Dear Helios and Calendar People
One can define nodetides for the lunar year of 12 months, using the same two formulae. For an eclipse cycle of M months and E eclipse seasons, setting M = 12*Y in the second formula gives: N = E - M/6 The shorter year makes a longer nodetide and this lunar nodetide is equal to exactly six eclimpiads and so consists of a number of six-month eclipse season intervals and six five-month eclipse season intervals (Pentalunex). This comes out at about 21.7 years (22.4 lunar years) on average. In https://www.staff.science.uu.nl/~gent0113/eclipse/eclipsecycles.htm the cycles less than 1000 years that are a multiple of 12 months are (with both years and nodetides lunar): Lunar year: 1 year, 0 nodetides Unidos: 67 years, 3 nodetides Trihex: 201 years, 9 nodetides (3 Unidos) Unnamed (327): 337 years, 15 nodetides Hexodeka: 402 years, 18 nodetides (6 Unidos) Grattan-Guinness: 403 years, 18 nodetides Basic Period: 537 years, 24 nodetides Tetradia: 604 years, 27 nodetides Hyper Exeligmos: 939 years, 42 nodetides. In all these examples, the number of lunar nodetides is divisible by 3. The longest cycle listed of a whole number of lunar years (Immobilis) has 84 lunar nodetides also divisible by 3. I expect a longer cycle will have a number of lunar nodetides not divisible by 3. The number of lunar nodetides in a cycle of A Inex cycles and B Saros cycle is equal to (4/3)*A + (5/6)*B. Karl 16(04(10 -----Original Message----- From: East Carolina University Calendar discussion List [mailto:[hidden email]] On Behalf Of Helios Sent: 27 October 2016 08:43 To: [hidden email] Subject: Nodetides Dear Calendar People, There's a basic period between when a node alignment and the solar year nearly coincide. This period averages about 9.3 years. I call this a nodetide. mean nodetide = 1/[ ( y/e ] - 2 ) Should we subtract from the year two eclipse seasons, the remaining 19 days is the portion of the year within which we can expect a node alignment to occur every nodetide. A luni-solar eclipse cycle will always contain a whole number of nodetides. For a luni-solar eclipse cycle, we can subtract from the eclipse seasons the number of years doubled to find the number of nodetides in the cycle. N = E - 2*Y Beginning from a node alignment at a certain time in year 0, we can predict the following nodetide years. I'm satisfied with the following accumulator function; ( 75*Y + 34 )MOD( 698 ) < 75 9, 19, 28, 37, 47, 56, 65, 74, 84, 93, 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, 354, 363, 372, 391, . . . The function seems to test whether or not a luni-solar cycle is an eclipse cycle. -- View this message in context: http://calndr-l.10958.n7.nabble.com/Nodetides-tp17237.html Sent from the Calndr-L mailing list archive at Nabble.com. |
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