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

[oz93666]

It's generally said that a moon walker should be able to jump six times higher on the moon, than he can on Earth, which sounds sensible since gravity is one sixth , but consider this...

Two men , one on the Earth, one on the moon, both bend their knees the same amount ready to do a 'standing' jump.
They both start , and accelerate upwards.
The Earth man is pushing against the inertia of his mass and the downward force of his Earth weight.
The moon man is pushing against the inertia of his mass and the downward force of his moon weight.

Clearly the moon man will have a higher velocity when his feet leave the ground than the Earth man, and so will travel OVER six times higher.

How much higher will depend on how long the acceleration zone is.

When I do this I squat right down to the floor and can jump 30cm high. the acceleration zone is 90cm.

The energy the Earth man expends against the extra gravity iis 0.9 x( 1 - 1/6)g

Equivalent to 0.9 x5/6 =75cm in Earth gravity total jump =30 +75 =105cm

This means he can jump 6.3meters on the moon! 21 times higher

 

[Vanadium 50]

Welcome to PF! I don't understand the purpose of your message. Is there a question in there? Or did you just want to show us an incorrect calculation?

 

[oz93666]

If you want a question then ... Is my logic flawed? Show me where the error is.

 

[sophiecentaur]

If you want a question then ... Is my logic flawed? Show me where the error is.

You are constructing a strange argument and it is based on arbitrary and irrelevant assumptions. There is no way you can get a valid answer that contradicts one that's based on Energy considerations. When you find (and it's very common for all of us) a nonsense answer, you need to scrutinise what you've done and look for the flaw.
If the available energy (mgh) is the same in each case, and g is one sixth, the maximum achievable h will be six times what it is on Earth*. But there will be a problem in achieving this in practice. The impulse that the legs can deliver is matched to Earth conditions. On the Moon, it would be limited, without some help from a leverage system of some sort. What you seem to have done is to build a model that just gives the wrong answer. Time to think again and to reconsider what you mean, for instance, by "acceleration zone".
* Bear in mind that millions of years of development of the muscular / skeletal system had produced a near-optimum utilisation of the muscles on Earth. The Energy based estimate will be surely over optimistic regarding possible performance on the Moon.

 

[oz93666]

Consider these two men , both with knees bent ready to push themselves into the air/space.
Are you suggesting the man on the moon can't exert the same force as his twin on Earth?

 

[sophiecentaur]

Consider these two men , both with knees bent ready to push themselves into the air/space.
Are you suggesting the man on the moon can't exert the same force as his twin on Earth?

Of course he can - but he is only applying that force for a shorter time as he will very soon be going as fast as his muscles will allow and as far as his skeleton will let him. That is why I introduced the term 'Impulse'. To get the full height, he must be able to apply his maximum force for a longer time - that would be by using a lever system (distance magnifier).
Your argument is a bit along the lines of an argument that you should be able to throw a small ball bearing ten times further on Earth than you can throw a cricket ball (easy to disprove experimentally). Except with massive objects, your throwing ability is only limited by your arms and muscles and not by the object you are doing work on. That's why we use a rod to cast a fishing weight a lot further than we can throw it.

 

[jbriggs444]

The result I get, based on the assumptions that I believe that oz93666 is using, matches his.

Neglect air resistance. Assume that the force the man's legs generate is constant throughout the jump and does not depend on the magnitude of the man's acceleration. Assume that the jump from the fully crouched posture to when the feet leave the ground is 90 cm. Measure the height of the resulting jump from the point at which the feet leave the ground.

On earth, we have that the 90 cm take-off results in a 30 cm jump. It is then clear that the take-off was at 1/3 g. One can obtain that result by equating the work done by the legs + gravity over the 90 cm take off and the negative work done by gravity alone over the resulting 30 cm jump.

This means that the legs alone are producing a force equal to 1 1/3 of the man's Earth-weight.

