On 7/12/15, WRSomsky <wrsomsky@speakeasy.net> wrote:

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.

--that would indeed be what I was claiming. You do not believe?

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.

--Oh dear. So it is kind of an alternating sum, and I omitted the minus signs every other term? Shit. The exterior "turn" angle and interior polygon corner angles are exact negatives of each other (mod 180) so it does not matter in my argument which one you use. (I prefer the former.) However, omitting minus signs in every other term, is something that really does matter! You have refuted my proof. However, G is a bipartite graph, i.e. has two classes of vertices ("pink" and "blue") and as we travel round any cycle in G, we alternately hit pink and blue vertices. So for each cycle, there really are two sums we need to worry about: the pink-angles sum p and the blue-angles sum b. My old argument was that p+b=0 (mod 360) which was automatic for Euclidean polygons. What we really want is that p-b=rational (mod 360). That sadly is not automatic. But it is implied by, for example, p=rational. However, the pink vertices happen to be the sun and outer-ring gear (which I'll call the "antisun") in planetary systems; the planets then are the blue vertices. So long as the planets viewed from the sun are spaced at rational angles apart (measured in degrees) and ditto for the antisun, we are ok. Indeed it suffices if each sun-viewed angle plus its corresponding antisun-viewed angle, is a rational. These demands are certainly true in Somsky's all-collinear case where p=0, or any distorted version of it where, e.g, his angle "Omega" is rational.