Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It seems pretty well established that organisms grow according to a 3/4-power law. For example, Niklas and Enquist, in their paper "Invariant scaling relationships for interspecific plant biomass production rates and body size," PNAS 2001, 98(5):2922 -2927, say:

Annualized rates of growth $G$ scale as the 3/4-power of body mass $M$ over 20 orders of magnitude of $M$ (i.e., $G \propto M^{\frac{3}{4}}$).

Does anyone know if there is some geometric reason to expect such a growth-rate law?

$$\frac{d M}{d t} \sim M^{\frac{3}{4}}$$

Apparently attempts to derive this growth-rate law from Kleiber's Law, which claims that metabolic rate scales as $M^{\frac{3}{4}}$, are controversial. So I was wondering if there might be some geometric viewpoint that makes growth proportional to $M^{\frac{3}{4}}$ not unexpected.

share|cite|improve this question
Interesting topic! There's an accessible discussion of the geometry here: – minopret Feb 28 '12 at 1:52
@minopret: Thanks for the useful link! Note that those pages focus on metabolism rate, rather than growth rate, but there is a (conjectured) connection. – Joseph O'Rourke Feb 28 '12 at 12:03
The literature of allometric growth rates is littered with geometric descriptions of rates that ended up not being supported by data. It looks like it's easy to come up with an explanation but hard to know that your explanation is right. – Charles Feb 28 '12 at 16:12

h=height, M=mass, A=cross-sectional area of stem d=density of stem V=volume of plant

The total cross-sectional area of a plant remains roughly the same at the top as at the bottom. (Da Vinci's rule).

Plants collect light for photosynthesis along their cross-sectional area. The thickness of stem A needed to support a plant is proportional to $h^3$.

$A\propto h^3$

$M = Ahd$

$A \propto M^{3/4}$

growth rate $\propto$ area exposed to the sun for photosynthesis $\propto \frac{V}h \propto A$

share|cite|improve this answer
Nice! $\mbox{}$ – Joseph O'Rourke Feb 29 '12 at 0:39
Why is the thickness of stem $A$ needed to support a plant proportional to $h^3$? That seems to be the key step, but I am not seeing the logic behind it... – Joseph O'Rourke Feb 29 '12 at 13:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.