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

[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.

 

[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.