Gentlemen, Whiles I am still unsure whether there is agreement about Tom Rokicki's analysis that all Somsky Gears can be offset, nor can I follow that various mathematical reasonings, so I did an experiment with regular 120-degrees planetary gears. The attached sketch shows that a 26-8-10 fits as well as the 27-9-9 that it was offset from. Of course this does not prove anything, but at least it is consistent with Tom's analysis. Oskar -----Original Message----- From: Warren D Smith Sent: Tuesday, July 14, 2015 8:42 PM To: Fred Lunnon Cc: math-fun ; Tom Rokicki ; Bill Gosper ; M. Oskar van Deventer ; Julian Ziegler Hunts Subject: Re: [math-fun] New challenge: Offset Sonsky Gears WRSomsky 14 July 2015: I believe the following eight-planet system to be exact: ring = 53, sun = 27, offset = 20.00, planets = 3, 5, 5, 13, 13, 21, 21, 23 Here is an animation w/ the 23-tooth gear omitted (as it overlaps): [gif omitted] WDSmith: I compute RingCenter-to-SunCenter angles (in degrees) as viewed from planets.
From 3: arccos( ((27+3)^2+(53-3)^2-20^2)/(2*(27+3)*(53-3)) ) = arccos(1) = 0
From 5: arccos( ((27+5)^2+(53-5)^2-20^2)/(2*(27+5)*(53-5)) ) = arccos(61/64) = 17.6124390703503806061448260043641699743420118772169176197494099557435320963627421627054510333214083707090019564325635310... = X = 360/20.4400991... = [17; 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 19, 1, 4, 9, 5, 1, 1, 3, 1, 3, 5, 1, 2, 1, 1, 2, ...] as a continued fraction Here [17; 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1] = 3681/209 is a close rational approx
From 13: arccos( ((27+13)^2+(53-13)^2-20^2)/(2*(27+13)*(53-13)) ) = arccos(7/8) = 28.9550243718598477575420695982543320102631952491132329521002360177025871614549031349178195866714366517163992174269745875 = Y = 360/12.433075... = [28; 1, 21, 4, 3, 1, 2, 1, 1, 2, 7, 1, 2, 1, 6, 1, 4, 3, 1, 3, 1, 2, 2, 1, 1, 1, 97, ...] Here [28; 1, 21, 4, 3, 1, 2, 1, 1, 2] = 184125/6359
From 21: arccos( ((27+21)^2+(53-21)^2-20^2)/(2*(27+21)*(53-21)) ) = arccos(61/64) we already encountered this angle X
From 23: arccos( ((27+23)^2+(53-23)^2-20^2)/(2*(27+23)*(53-23)) ) = arccos(1) = 0
Here the two angles X, Y that look irrational in degrees indeed are. These claims happen to be the same claim I made before re one of Somsky 6-planet animations, and they allegedly are proven in the same paper I cited before: JH Conway, C Radin, L Sadun: On angles whose squared trigonometric functions are rational, Discrete Comput. Geom. 22 (1999) 321-332 http://www.ma.utexas.edu/users/radin/papers/geodetic.pdf and the proof ultimately derives from Alan Baker's theorem about "linear forms in logarithms" https://en.wikipedia.org/wiki/Baker%27s_theorem Now previously I had claimed not only that, but indeed X,Y were independent transcendentals, that is a*X+b*Y+c is always irrational and nonalgebraic if a,b,c rational with a*b nonzero. This, I thought also followed from the same papers and theorems. But I now see this stronger claim by me must have been WRONG because in fact we have the miraculous identity 2*X + 5*Y = 2*arccos(61/64) + 5*arccos(7/8) = 180 degrees, which I now have confirmed by high precision computation to over a thousand decimal places. My old claim about aX+bY+c always being irrational, made me believe it was IMPOSSIBLE for the variously-named "belt condition" or "alternating sum" to be 0 or any rational if the 5 and 21 toothed gears are involved in the path. Therefore, I believed this -- as well as Somsky's 6-planet animation with sun=25, antisun=55, planets in cyclic order 25,15,7,5,7,15 which had sun-antisun angle X as viewed from 7 and Y as viewed from 15 -- both were NOT exact solutions, they are merely good approximate solutions exhibiting slight phase mismatches. However, in view of the discovery of the miraculous identity 2*X+5*Y=180 degrees, holy cow, does this change that verdict? It doesn't seem to because the gears invovled do not have 5:2 radius ratio? In case you are wondering -- How can 2*X+5*Y=180 degrees be proven? -- Here is a proof. We can use the cosine multiple angle and angle-sum identities to see that Cos[2*X+5*Y] = Cos[X]^2*Cos[Y]^5-Cos[Y]^5*Sin[X]^2-10*Cos[X]*Cos[Y]^4*Sin[X]*Sin[Y] -10*Cos[X]^2*Cos[Y]^3*Sin[Y]^2+10*Cos[Y]^3*Sin[X]^2*Sin[Y]^2 +20*Cos[X]*Cos[Y]^2*Sin[X]*Sin[Y]^3+5*Cos[X]^2*Cos[Y]*Sin[Y]^4 -5*Cos[Y]*Sin[X]^2*Sin[Y]^4-2*Cos[X]*Sin[X]*Sin[Y]^5 for general X,Y and then substitute in our values Cos[X]=61/64, Cos[Y]=7/8, Sin[X]=sqrt(375/4096), Sin[Y]=sqrt(15/64) to get cos(2*X+5*Y) = (61/64)^2*(7/8)^5-(7/8)^5*sqrt(375/4096)^2-10*(61/64)*(7/8)^4*sqrt(375/4096)*sqrt(15/64)-10*(61/64)^2*(7/8)^3*sqrt(15/64)^2+10*(7/8)^3*sqrt(375/4096)^2*sqrt(15/64)^2 +20*(61/64)*(7/8)^2*sqrt(375/4096)*sqrt(15/64)^3+5*(61/64)^2*(7/8)*sqrt(15/64)^4-5*(7/8)*sqrt(375/4096)^2*sqrt(15/64)^4-2*(61/64)*sqrt(375/4096)*sqrt(15/64)^5 = -56440307/33554432 + 22885875/33554432 = -1 thus proving the miracle identity.