On the moon, we have legs producing a force of 1 1/3 of the man's Earth-weight and the moon pulling back with a force of 1/6 of the man's Earth-weight. The net is 7/6 of the man's Earth weight. This produces an acceleration of 7/6 of a g.

Again, equating the work done by 7/6 of a g through a 90 cm stroke with the negative work done by 1/6 of a g through a jump of unknown height we get that the moon jump is 7 times 90 cm = 630 cm.

Divide by 630 cm by 30 cm and one sees that the moon jump is indeed 21 times the height of the Earth jump -- under the stated assumptions.

Sopie used a different implicit assumption (in particular that the height of the jump is measured from the bottom of the crouch rather than from the top) to reach an apparently contradictory conclusion.

 

[sophiecentaur]

You are assuming that there is no operational speed limit to the legs and muscles. Do you really think that you could throw a small (0.5kg ) mass as much as 21 times as high as you could make a standing jump, with a straight vertical push with your legs (with your body in a suitable upside down cradle arrangement)? You need to match your body to the required task if you want to approach the same amount of work done - and it is unlikely that you will be able to do more work than you can on Earth. If you could, then you would have to blame Mr Darwin for doing such a sloppy job with Earthbound body design! Your argument is missing out a very important consideration.

 

[oz93666]

Of course he can - but he is only applying that force for a shorter time as he will very soon be going as fast as his muscles will allow

So you concede on the moon he will be going faster than on earth, so ' take off velocity' will be higher than on Earth hence he can reach higher than 6 times Earth jump, we just have to agree on how much higher.

Yes , you're right , this is all dependent on being able to exert the same force at higher speed, I've a feeling this will be possible, as the speed is very moderate, and this can all be measured here on Earth in a gym.

 

[oz93666]

You are assuming that there is no operational speed limit to the legs and muscles. Do you really think that you could throw a small (0.5kg ) mass as much as 21 times as high...

Are you referring here to throwing a cricket ball or similar with your arm?

No there would be very little difference here because the 'acceleration zone' is short compared to the overall distance involve, it would be only slightly higher than six times Earth hight. perhaps 6.1

 

[sophiecentaur]

Are you referring here to throwing a cricket ball or similar with your arm?

No there would be very little difference here because the 'acceleration zone' is short compared to the overall distance involve, it would be only slightly higher than six times Earth hight. perhaps 6.1

I am pointing out the fact that, below a certain mass, there is very little, if any, advantage in reducing the load. In the case of throwing upwards with a push of the feet (like a standing jump on the Moon) there would be little chance of even achieving six times the height. Your 'acceleration zone' idea will only work if, in fact, the acceleration can be sustained over the whole distance. What evidence would you have of this?

So you concede on the moon he will be going faster than on earth, so ' take off velocity' will be higher than on Earth hence he can reach higher than 6 times Earth jump, we just have to agree on how much higher.

Yes , you're right , this is all dependent on being able to exert the same force at higher speed, I've a feeling this will be possible, as the speed is very moderate, and this can all be measured here on Earth in a gym.

Of course he can and will be traveling faster. I maintain that he will be Energy Limited, though. There are enough instances on Earth where we find we are limited in how we can actually expend energy in a desired task and this can only be dealt with by using gears of some kind. A bicycle with gears can go much faster than a skateboarder on the flat, for this reason. The skateboarder can't actually add to his velocity once his legs are going as fast as possible (ignoring fancy leaning moves etc - which constitute another form of 'matching') Running down hill will result in an accident if you don't actually control your motion and the actual (safe and controlled) speed increase is never very great.

I am taking on board the points you are making, though, and trying to reconcile the differences between our two models. The bit about the extra Kinetic Energy of the jumper when there is less weight force, makes sense. Certainly he is doing less work (1/6) in raising his body before he takes off and the remaining energy could be available as extra KE at the end of the muscle action. I am still questioning whether he could actually achieve the suggested higher speed at the end of the action. I imagine it would be easy to do some measurements with suitable gymn equipment. A simple test could be to do the experiment I suggested, involving a straight 'push' throw into the air with the feet. The height reached would indicate a maximum take off speed, on Moon or anywhere.

