feet 9 inches or 1.7526 meters. The crouching height is as you can see 3 feet 7 inches or 1.0922 meters. When subtracted from the standing height, this gives a distance of: .6604 m. I'll assume that the energy necessary to raise the subject from the crouching to the standing height is the same energy as is required for vertical displacement, minus the lower legs and feet and knees, which aren't displaced significantly (I'm assuming that my jumping subjects will have their legs bent like the figure's left leg, which is most natural for a vertical jump). I found a document online entitled: ”Weight, Volume, and Centers of Mass of Segments of the Human Body”, put out by the US Air Force, and on pg. 13 of the pdf I came across a table that mentions various matter-of-fact weight measurements for just about any butcher's cut you please, including the ones we're after. With a "calf + foot, right" weight of 3.970 kg (which is meta-averaged from various sources), we just double this number to figure out how many kg need to be subtracted in the energy calculation to follow. And not to forget the knees: from this site I discovered that knees themselves weigh 161.71 g for men. Double and subtract them as well. Note that I didn't take away some from the thigh on account of its rotation and not vertical displacement (which would have involved trigonometric calculations of which I'm rusty with - if someone wanted to help with that, it would be appreciated).

Taking a hypothetical subject weighing 61.14 kg, I then subtract the aforelisted anatomical components, to get the total mass which will be vertically displaced:
$$ 61.14_{\text{kg}} - .32342_{\text{kg(knees)}} - 7.940_{\text{kg(calf & foot)}} = 52.87658_{\text{kg}}$$
Then multiplying this mass with the gravity on Earth and the aforementioned vertical displacement, we get the total energy necessary to bring the subject up to a standing position on Earth during the course of his jump:
$$ (52.87658_{\text{kg}})(9.807_{\frac{m}{s^2}})(.6604_m) = 342.457433488_{\text{Joules}}$$
Taking this number and multiplying it by the quotient of lunar gravity over Earth's gravity already arrived at, I get:
56.6397427464 Joules. This is how much energy would be necessary on the moon to perform the same act of standing during the jump. I subtract this from the calculated total number of Joules previously arrived at, and get: 285.817690742 Joules. This is the total amount of energy in the act of standing during the jump which is available on the moon for conversion into vertical displacement. How far could the subject go with this much energy on the moon? Using the simple formula of U=m*g*height, I algebraically isolate height, plug in the relevant values and solve:
$$\text{height} = \frac{285.817690742_J}{(52.87658_{\text{kg}})(1.622_{\frac{m}{s^2}})} = 3.33253637485_m$$
Now bear in mind that this is only the component involved in the standing action portion of the jump. It's a substantial distance!

 

[litup]

You could test all this on Earth: using an elevator going from top floor to bottom, accelerate the elevator downwards at 5/6 g, leaving 1/6 g for someone say, on the top of the elevator where there would be plenty of room for the jumper to go vertical, say from the top floor of a 100 story building or whatever the max limit is for the number of floors an elevator can go in one flight. That would be the elevator accelerating downwards at 26.666 f/sec^2, so for a fall of 333 feet you would get a flight time of 5 seconds. Maybe enough to perform the jump test. That's all the further I took the math. Anyone else see a problem with this? So 6 seconds=479 ft, 7 sec=653 ft, 8 sec=853 feet needed. Can an elevator do 853 feet in one trip?

 

[5P@N]

Just make sure the elevator has a high ceiling...this is a bad way to get down with headbangers (ouch)!

 

[sophiecentaur]

There are so many unknowns in this problem - too many for a good estimage, I think. But it would be easy enough to simulate a zero g situation by having a 'jumper' working horizontally on a trolley or on ice. That would establish the maximum velocity achievable with his or her particular muscle / skeleton proportions. The limitations on speed will be imposed by the 'gearing' and velocity ratio of all the joints. The takeoff velocity could then be used to work out a trajectory. You could modify it by usin an inclined plane or a suitable weight on a pulley to simulate Moon weight. I would imagine the '21 times' figure would be a wild overestimate without using some arrangement for 'gearing' the athlete and matching the muscles to the task.

 

[oz93666]

.Pounds is a unit of mass, it is independent of the position.

Not true. The Slug is the unit of mass ... It exerts a force of 1lb in standard Earth gravity... this mass of 1 slug on the moon 'weighs' 1/6 ponds.

Although I'm British , I hate the imperial system, so messy and hard work , metric is so much better.

 

[jbriggs444]

Not true. The Slug is the unit of mass ... It exerts a force of 1lb in standard Earth gravity... this mass of 1 slug on the moon 'weighs' 1/6 ponds.

