### The Mathematics of Honeycomb: Part III

In the first two parts of the article, I have given a historical review of the honeycomb problem and a calculus-based technique to compute the honeycomb rhombic angle. In this article, I will try to give a crude analysis on the results we obtained in Part II.

For a closed curve in two-dimensional space, there exists an inequality to govern its perimeter \(L\) and area \(A\): $$A \le \tfrac{1}{4\pi}L^2$$ This inequality is called the isoperimetric inequality. It simply means that the ratio of \(A/L^2\) cannot be greater than \(\frac{1}{4\pi}\). This result can be generalized to three-dimensional space to give: $$V \le \tfrac{1}{6\sqrt{\pi}}S^{3/2}$$ This shows that the theoretical maximum \(V/S^{3/2}\) (when you have the least surface for a given volume) is \(\frac{1}{6\sqrt{\pi}} = 0.0940\). For honeycomb cells, the ratio of \(V\) to \(S^{3/2}\) is: $$\frac{V_{\rm honeycomb}}{(S_{\rm honeycomb})^{3/2}} = \frac{\frac{3\sqrt{3}}{2}a^2 b}{\left(6ab + \frac{6}{\sqrt{8}}a^2\right)^{3/2}}= \frac{\sqrt{2}\lambda}{(\sqrt{2}+4\lambda)^{3/2}}$$ where \(a = \lambda b\). For honeycomb, the value of \(V/S^{3/2}\) is approximately \(0.0901\) if \(\lambda_{\rm drone} = 2.64\), which is pretty close to the theoretical maximum. On the other hand, the same ratio is approximately \(0.0866\) for a prism with hexagonal top and hexagonal bottom.

This result is not too surprising from an evolutionary perspective. Suppose initially all of the ancestors of the honeybees were building their cells the easier way with hexagonal top and hexagonal bottom. Owing to genetic mutation, a new group of honeybees evolved the ability to construct cells with rhombic bottom. Over time, this group of honeybees gained evolutionary advantage for able to economize their resources, they were able to reproduce more efficiently with the same amount of food resources, and slowly displaced all of the old honeybee groups.

Note that a small difference of 4 percent in this ratio is sufficient for Nature to decide which honeybee species were to be eliminated. Nature is very calculative in her rule of natural selection.

For a closed curve in two-dimensional space, there exists an inequality to govern its perimeter \(L\) and area \(A\): $$A \le \tfrac{1}{4\pi}L^2$$ This inequality is called the isoperimetric inequality. It simply means that the ratio of \(A/L^2\) cannot be greater than \(\frac{1}{4\pi}\). This result can be generalized to three-dimensional space to give: $$V \le \tfrac{1}{6\sqrt{\pi}}S^{3/2}$$ This shows that the theoretical maximum \(V/S^{3/2}\) (when you have the least surface for a given volume) is \(\frac{1}{6\sqrt{\pi}} = 0.0940\). For honeycomb cells, the ratio of \(V\) to \(S^{3/2}\) is: $$\frac{V_{\rm honeycomb}}{(S_{\rm honeycomb})^{3/2}} = \frac{\frac{3\sqrt{3}}{2}a^2 b}{\left(6ab + \frac{6}{\sqrt{8}}a^2\right)^{3/2}}= \frac{\sqrt{2}\lambda}{(\sqrt{2}+4\lambda)^{3/2}}$$ where \(a = \lambda b\). For honeycomb, the value of \(V/S^{3/2}\) is approximately \(0.0901\) if \(\lambda_{\rm drone} = 2.64\), which is pretty close to the theoretical maximum. On the other hand, the same ratio is approximately \(0.0866\) for a prism with hexagonal top and hexagonal bottom.

This result is not too surprising from an evolutionary perspective. Suppose initially all of the ancestors of the honeybees were building their cells the easier way with hexagonal top and hexagonal bottom. Owing to genetic mutation, a new group of honeybees evolved the ability to construct cells with rhombic bottom. Over time, this group of honeybees gained evolutionary advantage for able to economize their resources, they were able to reproduce more efficiently with the same amount of food resources, and slowly displaced all of the old honeybee groups.

Note that a small difference of 4 percent in this ratio is sufficient for Nature to decide which honeybee species were to be eliminated. Nature is very calculative in her rule of natural selection.

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