What's wrong with Arnold's scaling argument on jumping height?

Question

The following question was put on hold: Is it possible to prove that an elephant and a human could jump to the same height?

It reminded me of an exercise (24a) on that exact topic in Arnold's "Mathematical Methods of Classical Mechanics". The solution he gives goes like that:

For a jump of height h one needs energy proportional to $L^3h$, and the work accomplished by muscular strength $F$ is proportional to $FL$. The force $F$ is proportional to $L^2$ (since the strength of the bones is proportional to their section). Therefore, $L^3h$~$L^2L$, i.e. the height of a jump does not depend on the size of the animal. In fact, a jerboa and a kangaroo can jump to approximately the same height.

The comments of the above question tended to dismiss that argument. What's wrong with it?

Addendum: It seems obvious that not all animals jump exactly to the same height, given their different physiologies/shapes. Some of them can't jump at all.

The question is to be understood in the following spirit: if we plotted jumping height vs animal size for a lot of different species, would there be a correlation? I don't mean there is no spread; I totally expect a big spread due to the other factors involved.

Second addendum: Some interesting points have been raised in comments and answers. I will accept an answer that incorporates: the domain of validity of Arnold's argument (or explain why it is never valid), the effect of air drag on very small jumpers and the impossibility of very large animals.

Bonus points for documenting the yet elusive jumping elephant and plotting jumping height vs size for different species ;)

Answer 1

There is nothing wrong with the argument. The mathematics are quite simple and the conclusion is sound - scale cancels out.

Let's consider the essence of the question; How does the scale size of an animal affect the absolute height it can jump?

Let's assume an on-the-spot spring jump so we exclude a run-up. Now consider an arbitrary animal (let's call it a ballerina). It crouches, extends itself and leaps in the air. The difference in height of, say, its head between crouch position and when it leaves the ground is its leg extension ($s$). The difference in height between leaving the ground and top of the jump is the jump height ($h$).

As the ballerina jumps, its muscles provide an upwards acceleration of $a$ over the extension length, $s$ of its legs. This leads to an initial upwards velocity, $v$ that is given by: $$ v^2 = 2as $$

As it flies upwards, it is decelerated by gravity until it stops at the top of its leap. The height reached is given by the same equation where the acceleration is that due to gravity, $g$: $$ v^2 = 2gh $$ hence $$ h = {v^2\over{2g}} $$

Since $v$ is the same in both equations, we can equate them, giving us: $$ h = {{as}\over{g}} $$

Now, $a = {F \over m}$, so: $$ h = {{Fs}\over{mg}} $$

We now have an equation for how high a ballerina can jump that depends on the force in its muscles, $F$, the length of its leg extension, $s$, its mass, $m$, and $g$.

What happens if we scale up the ballerina by a factor $x$? Leg extension simply goes up by $x$. Mass, which depends on its volume, goes up by $x^3$. Interestingly, muscle strength goes up by $x^2$ because it is the cross-sectional area of a muscle that gives it its strength (not its volume!). Plugging in the scale factors:

$$\begin{equation}\begin{aligned} h &= {{x^2 F \cdot xs}\over{x^3m \cdot g}} \\ &= {{{x^3}\over{x^3}} \cdot {{Fs}\over{mg}}} \\ &= {{Fs}\over{mg}} \end{aligned}\end{equation}$$

So we get the same height.

The conclusion is that the scale size of a creature is not a factor in calculating how high it can jump. To put it another way, the graph of jump-height versus scale size is a horizontal line.

Incidentally, elephants can actually jump - there is a circus stunt in which they stand on their two hind legs and hop slightly. They go up 10-20 cm, which is about the same as a flea.

Answer 2

What's wrong is :

For a jump of height h one needs energy proportional to $L^3/h$

Taking L as a measure of animal size then we should actually have

$$E \approx Mgh \propto L^3h $$

So not divided by $h$ but multiplied by $h$ !

And a little thought would show that dividing by $h$ would make no sense, as it implies you need less energy to jump higher. I don't know if that's a typo by the OP or in Arnold.

, and the work accomplished by muscular strength F is proportional to FL. The force F is proportional to $L^2$ (since the strength of the bones is proportional to their section). Therefore, $L^3/h \sim L^2L$

A serious flaw here is assuming all animals are built the same way. We're implicitly assuming that animals have both the same ways of storing energy for activities like jump and that the proportion of muscle mass available for jumping is the same, which is not the case.

Now I did a bit of back of the envelope modeling of this, as I think it's fair to say that the proportion of an elephant's volume that is leg is smaller than that of e.g. a human. Whatever the case we might modify the relationship to produce :

$$L^3h \sim FL \propto \frac {V_{legs}}{V_{total}}L^3$$

So at the very least we get

$$h \sim \frac {V_{legs}}{V_{total}}$$

Which casts a very different view of this idea that all animals jump to the same height. It now becomes a function of the design of the animal. It even tells me that other things being equal a fat me will not jump as high as a thin me, which sounds like a better model than we all jump the same height !

I see this is another answer :

all animals jump about the same height to within an order of magnitude - from about 20cm to 2m.

I really think of this as a cop-out argument. I've seen a similar argument in another place online and it boils down to "can't be bothered to work out a better model so ignore the order of magnitude difference". I just don't see the point of trying to model something if you're going to do that.

Finally

In fact, a jerboa and a kangaroo can jump to approximately the same height.

Do we know this to be true ? How do we even define the jump height for the case of two such different animals (which strikes me as a non-trivial issue) ? But is this another case of ignoring order of magnitude differences and calling it "the same" ?

And what if I replace "jerboa" with "elephant" - does it work out then ?

This sounds like very sloppy logic.