 

[Nugatory]

So you concede on the moon he will be going faster than on earth, so ' take off velocity' will be higher than on Earth hence he can reach higher than 6 times Earth jump, we just have to agree on how much higher.

That does not follow, because there are serious limits to how quickly the jumper will be able to rotate his limbs and hence how effectively he will be able to take advantage of the (conjectured) opportunity to generate more energy during the jumping motion.

There is a related statement that is true: It is possible to design a device that will reach a height greater than six times what it reaches on the earth. But it is far from clear that the human body is such a device.

 

[zoki85]

21 times? Air Jordan could jump 20 meters high on the moon then. No way .

 

[oz93666]

That does not follow, because there are serious limits to how quickly the jumper will be able to rotate his limbs and hence how effectively he will be able to take advantage of the (conjectured) opportunity to generate more energy during the jumping motion.

There is no suggestion the moon jumper is generating more energy than the Earth jumper. If the force exerted by both jumpers is the same and the distance through which it operates is the same, then the energy is the same.
The only question is will the fact that on the moon the feet of the jumper are moving away from the body at higher speed , effect the jumpers ability to exert the same force.

Lets look at the speed at which the feet are traveling away from the body, at the point of maximum velocity, just before they leave the surface.

on the Earth the jump is 30cm.... m g h = 1/2 m v2 .3 x 9.81 x m = 1/2 m v2 v =2.426

on moon IF jump if 21 times higher ...m g h = 1/2 m v2 .3 x 21 x9.81/6 x m= 1/2 m v2 v =4.538

So even this higher speed is still very modest at 10mph.
We can get an idea of the legs ability to push at high speeds by looking at martial arts kicks. Swinging kicks(roundhouse)have been recorded at over 120mph, but they are not applicable to our case ,we need a side kick, where the foot is pushed away from the body, it seems these have speeds of over 40mph which shows the muscles ability to act quickly, and exert force at much higher speeds than our case.

 

[oz93666]

21 times? Air Jordan could jump 20 meters high on the moon then. No way .

Do you have any science or figures to support your position or is this just a 'feeling'.

If Air Jordan could jump 95 cm in a standing jump , he could indeed get to 20meters. Thats very high jump on earth, are you sure he's not taking a run up? I haven't considered a running jump , makes things more complex...

 

[zoki85]

Do you have any science or figures to support your position or is this just a 'feeling'.

If Air Jordan could jump 95 cm in a standing jump , he could indeed get to 20meters. Thats very high jump on earth, are you sure he's not taking a run up? I haven't considered a running jump , makes things more complex...

Kinesiology, I think that's the name of the science

 

[jbriggs444]

Do you have any science or figures to support your position or is this just a 'feeling'.

If Air Jordan could jump 95 cm in a standing jump , he could indeed get to 20meters. Thats very high jump on earth, are you sure he's not taking a run up? I haven't considered a running jump , makes things more complex...

Your calculations do not support this assertion. Let us re-run the numbers using the same [questionable] assumptions that went into the original 21 to 1 estimate.

"Air Jordan" crouches 90 cm and manages a jump of 95 cm. That means that his acceleration during the jump must be 95/90 of a g. The force output by his legs must therefore be 1 g + 95/90 g or about about 2.06 g. On the moon this nets him about 1.9 g of vertical acceleration. 1.9 g over 90 cm during takeoff equates to 1/6 g over about 10.3 meters.

He does not get to 20 meters. He only gets to 10 meters. The original claim of a 21 to 1 factor applies only to the special case of Joe Schmoe, not to the case of Air Jordan,.

 

[A.T.]

You are assuming that there is no operational speed limit to the legs and muscles.

While the common claim, of 6 times higher jumps on Moon, assumes those limits to be so small, that the take-off speeds will be identical on Moon and Earth. Both assumptions seem unrealistic and the truth is probably somewhere in the middle. It might not be 21 times higher, but could be more than those commonly stated 6 times.

