Wait... In your Theorem 3, in regards to the gear angle/phases, you seem to be making the claim that any set of abutted gears will automatically mesh properly? That if, say, I have four gears of radius 1 with, say 8 teeth each, that as long as I place their axles such that the A-B, B-C, C-D, and D-A distances are all two, that they will always mesh properly? I don't believe that is true. Ah, I see where your reasoning is faulty. You're measuring the exterior angles. However, for measuring the phase-angle of the gears, you need to "follow the flow" and you need to measure the *interior* angle of every other gear. Imagine threading a timing belt around and thru the gears that will flow along as the gears rotate. That belt will go (wlog) on the outside of the gears that turn clockwise, and on the inside of the gears that turn counter-clockwise. Hopefully you see what I mean. It's a bit hard to explain w/out a diagram, and I'm not set up at the moment to create one. - Bill On 2015-07-10 05:29, Warren D Smith wrote:
I. If we are not interested in movable gear-axes, then it seems to me that Somsky's sum-of-radii condition is not needed, which makes my theorem 2 work without needing to choose any "mates" obeying Somsky's condition, and then for any N>1, including odd N, number of planets, in Deventer's "unsymmetric planetary gears."
II. Also as I remarked before, after unconfusion supplied with aid by Rokicki, my theorem 1 was false, or more precisely: it is true that Somsky's type of Omega-altering motion has bounded travel, but the other kind of motion with all gear-axis locations fixed, permits infinite turning.
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III. I now believe I can prove this highly general THEOREM 3: Let G be _any_ planar bipartite graph. ( https://en.wikipedia.org/wiki/Planar_graph , https://en.wikipedia.org/wiki/Bipartite_graph .) * With each vertex of G we can associate a circular gear in the plane. * With each edge (a,b) of G we can demand the gears associated with vertices a and b be perfectly meshed. * With each non-edge (a,b) of G we can demand the gears a,b do not mesh and do not conflict by overlapping. * Any one of the vertices of G can be demanded to be an "outer ring" type gear with all the others being conventional filled-circle type gears. * All gear radii can be chosen to be integers in a countably infinite number of scaling-inequivalent ways. * The gears will all rotate in sync with their centers (axes) fixed at the specified vertex locations, with infinite angle of turning permitted. And suitable initial angles exist such that all gear teeth mesh.
Note, for non-bipartite G, there would be an odd-cycle in the graph, and then gear rotation would be impossible because of a sign problem. E.g. 3 gears meshed in a triangle are jammed and cannot rotate, but 4 gears meshed according to a quadrilateral can rotate.
Also for non-planar G, no such configuration is possible in the sense that some gears would need to overlap. (However, with offset parallel planes some such overlaps could be permissible, in which case nonplanar G could be permissible, and I have not attempted to determine which nonplanar graphs G then would be permissible. It is easy to see that some such nonplanar G do exist.) Thus in these senses THEOREM 3 is "best possible." In particular my theorem 2 about the unsymmetrical planetary gears is only a very special case of it.
HOW TO PROVE IT: Many of the underlying ideas were described in a paper I wrote many years ago, which you can read here: paper #6 here: http://rangevoting.org/WarrenSmithPages/homepage/works.html the new ideas beyond that paper involve showing it can always be done with integer radii, in a countably infinite number of ways, and that suitable angles exist. Wlog let G be "polyhedral" having V vertices, E edges, F faces, where V-E+F=2 by Euler theorem. (Basically because if G's edge-set is a subset of such a polyhedral graph the proof will only get easier.) Each face is a 2k-gon for integers k>1. This enables seeing that F<=V-2, and then that E<=2V-4. Now consider supplementing G's edge-set with extra "red" edges to convert it to a maximal (fully triangulated polyhedral) graph H. The number of extra red edges will be at least F (if all faces are quadrilaterals) and indeed at least 2E-3F=2(V-2)-F=2V-F-4>=V-2. Embed H using the Andreev-Thurston theorem (or my paper's polynomial time algorithmic version of it). By an appropriate inversive transformation and real-rescaling, any particular G-vertex can be demanded to be "outer" i.e. an outer-ring type gear, of specified radius, let us say radius=1, with all the other G-vertex circles lying inside it, and two of these interior circles also have specified sizes which I can and will demand to be rational numbers, by using the 3 continuous degrees of freedom of the 3 parameters specifying the inversive transformation and rescaling. Now, there are V-3 gears (G vertices) whose radii have not yet been specified. These are variable because the red edges of H are not present in G, and for each one we remove from H there is a new degree of freedom of motion of the G-vertices while still satisfying the conditions that all the Andreev-Thurston circles associated with the G-vertices are tangent if and only if there is a G-edge between those two vertices, otherwise they do not touch. These motions are permissible within some perhaps small, but nonzero range. We keep removing the red edges of H one by one, using the resulting freedom to specify one more gear-radius associated with 1 more G-vertex, to be (again each time) a rational number. This can be done because the rationals are everywhere dense in the reals, so no matter how small the range of motion we get by removing a red H-edge, there always is a rational radius inside that range. We can do this for at least the 3 initially specified rationals, plus at least V-2 more. And since V-2+3>V, we clearly have enough degrees of embedding freedom to make all gear radii rational in this way, indeed with at least one degree of freedom left over. Now multiply everything by LCM(denominators) to make all gear radii be integers.
Now at this point we need to associate an "initial angle of rotation" with each gear. If you have a gear, at a specified rotation angle A to the x-axis-direction, then any gear that is meshed with it has a rotation angle that then is determined (up to rotations by multiples of 360/T degrees if T is its number of teeth) by A and by the relative locations of the two gears. If we go round a polygonal cycle in the graph G, fortunately the total sum of exterior "bend" angles in this polygon must sum to 360 degrees, which means we never get a conflict at the end of the cycle when we get back to the original gear; the final demand is always the same modulo 360 as the initial demand. So we can color "chains" of edges of G "blue" one chain at a time, each chain linking two previously-blued vertices as its endpoints. Each chain we add will not generate a conflict. Eventually the entirety of G is colored blue, i.e. we have chosen acceptable initial angles of rotation for every gear, without ever suffering a conflict.
In that argument, you might worry that there are E demands about angles, but only V angles which is too many equations for too few unknowns. However, the "sum of bends in polygon=360" Euclidean theorem means that really, many of these E demands are redundant, i.e. are consequences of others, not independent. The number of independent demands is F, due to the well known theorem that G's faces form a "cycle basis"; and since F<=V-2 this worry is obliterated.
At this point we have specified the integer radii, the center (axis) locations, and the initial rotation angles, for every gear. Finally, we can turn one of the gears, and the others will all turn too with the invariant that the rim-speed is equal for every gear. (And due to bipartiteness, the speeds all have the correct signs.) This invariant shows that there is no "jam" and it turns freely forever. QED.
3+F>2+F<=V