schoolyard subversion

2000-10-10 [< * >]

Kragen Sitaker informs me on the surface area/volume controversy (1 2), saying that it does work with real units, but only if you divide:

On the surface-to-volume ratio: I have several tens of feet of intestines in my belly, with extremely rough surfaces, to solve this exact problem. You may note that slime molds, which are extremely large multinuclear cells, adopt shapes with a very high surface-to-volume ratio so they can feed.

The problem, as you correctly saw, is that it is not a dimensionless quantity. The volume-to-surface-area ratio of a one-foot cube is 1 cubic foot divided by six square feet: one sixth of a foot, or 2 inches. It stays the same if you change units: 1728 cubic inches divided by 6*144 = 864 square inches is two inches. 28317 cubic centimeters divided by 6 * 929 square centimeters is 5.08 centimeters -- two inches, again.

Dimensional analysis is, for some reason, traditionally treated as an arcane topic in elementary, middle, and high schools. That's why I had such a hard time with E = mc^2 for many years, before I knew enough physics to understand why the units worked out.

Anyway, you can measure how much power a particular kind of cell needs per cubic micron or whatever, divide that by how much power it can consume from the environment per square micron or whatever, and get a maximum survivable surface-to-volume ratio for the cell, denominated in microns (or whatever units of length you used). For a given cell shape, the actual surface-to-volume ratio will increase linearly with the length of the cell, and so, above a certain point, the cell won't be able to eat enough to survive.

Thanks! Now things are making a lot more sense.

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