 

[oz93666]

Your calculations do not support this assertion. Let us re-run the numbers using the same [questionable] assumptions that went into the original 21 to 1 estimate.

"Air Jordan" crouches 90 cm and manages a jump of 95 cm. That means that his acceleration during the jump must be 95/90 of a g. The force output by his legs must therefore be 1 g + 95/90 g or about about 2.06 g. On the moon this nets him about 1.9 g of vertical acceleration. 1.9 g over 90 cm during takeoff equates to 1/6 g over about 10.3 meters.

He does not get to 20 meters. He only gets to 10 meters. The original claim of a 21 to 1 factor applies only to the special case of Joe Schmoe, not to the case of Air Jordan,.

Yes ...your right , it's all got to do with the ratio of the acceleration zone to the hight jumped.
Who's Air Jordan ? One of those lanky basketball players? I find it hard to imagine anyone from a squat could jump 95cm. If he's very tall then the acceleration zone , the amount he reduces his hight by squatting should be more, which means he can jump higher on moon.
The figures I gave were for me, 5 foot 10 inches tall jump .3m acc zone .9m

 

[A.T.]

I find it hard to imagine anyone from a squat could jump 95cm.

If you start standing, then squat and jump, the record is 117 cm:
http://www.topendsports.com/testing/records/vertical-jump.htm

 

[zoki85]

Who's Air Jordan ? One of those lanky basketball players?

I think this could be him:

 

[bahamagreen]

I tried to simplify this in order to find what aspects of the mechanics determine the height... have run into a snag.

Here is how I'm thinking...

Consider a ball resting on a latched compressed spring, and then released. The max ball flight height is a function of initial velocity of the ball at separation from the uncompressed spring with respect to the gravitational acceleration acting during uncompression and after separation.

That velocity at separation has two contributing components:
- a positive component of the impulse from the spring acting on the inertial mass of the ball
- a negative component of gravitational acceleration on the ball during the impulse period

The max flight height of the ball will be based on:
- the separation velocity
- subsequent accumulation of negative acceleration

My wording is clumsy, but correct so far? If so, then the question I have at this point is:

It looks to me like the period of the impulse is going to be longer on Earth because of the second contribution component to the separation velocity, but does this mean that the separation velocities of the balls on Earth and Moon will be equal (with the max flight height determined only subsequently by gravity), or different (with gravity acting during the impulse period to effect the separation velocity)?

Hunting around I find:
- potential energy a la Hooke's Law is (1/2)k(x^2) without regard to time
- impulse is the integral of the force during the uncompression interval
- the mass of the ball is unchanged so the impulse is just change in momentum
- but I find this "The force is the gradient of the potential energy, which involves DERIVATIVES WRT TO SPACE COORDINATES AND NOT WRT TO TIME...So there is no simple connection between momentum and potential energy." at post #5 here - https://www.physicsforums.com/threads/momentum-is-potential-energy.58521/)...

Maybe I'm going in circles... :)
:

 

[sophiecentaur]

"Acceleration Zone"
This term is being used as if it's a universally applicable concept for such problems. Unless you can achieve unlimited acceleration over the whole of this distance, it has no meaning. There are far too many assumptions being made about how the body would work in 1/6 gravity.
The one really valid point that has been made - and that could affect the 'better than 6 times' argument - is the extra force available for accelerating during the jump. Without some actual evidence about the achievable speed of the legs, unloaded, there is no answer.

 

[A.T.]

Without some actual evidence about the achievable speed of the legs, unloaded, there is no answer.

I did some quick tests, with a high speed camera. As a very rough estimate, extending unloaded legs was about twice as fast as extending them in vertical jump (0.2s vs 0.4s). So it seems like humans could achieve higher take-off speeds in a low-g environment.

 

[Dale]

Of course he can - but he is only applying that force for a shorter time as he will very soon be going as fast as his muscles will allow and as far as his skeleton will let him.

