Dear Helios and Calendar People
---------------------------------------------------------- by the intersection of the two accumulators, ( 75*Y + 37 )MOD( 698 ) < 75 ( 253*Y + 126 )MOD( 687 )< 253 ---------------------------------------------------------- The first accumulator is for nodetide years and the second is for lunisolar leap month years for a 687-year cycle formed from one 334-year cycle and one 353-year cycle. Finding the years is less laborious if you exploit the patterns in the accumulators. For ( 75*Y + 37 )MOD( 698 ) < 75, the intervals are 9 or 10 years, the respect accumulator changes are -23 and +52 as I showed in an earlier note and in that note I constructed a complete first half of this cycle -009, 000, +009, +019, +028, +037, +047, +056, +065, +074, +084, +093, +102, +112, +121, +130, +140, +149, +158, +168, +177, +186, +195, +205, +214, +223, +233, +242, +251, +261, +270, +279, +289, +298, +307, +316, +326, +335, +344, The other half can be found by symmetry. The respective accumulators are 60, 37, 14, 66, 43, 20, 72, 49, 26, 03 55, 32, 09, etc.. See how the accumulators go up 6 after each short row and down 17 after each long row. This enables you the quickly produce later rows. Each row has intervals of 9 years within and intervals of 10 years between. Short rows have 28 years long rows have 37 years. The second accumulator ( 253*Y + 126 )MOD( 687 ) < 253 has complexity 5 rather than 4 of the first accumulator expression so its patterns are more complicated. Also the leap years are much closer together. The intervals are 2 or 3 years with accumulator changes of -181 and +72 respectively (compared to -23 and +52). However there is a worse criticism of this second accumulator. It allows lunisolar cycles that are several days out. Instead, I suggest going to the cycle created by the first accumulator and checking the remainder its years divided by 19. For years -698/4 to +698/4 (-174 to +174) select if that remainder is 0. For years 689/4 to 698*(3/4) (175 to 523) select if that remainder is 11. For years 698*(3/4) to 698*(5/4) (524 to 872) select if remainder is 3. etc. This is equivalent to selecting year Y if ( 7*Y - NINT(Y/349) ) MOD ( 19 ) = 0 NB: Use (x)MOD(y) = x - y*FLOOR(x/y). The NINT(Y/349) corrects the 19-year Metonic cycle. NINT(x) is the nearest integer to x and is well defined, because x has an odd denominator of 349. I chose NINT rather than FLOOR, because it gives symmetry about year 0 as in the 698-year cycle. One could use some other number of the magnitude of 349 instead of 349 such as 343.5, but 349 has the advantage of interlocking with the 698-year cycle and so gives each year of the 698-year cycle one place in a range of years -6631 to +6631, which is determined by its remainder when divided by 19 and which part of the cycle (0 to 174, 175 to 523 or 523 to 697) it belongs. I reckon the series starts 0, 19, 372, 391, 763, 782, ... I may extend this series in a later note. Karl 16(03(22 -----Original Message----- From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Helios Sent: 21 November 2016 06:31 To: [hidden email] Subject: Re: Structural complexity of Nodetide-year cycles RE: Nodetides Dear Karl and Calendar People, I now see a working method that will generate a list of lunisolar eclipse cycles, or ec-lunisolar cycles. ---------------------------------------------------------- by the intersection of the two accumulators, ( 75*Y + 37 )MOD( 698 ) < 75 ( 253*Y + 126 )MOD( 687 )< 253 ---------------------------------------------------------- this series begins 0, 19, 65, 84, 130, 149, 177, 223, 242, 261, 307, 326, 372, 391, . . . It's laborious to find these years and a computer program could make a longer list. -- View this message in context: http://calndr-l.10958.n7.nabble.com/Structural-complexity-of-Nodetide-year-cycles-RE-Nodetides-tp17264p17286.html Sent from the Calndr-L mailing list archive at Nabble.com. |
Dear Helios and Calendar People
I've worked out the number of years Y in all 38 the lunisolar eclipse cycles that satisfy both ( 75*Y + 37 )MOD( 698 ) < 75 and ( 7*Y - NINT(Y/349) )MOD( 19 ) = 0 up to Y = 349*19 = 6631. If anyone shows interest in how I worked them out, I'll show this in a later note. I also show the number of nodetides in () and each row corresponds to the number of corrections of the Metonic cycle. 