Optimal Gear Trains - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:29:49Z http://mathoverflow.net/feeds/question/109343 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109343/optimal-gear-trains Optimal Gear Trains Joseph O'Rourke 2012-10-11T00:28:05Z 2012-10-11T01:07:58Z <p>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: <br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <img src="http://www.ams.org/featurecolumn/images/december2008/gears.1.gif" width="400" alt="Gears" /> <br /> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <sub>(Image from <a href="http://www.ams.org/samplings/feature-column/fcarc-stern-brocot" rel="nofollow">AMS Feature Column</a>.)</sub> <br /></p> <p>There are two natural <em>costs</em> to a gear train: the <em>teeth cost</em>, the total number of teeth (assuming the manufacturing costs are dominated by the teeth), and the <em>metal cost</em>, 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 $(10/20,10/20,10/150)$&mdash;$\frac{1}{2} \frac{1}{2} \frac{1}{15}$&mdash;is also more costly.</p> <p><b>Q1</b>. 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 $n$?</p> <p><b>Q1a</b>. 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.</p> <p><b>Q2</b>. Same question to achieve a $\frac{a}{b}$ slow-down.</p> <p><b>Q3</b>. 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.)</p> <p>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!</p>