Re: [math-fun] Offset Somsky Gears, coaxial offset
Dear Bill, Tom, all, Thanks to Tom's fantastic tool and your explanation, I now understand that any set of Somsky Gears can be offset to produce an new set. I tested this with a random set from Tom's program (26-9-7-5) and I could see that its offset (25-10-6-4) is a Somsky set too. However, those two Somsky sets are not coaxial. I used some crude editing to overlay the two, see the attached figure. It is clear to see that the two are not coaxial. Tom has already provided me with a set of Somsky Gears, which can be coaxially offset into another set of Somsky Gears. However, not all sets of Somsky Gears can be coaxially offset like this. So what is the additional requirement that enables this? Is there a quick way how I can determine whether a Somsky set can be coaxially offset? Best regards, Oskar -----Original Message----- From: wrsomsky@gmail.com Sent: Friday, July 31, 2015 6:44 AM
From some thinking I've done, the answer is yes. Ignoring overlap, any set of Somsky Gears can be offset to produce an new set. You can prove it by noting that for a given ring-sun-planet, the strap-length changes by pi (one half a tooth) if you increase the ring and planet radii by one, and decrease the sun radius by one (or vice versa).
On 2015-07-21 13:12, M. Oskar van Deventer wrote:
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
No, ignoring overlap, they would all be, or CAN all be, coaxial... The one thing you need to be careful with though, is that, in general, there are several different sun-ring offsets that work for a given set of gears. If you pick one offset for one set of gears and a different one for the shifted set, they won't be coaxial. But if you pick the same offset, they are. On 08/03/15 01:46, M. Oskar van Deventer wrote:
Dear Bill, Tom, all,
Thanks to Tom's fantastic tool and your explanation, I now understand that any set of Somsky Gears can be offset to produce an new set. I tested this with a random set from Tom's program (26-9-7-5) and I could see that its offset (25-10-6-4) is a Somsky set too. However, those two Somsky sets are not coaxial. I used some crude editing to overlay the two, see the attached figure. It is clear to see that the two are not coaxial.
Tom has already provided me with a set of Somsky Gears, which can be coaxially offset into another set of Somsky Gears. However, not all sets of Somsky Gears can be coaxially offset like this. So what is the additional requirement that enables this? Is there a quick way how I can determine whether a Somsky set can be coaxially offset?
Best regards,
Oskar
-----Original Message----- From: wrsomsky@gmail.com Sent: Friday, July 31, 2015 6:44 AM
From some thinking I've done, the answer is yes. Ignoring overlap, any set of Somsky Gears can be offset to produce an new set. You can prove it by noting that for a given ring-sun-planet, the strap-length changes by pi (one half a tooth) if you increase the ring and planet radii by one, and decrease the sun radius by one (or vice versa).
On 2015-07-21 13:12, M. Oskar van Deventer wrote:
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
In my tabulations, the set you call 26-9-7-5 is 26-9-5,12 (ring-sun-list-planets) where the planets are in increasing size and all from one side of the centerline. (7 & 12 form a complementary "somsky-set", if you want to call it that, but the 7 is on the opposite side of the 5) For 26-9-5,12 you get the solutions (ring, sun, planet, planet, offset, sun-phase): 26 9 5 12 16.048313 0.500000 26 9 5 12 13.388239 1.000000 26 9 5 12 11.624141 0.500000 26 9 5 12 10.372542 0.000000 26 9 5 12 9.448488 0.500000 26 9 5 12 8.750000 1.000000 26 9 5 12 8.215897 0.500000 26 9 5 12 7.807356 1.000000 26 9 5 12 7.498839 0.500000 26 9 5 12 7.273254 1.000000 26 9 5 12 7.119219 0.500000 26 9 5 12 7.029480 1.000000 26 9 5 12 7.000000 0.500000 For 25-10-4,11 you get: 25 10 4 11 13.388239 0.500000 25 10 4 11 11.624141 1.000000 25 10 4 11 10.372542 0.500000 25 10 4 11 9.448488 1.000000 25 10 4 11 8.750000 0.500000 25 10 4 11 8.215897 0.000000 25 10 4 11 7.807356 0.500000 25 10 4 11 7.498839 0.000000 25 10 4 11 7.273254 0.500000 25 10 4 11 7.119219 0.000000 25 10 4 11 7.029480 0.500000 25 10 4 11 7.000000 0.000000 If you use the same offset, you get concentric matches. "But you said that they can always be concentric, barring overlap? What about the 16.04?" you ask. Well, in the 25-10-4,11 case, an offset of 16.04 causes the sun and ring to overlap, and my program (and Tom's I certainly believe) doesn't even consider those cases. If you used Tom's program, I expect it might be using the greatest possible offset, which would be 16.04 for one case, and 13.38 for the other. (Also, there may be scaling problems, as I don't know if Tom scales his image to the size of the ring gear -- mine normally does -- otherwise a set w/ a ring of 7 would look tiny, while a ring of 57 would go off the image area.) Bill On 08/03/15 10:59, William R Somsky wrote:
No, ignoring overlap, they would all be, or CAN all be, coaxial...
The one thing you need to be careful with though, is that, in general, there are several different sun-ring offsets that work for a given set of gears. If you pick one offset for one set of gears and a different one for the shifted set, they won't be coaxial. But if you pick the same offset, they are.
On 08/03/15 01:46, M. Oskar van Deventer wrote:
Dear Bill, Tom, all,
Thanks to Tom's fantastic tool and your explanation, I now understand that any set of Somsky Gears can be offset to produce an new set. I tested this with a random set from Tom's program (26-9-7-5) and I could see that its offset (25-10-6-4) is a Somsky set too. However, those two Somsky sets are not coaxial. I used some crude editing to overlay the two, see the attached figure. It is clear to see that the two are not coaxial.
Tom has already provided me with a set of Somsky Gears, which can be coaxially offset into another set of Somsky Gears. However, not all sets of Somsky Gears can be coaxially offset like this. So what is the additional requirement that enables this? Is there a quick way how I can determine whether a Somsky set can be coaxially offset?
Best regards,
Oskar
-----Original Message----- From: wrsomsky@gmail.com Sent: Friday, July 31, 2015 6:44 AM
From some thinking I've done, the answer is yes. Ignoring overlap, any set of Somsky Gears can be offset to produce an new set. You can prove it by noting that for a given ring-sun-planet, the strap-length changes by pi (one half a tooth) if you increase the ring and planet radii by one, and decrease the sun radius by one (or vice versa).