EDIT: I messed up on some of the math, and went back and corrected.

Sorry to come late to this thread, but this struck me as quite interesting. I looked around for accepted models of muscle mechanics that describe the force as a function of the contraction velocity.

I found that one common model is the so-called Hill's muscle model. I solved the model for the force and got:
$$F=\frac{F_0 k (v_0 - v)}{v+k v_0}$$

Where $F_0$ is the maximum static force, $v_0$ is the maximum (unloaded) contraction velocity, and $k$ is the dimensionless Hill's parameter.

Using literature values of $k=.24$ and $F_0 = 3000~N$ and setting $v_0=21~m/s$ and assuming an 80 cm extension of the legs during takeoff we get a realistic "high performance" jump of 63 cm taking .37 s to reach a peak velocity of 3.5 m/s.

Using the same muscle model with the same biological parameters on the moon we get a jump of 6.9 m taking .29 s to reach a peak velocity of 4.7 m/s.

I don't know how realistic the Hill's model is for this type of thing, but a 11x higher jump on the moon certainly seems more plausible than 21x, and to me it even seems more plausible than the naive figure of 6x higher.

 

[A.T.]

I found that one common model is the so-called Hill's muscle model. I solved the model for the force and got:
$$F=\frac{F_0 k (v_0 - v)}{v+k v_0}$$

Where $F_0$ is the maximum static force, $v_0$ is the maximum (unloaded) contraction velocity, and $k$ is the dimensionless Hill's parameter.

Using literature values of $k=.24$ and $F_0 = 300 ~9.8$ and setting $v_0=21$ and assuming an 80 cm extension of the legs during takeoff we get a realistic "high performance" jump of 62 cm taking .37 s to reach a peak velocity of 3.5 m/s.

How did you calculate the contraction velocity of the muscles $v$ ?

 

[zoki85]

Using literature values of $k=.24$ and $F_0 = 300 ~9.8$ and setting $v_0=21$ and assuming an 80 cm extension of the legs during takeoff we get a realistic "high performance" jump of 62 cm taking .37 s to reach a peak velocity of 3.5 m/s.

Using the same muscle model with the same biological parameters on the moon we get a jump of 1.9 m taking .30 s to reach a peak velocity of 4.7 m/s.

I don't know how realistic the Hill's model is for this type of thing, but a 3x higher jump on the moon certainly seems more plausible than 21x, and to me it even seems more plausible than the naive figure of 6x higher.

If I understood correct that would mean 10-11x higher jump on Moon? Moon's gravity is aprox 6x weaker than Earth gravity.
h=v ² /(2g)

 

[Nick O]

There is no suggestion the moon jumper is generating more energy than the Earth jumper. If the force exerted by both jumpers is the same and the distance through which it operates is the same, then the energy is the same.
The only question is will the fact that on the moon the feet of the jumper are moving away from the body at higher speed , effect the jumpers ability to exert the same force.

If I understand correctly, you are saying that the energy in both cases is the same, but both the initial velocity and the peak height are greater on the moon. If this is not what you are saying, please correct me.

For energy to be equal in both cases, assuming ideal conditions and negligible acceleration of the earth/moon, the kinetic energy when the feet leave the ground should be equal in both cases, as should gravitational potential energy at the highest point. This is not the case in your setup.

 

[jbriggs444]

If I understand correctly, you are saying that the energy in both cases is the same, but both the initial velocity and the peak height are greater on the moon. If this is not what you are saying, please correct me.

For energy to be equal in both cases, assuming ideal conditions and negligible acceleration of the earth/moon, the kinetic energy when the feet leave the ground should be equal in both cases, as should gravitational potential energy at the highest point. This is not the case in your setup.

Energy is the same if one measures from the bottom of the crouch rather than at the instant the feet leave the ground. Kinetic energy when the feet leave the ground need not be equal in the two cases because gravitational potential energy (measured from the bottom of the crouch) is not the same when the feet leave the ground.