Although I'm British , I hate the imperial system, so messy and hard work , metric is so much better.

There are multiple systems of measurement using pound as the name of a unit of measurement. It is not a uniquely a unit of force in all of them.

There is the foot/pound/poundal/second scheme in which the pound is a unit of mass and the poundal (one pound foot per second squared) is a unit of force.

There is the foot/slug/pound/second scheme in which the slug is a unit of mass and the pound (one slug foot per second squared) is a unit of force.

There is the gravitational foot/pound/pound/second scheme in which the pound is a unit of mass the pound-force (one gee-pound) is a unit of force. This is the customary system of units in the U.S. For purposes of commerce, when the "pound" is used as a unit of measurement for the "weight" of goods bought and sold, it is legally defined as a unit of mass and the weight is a mass measurement.

Schools in the U.S. (at least when I attended) went to great pains to teach that the pound is a unit of force and that weight refers to gravitational attraction. The truth is more nuanced than that. Words do not always have a single meaning.

 

[sophiecentaur]

There are multiple systems of measurement using pound as the name of a unit of measurement. It is not a uniquely a unit of force in all of them.

There is the foot/pound/poundal/second scheme in which the pound is a unit of mass and the poundal (one pound foot per second squared) is a unit of force.

There is the foot/slug/pound/second scheme in which the slug is a unit of mass and the pound (one slug foot per second squared) is a unit of force.

There is the gravitational foot/pound/pound/second scheme in which the pound is a unit of mass the pound-force (one gee-pound) is a unit of force. This is the customary system of units in the U.S. For purposes of commerce, when the "pound" is used as a unit of measurement for the "weight" of goods bought and sold, it is legally defined as a unit of mass and the weight is a mass measurement.

Schools in the U.S. (at least when I attended) went to great pains to teach that the pound is a unit of force and that weight refers to gravitational attraction. The truth is more nuanced than that. Words do not always have a single meaning.

I think you just proved that the fps system(s) are just total confusion. SI avoids many of those problems although it isn't without a few.

 

[oz93666]

There are multiple systems of measurement using pound as the name of a unit of measurement. It is not a uniquely a unit of force in all of them.

There is the foot/pound/poundal/second scheme in which the pound is a unit of mass and the poundal (one pound foot per second squared) is a unit of force.

There is the foot/slug/pound/second scheme in which the slug is a unit of mass and the pound (one slug foot per second squared) is a unit of force.

There is the gravitational foot/pound/pound/second scheme in which the pound is a unit of mass the pound-force (one gee-pound) is a unit of force. This is the customary system of units in the U.S. For purposes of commerce, when the "pound" is used as a unit of measurement for the "weight" of goods bought and sold, it is legally defined as a unit of mass and the weight is a mass measurement.

Schools in the U.S. (at least when I attended) went to great pains to teach that the pound is a unit of force and that weight refers to gravitational attraction. The truth is more nuanced than that. Words do not always have a single meaning.

Well done, I stand corrected ... I had no idea things were as insane as all that ! Of course the pound is also the UK unit of currency , I think it goes back to when it was worth a pound of gold... Please petition politicians to change things to the metric system to protect future generations from this madness.

 

[sophiecentaur]

, I think it goes back to when it was worth a pound of gold.

HAha. I know the Pound Stirling (GBP) has been doing quite well recently and at times in the past but a pound of gold? At the present time, scrap gold (high carrat) is about £20.00 per gram. The Pound (livre) was originally based on a pound (troy) of silver and that's bought for scrap at about £0.25 per gram at the present time.

 

[Art Vanderlay]

The answer is more like a ratio of 10. Here's the analysis:

Equate k.e. at take-off point with p.e.:

1/2 mv^2 = mgh => h = v^2/2g

where m is the mass, h is the peak height.

Assume force applied by legs F is constant and the same in both gravities (will come back to this). Let d me the crouching distance (i.e. the distance over which the upwards force will be applied) and t be the time the force is applied

F - mg = ma

a = v/t and d = at^2 / 2. Using these and the formula for h above, you get a = gh/d (there's probably a quicker way to get to that result)

So F - mg = mgh/d

let F = kmg (i.e. F as the ratio of the weight of the person). Then

(k - 1) = h/d

On the moon, let the gravity g' be rg and h' be the peak height. So on the moon:

(k - r) = rh'/d

so h'/h = (k-r)/r(k-1)

Since k = 1 + h/d (above),

h'/h = (1 + h/d - r)/(rh/d)

Putting in numbers, assume a jump height on Earth of .5m and a crouch distance of .4m and r is about 1/6, then:

h'/h = (1 + 5/4 - 1/6)/(1/6 * 5/4) = 10

The crouch distance, d is going to be between .3 and .5, so at these extremes (but assuming a jump height still of .5 you get 9 and 11 respectively for the ratio of heights.