On 2015-07-21 13:12, M. Oskar van Deventer wrote:
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
WHOOPS! Sorry, went to plot it, and found I'd done the wrong set. The proper match to 26-9-7-5 is 26-9-5,10: 26 9 5 10 13.702903 0.072038 26 9 5 10 11.174443 0.338222 26 9 5 10 9.588008 0.585295 26 9 5 10 8.527280 0.807389 26 9 5 10 7.805237 0.997014 26 9 5 10 7.330664 0.144455 26 9 5 10 7.064324 0.236996 And 25-10-6-4 is 25-10-4,9: 25 10 4 9 13.702903 0.572038 25 10 4 9 11.174443 0.838222 25 10 4 9 9.588008 0.085295 25 10 4 9 8.527280 0.307389 25 10 4 9 7.805237 0.497014 25 10 4 9 7.330664 0.644455 25 10 4 9 7.064324 0.736996 Also 24-11-3,8: 24 11 3 8 11.174443 0.338222 24 11 3 8 9.588008 0.585295 24 11 3 8 8.527280 0.807389 24 11 3 8 7.805237 0.997014 24 11 3 8 7.330664 0.144455 24 11 3 8 7.064324 0.236996 So, using the 7.805 displacement you get the attached diagram, (also available as https://drive.google.com/open?id=0B2889vNnzpsTWVJzTTZ6TUNod0U) which has concentric gears. WRSomsky On 08/03/15 11:23, William R Somsky wrote:
In my tabulations, the set you call 26-9-7-5 is 26-9-5,12 (ring-sun-list-planets) where the planets are in increasing size and all from one side of the centerline. (7 & 12 form a complementary "somsky-set", if you want to call it that, but the 7 is on the opposite side of the 5)
For 26-9-5,12 you get the solutions (ring, sun, planet, planet, offset, sun-phase):
26 9 5 12 16.048313 0.500000 26 9 5 12 13.388239 1.000000 26 9 5 12 11.624141 0.500000 26 9 5 12 10.372542 0.000000 26 9 5 12 9.448488 0.500000 26 9 5 12 8.750000 1.000000 26 9 5 12 8.215897 0.500000 26 9 5 12 7.807356 1.000000 26 9 5 12 7.498839 0.500000 26 9 5 12 7.273254 1.000000 26 9 5 12 7.119219 0.500000 26 9 5 12 7.029480 1.000000 26 9 5 12 7.000000 0.500000
For 25-10-4,11 you get:
25 10 4 11 13.388239 0.500000 25 10 4 11 11.624141 1.000000 25 10 4 11 10.372542 0.500000 25 10 4 11 9.448488 1.000000 25 10 4 11 8.750000 0.500000 25 10 4 11 8.215897 0.000000 25 10 4 11 7.807356 0.500000 25 10 4 11 7.498839 0.000000 25 10 4 11 7.273254 0.500000 25 10 4 11 7.119219 0.000000 25 10 4 11 7.029480 0.500000 25 10 4 11 7.000000 0.000000
If you use the same offset, you get concentric matches.
"But you said that they can always be concentric, barring overlap? What about the 16.04?" you ask. Well, in the 25-10-4,11 case, an offset of 16.04 causes the sun and ring to overlap, and my program (and Tom's I certainly believe) doesn't even consider those cases.
If you used Tom's program, I expect it might be using the greatest possible offset, which would be 16.04 for one case, and 13.38 for the other. (Also, there may be scaling problems, as I don't know if Tom scales his image to the size of the ring gear -- mine normally does -- otherwise a set w/ a ring of 7 would look tiny, while a ring of 57 would go off the image area.)
Bill
Are these approximate x and y numbers roots of polynomials? If not, thought to be algebraic anyway? If the latter, and available to high precision, I could run them through RootApproximant. --rwg On Mon, Aug 3, 2015 at 12:02 PM, William R Somsky <wrsomsky@gmail.com> wrote:
WHOOPS! Sorry, went to plot it, and found I'd done the wrong set.
The proper match to 26-9-7-5 is 26-9-5,10:
26 9 5 10 13.702903 0.072038 26 9 5 10 11.174443 0.338222 26 9 5 10 9.588008 0.585295 26 9 5 10 8.527280 0.807389 26 9 5 10 7.805237 0.997014 26 9 5 10 7.330664 0.144455 26 9 5 10 7.064324 0.236996
And 25-10-6-4 is 25-10-4,9:
25 10 4 9 13.702903 0.572038 25 10 4 9 11.174443 0.838222 25 10 4 9 9.588008 0.085295 25 10 4 9 8.527280 0.307389 25 10 4 9 7.805237 0.497014 25 10 4 9 7.330664 0.644455 25 10 4 9 7.064324 0.736996
Also 24-11-3,8:
24 11 3 8 11.174443 0.338222 24 11 3 8 9.588008 0.585295 24 11 3 8 8.527280 0.807389 24 11 3 8 7.805237 0.997014 24 11 3 8 7.330664 0.144455 24 11 3 8 7.064324 0.236996
So, using the 7.805 displacement you get the attached diagram, (also available as https://drive.google.com/open?id=0B2889vNnzpsTWVJzTTZ6TUNod0U) which has concentric gears.
WRSomsky
On 08/03/15 11:23, William R Somsky wrote:
In my tabulations, the set you call 26-9-7-5 is 26-9-5,12 (ring-sun-list-planets) where the planets are in increasing size and all from one side of the centerline. (7 & 12 form a complementary "somsky-set", if you want to call it that, but the 7 is on the opposite side of the 5)
For 26-9-5,12 you get the solutions (ring, sun, planet, planet, offset, sun-phase):
26 9 5 12 16.048313 0.500000 26 9 5 12 13.388239 1.000000 26 9 5 12 11.624141 0.500000 26 9 5 12 10.372542 0.000000 26 9 5 12 9.448488 0.500000 26 9 5 12 8.750000 1.000000 26 9 5 12 8.215897 0.500000 26 9 5 12 7.807356 1.000000 26 9 5 12 7.498839 0.500000 26 9 5 12 7.273254 1.000000 26 9 5 12 7.119219 0.500000 26 9 5 12 7.029480 1.000000 26 9 5 12 7.000000 0.500000
For 25-10-4,11 you get:
25 10 4 11 13.388239 0.500000 25 10 4 11 11.624141 1.000000 25 10 4 11 10.372542 0.500000 25 10 4 11 9.448488 1.000000 25 10 4 11 8.750000 0.500000 25 10 4 11 8.215897 0.000000 25 10 4 11 7.807356 0.500000 25 10 4 11 7.498839 0.000000 25 10 4 11 7.273254 0.500000 25 10 4 11 7.119219 0.000000 25 10 4 11 7.029480 0.500000 25 10 4 11 7.000000 0.000000
If you use the same offset, you get concentric matches.
"But you said that they can always be concentric, barring overlap? What about the 16.04?" you ask. Well, in the 25-10-4,11 case, an offset of 16.04 causes the sun and ring to overlap, and my program (and Tom's I certainly believe) doesn't even consider those cases.
If you used Tom's program, I expect it might be using the greatest possible offset, which would be 16.04 for one case, and 13.38 for the other. (Also, there may be scaling problems, as I don't know if Tom scales his image to the size of the ring gear -- mine normally does -- otherwise a set w/ a ring of 7 would look tiny, while a ring of 57 would go off the image area.)
Bill
How many bits do you need? I can give you an equation. It is not very complicated. I think they are algebraic in some cases but not all. On Aug 3, 2015 9:33 AM, "Bill Gosper" <billgosper@gmail.com> wrote:
Are these approximate x and y numbers roots of polynomials? If not, thought to be algebraic anyway? If the latter, and available to high precision, I could run them through RootApproximant. --rwg
On Mon, Aug 3, 2015 at 12:02 PM, William R Somsky <wrsomsky@gmail.com> wrote:
WHOOPS! Sorry, went to plot it, and found I'd done the wrong set.
The proper match to 26-9-7-5 is 26-9-5,10:
26 9 5 10 13.702903 0.072038 26 9 5 10 11.174443 0.338222 26 9 5 10 9.588008 0.585295 26 9 5 10 8.527280 0.807389 26 9 5 10 7.805237 0.997014 26 9 5 10 7.330664 0.144455 26 9 5 10 7.064324 0.236996
And 25-10-6-4 is 25-10-4,9:
25 10 4 9 13.702903 0.572038 25 10 4 9 11.174443 0.838222 25 10 4 9 9.588008 0.085295 25 10 4 9 8.527280 0.307389 25 10 4 9 7.805237 0.497014 25 10 4 9 7.330664 0.644455 25 10 4 9 7.064324 0.736996
Also 24-11-3,8:
24 11 3 8 11.174443 0.338222 24 11 3 8 9.588008 0.585295 24 11 3 8 8.527280 0.807389 24 11 3 8 7.805237 0.997014 24 11 3 8 7.330664 0.144455 24 11 3 8 7.064324 0.236996
So, using the 7.805 displacement you get the attached diagram, (also available as https://drive.google.com/open?id=0B2889vNnzpsTWVJzTTZ6TUNod0U) which has concentric gears.