Of course, this ignores DaleSpam's refinement to more closely model real world muscle behavior.

 

[oz93666]

If I understand correctly, you are saying that the energy in both cases is the same, but both the initial velocity and the peak height are greater on the moon.

Yes , I'm saying the energy put into the jumps is the same. The force exerted by the leg muscles is the same and squat length (acc. zone) is same.

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)

The above cannot be disputed , it's hard basic science... so if the take off velocity is higher on moon this MUST result in a greater than six times jump, remember , if take off velocities are equal then the jump would be six times higher.

So we must agree ALL jumps on the moon are OVER six times higher.

The only question is, how much does the increase velocity with which the feet are pushed away from the body on the moon, affect the muscles ability to apply the same force as on Earth.
This could be accurately measured with gym equipment , but to get an idea about this , we can note that martial artists kick at 4 times the speed we are talking about, so I don't imagine there would be too much reduction in force.

In my case... I squat .9m and jump .3m so on moon my jump should be nearly 21 times higher = 6.3m

An athlete.. with squat .9m and jump .75m will on the moon only jump around 12 times higher = 9 m

 

[Nick O]

Is the force applied during the "squat" constant? My gut tells me that it's probably linear, but not constant.

My gut has been known to lie on occasion, though.

Edit: Actually, linear force would necessitate two discontinuities: one at the start of application, and one at the end. It's probably more of a rounded, bell-shaped curve.

Whatever the case, I think this can screw up the calculations quite a bit.

 

[sgphysics]

I think bahamagreen's comparision to a compressed spring would be an ok representation of the problem. The energy of the compressed spring $E_c$ must be spent on partly lifting an object with a mass while extending the spring $E_{ext}=mg\delta$ and partly on the objects flight $E_f=mgh$. The only difference between moon and Earth here are the value of g, beeing 6 times larger for earth. (I neglect the possible difference in $\delta$).
In other words we get:
$E_{f}=E_{c}-E_{ext}$
so from this you can show that the ratio of jumping heights are:

$\frac{h_m}{h_e}=\frac{g_e}{g_m} \frac{E_c-mg_m\delta}{E_c - mg_e\delta}$ (index m=moon, e=earth)

Now you see that if the extension part is neglected ($\delta=0$), the ratio of jump heights are 6. When including the extension energy the ratio will be larger. I have not checked if it can be 21 for a man's jump.

 

[zoki85]

I think bahamagreen's comparision to a compressed spring would be an ok representation of the problem.

I think a direct comparison with spring is not very good

 

[sophiecentaur]

I think a direct comparison with spring is not very good

I agree. The muscles are energy converters - motors - and not simple energy storage Springs / Capacitors. The force (often in the form of torque) available is limited by the Power available:
F=P/v
If the speed is greater then the Force will be less. Under less load, the speed of travel will be greater and both the available Force and the time applied will be less. There is no reason to conclude that the total Energy delivered will be the same under all circumstances. As I commented near the top of this thread, the only way to achieve exceptional heights would be by using some form of gearing, lever system. This would give the muscles the same time to act.

There will be reasons to expect the achievable jump height to be greater but the '21' factor is not reasonable as it is based on a faulty model. That 'trojan horse' snuck in at the beginning of many of the arguments and still seems to keep going. DaleSpam introduced the reference to a possible, simple model of muscular behaviour but people seem to have ignored the need for it.

 

[sgphysics]

I agree. The muscles are energy converters - motors - and not simple energy storage Springs / Capacitors. The force (often in the form of torque) available is limited by the Power available:

My main point was to show that the problem can be best understood, and formulated, by energy consideration.

 

[zoki85]

My main point was to show that the problem can be best understood, and formulated, by energy consideration.

With small problem that this was unrealistic...

 

[sophiecentaur]

With small problem that this was unrealistic...

Exactly, because it is not known how much energy the muscles can supply. The same considerations apply to and motor as the load and speed change. The constant energy model is a first stab, perhaps, but is no good for making a useful prediction. DaleSpam shows the direction to go.

