Optimal Gear Trains - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T22:29:49Zhttp://mathoverflow.net/feeds/question/109343http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/109343/optimal-gear-trainsOptimal Gear TrainsJoseph O'Rourke2012-10-11T00:28:05Z2012-10-11T01:07:58Z
<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 />
<img
src="http://www.ams.org/featurecolumn/images/december2008/gears.1.gif"
width="400"
alt="Gears" />
<br />
<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
$(10/20,10/20,10/150)$—$\frac{1}{2} \frac{1}{2} \frac{1}{15}$—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>