Coming back to the assumption of the force, let's check the time to execute the jump implied by these results. From d = 1/2 at^2 and a = gh/d:

t^2 = 2d^2 / gh

So assuming d = .4 and h = .5, t = .26s, which seems reasonable. In the reduced gravity situation, the corresponding figure is this divided by sqrt(rh'/h), which is .2s. I would assume therefore that the speed limit of contracting the muscles is not being reached, since we are only asking them to contract slightly faster. If speed of contraction is actually a limiting factor, then the effect would be a reduction in the force applied and therefore slightly reduce the height ratio.

The additional assumption about the force is that it is constant and (implicitly) that the way it is created (the physiology) is not affected by the reduced gravity. That would have to be verified experimentally, but my guess is that it has little effect.

 

[sophiecentaur]

I would assume therefore that the speed limit of contracting the muscles is not being reached,

It would be easy to estimate this effect by seeing how far you can launch a mass of man/6 vertically when laying on your back. This would give a launch speed. It would be useful to do this with a range of masses. I realize that both mg and ma are relevant. The ma could be examined in a similar way with masses on a pendulum and seeing what height they reach. You could find the relationship between ma and mg with this sort of experiment.
Not to denigrate the theoretical approach.

 

[A.T.]

I would assume therefore that the speed limit of contracting the muscles is not being reached, since we are only asking them to contract slightly faster. If speed of contraction is actually a limiting factor, then the effect would be a reduction in the force applied and therefore slightly reduce the height ratio.

Faster contraction speed always reduces the force, even if the speed limit isn't reached.

https://en.wikipedia.org/wiki/Muscle_contraction#Force-velocity_relationships

 

[Art Vanderlay]

Faster contraction speed always reduces the force, even if the speed limit isn't reached.

OK, that's good. To account for this, let's say the force is reduced to x of it's value on Earth, then h'/h = (x(1+h/d)-r)/(rh/d). Putting in some numbers, you get x=.9, h'/h = 8.9 and x = .8, h'/h = 7.8.

The other issue with the model of course is that the contraction force is not constant, so you actually want ∫F over the contraction period. The factor x above is more accurately the reduced impulse due to the different force function.

As a next approximation you could use the force/velocity relationship to introduce a velocity dependent resistive force, i.e. the EoM becomes F - mg - βv where β is effectively a drag coefficient. If you looked at standard drag theory (e.g. sinking object in a viscous fluid) then it should be simple to apply. But given that we don't know how the force varies kinematically due to the way the legs generate it (i.e. both the physiology and the machinery - there are at least 3 pivot points in the joints and the force is generated by effectively pulling cables to straighten the joins... non-trivial) it's probably too much model and too little data.

The main point is, assuming no other advantage for the legs in lower gravity it can't be more than 10 (for d=.4, h=.5) and it can be more than 6 (which is based on flawed analysis that ignores that fact that you are able to apply more net force due to less gravity).

 

[sophiecentaur]

Humans took a million years to get their leg mechanics right for the local g. When (not if) the colonisation of the Moon has produced a big enough enclosure to do the test for real, I think that the record height will be achieved using a lever system to match the load on the legs to something nearer weight on Earth. I'm not suggesting energy storage (trampolining would break the rules) but a simple impedance transformation. (No idea how it could be done, to extend all those levers in the leg.)

 

[Art Vanderlay]

Humans took a million years to get their leg mechanics right for the local g. When (not if) the colonisation of the Moon has produced a big enough enclosure to do the test for real, I think that the record height will be achieved using a lever system to match the load on the legs to something nearer weight on Earth. I'm not suggesting energy storage (trampolining would break the rules) but a simple impedance transformation. (No idea how it could be done, to extend all those levers in the leg.)

Just because the mechanics may be optional for Earth, it doesn't mean that it is not also optimal for lower g. I say may, because that's not how evolution works - it doesn't optimise everything, only the things that provide a selective advantage. That's why we aren't as strong as most of the primate family even though we are descended from a common ancestor.

That aside, here's a much quicker way of deriving the result (the first one was just writing down stuff as I thought of it!):

Work done in jumping phase = Fd (=kmg where k in units of weight of person)
mgd of this is used against gravity, so net remaining energy = (k-1)mgd
This is in the form of k.e. and converted to p.e., so mgh = (k-1)mgd => h = (k-1)d
On moon: g'h' = (k-g'/g)d
So h'/h = (k-r)/g(k-1)

The rest is the same as above

 

[sophiecentaur]

Just because the mechanics may be optional for Earth, it doesn't mean that it is not also optimal for lower g. I say may, because that's not how evolution works - it doesn't optimise everything, only the things that provide a selective advantage.