 

[Dale]

How did you calculate the contraction velocity of the muscles $v$ ?

I used the "guess and try" method :-)

I guessed several values until I got one that resulted in ~60 cm vertical jump on earth. 60 cm is considered very good for a young athletic individual, but not superhuman.

The $v_0$ was the only value I couldn't find from the literature, but given the other literature values and literature values for a good vertical jump, I felt OK using this rough method.

EDIT: I just noticed that you asked about $v$ and not about $v_0$. For $v$ I just plugged it into Mathematica and used the NDSolve function to solve the numerical differential equation:
$m v'(t)=F- m g$
Where F is the Hill's model force given above which is a function of the constants $k,~F_0,~v_0$ as well as $v$.

 

[Dale]

If I understood correct that would mean 10-11x higher jump on Moon? Moon's gravity is aprox 6x weaker than Earth gravity.
h=v ² /(2g)

That is correct. I messed up on the math, specifically the translation of the final velocity to height. I have gone back and corrected it.

 

[bahamagreen]

If the speed is greater then the Force will be less. Under less load, the speed of travel will be greater and both the available Force and the time applied will be less. There is no reason to conclude that the total Energy delivered will be the same under all circumstances.

I was thinking that both contributing factors to the separation velocity include the variation in load, speed of travel, and force and time applied... in a simple way for the spring and a more complicated way for the extending legs. I think you are saying that mechanically because of the geometry of lever actions in the legs these variations will present a different force application profile through the impulse period than that of a simple spring and therefore may deliver a different energy. I agree.

But it is not clear to me why the spring is not a suitable model of the leg - it looks to me like the resultant velocity at separation is due to the accrual and effect of both contributing components in both spring and leg systems which include the actions of these variations instantaneously throughout the extension.subject to the same interacting variations throughout the extension. I mean that both the spring and the leg are varying their force application instantaneously in response to the variations... I guess I'm not see which variations and how they act during the impulse can have any other resultant effect than presenting the ball with a separation velocity - given that the spring and leg will do this differently with a different result, why is the spring an inadequate model for the leg when the object of the model is to evaluate only the flight height of the ball?

I'm thinking that the impulse includes all these variation results at the time of separation so the ball has only a velocity, all prior force profiles through the extension period are irrelevant to the ball at separation...if we want to know is how high the ball or jumper will go?

"That velocity at separation has two contributing components:
- a positive component of the impulse from the spring acting on the inertial mass of the ball
- a negative component of gravitational acceleration on the ball during the impulse period"

 

[sophiecentaur]

why is the spring an inadequate model for the leg when the object of the model is to evaluate only the flight height of the ball?

The spring model is not the same as an engine or a motor or a muscle. A spring has so much Potential Energy stored in it and all of that energy will be converted to gpe plus KE, whatever load you put on it - assuming there is some KE left at the top of the stroke. (I refer to an ideal spring and not one with some power limiter, like a dashpot or friction device). With a decreasing load, an ideal spring can deliver unlimited power (or at least, only limited by its own mass).
If you have an engine or a muscle (Chemical energy in / mechanical energy out) then the amount of energy expended will depend upon the Power available. There is always an upper power limit for an engine and that power can only be delivered under certain circumstances (the right gearing / slope etc) and the Power equals the Force times the speed. Despite immense engine power, a Ferrari cannot accelerate to 100mph without changing up through the gears and nor can a leg muscle produce more than a certain power, un aided by some gears or a lever. You have to match the engine to the task if you want to get the most out of it. There is no reason to assume that jumping on the Moon is a task that is matched to your legs. They evolved for running and jumping on Earth and will work optimally (velocity ratio of the joints and chemical reaction rates, for instance)
Do you really not see the difference between the two cases?

It will be be true that a human could probably project a human mass to an impressive height using the right form of lever / gear system but that is not what the OP is suggesting.

 

[A.T.]

