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Suppose you need to slow down a turning motor so that a gear turns at an angular velocity $\frac{a}{b}$ of that of the motor shaft, where $a$ and $b$ are natural numbers. For example, this set of four gears achieves a slowdown of $\frac{1}{60}$: the $10$ teeth of the motor green gear are slowed to $\frac{1}{6}$ by the $60$ teeth of the blue gear, and the $10$ teeth of the red gear are slowed another $\frac{1}{10}$ by the $100$ teeth of the gray gear:
           Gears
           (Image from AMS Feature Column.)

There are two natural costs to a gear train: the teeth cost, the total number of teeth (assuming the manufacturing costs are dominated by the teeth), and the metal cost, the total area of the gears (assuming the cost of the metal dominates). Since the area is determined by the radii, which are proportional to the number of teeth, let us say the metal cost is the sum of the squares of the number of teeth. So the gear train above has teeth cost $10 + 60 + 10 + 100 = 180$ and metal cost $10^2 + 60^2 + 10^2 + 100^2 = 13,800$. If we insist no gear has fewer than $10$ teeth, then another alternative would be to have just one gear with $600$ teeth attached to $10$, which would have much larger teeth and metal costs. The gear train $(10/20,10/20,10/150)$—$\frac{1}{2} \frac{1}{2} \frac{1}{15}$—is also more costly.

Q1. If no gear has fewer than $t$ teeth, characterize the optimal gear train (under either cost measure) to slow down rotation to exactly $\frac{1}{n}$. Short of a characterization, is there an algorithm more efficient than trying all trains compatible with the factorizations of $n$?

Q1a. The case $t=2^k$ and $n=2^m$ seems especially approachable. For example, for $t=2^3=8$ and $n=2^8=256$, it seems the optimal teeth cost is achieved by four $\frac{1}{4}$ slowdowns, that is, a cascade of four $8/32$ slowdowns, achieving a cost of $160$, whereas eight $8/16$ slowdowns has a cost of $192$. For $t=8$ but $n=2^{16}=65,536$, again it seems that a series of $8/32$ slowdowns is optimal: eight yield a teeth cost of $320$. But I have not proven that these are optimal.

Q2. Same question to achieve a $\frac{a}{b}$ slow-down.

Q3. Is there a clear case where the optimal train under the teeth cost is not the same as that for the metal cost? (I haven't explored the metal model.)

It seems likely that Q1 might map into a well-studied number theory problem, somehow balancing the size of factors. If anyone knows, I'd appreciate learning of the connection. Thanks!

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I don't think even trying all trains compatible with the factorizations of $n$ is sufficient. e.g. take $n=17$, $t=4$, then $(17:4)(16:4)$ beats $(68:4)$. –  Will Sawin Oct 11 '12 at 0:56
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For $Q3$, take $n = 4$, $t = 2$. It seems to me (no proof, and Will's example might mean that I'm wrong), that the only relevant slowdowns will be either $(2/8)$ or $(2/4,2/4)$. The first has teeth cost $10$ and metal cost $68$, while the second has teeth cost $12$ and metal cost $40$. –  Michael Biro Oct 11 '12 at 1:41
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You might consider the friction belt analogue first, where the ratio can be any real number, and pretend the cost is proportional to sums of radii or areas as appropriate. You can then use calculus to come up with a minimum, and then continued fractions or rational approximation to pull the results back to the tooth case. Gerhard "Deciding Between Bridge Or Implants" Paseman, 2012.10.10 –  Gerhard Paseman Oct 11 '12 at 2:51
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There is a problem with units. Namely, I don't understand why $\pi$ square inches of metal should cost the same as however many teeth fill $2\pi$ linear inches. Presumably, a best answer would explain how to minimize any positive linear combination of #(teeth) and (#teeth)^2. –  Theo Johnson-Freyd Oct 11 '12 at 3:03
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See ams.org/samplings/feature-column/fcarc-stern-brocot for a discussion of the Stern-Brocot tree which has been used to design optimal gear trains. Tables of optimal gear trains based on this method have been around for a long time. –  Brian Borchers Oct 11 '12 at 3:42
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