WRSomsky
On 08/03/15 11:23, William R Somsky wrote:
In my tabulations, the set you call 26-9-7-5 is 26-9-5,12 (ring-sun-list-planets) where the planets are in increasing size and all from one side of the centerline. (7 & 12 form a complementary "somsky-set", if you want to call it that, but the 7 is on the opposite side of the 5)
For 26-9-5,12 you get the solutions (ring, sun, planet, planet, offset, sun-phase):
26 9 5 12 16.048313 0.500000 26 9 5 12 13.388239 1.000000 26 9 5 12 11.624141 0.500000 26 9 5 12 10.372542 0.000000 26 9 5 12 9.448488 0.500000 26 9 5 12 8.750000 1.000000 26 9 5 12 8.215897 0.500000 26 9 5 12 7.807356 1.000000 26 9 5 12 7.498839 0.500000 26 9 5 12 7.273254 1.000000 26 9 5 12 7.119219 0.500000 26 9 5 12 7.029480 1.000000 26 9 5 12 7.000000 0.500000
For 25-10-4,11 you get:
25 10 4 11 13.388239 0.500000 25 10 4 11 11.624141 1.000000 25 10 4 11 10.372542 0.500000 25 10 4 11 9.448488 1.000000 25 10 4 11 8.750000 0.500000 25 10 4 11 8.215897 0.000000 25 10 4 11 7.807356 0.500000 25 10 4 11 7.498839 0.000000 25 10 4 11 7.273254 0.500000 25 10 4 11 7.119219 0.000000 25 10 4 11 7.029480 0.500000 25 10 4 11 7.000000 0.000000
If you use the same offset, you get concentric matches.
"But you said that they can always be concentric, barring overlap? What about the 16.04?" you ask. Well, in the 25-10-4,11 case, an offset of 16.04 causes the sun and ring to overlap, and my program (and Tom's I certainly believe) doesn't even consider those cases.
If you used Tom's program, I expect it might be using the greatest possible offset, which would be 16.04 for one case, and 13.38 for the other. (Also, there may be scaling problems, as I don't know if Tom scales his image to the size of the ring gear -- mine normally does -- otherwise a set w/ a ring of 7 would look tiny, while a ring of 57 would go off the image area.)
Bill
Assuming I've made no typos, the core is the following: Define Ph[a,c,d,b] = b pi + (a+c) arccos [ ( (a-b)^2 + d^2 - (c+b)^2 ) / ( 2 (a-b) d ) ] + (b+c) arccos [ ( (a-b)^2 + (c+b)^2 - d^2 ) / ( 2 (a-b) (c+b) ) ] (a is the ring radius, c is the sun radius, b is the planet radius, and d is the offset of the sun from the center of the ring) Find {a,c,d,b,b,...} w/ d non-negative real, a,c,b... positive integers, a > c+d, a-d-c < 2b < a+d-c such that for all pairs of b1,b2: Ph[a,c,d,b1] - Ph[a,c,d,b2] is a multiple of 2pi Once you have that, it's just solving: bx^2+by^2 == (a-b)^2 && (bx+d)^2+by^2 == (c+b)^2 On 08/03/15 12:38, Tom Rokicki wrote:
How many bits do you need?
I can give you an equation. It is not very complicated.
I think they are algebraic in some cases but not all.
On Aug 3, 2015 9:33 AM, "Bill Gosper" <billgosper@gmail.com <mailto:billgosper@gmail.com>> wrote:
Are these approximate x and y numbers roots of polynomials? If not, thought to be algebraic anyway? If the latter, and available to high precision, I could run them through RootApproximant. --rwg
On Mon, Aug 3, 2015 at 12:02 PM, William R Somsky <wrsomsky@gmail.com <mailto:wrsomsky@gmail.com>> wrote:
WHOOPS! Sorry, went to plot it, and found I'd done the wrong set.
The proper match to 26-9-7-5 is 26-9-5,10:
26 9 5 10 13.702903 0.072038 26 9 5 10 11.174443 0.338222 26 9 5 10 9.588008 0.585295 26 9 5 10 8.527280 0.807389 26 9 5 10 7.805237 0.997014 26 9 5 10 7.330664 0.144455 26 9 5 10 7.064324 0.236996
And 25-10-6-4 is 25-10-4,9:
25 10 4 9 13.702903 0.572038 25 10 4 9 11.174443 0.838222 25 10 4 9 9.588008 0.085295 25 10 4 9 8.527280 0.307389 25 10 4 9 7.805237 0.497014 25 10 4 9 7.330664 0.644455 25 10 4 9 7.064324 0.736996
Also 24-11-3,8:
24 11 3 8 11.174443 0.338222 24 11 3 8 9.588008 0.585295 24 11 3 8 8.527280 0.807389 24 11 3 8 7.805237 0.997014 24 11 3 8 7.330664 0.144455 24 11 3 8 7.064324 0.236996
So, using the 7.805 displacement you get the attached diagram, (also available as https://drive.google.com/open?id=0B2889vNnzpsTWVJzTTZ6TUNod0U) which has concentric gears.
WRSomsky
On 08/03/15 11:23, William R Somsky wrote:
In my tabulations, the set you call 26-9-7-5 is 26-9-5,12 (ring-sun-list-planets) where the planets are in increasing size and all from one side of the centerline. (7 & 12 form a complementary "somsky-set", if you want to call it that, but the 7 is on the opposite side of the 5)
For 26-9-5,12 you get the solutions (ring, sun, planet, planet, offset, sun-phase):
26 9 5 12 16.048313 0.500000 26 9 5 12 13.388239 1.000000 26 9 5 12 11.624141 0.500000 26 9 5 12 10.372542 0.000000 26 9 5 12 9.448488 0.500000 26 9 5 12 8.750000 1.000000 26 9 5 12 8.215897 0.500000 26 9 5 12 7.807356 1.000000 26 9 5 12 7.498839 0.500000 26 9 5 12 7.273254 1.000000 26 9 5 12 7.119219 0.500000 26 9 5 12 7.029480 1.000000 26 9 5 12 7.000000 0.500000
For 25-10-4,11 you get:
25 10 4 11 13.388239 0.500000 25 10 4 11 11.624141 1.000000 25 10 4 11 10.372542 0.500000 25 10 4 11 9.448488 1.000000 25 10 4 11 8.750000 0.500000 25 10 4 11 8.215897 0.000000 25 10 4 11 7.807356 0.500000 25 10 4 11 7.498839 0.000000 25 10 4 11 7.273254 0.500000 25 10 4 11 7.119219 0.000000 25 10 4 11 7.029480 0.500000 25 10 4 11 7.000000 0.000000
If you use the same offset, you get concentric matches.
"But you said that they can always be concentric, barring overlap? What about the 16.04?" you ask. Well, in the 25-10-4,11 case, an offset of 16.04 causes the sun and ring to overlap, and my program (and Tom's I certainly believe) doesn't even consider those cases.
If you used Tom's program, I expect it might be using the greatest possible offset, which would be 16.04 for one case, and 13.38 for the other. (Also, there may be scaling problems, as I don't know if Tom scales his image to the size of the ring gear -- mine normally does -- otherwise a set w/ a ring of 7 would look tiny, while a ring of 57 would go off the image area.)
