



 
Hi Grant, I redrew the diagram you sent and added some labels. A few years ago I answered a similar question for a system involving two pulleys. You can use the same technique to find the length of the belt between $P$ and $Q.$ Using your notation I get \[PQ = \sqrt{c^2  \left(\frac{Dd}{2}\right)^2}.\] Again from the technique in the earlier problem \[\angle PRU = \angle QVU = tan^{1}\left(\frac{PQ}{\left(\frac{D  d}{2}\right)}\right).\] If you measure the angles in radians then $\angle SRP = \pi  \angle PRU$ and you can then find the lenghs of the belt sections from $S$ to $P$ and $Q$ to $U$ using the fact that the length of an arc of a circle is radius of the circle times the measure of the angle in radians. Next look at the section between the large pulley and the tension pulley. $R$ and $W$ are the centers of the two circles and $XY$ is parallel to$RW.$ Triangle $ZRW$ is a right triangle, $WZ = y \mbox{ and } ZR = c  x$ and hence Pythagoras theorem can be used to find the distance between the centers, $RW.$ Triangle $XAY$ is also a right triangle, $YX = RW$ and $XA = \frac{D + T}{2}$ and hence the length of belt $AY$ can be found using Pythagoras theorem. Since triangle $ZRW$ is a right triangle the measure of angle $ZRW$ can be found fron the inverse tangent function. Likewise $\angle YXA = \angle WRA$ can be found from the inverse tangent function. Thus we have $\angle ARS = \pi  \angle ZRA$ and we can find the length of belt from $S$ to $A.$ Furthermore $\angle YWC = \angle ARB = \frac{\pi}{2}  \angle ZRA$ and thus we can find the length of belt from $Y$ to $C.$ A similar construction can be used to find the length of the remaining piece of the belt from $C$ to $U.$ Harley  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 