I would have said that efficient use of muscles in running and lifting could be a massive evolutionary advantage. But I have no direct evidence one way or another, of course. It's not the sort of thing that can be resolved by Physics. The "why"s about this sort of thing are impossible to be certain about but humans have been making use of tools for a long time and intelligent use of tools reduces the need for sheer strength, compared with the other apes. As the astronauts have shown, muscles tend to adopt the appropriate size to match demand. I guess that implies the need to do the moon experiment with recently arrived athletes.

 

[Art Vanderlay]

I would have said that efficient use of muscles in running and lifting could be a massive evolutionary advantage. But I have no direct evidence one way or another, of course. It's not the sort of thing that can be resolved by Physics. The "why"s about this sort of thing are impossible to be certain about but humans have been making use of tools for a long time and intelligent use of tools reduces the need for sheer strength, compared with the other apes. As the astronauts have shown, muscles tend to adopt the appropriate size to match demand. I guess that implies the need to do the moon experiment with recently arrived athletes.

Yep, exactly right, it isn't likely that there has been an advantage to further optimising the muscles since we have relied more heavily on intelligence. Also, the fact that there is no advantage means that there is no disadvantage to it becoming less efficient. It will do so due to natural drift and also if by becoming less efficient it adds another advantage, e.g. in our case, not having fuel-hungry large muscles enables us to survive with less food.

When we were developing in this direction I think it is highly likely we were evolving for speed and stamina rather than jump height.

 

[A.T.]

The "why"s about this sort of thing are impossible to be certain about but humans have been making use of tools for a long time and intelligent use of tools reduces the need for sheer strength, compared with the other apes.

Here a study about vertical jump performance of apes, suggesting that it requires muscle properties significantly different than those of human muscles.

http://rspb.royalsocietypublishing.org/content/273/1598/2177

 

[sophiecentaur]

Here a study about vertical jump performance of apes, suggesting that it requires muscle properties significantly different than those of human muscles.

http://rspb.royalsocietypublishing.org/content/273/1598/2177

That's interesting and appears to be a pretty thorough bit of investigation. I am surprised that I couldn't find a comment on the obvious difference between ape and early human lifestyle - humans were runners and not arboreal - probably some of the most effective running hunters ever and seemed to have managed to bring down massive prey by simply exhausting them by running them into the ground. I heard (unspecified radio programme) that it was assumed that both sexes would have needed the same abilities in order for the mums with children could be present at the kill in order to eat the stuff before other predators arrived to steal it.
But I am talking well above my pay grade on this topic, of course. (Standard PF practice, so no apology. )
On the topic of evolution and gravity, whatever the lifestyle of an Earth organism, it will have developed with a fixed value of g and, just as with all other abilities, there is every reason to suspect that some degree of optimisation of all abilities must have taken place. 'Nature' seems to do no more than absolutely necessary, when it comes to abilities. We really don't preform well outside a narrow range of temperatures (without clothes etc.), in non-standard proportions of atmospheric gases or in the presence of unfamiliar microbes. A local g of g/6 is not a trivial difference and I don't see how it can be assumed that we would be well adapted well to it.

 

[Art Vanderlay]

A local g of g/6 is not a trivial difference and I don't see how it can be assumed that we would be well adapted well to it.

Because there is nothing significant about the physics to suggest otherwise. Go and check out a gym and see people doing leg press (any g) or squat (greater g). there are many examples of where we would use our legs to push less than our own weight. A simple experiment on Earth would be to suspend someone from a tether providing a constant upwards force of 5/6g... that's an identical environment to the moon. I'd be surprised if this has not already been done, e.g. in preparing for the moon landings.

Also, I think it is a pretty trivial difference. Do the mechanics of jumping change when you have a heavy bag on you bag? Why should it be so much different in lower g?

 

[sophiecentaur]

Because there is nothing significant about the physics to suggest otherwise.

If you were doing any other experiment and suddenly introduced a factor of 1/6 into the variables then would it be good practice to 'assume' that it would would make no difference? You would just have to consider this as a major factor until you could prove otherwise.
We evolved with our weight and mass being relegated by a constant reducing one but not the other by such a large factor is not trivial until it's proved to be.

a tether providing a constant upwards force of 5/6g.

Why not find out about it and prove me wrong then? Unfortunately, I can't imagine NASA having built a test rig tall enough to cope with high jump records but you never know.
Actually, you wouldn't need NASA facilities. If you could find a high bridge with access top and bottom you could attach a pulley with a 5/6 body weight mass on a rope . .etc.