Bill
Right. And if you rewrite things with a small change of variables: b' = d - b a' = a + b both of which are invariant under offsets, then you'll see that offsetting just works (since offsets don't change b' or a', and the final result multiplies d by 2pi so that drops out as well.) On Mon, Aug 3, 2015 at 1:37 PM, William R Somsky <wrsomsky@gmail.com> wrote:
Assuming I've made no typos, the core is the following:
Define Ph[a,c,d,b] = b pi + (a+c) arccos [ ( (a-b)^2 + d^2 - (c+b)^2 ) / ( 2 (a-b) d ) ] + (b+c) arccos [ ( (a-b)^2 + (c+b)^2 - d^2 ) / ( 2 (a-b) (c+b) ) ]
(a is the ring radius, c is the sun radius, b is the planet radius, and d is the offset of the sun from the center of the ring)
Find {a,c,d,b,b,...} w/ d non-negative real, a,c,b... positive integers, a > c+d, a-d-c < 2b < a+d-c such that for all pairs of b1,b2: Ph[a,c,d,b1] - Ph[a,c,d,b2] is a multiple of 2pi
Once you have that, it's just solving: bx^2+by^2 == (a-b)^2 && (bx+d)^2+by^2 == (c+b)^2
On 08/03/15 12:38, Tom Rokicki wrote:
How many bits do you need?
I can give you an equation. It is not very complicated.
I think they are algebraic in some cases but not all.
On Aug 3, 2015 9:33 AM, "Bill Gosper" <billgosper@gmail.com> wrote:
Are these approximate x and y numbers roots of polynomials? If not, thought to be algebraic anyway? If the latter, and available to high precision, I could run them through RootApproximant. --rwg
On Mon, Aug 3, 2015 at 12:02 PM, William R Somsky <wrsomsky@gmail.com> wrote:
WHOOPS! Sorry, went to plot it, and found I'd done the wrong set.
The proper match to 26-9-7-5 is 26-9-5,10:
26 9 5 10 13.702903 0.072038 26 9 5 10 11.174443 0.338222 26 9 5 10 9.588008 0.585295 26 9 5 10 8.527280 0.807389 26 9 5 10 7.805237 0.997014 26 9 5 10 7.330664 0.144455 26 9 5 10 7.064324 0.236996
And 25-10-6-4 is 25-10-4,9:
25 10 4 9 13.702903 0.572038 25 10 4 9 11.174443 0.838222 25 10 4 9 9.588008 0.085295 25 10 4 9 8.527280 0.307389 25 10 4 9 7.805237 0.497014 25 10 4 9 7.330664 0.644455 25 10 4 9 7.064324 0.736996
Also 24-11-3,8:
24 11 3 8 11.174443 0.338222 24 11 3 8 9.588008 0.585295 24 11 3 8 8.527280 0.807389 24 11 3 8 7.805237 0.997014 24 11 3 8 7.330664 0.144455 24 11 3 8 7.064324 0.236996
So, using the 7.805 displacement you get the attached diagram, (also available as https://drive.google.com/open?id=0B2889vNnzpsTWVJzTTZ6TUNod0U) which has concentric gears.
WRSomsky
On 08/03/15 11:23, William R Somsky wrote:
In my tabulations, the set you call 26-9-7-5 is 26-9-5,12 (ring-sun-list-planets) where the planets are in increasing size and all from one side of the centerline. (7 & 12 form a complementary "somsky-set", if you want to call it that, but the 7 is on the opposite side of the 5)
For 26-9-5,12 you get the solutions (ring, sun, planet, planet, offset, sun-phase):
26 9 5 12 16.048313 0.500000 26 9 5 12 13.388239 1.000000 26 9 5 12 11.624141 0.500000 26 9 5 12 10.372542 0.000000 26 9 5 12 9.448488 0.500000 26 9 5 12 8.750000 1.000000 26 9 5 12 8.215897 0.500000 26 9 5 12 7.807356 1.000000 26 9 5 12 7.498839 0.500000 26 9 5 12 7.273254 1.000000 26 9 5 12 7.119219 0.500000 26 9 5 12 7.029480 1.000000 26 9 5 12 7.000000 0.500000
For 25-10-4,11 you get:
25 10 4 11 13.388239 0.500000 25 10 4 11 11.624141 1.000000 25 10 4 11 10.372542 0.500000 25 10 4 11 9.448488 1.000000 25 10 4 11 8.750000 0.500000 25 10 4 11 8.215897 0.000000 25 10 4 11 7.807356 0.500000 25 10 4 11 7.498839 0.000000 25 10 4 11 7.273254 0.500000 25 10 4 11 7.119219 0.000000 25 10 4 11 7.029480 0.500000 25 10 4 11 7.000000 0.000000
If you use the same offset, you get concentric matches.
"But you said that they can always be concentric, barring overlap? What about the 16.04?" you ask. Well, in the 25-10-4,11 case, an offset of 16.04 causes the sun and ring to overlap, and my program (and Tom's I certainly believe) doesn't even consider those cases.
If you used Tom's program, I expect it might be using the greatest possible offset, which would be 16.04 for one case, and 13.38 for the other. (Also, there may be scaling problems, as I don't know if Tom scales his image to the size of the ring gear -- mine normally does -- otherwise a set w/ a ring of 7 would look tiny, while a ring of 57 would go off the image area.)
Bill
-- -- http://cube20.org/ -- [Golly link suppressed; ask me why] --
Oops, I notice you used "your" abcd, where I use a=sun, d=outer, b/c are planets. The change of variables should be r'_planet = r_ring - r_planet r'_sun = r_sun + r_planet On Mon, Aug 3, 2015 at 1:47 PM, Tom Rokicki <rokicki@gmail.com> wrote:
Right. And if you rewrite things with a small change of variables:
b' = d - b a' = a + b
both of which are invariant under offsets, then you'll see that offsetting just works (since offsets don't change b' or a', and the final result multiplies d by 2pi so that drops out as well.)
On Mon, Aug 3, 2015 at 1:37 PM, William R Somsky <wrsomsky@gmail.com> wrote:
Assuming I've made no typos, the core is the following:
Define Ph[a,c,d,b] = b pi + (a+c) arccos [ ( (a-b)^2 + d^2 - (c+b)^2 ) / ( 2 (a-b) d ) ] + (b+c) arccos [ ( (a-b)^2 + (c+b)^2 - d^2 ) / ( 2 (a-b) (c+b) ) ]
(a is the ring radius, c is the sun radius, b is the planet radius, and d is the offset of the sun from the center of the ring)
Find {a,c,d,b,b,...} w/ d non-negative real, a,c,b... positive integers, a > c+d, a-d-c < 2b < a+d-c such that for all pairs of b1,b2: Ph[a,c,d,b1] - Ph[a,c,d,b2] is a multiple of 2pi
Once you have that, it's just solving: bx^2+by^2 == (a-b)^2 && (bx+d)^2+by^2 == (c+b)^2
On 08/03/15 12:38, Tom Rokicki wrote:
How many bits do you need?
I can give you an equation. It is not very complicated.
I think they are algebraic in some cases but not all.
On Aug 3, 2015 9:33 AM, "Bill Gosper" <billgosper@gmail.com> wrote:
Are these approximate x and y numbers roots of polynomials? If not, thought to be algebraic anyway? If the latter, and available to high precision, I could run them through RootApproximant. --rwg
On Mon, Aug 3, 2015 at 12:02 PM, William R Somsky <wrsomsky@gmail.com> wrote:
WHOOPS! Sorry, went to plot it, and found I'd done the wrong set.