For $v$ I just plugged it into Mathematica and used the NDSolve function to solve the numerical differential equation:
$m v'(t)=F- m g$
Where F is the Hill's model force given above which is a function of the constants $k,~F_0,~v_0$ as well as $v$.

If I understand this correctly, you are assuming the upwards velocity of the human body (v in left side of eq.) equals the contraction velocity of the muscles (v in Hill model).

If that was your assumption, I don't think it's correct. The moment arm of the quadriceps (r) is about 4-5cm, while the length of the femur (R) is 40-50cm:
http://www.gla.ac.uk/t4/~fbls/files/fab/tutorial/biomech/akp2.html
http://www.scielo.org.za/scielo.php?pid=S0038-23532008000200010&script=sci_arttext

So for example at 90° knee flexion, the distance hip-ankle (c) changes about 7 times faster than the quadriceps length (q). I think the relationship can be approximated as:

c = R * 2 * sin($\Delta$q / (2 * r)) where $\Delta$q is the change of q compared to c = 0 state (hypothetical 180° flexion)

But note that c'(t) is still not the upwards velocity of the hip, because the ankle is moving upwards too, driven by the plantar flexion of the foot, which in turn is partly driven by the knee extension via the tension in the gastrocnemius muscles that cross both joints.

Bottom line is, you need a musculo skeletal model to computationally estimate the maximal jump height on the Moon. Something similar the one used in this study:
http://www.sciencedirect.com/science/article/pii/S0306452213000997

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). From this you could get the take-off speed at 1/6 g.

 

[Dale]

If I understand this correctly, you are assuming the upwards velocity of the human body (v in left side of eq.) equals the contraction velocity of the muscles (v in Hill model).

If that was your assumption, I don't think it's correct.

I am sure it is not "correct", I was deliberately sacrificing accuracy for simplicity.

I was assuming that the net effect of all of the different lever arms and joint rotations would be to scale the Hill model as a first approximation. I would have preferred an equation which simply described the movement force or even joint torque, but I didn't find any such formulas. Papers which measured the jump force or the knee extension force did not attempt to fit their data to a model, so they could not be applied elsewhere.

Bottom line is, you need a musculo skeletal model to computationally estimate the maximal jump height on the Moon. Something similar the one used in this study:
http://www.sciencedirect.com/science/article/pii/S0306452213000997

Yes, but I don't have one of those here :-)

 

[zoobyshoe]

An astronaut tries to set the moon high jump record:



There's a lot of real-life obstacles to reaching any calculated ideal. With space suit and backpack he says he weighs 380 pounds? I assume that's Earth pounds.

 

[A.T.]

With space suit and backpack...

We are or course talking about jumping within the Moon base that Newt Gingrich will build.

 

[zoobyshoe]

We are or course talking about jumping within the Moon base that Newt Gingrich will build.

I can't wait!

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.

 

[A.T.]

I am sure it is not "correct", I was deliberately sacrificing accuracy for simplicity.

I fear that simply ignoring the up to 10:1 leverage the muscles have is really too much of a sacrifice for the result to be useful. But a practical test on an inclined rowing machine (or similar device) would probably be more conclusive than even a complex musculo skeletal model.

 

[Dale]

But a practical test on an inclined rowing machine (or similar device) would probably be more conclusive than even a complex musculo skeletal model.

I agree.

I was actually fairly surprised to find the lack of simple models for functional movements in the literature. The literature seems to focus on either simple models of non-functional movement, highly complex models of functional movements, or non-modeled descriptions of functional movements. I have an inherent distrust of complex models since their great freedom provides too much opportunity to fit the noise.

 

[mfb]

Even better than in inclined rowing machine would be a moveable floor (accelerating downwards) or an airplane on an appropriate flight path to simulate 1/6g.

I assume that's Earth pounds.

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

 

[A.T.]

Even better than in inclined rowing machine would be a moveable floor (accelerating downwards) or an airplane on an appropriate flight path to simulate 1/6g...

... or an actual Moon base. But how about cheaper and easier to realize.