I don't quite understand this quote from Stephen J. Gould's Ever since Darwin, where he talks about the compensating physical characteristics of organisms for their size.

Other essential features of organisms change even more rapidly with increasing size than the ratio of surface to volume. Kinetic energy, in some situations, increases as length raised to the fifth power. If a child half your height falls down, its head will hit with not half, but only 1/32 the energy of yours in a similar fall. In return, we are protected from the physical force of its tantrums, for the child can strike with, not half, but only 1/32 of the energy we can muster.

Mass increases with the third power of length, and even if we allow for a shorter fall (even though that is not clear from the wording) for the child (who is half as tall), I can only come up with the fourth power.

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    $\begingroup$ Your question will be more clear if You present some of Your calculations. $\endgroup$
    – Wojciech
    Commented Mar 24, 2014 at 15:43
  • $\begingroup$ My calculations are simple: m increases as the third power and h is linear => mgh increases with the fourth power. Btw, I expanded the quote to include the next sentence that may shed some light on Gould's thinking. $\endgroup$
    – Innuo
    Commented Mar 24, 2014 at 16:29
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    $\begingroup$ I doubt the author actually sat down and calculated the physics rigorously. He may just have casually noted that mass scales as $l^3$ and that kinetic energy is $mv^2$, and (incorrectly) added the powers 2 and 3 to arrive at $l^5$. Just speculation of course, but my point is that it is easy to believe a biologist/paleontologist could make this kind of physics mistake. (Not to discredit biologists or paleontologists at all!) $\endgroup$
    – TypeIA
    Commented Mar 24, 2014 at 20:56
  • $\begingroup$ It's just that the book is quite old (1979) and the essay is probably even older. I am surprised that this was not caught and corrected in my much later edition. $\endgroup$
    – Innuo
    Commented Mar 25, 2014 at 14:39

2 Answers 2


Model the child as a rod of mass $m$ and length $l$ standing on the ground vertically, with center of mass at height $l/2$, with feet glued to the ground but the rest of the body able to rotate.

When upright, the potential energy is $mgl/2$. When lying on the ground, the potential is 0. So when falling, the child hits the ground with energy $mgl/2$. Equating this with $\frac{1}{2}I\omega^2$, with $I=\frac{m l^2}{3}$ (the moment of inertia of a rigid rod rotating about its end) one gets $$\omega=\sqrt{\frac{3g}{l}}.$$ The head hits the ground with velocity $l\omega$. Letting the mass of the head be $m_h$, the kinetic energy of the head hitting the ground is $$T=\frac{1}{2}m_h(l\omega)^2=\frac{3}{2} g l m_h.$$ Since $l\rightarrow \lambda l$ and $m_h\rightarrow\lambda^3m_h$ under a scale transformation $\lambda$, we have $$T\rightarrow\lambda^4T.$$ So unless I'm overlooking something, I can't quite see where Gould's extra factor of $\lambda$ comes from. Assuming this is right and Gould is mistaken, I suspect he got the $\lambda^5$ figure by looking at $E=\frac{1}{2}I\omega^2$ and noting that $I\rightarrow\lambda^5I$, but forgot to look in detail as to how $\omega$ transforms.

NOTE: If it makes it easier to swallow, this analysis can also be carried out by assuming the child is a cylinder of radius $R$, length $L$ and constant density $\rho$ which is free to rotate about its base. The inertia tensor is $$\mathbf{I}=\left( \begin{array}{ccc} \frac{1}{12} L \pi R^2 \left(4 L^2+3 R^2\right) \rho & 0 & 0 \\ 0 & \frac{1}{12} L \pi R^2 \left(4 L^2+3 R^2\right) \rho & 0 \\ 0 & 0 & \frac{1}{2} L \pi R^4 \rho \\ \end{array} \right)$$ and repeating the previous analysis using $I=\mathbf{I}_{xx}$ and $m=\pi L \rho R^2$ gives a head energy which scales as $$T=m_h(L\omega)^2=\frac{6 g L^3 m_h}{4 L^2+3 R^2}\rightarrow\lambda^4T$$ under scale transformation.

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    $\begingroup$ Shouldn't the head's mass scale by an additional factor of $\lambda$? $\endgroup$ Commented Mar 24, 2014 at 16:05
  • $\begingroup$ Indeed, keeping the density $\rho$ of your rod-child constant, the mass will add another factor of $\lambda$: $ m = \rho \cdot l \rightarrow \rho \cdot \lambda l$. $\endgroup$
    – Neuneck
    Commented Mar 24, 2014 at 16:07
  • $\begingroup$ @JerrySchirmer: Do you mean $m_h\rightarrow\lambda^4m_h$? I'm not sure I understand. Modeling the head as a cube attached to the top of the rod, shouldn't it's mass scale cubically like a volume element? $\endgroup$ Commented Mar 24, 2014 at 16:29
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    $\begingroup$ @Neuneck: While I am using a rod for geometric simplicity, I am implicitly assuming that the child's body is actually a volume element which scales cubically, but which has been compressed into a rod, so $\rho\rightarrow\lambda^2\rho$. This ensures that the child's mass scales correctly like a volume should. $\endgroup$ Commented Mar 24, 2014 at 16:31

It appears that the statement is half-right....

As answered by @DumpsterDoofus, the factor for a scaled object falling a scaled height, is the fourth power.

However, if we scale an object that is traveling at a velocity of $x \frac{\text{body-lengths}}{\text{second}}$, then a fifth power for the kinetic energy is correct. The size and thus mass in the kinetic energy equation scales as the cube, and the velocity term in the kinetic energy scales as the square.

The difference in the two scenarios comes from correctly not scaling the distance factor in the value of $g$.


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