The proper match to 26-9-7-5 is 26-9-5,10:
26 9 5 10 13.702903 0.072038 26 9 5 10 11.174443 0.338222 26 9 5 10 9.588008 0.585295 26 9 5 10 8.527280 0.807389 26 9 5 10 7.805237 0.997014 26 9 5 10 7.330664 0.144455 26 9 5 10 7.064324 0.236996
And 25-10-6-4 is 25-10-4,9:
25 10 4 9 13.702903 0.572038 25 10 4 9 11.174443 0.838222 25 10 4 9 9.588008 0.085295 25 10 4 9 8.527280 0.307389 25 10 4 9 7.805237 0.497014 25 10 4 9 7.330664 0.644455 25 10 4 9 7.064324 0.736996
Also 24-11-3,8:
24 11 3 8 11.174443 0.338222 24 11 3 8 9.588008 0.585295 24 11 3 8 8.527280 0.807389 24 11 3 8 7.805237 0.997014 24 11 3 8 7.330664 0.144455 24 11 3 8 7.064324 0.236996
So, using the 7.805 displacement you get the attached diagram, (also available as https://drive.google.com/open?id=0B2889vNnzpsTWVJzTTZ6TUNod0U) which has concentric gears.
WRSomsky
On 08/03/15 11:23, William R Somsky wrote:
In my tabulations, the set you call 26-9-7-5 is 26-9-5,12 (ring-sun-list-planets) where the planets are in increasing size and all from one side of the centerline. (7 & 12 form a complementary "somsky-set", if you want to call it that, but the 7 is on the opposite side of the 5)
For 26-9-5,12 you get the solutions (ring, sun, planet, planet, offset, sun-phase):
26 9 5 12 16.048313 0.500000 26 9 5 12 13.388239 1.000000 26 9 5 12 11.624141 0.500000 26 9 5 12 10.372542 0.000000 26 9 5 12 9.448488 0.500000 26 9 5 12 8.750000 1.000000 26 9 5 12 8.215897 0.500000 26 9 5 12 7.807356 1.000000 26 9 5 12 7.498839 0.500000 26 9 5 12 7.273254 1.000000 26 9 5 12 7.119219 0.500000 26 9 5 12 7.029480 1.000000 26 9 5 12 7.000000 0.500000
For 25-10-4,11 you get:
25 10 4 11 13.388239 0.500000 25 10 4 11 11.624141 1.000000 25 10 4 11 10.372542 0.500000 25 10 4 11 9.448488 1.000000 25 10 4 11 8.750000 0.500000 25 10 4 11 8.215897 0.000000 25 10 4 11 7.807356 0.500000 25 10 4 11 7.498839 0.000000 25 10 4 11 7.273254 0.500000 25 10 4 11 7.119219 0.000000 25 10 4 11 7.029480 0.500000 25 10 4 11 7.000000 0.000000
If you use the same offset, you get concentric matches.
"But you said that they can always be concentric, barring overlap? What about the 16.04?" you ask. Well, in the 25-10-4,11 case, an offset of 16.04 causes the sun and ring to overlap, and my program (and Tom's I certainly believe) doesn't even consider those cases.
If you used Tom's program, I expect it might be using the greatest possible offset, which would be 16.04 for one case, and 13.38 for the other. (Also, there may be scaling problems, as I don't know if Tom scales his image to the size of the ring gear -- mine normally does -- otherwise a set w/ a ring of 7 would look tiny, while a ring of 57 would go off the image area.)
Bill
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-- -- http://cube20.org/ -- [Golly link suppressed; ask me why] --
I'm not comprehending some of what you all are saying. I had recently presented an argument algebraic numbers came up in Somsky gears -- and also my general purpose older argument about general bipartite planar graphs as gear configurations, will always yield algebraic numbers whenever configuration rigid. To actually find out the minimal polynomial of a given algebraic number, though, you tend to need a lot of decimals, but if you have them there are general tools to do it (using, e.g. Lovasz lattice reduction) which I presume Gosper was referring to and already knows about. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
I cannot say I always comprehend your emails, Warren. For this I apologize. On Mon, Aug 3, 2015 at 1:19 PM, Warren D Smith <warren.wds@gmail.com> wrote:
I'm not comprehending some of what you all are saying. I had recently presented an argument algebraic numbers came up in Somsky gears -- and also my general purpose older argument about general bipartite planar graphs as gear configurations, will always yield algebraic numbers whenever configuration rigid. To actually find out the minimal polynomial of a given algebraic number, though, you tend to need a lot of decimals, but if you have them there are general tools to do it (using, e.g. Lovasz lattice reduction) which I presume Gosper was referring to and already knows about.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Boy am I confused. You have non-algebraic equations with suspected algebraic roots? If they're algebraic, I don't need any decimal places at all. --Bill On Mon, Aug 3, 2015 at 1:19 PM, Warren D Smith <warren.wds@gmail.com> wrote:
I'm not comprehending some of what you all are saying. I had recently presented an argument algebraic numbers came up in Somsky gears -- and also my general purpose older argument about general bipartite planar graphs as gear configurations, will always yield algebraic numbers whenever configuration rigid. To actually find out the minimal polynomial of a given algebraic number, though, you tend to need a lot of decimals, but if you have them there are general tools to do it (using, e.g. Lovasz lattice reduction) which I presume Gosper was referring to and already knows about.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Boy am I confused. You have non-algebraic equations with suspected algebraic roots? If they're algebraic, I don't need any decimal places at all. --Bill Gosper
--Answer (WDS): I showed how the non-algebraic equations involving arctrig, could be restated in a way involving integer powers of complex numbers (with coordinates inside the complexes involving square roots). So then really, they are algebraic equations in disguise. And then yes, you do not need any decimals at all to know algebraic. Furthermore in an early post I explored general bipartite-planar gear-circle-contact graphs, and the equations there are just about distances, which are of course algebraic functions. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Before this intrusion of ineptitude risks the moral endangerment of any further tender young minds, I feel obliged to point out the logical inequivalence between the statements cos(rational * 2pi) = algebraic ! and arccos(rational) = algebraic * 2pi ? --- the second of which is in general (to put it mildly) problematic. Those remaining in doubt about the matter should reflect upon the kindergarten fact arccos(1/3) = dihedral angle of regular tetrahedron, the expression of which as an algebraic angle ought to be worth a Fields at least! Fred Lunnon On 8/3/15, Warren D Smith <warren.wds@gmail.com> wrote:
Boy am I confused. You have non-algebraic equations with suspected algebraic roots? If they're algebraic, I don't need any decimal places at all. --Bill Gosper
--Answer (WDS): I showed how the non-algebraic equations involving arctrig, could be restated in a way involving integer powers of complex numbers (with coordinates inside the complexes involving square roots). So then really, they are algebraic equations in disguise. And then yes, you do not need any decimals at all to know algebraic.
Furthermore in an early post I explored general bipartite-planar gear-circle-contact graphs, and the equations there are just about distances, which are of course algebraic functions.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
In fact, the only solutions to: cos(algebraic * pi) = algebraic are of the form: cos(rational * pi) = algebraic Proof: -------- It is clear that cos(algebraic * pi) and exp(algebraic * (pi i)) are algebraically dependent, so we can consider the equivalent problem to find solutions to: exp(algebraic * (pi i)) = different algebraic which rearranges to: (-1)^algebraic = different algebraic By Gelfond-Schneider, this occurs only when the exponent is rational. -------- Clearly the converse is also true, so the algebraic numbers x for which cos(x * pi) is also algebraic are precisely the rationals. Sincerely, Adam P. Goucher
Sent: Tuesday, August 04, 2015 at 8:23 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Offset Somsky Gears, coaxial offset
Before this intrusion of ineptitude risks the moral endangerment of any further tender young minds, I feel obliged to point out the logical inequivalence between the statements cos(rational * 2pi) = algebraic ! and arccos(rational) = algebraic * 2pi ? --- the second of which is in general (to put it mildly) problematic.
