Teaching Physics with the Physics Suite
(a) A sculptor builds a model for a statue of a terrapin to replace Testudo.* She discovers that to cast her small scale model she needs 2 kg of bronze. When she is done, she finds that she can give it two coats of finishing polyurethane varnish using exactly one small can of varnish. The final statue is supposed to be 5 times as large as the model in each dimension. How much bronze will she need? How much varnish should she buy? (Hint: If this seems difficult, you might start by writing a simpler question that is simpler to work on before tackling this one.)
(a) The simpler problem we might tackle is to consider her making a simpler sculpture than Testudo -- say a cube. (We could imagine that she is doing a modernistic version and incising small lines on the cube so as to give it the apparent shape of a terrapin. Perhaps she is "making a statement.") For a cube multiplying each dimension by 5 means that the volume increases by a factor of 5x5x5=125. Since the bronze is bought by volume, she must buy 125x2 kg = 250 kg.
For the varnish, she has to cover 6 sides of the square. If the original square had a side a, the area she has to cover is 6a2. This requires 2 cans of varnish so each can can cover an area 3a2. When she multiplies her dimensions by 5 each side has an area (5 a)x(5 a) = 25 a2. There are 6 sides, so the total area is 6 x 25 a2 = 150 a2. Since she needs one can to cover an area 3 a2 with two coats she will need 25 cans -- 25 times as many as before. If we assume she only gives it one coat, she will need 12.5 cans worth of varnish so she will have to buy 13 cans.
Now what about a more complex terrapin shape? If you imagine the terrapin now made up of small cubes, it's clear that when you scale up, the amount of bronze for each little cube scales up by a factor of 125 and the area of each side of the little cube scales up by a factor of 25. We can choose our cubes smaller and smaller until we have essentially a perfect terrapin and the factors are still 125 and 25.
Of course you can see this much more easily by using dimensional analysis. We know that a volume has the dimensions of the cube of a length ([volume] = L 3). If we scale the length by some factor (say 5), then the volume scales by the cube of that factor, or 125. We know that an area has the dimensions of the square of a length ([area] = L2). If we scale the length by some factor (say 5), then the volume scales by the square of that factor, or 25. And that will be true for any shape at all.
(b) To decide what the effect of convolution is, let's suppose that the human brain were not convoluted. How much bigger would it's surface area be than a mouse's brain? We can then suppose that the remaining factor is due to convolution, What we learned from the previous problem is that because of its dimension (L2), area scales as the square of the linear dimension. That is, if we increase all linear dimensions of an object by a factor x, then its area increases by a factor x2 (and its volume increases by a factor x3).
We therefore only need to know how much the linear dimension is increased in going from a mouse to a human brain. (Assuming they otherwise have about the same shape.) This really depends on which mouse and which human you choose. All of them are not the same. The statement in the problem came directly from Science magazine, but they did not specify anything more. To really solve this problem, we would need to know not only the average size of human and mouse brains, but how much they vary from sample to sample (the expected range of variation). Since we don't have that info, let's try to do the best estimate we can.
Measuring my head from front to back, I expect my brain has a length of about 15 cm. Thinking about the size of a mouse's head and looking at my ruler, I estimate that a mouse's brain has a length of about 1 cm. This says that the linear size of a human brain is about 15 times that of a mouse's so x = 15 and the human brain area would be about x2 = 15x15 = 225 times the mouse brain area. This would say that the convolutions account for a factor of 1000/225 or about 4. That is, the convolutions of the human brain give it about four times the area one might expect from size alone.
How good are these numbers? I just made them up, after all from looking at a ruler and thinking about my head and a mouse. Looking on the web, we find data to suggest that the average mass of a human brain is about 1500 grams and the average mass of a mouse brain is about 0.4 grams. Since the mass should be proportional to the volume with a constant (the density -- assuming that all brains are made of about the same stuff and therefore have the same density) the mass should scale like x3 where x is the linear increase in size from human to mouse brain. That is,
Solving this for x gives
Very close to the number from my estimate, so my estimate is probably pretty good! For more info on this stuff, check out the website at http://serendip.brynmawr.edu/bb/kinser/Home1.html.
Page last modified August 29, 2004: G08