Re: [math-fun] New challenge: Offset Sonsky Gears
Numerically, it just comes too close to feel like an accident. It's very similar to the six-planet systems that symmetry arguments say are exact. And I *think* I'm starting to see hints of a pattern to this, but I'm not sure. I'm going to try to examine the specific numbers of this one and see if it can be shown to be exact. (I'm hoping maybe it has a lot of serendipitous things like 'cos alpha = 1/2' in it, which we can put nice, clean values to, rather that stuff like 'cos alpha = 137/293') When I get a bit more time, I'll see about posting my algorithm for generating these numbers. On 07/13/15 23:06, Tom Rokicki wrote:
Can you share any support for your belief, or is this a puzzle for us to resolve?
-tom
On Mon, Jul 13, 2015 at 10:50 PM, William Somsky <wrsomsky@gmail.com <mailto:wrsomsky@gmail.com>> wrote:
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
It would be nice to see links to these animations (Dropbox? YouTube?), so that the math-fun list could view them. Failing that, would somebody email a copy directly to me please? Fred Lunnon On 7/14/15, William Somsky <wrsomsky@gmail.com> wrote:
Numerically, it just comes too close to feel like an accident. It's very similar to the six-planet systems that symmetry arguments say are exact. And I *think* I'm starting to see hints of a pattern to this, but I'm not sure. I'm going to try to examine the specific numbers of this one and see if it can be shown to be exact. (I'm hoping maybe it has a lot of serendipitous things like 'cos alpha = 1/2' in it, which we can put nice, clean values to, rather that stuff like 'cos alpha = 137/293')
When I get a bit more time, I'll see about posting my algorithm for generating these numbers.
On 07/13/15 23:06, Tom Rokicki wrote:
Can you share any support for your belief, or is this a puzzle for us to resolve?
-tom
On Mon, Jul 13, 2015 at 10:50 PM, William Somsky <wrsomsky@gmail.com <mailto:wrsomsky@gmail.com>> wrote:
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
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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.
On 07/14/15 11:42, Warren D Smith wrote:
[...]
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?
[...]
Ah, but (5+27):(27+53) = 32:80 = 2:5, and those are the coefficients needed to get the "strap length" around the 5-tooth gear...
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.
participants (5)
-
Fred Lunnon -
M. Oskar van Deventer -
Warren D Smith -
William R Somsky -
William Somsky