Those remaining in doubt about the matter should reflect upon the kindergarten fact arccos(1/3) = dihedral angle of regular tetrahedron, the expression of which as an algebraic angle ought to be worth a Fields at least!
Fred Lunnon
On 8/3/15, Warren D Smith <warren.wds@gmail.com> wrote:
Boy am I confused. You have non-algebraic equations with suspected algebraic roots? If they're algebraic, I don't need any decimal places at all. --Bill Gosper
--Answer (WDS): I showed how the non-algebraic equations involving arctrig, could be restated in a way involving integer powers of complex numbers (with coordinates inside the complexes involving square roots). So then really, they are algebraic equations in disguise. And then yes, you do not need any decimals at all to know algebraic.
Furthermore in an early post I explored general bipartite-planar gear-circle-contact graphs, and the equations there are just about distances, which are of course algebraic functions.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Unfortunately, my previous carefully-honed missile was directed at an obsolete target (arc-lengths of belts), rather than the current one (offset of sun). For offset, we want to solve the equation in Z with rational A,B,C,D A arccos( B Z^2 + C) + ... = D , so that cos D is a polynomial with rational coefficients in cos(Z^2) and sin(Z^2) , which can be further rationalised to a polynomial in cos^2(Z^2) alone. Solving this yields Z^2 = arccos(algebraic) ; and APG reminds us that this never has properly algebraic solutions, only rational or (more usually) transcendental! Fred Lunnon On 8/4/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, the only solutions to:
cos(algebraic * pi) = algebraic
are of the form:
cos(rational * pi) = algebraic
Proof: --------
It is clear that cos(algebraic * pi) and exp(algebraic * (pi i)) are algebraically dependent, so we can consider the equivalent problem to find solutions to:
exp(algebraic * (pi i)) = different algebraic
which rearranges to:
(-1)^algebraic = different algebraic
By Gelfond-Schneider, this occurs only when the exponent is rational.
--------
Clearly the converse is also true, so the algebraic numbers x for which cos(x * pi) is also algebraic are precisely the rationals.
Sincerely,
Adam P. Goucher
Sent: Tuesday, August 04, 2015 at 8:23 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Offset Somsky Gears, coaxial offset
Before this intrusion of ineptitude risks the moral endangerment of any further tender young minds, I feel obliged to point out the logical inequivalence between the statements cos(rational * 2pi) = algebraic ! and arccos(rational) = algebraic * 2pi ? --- the second of which is in general (to put it mildly) problematic.
Those remaining in doubt about the matter should reflect upon the kindergarten fact arccos(1/3) = dihedral angle of regular tetrahedron, the expression of which as an algebraic angle ought to be worth a Fields at least!
Fred Lunnon
On 8/3/15, Warren D Smith <warren.wds@gmail.com> wrote:
Boy am I confused. You have non-algebraic equations with suspected algebraic roots? If they're algebraic, I don't need any decimal places at all. --Bill Gosper
--Answer (WDS): I showed how the non-algebraic equations involving arctrig, could be restated in a way involving integer powers of complex numbers (with coordinates inside the complexes involving square roots). So then really, they are algebraic equations in disguise. And then yes, you do not need any decimals at all to know algebraic.
Furthermore in an early post I explored general bipartite-planar gear-circle-contact graphs, and the equations there are just about distances, which are of course algebraic functions.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Here's one solution based on one of Warren's miracles. Plot (39, 21, 3, 3) (ring, sun, planet, planet). This gives an offset of exactly 15. This is due to the odd equation 2 * acos(-11/16) - 3 acos(7/8) = pi Integral offsets such as these are rare except in certain degenerate cases. On Tue, Aug 4, 2015 at 4:22 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Unfortunately, my previous carefully-honed missile was directed at an obsolete target (arc-lengths of belts), rather than the current one (offset of sun).
For offset, we want to solve the equation in Z with rational A,B,C,D A arccos( B Z^2 + C) + ... = D , so that cos D is a polynomial with rational coefficients in cos(Z^2) and sin(Z^2) , which can be further rationalised to a polynomial in cos^2(Z^2) alone.
Solving this yields Z^2 = arccos(algebraic) ; and APG reminds us that this never has properly algebraic solutions, only rational or (more usually) transcendental!
Fred Lunnon
On 8/4/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, the only solutions to:
cos(algebraic * pi) = algebraic
are of the form:
cos(rational * pi) = algebraic
Proof: --------
It is clear that cos(algebraic * pi) and exp(algebraic * (pi i)) are algebraically dependent, so we can consider the equivalent problem to find solutions to:
exp(algebraic * (pi i)) = different algebraic
which rearranges to:
(-1)^algebraic = different algebraic
By Gelfond-Schneider, this occurs only when the exponent is rational.
--------
Clearly the converse is also true, so the algebraic numbers x for which cos(x * pi) is also algebraic are precisely the rationals.
Sincerely,
Adam P. Goucher
Sent: Tuesday, August 04, 2015 at 8:23 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Offset Somsky Gears, coaxial offset
Before this intrusion of ineptitude risks the moral endangerment of any further tender young minds, I feel obliged to point out the logical inequivalence between the statements cos(rational * 2pi) = algebraic ! and arccos(rational) = algebraic * 2pi ? --- the second of which is in general (to put it mildly) problematic.
Those remaining in doubt about the matter should reflect upon the kindergarten fact arccos(1/3) = dihedral angle of regular tetrahedron, the expression of which as an algebraic angle ought to be worth a Fields at least!
Fred Lunnon
On 8/3/15, Warren D Smith <warren.wds@gmail.com> wrote:
Boy am I confused. You have non-algebraic equations with suspected algebraic roots? If they're algebraic, I don't need any decimal places at all. --Bill Gosper
--Answer (WDS): I showed how the non-algebraic equations involving arctrig, could be restated in a way involving integer powers of complex numbers (with coordinates inside the complexes involving square roots). So then really, they are algebraic equations in disguise. And then yes, you do not need any decimals at all to know algebraic.
Furthermore in an early post I explored general bipartite-planar gear-circle-contact graphs, and the equations there are just about distances, which are of course algebraic functions.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Bang on cue! This baby reduces to a "grandfather" rational train as follows: (39, 21, 3, 3) ; Replacing one planet by its Somsky opposite, then completing a third planet on the same side as the other two -> [39, 21; 15, 9, 3] ; Scaling down by factor 3 -> [13, 7; 5, 3, 1] ; Increasing concentrically -> [14, 6; 6, 4, 2] ; Scaling down by factor 2 -> [7, 3; 3, 2, 1] ; Final offset = 5/2 . I conjecture that any rational train is reducible to the grandfather in a similar fashion. Fred Lunnon On 8/5/15, Tom Rokicki <rokicki@gmail.com> wrote:
Here's one solution based on one of Warren's miracles.
Plot (39, 21, 3, 3) (ring, sun, planet, planet).
This gives an offset of exactly 15.
This is due to the odd equation
2 * acos(-11/16) - 3 acos(7/8) = pi
Integral offsets such as these are rare except in certain degenerate cases.
On Tue, Aug 4, 2015 at 4:22 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Unfortunately, my previous carefully-honed missile was directed at an obsolete target (arc-lengths of belts), rather than the current one (offset of sun).
For offset, we want to solve the equation in Z with rational A,B,C,D A arccos( B Z^2 + C) + ... = D , so that cos D is a polynomial with rational coefficients in cos(Z^2) and sin(Z^2) , which can be further rationalised to a polynomial in cos^2(Z^2) alone.
Solving this yields Z^2 = arccos(algebraic) ; and APG reminds us that this never has properly algebraic solutions, only rational or (more usually) transcendental!