0(0), 19(2), 372(40), 391(42), 763(82), 782(84), 1135(122), 1154(124), 1173(126), 1526(164), 1545(166), 1564(168), 1917(206), None with 6, 7 or 8 corrections of Metonic cycle 2987(321), 3006(323), 3025(325), 3378(363), 3397(365), 3416(367), 3750(403), 3769(405), 4141(445), 4160(447), 4179(449), 4532(487), 4551(489), 4570(491), 4923(529), 4942(531), 5295(569), 5314(571), 5333(573), 5686(611), 5705(613), 5724(615), 6077(653), 6096(655). More can be found by subtracting these from 13262(1425). For example 13262(1425) - 6096(655) = 7166(770), which is the next one after the 6096-year cycle. They are optimised to a mean nodetide of 9 & 23/75 = 9.306667 years and a mean year of 12 & 2442/6631 = 12.368270 synodic months. This may cease to be accurate over a very long cycle. Karl 16(03(23 -----Original Message----- From: Palmen, Karl (STFC,RAL,ISIS) Sent: 21 November 2016 13:22 To: 'East Carolina University Calendar discussion List' Subject: Lunisolar Eclipse Cycles RE: Structural complexity of Nodetide-year cycles RE: Nodetides Dear Helios and Calendar People ---------------------------------------------------------- by the intersection of the two accumulators, ( 75*Y + 37 )MOD( 698 ) < 75 ( 253*Y + 126 )MOD( 687 )< 253 ---------------------------------------------------------- The first accumulator is for nodetide years and the second is for lunisolar leap month years for a 687-year cycle formed from one 334-year cycle and one 353-year cycle. Finding the years is less laborious if you exploit the patterns in the accumulators. For ( 75*Y + 37 )MOD( 698 ) < 75, the intervals are 9 or 10 years, the respect accumulator changes are -23 and +52 as I showed in an earlier note and in that note I constructed a complete first half of this cycle -009, 000, +009, +019, +028, +037, +047, +056, +065, +074, +084, +093, +102, +112, +121, +130, +140, +149, +158, +168, +177, +186, +195, +205, +214, +223, +233, +242, +251, +261, +270, +279, +289, +298, +307, +316, +326, +335, +344, The other half can be found by symmetry. The respective accumulators are 60, 37, 14, 66, 43, 20, 72, 49, 26, 03 55, 32, 09, etc.. See how the accumulators go up 6 after each short row and down 17 after each long row. This enables you the quickly produce later rows. Each row has intervals of 9 years within and intervals of 10 years between. Short rows have 28 years long rows have 37 years. The second accumulator ( 253*Y + 126 )MOD( 687 ) < 253 has complexity 5 rather than 4 of the first accumulator expression so its patterns are more complicated. Also the leap years are much closer together. The intervals are 2 or 3 years with accumulator changes of -181 and +72 respectively (compared to -23 and +52). However there is a worse criticism of this second accumulator. It allows lunisolar cycles that are several days out. Instead, I suggest going to the cycle created by the first accumulator and checking the remainder its years divided by 19. For years -698/4 to +698/4 (-174 to +174) select if that remainder is 0. For years 689/4 to 698*(3/4) (175 to 523) select if that remainder is 11. For years 698*(3/4) to 698*(5/4) (524 to 872) select if remainder is 3. etc. This is equivalent to selecting year Y if ( 7*Y - NINT(Y/349) ) MOD ( 19 ) = 0 NB: Use (x)MOD(y) = x - y*FLOOR(x/y). The NINT(Y/349) corrects the 19-year Metonic cycle. NINT(x) is the nearest integer to x and is well defined, because x has an odd denominator of 349. I chose NINT rather than FLOOR, because it gives symmetry about year 0 as in the 698-year cycle. One could use some other number of the magnitude of 349 instead of 349 such as 343.5, but 349 has the advantage of interlocking with the 698-year cycle and so gives each year of the 698-year cycle one place in a range of years -6631 to +6631, which is determined by its remainder when divided by 19 and which part of the cycle (0 to 174, 175 to 523 or 523 to 697) it belongs. I reckon the series starts 0, 19, 372, 391, 763, 782, ... I may extend this series in a later note. Karl 16(03(22 -----Original Message----- From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Helios Sent: 21 November 2016 06:31 To: [hidden email] Subject: Re: Structural complexity of Nodetide-year cycles RE: Nodetides Dear Karl and Calendar People, I now see a working method that will generate a list of lunisolar eclipse cycles, or ec-lunisolar cycles. ---------------------------------------------------------- by the intersection of the two accumulators, ( 75*Y + 37 )MOD( 698 ) < 75 ( 253*Y + 126 )MOD( 687 )< 253 ---------------------------------------------------------- this series begins 0, 19, 65, 84, 130, 149, 177, 223, 242, 261, 307, 326, 372, 391, . . . It's laborious to find these years and a computer program could make a longer list. -- View this message in context: http://calndr-l.10958.n7.nabble.com/Structural-complexity-of-Nodetide-year-cycles-RE-Nodetides-tp17264p17286.html Sent from the Calndr-L mailing list archive at Nabble.com. |
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