 

[Art Vanderlay]

If you were doing any other experiment and suddenly introduced a factor of 1/6 into the variables then would it be good practice to 'assume' that it would would make no difference? You would just have to consider this as a major factor until you could prove otherwise.
We evolved with our weight and mass being relegated by a constant reducing one but not the other by such a large factor is not trivial until it's proved to be.

Why not find out about it and prove me wrong then? Unfortunately, I can't imagine NASA having built a test rig tall enough to cope with high jump records but you never know.
Actually, you wouldn't need NASA facilities. If you could find a high bridge with access top and bottom you could attach a pulley with a 5/6 body weight mass on a rope . .etc.

High enough? The amount someone can jump from a standing start is about half a meter. Do you think this is high?

This has become a pointless debate with no new physics. The only unknown is whether the mechanics of the human legs dramatically changes in lower g. Well people have different power to weight ratios here on Earth also - do they jump differently?

Just saying some assumption is not valid with no reasoning is not science. There's no reason to believe it is either more or less efficient to jump on the moon - but it could be either way.

Reply if you wish, but I'm no longer responding on this thread since it appears to be just point-scoring based on blind objections with no physical reasoning.

I've given a reasonable model for the system which I hope others will find valuable.

 

[sophiecentaur]

Just saying some assumption is not valid with no reasoning is not science.

It's the other way round, surely. If you make an assumption then you should validate it. I was introducing a note of caution into making such a simple assumption - after all, there have been more than one approach to the theory, even on this thread.
Of course different people perform differently, under the same conditions. The issue is how differently one particular individual would perform under different conditions. Unless you can justify ignoring some parameter then you should really include it - that's a good scientific practice, isn't it?
I think the scientific reasoning behind my doubt would probably come from reference to the Power / Force graph, which doesn't show a simple relationship. I should have thought that evolution would have optimised the way the legs propel and lift the individual (running and hunting - not in a gym exercise).
It would be an interesting experiment to do before actually going all the way to the Moon.

 

[Art Vanderlay]

It's the other way round, surely. If you make an assumption then you should validate it. I was introducing a note of caution into making such a simple assumption - after all, there have been more than one approach to the theory, even on this thread.
Of course different people perform differently, under the same conditions. The issue is how differently one particular individual would perform under different conditions. Unless you can justify ignoring some parameter then you should really include it - that's a good scientific practice, isn't it?
I think the scientific reasoning behind my doubt would probably come from reference to the Power / Force graph, which doesn't show a simple relationship. I should have thought that evolution would have optimised the way the legs propel and lift the individual (running and hunting - not in a gym exercise).
It would be an interesting experiment to do before actually going all the way to the Moon.

You are just providing reasons why others' analysis is "wrong", rather than offering any yourself. Making assumptions is the way we simplify a problem enough to solve it. You could argue that g isn't constant in either case due to height, you could argue that Newtonian physics is invalid since general relativity more correctly describes reality.

I assume the reason why one might ask this question is to understand how the mechanics of the problem affect the outcome, not to understand if the action of jumping itself is changed. So if you want to continue to speculate about this point, carry on, I'll split the problem into 2:

1/ What height will a mass propelled with the same force over the same distance vertically reach on the moon vs the Earth?
2/ Is the way a person jumps affected on the moon?

Answers:

1/ h'/h = (1 + h/d - r)/(rh/d) where h' is the height on the moon, h is the height on the Earth, d is the distance the force is applied over and r is the ratio of the moon's to Earth's gravity. Anyone looking on this forum for the answer to that question, use the parameters you want for d, h and r and you have h'
2/ Feel free to answer yourself or continue to speculate. It doesn't interest me because it can't be tackled theoretically, it requires empirical evidence. My guess is that the only significant factor is the fact the the contraction speed would need to increase (as I have already analysed). This could in principal change the total force integral. I can't see that the contraction mechanism itself would be affected (you should be able to work out why that is yourself)

Maybe someone else can pick up the debate with you on 2. Good luck.

 

[sophiecentaur]

I assume the reason why one might ask this question is to understand how the mechanics of the problem affect the outcome, not to understand if the action of jumping itself is changed. So if you want to continue to speculate about this point, carry on,

My speculation is totally based on the fact that humans who use machines of any sort tend to use Gears. If you have ever tried to cycle fast in a low gear, you will remember that you are speed limited. The same thing applies to motor cars. Engines and (from the graph in that previous post) muscles have a definite optimum speed for operation. Why should this not apply when trying to jump as high as possible in low g?