Fred Lunnon
On 8/4/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, the only solutions to:
cos(algebraic * pi) = algebraic
are of the form:
cos(rational * pi) = algebraic
Proof: --------
It is clear that cos(algebraic * pi) and exp(algebraic * (pi i)) are algebraically dependent, so we can consider the equivalent problem to find solutions to:
exp(algebraic * (pi i)) = different algebraic
which rearranges to:
(-1)^algebraic = different algebraic
By Gelfond-Schneider, this occurs only when the exponent is rational.
--------
Clearly the converse is also true, so the algebraic numbers x for which cos(x * pi) is also algebraic are precisely the rationals.
Sincerely,
Adam P. Goucher
Sent: Tuesday, August 04, 2015 at 8:23 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Offset Somsky Gears, coaxial offset
Before this intrusion of ineptitude risks the moral endangerment of any further tender young minds, I feel obliged to point out the logical inequivalence between the statements cos(rational * 2pi) = algebraic ! and arccos(rational) = algebraic * 2pi ? --- the second of which is in general (to put it mildly) problematic.
Those remaining in doubt about the matter should reflect upon the kindergarten fact arccos(1/3) = dihedral angle of regular tetrahedron, the expression of which as an algebraic angle ought to be worth a Fields at least!
Fred Lunnon
On 8/3/15, Warren D Smith <warren.wds@gmail.com> wrote:
Boy am I confused. You have non-algebraic equations with suspected algebraic roots? If they're algebraic, I don't need any decimal places at all. --Bill Gosper
--Answer (WDS): I showed how the non-algebraic equations involving arctrig, could be restated in a way involving integer powers of complex numbers (with coordinates inside the complexes involving square roots). So then really, they are algebraic equations in disguise. And then yes, you do not need any decimals at all to know algebraic.
Furthermore in an early post I explored general bipartite-planar gear-circle-contact graphs, and the equations there are just about distances, which are of course algebraic functions.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Caveat: the accompanying gloss, while correct in a general sense, overlooks the mildly embarrassing fact that radius-3 planets of the original actually coincide on the axis! The radius-15 partner therefore lies opposite along the axis. The Somsky pair subsequently attached have (maximal) equal radius-9 . This illustrates a special case of a construction to which I want return shortly --- I had intended to delay until the discussion of coaxial / concentric families had finished, but events seem to be overtaking me! WFL On 8/5/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Bang on cue!
This baby reduces to a "grandfather" rational train as follows: (39, 21, 3, 3) ; Replacing one planet by its Somsky opposite, then completing a third planet on the same side as the other two -> [39, 21; 15, 9, 3] ; Scaling down by factor 3 -> [13, 7; 5, 3, 1] ; Increasing concentrically -> [14, 6; 6, 4, 2] ; Scaling down by factor 2 -> [7, 3; 3, 2, 1] ; Final offset = 5/2 .
I conjecture that any rational train is reducible to the grandfather in a similar fashion.
Fred Lunnon
On 8/5/15, Tom Rokicki <rokicki@gmail.com> wrote:
Here's one solution based on one of Warren's miracles.
Plot (39, 21, 3, 3) (ring, sun, planet, planet).
This gives an offset of exactly 15.
This is due to the odd equation
2 * acos(-11/16) - 3 acos(7/8) = pi
Integral offsets such as these are rare except in certain degenerate cases.
Please ignore previous (brainstorm) caveat: I was right first time, so was TR. However if instead grandad is simply scaled up by 4, the sunset becomes 10. There is now room along the axis to insert two more planets, yielding the 8-planet train [28, -12; 12, 8, 4, 3; 4, 8, 12, 13] . Scaling this up by 2 again then decreasing by 3 yields [53, -27; 21, 13, 5, 3; 5, 13, 21, 23] , the train reported by WRS on July 13th. I conjecture that (assuming nonzero sunset) every 8-planet train is thus constructable; and further that there are no trains with more than 8 planets. [ By the way, using "offset" to denote displacement of sun from ring centre turns out badly, for reasons best postponed. Similarly --- as previously hinted --- I should like from now on to change the sign of my sun radius: please be patient and indulge me! ] Fred Lunnon On 8/5/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Caveat: the accompanying gloss, while correct in a general sense, overlooks the mildly embarrassing fact that radius-3 planets of the original actually coincide on the axis! The radius-15 partner therefore lies opposite along the axis. The Somsky pair subsequently attached have (maximal) equal radius-9 .
This illustrates a special case of a construction to which I want return shortly --- I had intended to delay until the discussion of coaxial / concentric families had finished, but events seem to be overtaking me!
WFL
On 8/5/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Bang on cue!
This baby reduces to a "grandfather" rational train as follows: (39, 21, 3, 3) ; Replacing one planet by its Somsky opposite, then completing a third planet on the same side as the other two -> [39, 21; 15, 9, 3] ; Scaling down by factor 3 -> [13, 7; 5, 3, 1] ; Increasing concentrically -> [14, 6; 6, 4, 2] ; Scaling down by factor 2 -> [7, 3; 3, 2, 1] ; Final offset = 5/2 .
I conjecture that any rational train is reducible to the grandfather in a similar fashion.
Fred Lunnon
On 8/5/15, Tom Rokicki <rokicki@gmail.com> wrote:
Here's one solution based on one of Warren's miracles.
Plot (39, 21, 3, 3) (ring, sun, planet, planet).
This gives an offset of exactly 15.
This is due to the odd equation
2 * acos(-11/16) - 3 acos(7/8) = pi
Integral offsets such as these are rare except in certain degenerate cases.
With planets placed in cyclic order around the sun, and axis indicated by semicolon, the vector X = [ r, s; p, ..., q; u, ..., v ] of gear radii becomes essentially unique; and combined with sunset (offset) h , it specifies the train. While in a realisable train the radii (more properly, tooth counts) are positive integers, it becomes convenient to generalise them to be rationals. In particular, my sun radius s is normally negative! Vectors X, Y of different trains are related by some sequence of concentric (coaxial) increments and rescalings just when a X + b I = c Y + d I , where I = [1, 1; 1, ..., 1] with a,b,c,d integers. X can be reduced to a unique canonical form via Z = ( X - (r+s)/2 I ) / (r-s)/2 = [ 1, -1; p', ... ] ; to establish whether trains X, Y are related, we need only compare their reduced forms! Example: the smallest 8-planet train is X = [ 28, -12; 12, 8, 4, 3; 4, 8, 12, 13 ] , h = 10 . Scaling this up by c = 2 then incrementing by d = -3 yields the train reported earlier by Somsky, Y = [ 53, -27; 21, 13, 5, 3; 5, 13, 21, 23 ] , h = 20 . But if unaware of their relation, we could deduce it by reduction of both to Z = [ 1, -1; 1/5, 0, -1/5 -1/4; -1/5, 0, 1/5, 1/4 ] , h = 1/2 . Note that for this family, largest and smallest planets lie on the axis; also we can show that for any related vector representing realisable gears, it is impossible to prevent overlap between the largest three. Reduced forms have some pleasant properties: primarily, they save space in a train catalogue. Also a "Somsky pair" of planet radii p, v , on opposite sides with r + s = p + v , reduce to radii with equal size and opposite signs; if furthermore p = v then both reduce to zero. Although reduction apparently fails if r = s , this case is the null family with planets and sunset vanishing: we set Z = [0, 0; ..., 0] . Finally (as demonstrated elsewhere in inadvertant but convincing fashion) train coordinates as defined above labour under a dichotomy between radii (sic) X and centre h : the resulting nuisance involves not merely the scale factor 2pi , but varying behaviour under symmetry transformations. It is possible to eliminate this conceptual untidiness. One strategy ditches the centres, instead incorporating belt lengths around adjacent pairs of planets into the vector. Another uses Lie-sphere coordinates to unify radius and centre: rescaling and concyclic increment correspond to geometric transformations of an oriented circle, respectively dilation and (wait for it) offset (noooh!). Fred Lunnon
Umm.... Ah! You're setting X_2 = -(sun radius) so that I is all ones, rather than [1, -1; 1, ..., 1], yes? On 08/06/15 09:42, Fred Lunnon wrote:
With planets placed in cyclic order around the sun, and axis indicated by semicolon, the vector
X = [ r, s; p, ..., q; u, ..., v ]
of gear radii becomes essentially unique; and combined with sunset (offset) h , it specifies the train. While in a realisable train the radii (more properly, tooth counts) are positive integers, it becomes convenient to generalise them to be rationals. In particular, my sun radius s is normally negative!