Answer 3

The fact that animals are all very different aside, it ignores some important facts.

The first big problem is that it ignores the fact that an animal scaled up might not be able to stand at all. By the original argument, $F\propto x^2$ and $mg\propto x^3$, where $x$ is some scaling factor. So it should be obvious that, at some point, $mg>F$ and the animal is unable to stand, much less jump.

Connected to this issue is the definition of jump height that is used. It assumes that $mgh=Fs$, where $h$ is the height of the jump and $s$ is the extension of the legs. But this counts simply standing up as a "jump," even if the feet do not leave the ground. The proper expression would be:

$$ mgh=Fs-mgs$$

subtracting standing up from the height reached. But this no longer scales the same way:

$$ h=\frac{Fs-mgs}{mg}\sim\frac{x^2x-x^3x}{x^3}={x^3-x^4\over x^3} $$

So jump height is only approximately scale-invariant so long as the negative term in the numerator is negligible. Or in other words, only so long as $F\gg mgx$. Once this term is no longer negligible, jump height starts to decrease with scale.

An average human can leg press about twice their body weight once. So already $F\not\gg mg$. Scaling up a human only two times leaves them unable to stand under their own weight. A particular form for a biological organism is not remotely scale invariant- it only works at one particular scale.

In other words, all else being equal, an animal's jump height correlates inversely with its size. It is approximately constant only in the limit where an animal can already jump much higher than its own height. If there is not a negative correlation between size and jump height, it is because all else is not equal- the biochemistry involved is very non-trivial.

Answer 4

The force F is proportional to L² (since the strength of the bones is proportional to their section).

That is relevant only if strength of the bones is the limiting factor. If an animal with height L jumps a height h, then the acceleration is gL/h. For a tick of height 1 mm to jump 1 m, it would have to have an acceleration of 1000g. A human with height 2 m jumping 1 m would have to have an acceleration of only 0.5g. For small animals, the larger they are, the higher they can jump, generally. The height they can jump increases more slowly than size, so smaller animals can jump a larger number of "body lengths" than larger animals. As animals get larger, the slope of the curve decreases, and there's a point at which it goes negative.

Answer 5

Tl;dr: In general, and under ideal conditions of energy and power, smaller animals jump at a lower height than larger animals. And then, all 'big enough' animals jump at the same height.

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Energy

It was shown by several answers, and by the OP, that the energy content leads to a scale-independent height.

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Power

There's one thing that everyone here is ignoring. Power. It is true that, from the point of view of energy, scale cancels out, but, how about power? Can muscles deliver this energy content quickly enough for a successful jump? The answer is: no. From a power point of view, scale does matter.

This is a very good source, if you want: http://web.mit.edu/6.055/old/S2009/notes/jump-heights.pdf

This source says: Power production might also limit the jump height. In the preceding analysis, energy is the limiting reagent: The jump height is determined by the energy that an animal can store in its muscles. However, even if the animal can store enough energy to reach that height, the muscles might not be able to deliver the energy rapidly enough.

It then argues that: $$ P \sim \frac{E}{t} \sim \frac{E}{d/v} \sim \frac{L^3}{L^1/L^0} = L^2 $$

Therefore, the specific power: $$ \frac{P}{m}\sim L^{-1} $$

That is, smaller animals will need a higher muscle specific power if they plan to jump to their theoretical energy-limit height. Thus, if we assume that every animal has the same specific power [that is, if we assume specific power doesn't scale - because it depends on muscle design], then, smaller animals will jump at a lower height than larger animals, because quite simply, even if though are capable store the energy to jump higher, their muscles cannot release this energy fast enough.

In other words, a small-enough animal cannot jump to its energy-limited height.

=].

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Drag

The same document argues that: $$ \frac{E_{drag}}{E_{jump}} \sim \frac{\rho v^2 Ah}{mv^2} \sim \frac{L^2}{L^3} = L^{-1} $$

This means, the ratio between energies to overcome the drag, and the jump itself, scales with the inverse of size. Thus, the energy released by the jump, for smaller animals, will have a height portion consumed to drag, than to the actual gravitational potential energy to get to a certain height.

Answer 6

I've had second thoughts on the whole scale-invariance thing... Referring to my answer above, I think we're safe up to the equation for jump height: $$ h = {{Fs}\over{mg}} $$ In the argument above, I re-scaled the variables and showed that the scale factor cancels out - hence jumping is scale-invariant. This means that if real Matt Damon can jump 30cm, his character in Downsizing can also jump 30cm.

However, as @Chris et al. pointed out to me, the force, $F$, in the equation above isn't just the force available in the muscles (which simply scales with the square of the scale-factor). Rather, it is the force used to accelerate the animal - and that force is the net force after gravity has been subtracted. In other words: $$ F = F_{0} - mg $$ Now this force doesn't simply scale with size because it has $m$ in it... So the real jump height equation is: $$ h = {{(F_{0}-mg)s}\over{mg}} $$ Now if we scale this by $x$, we get: $$ h = {{(x^{2}F_{0}-x^{3}mg)xs}\over{x^{3}mg}} $$ which simplifies to: $$ h = \left( {{F_0}\over{mg}} - x \right) s $$ Note that all variables refer to the values for the original animal (not the scaled version).

It is apparent that when $x=1$ (original situation), $F_0$ must be greater than $mg$ to get off the ground at all. This is obvious so that's good.

If we scale down ($x \ll 1$), the subtraction of $x$ is not significant and the scale-invariant behaviour is approximately obtained.

But, if $x$ increases, $h$ shrinks until, when $x = {{F_0}\over{mg}}$, $h$ tends to zero and jumping becomes impossible. This chimes more with reality.