Vectors X, Y of different trains are related by some sequence of concentric (coaxial) increments and rescalings just when
a X + b I = c Y + d I , where I = [1, 1; 1, ..., 1]
with a,b,c,d integers. X can be reduced to a unique canonical form via
On 8/6/15, William R Somsky <wrsomsky@gmail.com> wrote:
Umm.... Ah!
You're setting X_2 = -(sun radius) so that I is all ones, rather than [1, -1; 1, ..., 1], yes?
There's a little more to it than that. My radius sign indicates direction of rotation (rather than of teeth, as one might naïvely expect). With this convention, concentric increment corresponds to the "offset" operation in spherical geometry, in which each oriented sphere (including points and hyperplanes) changes in radius by a constant amount. WFL
Incidentally, under a fairly mundane extension of mechanical realisability, the sun is allowed partially to overlap the ring, creating a new crescent-shaped region in which planets rotate counter to those remaining in the old crescent. The new planets require negative radii for previous formulae to work smoothly. WFL On 8/6/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 8/6/15, William R Somsky <wrsomsky@gmail.com> wrote:
Umm.... Ah!
You're setting X_2 = -(sun radius) so that I is all ones, rather than [1, -1; 1, ..., 1], yes?
There's a little more to it than that. My radius sign indicates direction of rotation (rather than of teeth, as one might naïvely expect). With this convention, concentric increment corresponds to the "offset" operation in spherical geometry, in which each oriented sphere (including points and hyperplanes) changes in radius by a constant amount.
WFL
What constraints determine whether a vector represents some realisable train? If sunset h = 0 the train is conventional with sun and ring concentric, and if h < 0 the axis may be reversed; so assume h > 0 . All radii will be nonzero integers, positive except for s . The sun and ring will not overlap; however, restricting h a priori to prevent planets overlapping does not currently appear feasible. Experimentally it appears that the overall structure of realisable X is quite simple. Each of the following cases is a subset of the case preceding; cases with unequal numbers of planets on each side can always be completed via Somsky's lemma. Proofs are mundane and available but omitted. ** Case 2 planets: Somsky's lemma shows X = [ r, s; p ; u ] with r + s = p + u and (say) p >= u is realisable for the continous interval of sunsets max( 0, 2 p - r - s, r + s - 2 u ) < h < r + s . ** Case 4 planets: X = [ r, s; p, q; u, v ] with r + s = p + u = q + v and p > q is realisable for some finite nonempty set of h , in the smaller interval deducible from previous case. ** Case 6 planets: X = [ r, s; p, t, u; u, t, p ] with r + s = p + u = 2 t and p > t > u is realisable for some finite nonempty set of h , in the previous interval. ** Case 8 planets: X = a Z + b I , where I = [ 1, 1; 1, 1, 1, 1; 1, 1, 1, 1 ] , Z = [ 1, -1; 1/5, 0, -1/5 -1/4; -1/5, 0, 1/5, 1/4 ] , h = 1/2 is realisable for integers a, b satisfying previous restrictions. A computer survey of trains with r <= 50 supports the ** Conjecture: Every realisable eccentric train falls under some case above! In particular, it would follow that there are no trains with 10 planets. Fred Lunnon On 8/6/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Incidentally, under a fairly mundane extension of mechanical realisability, the sun is allowed partially to overlap the ring, creating a new crescent-shaped region in which planets rotate counter to those remaining in the old crescent. The new planets require negative radii for previous formulae to work smoothly.
WFL
On 8/6/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
On 8/6/15, William R Somsky <wrsomsky@gmail.com> wrote:
Umm.... Ah!
You're setting X_2 = -(sun radius) so that I is all ones, rather than [1, -1; 1, ..., 1], yes?
There's a little more to it than that. My radius sign indicates direction of rotation (rather than of teeth, as one might naïvely expect). With this convention, concentric increment corresponds to the "offset" operation in spherical geometry, in which each oriented sphere (including points and hyperplanes) changes in radius by a constant amount.
WFL
The following may appear well off-topic from "coaxial" families (sic --- prefer "concentric"?); nonetheless there's a connection: Conjecture: a rational offset always takes the form (integer / 2) . WFL On 8/5/15, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Unfortunately, my previous carefully-honed missile was directed at an obsolete target (arc-lengths of belts), rather than the current one (offset of sun).
For offset, we want to solve the equation in Z with rational A,B,C,D A arccos( B Z^2 + C) + ... = D , so that cos D is a polynomial with rational coefficients in cos(Z^2) and sin(Z^2) , which can be further rationalised to a polynomial in cos^2(Z^2) alone.
Solving this yields Z^2 = arccos(algebraic) ; and APG reminds us that this never has properly algebraic solutions, only rational or (more usually) transcendental!
Fred Lunnon
On 8/4/15, Adam P. Goucher <apgoucher@gmx.com> wrote:
In fact, the only solutions to:
cos(algebraic * pi) = algebraic
are of the form:
cos(rational * pi) = algebraic
Proof: --------
It is clear that cos(algebraic * pi) and exp(algebraic * (pi i)) are algebraically dependent, so we can consider the equivalent problem to find solutions to:
exp(algebraic * (pi i)) = different algebraic
which rearranges to:
(-1)^algebraic = different algebraic
By Gelfond-Schneider, this occurs only when the exponent is rational.
--------
Clearly the converse is also true, so the algebraic numbers x for which cos(x * pi) is also algebraic are precisely the rationals.
Sincerely,
Adam P. Goucher
Sent: Tuesday, August 04, 2015 at 8:23 PM From: "Fred Lunnon" <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Offset Somsky Gears, coaxial offset
Before this intrusion of ineptitude risks the moral endangerment of any further tender young minds, I feel obliged to point out the logical inequivalence between the statements cos(rational * 2pi) = algebraic ! and arccos(rational) = algebraic * 2pi ? --- the second of which is in general (to put it mildly) problematic.
Those remaining in doubt about the matter should reflect upon the kindergarten fact arccos(1/3) = dihedral angle of regular tetrahedron, the expression of which as an algebraic angle ought to be worth a Fields at least!
Fred Lunnon
On 8/3/15, Warren D Smith <warren.wds@gmail.com> wrote:
Boy am I confused. You have non-algebraic equations with suspected algebraic roots? If they're algebraic, I don't need any decimal places at all. --Bill Gosper
--Answer (WDS): I showed how the non-algebraic equations involving arctrig, could be restated in a way involving integer powers of complex numbers (with coordinates inside the complexes involving square roots). So then really, they are algebraic equations in disguise. And then yes, you do not need any decimals at all to know algebraic.
Furthermore in an early post I explored general bipartite-planar gear-circle-contact graphs, and the equations there are just about distances, which are of course algebraic functions.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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participants (7)
-
Adam P. Goucher -
Bill Gosper -
Fred Lunnon -
M. Oskar van Deventer -
Tom Rokicki -
Warren D Smith -
William R Somsky