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« Return to Thread: 5515-Year Luni-Solar Cycle
Re: 5515-Year Luni-Solar Cycle
by Brillig
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Speaking of my 28/293 calendar, this month is the last in a series of
rectangular months because next month contains a special day. There won't be
another rectangular month until 7 more special days.
I define a rectangular month as a month where day 0 is on Sunday and the
last day of the month is on Saturday. Weeks are arranged Sunday through
Saturday as is the custom here in America.
The next special day is on -4-0-19, 09 Jan 2009 Gregorian. This is three
days after the troll in my yard solves a puzzle.
This isn't relevant to anything, but I find it interesting nonetheless. The
day number of the special day contains the same digits as the month and day
in the Gregorian calendar. Furthermore, the special day occurs on day 9 of
the Gregorian calendar and 9 days from the end of the month in the 28/293
calendar (the accumulator of the month is 9).
Victor
> -----Original Message-----
> From: East Carolina University Calendar discussion List [mailto:CALNDR-
> L@...] On Behalf Of Palmen, KEV (Karl)
> Sent: Monday, December 01, 2008 6:49 AM
> To: CALNDR-L@...
> Subject: Re: 5515-Year Luni-Solar Cycle
>
> Dear Helios, Victor and Calendar People
>
> So the Octaeteris new year would drift through all the seasons
> (spring->summer->fall->winter) three times in the course of one
> 5515-year cycle.
>
> The Octaeteris has eight years (an even-number) so cannot be made into a
> Helios cycle. This may be why Helios cannot find a leap-moon function
> for it. However the three leap months can be spread perfectly evenly
> through the 99 months of Octeateris exactly one every 33 months. To form
> a Helios cycle of months, select the 17th, 50th and 83rd month as leap
> months in the 99-month cycle.
>
> The 5515-year cycle has 2014311 = 3 * (13^2) * 29 * 137 days and 5512 =
> (2^3) * 13 * 53. So there is a common divisor of 13 for the 5512
> Octeateris years. So the octaeteris cycle would repeat once every 424
> octaeteris years = 53 octaeterides = three 1749-month cycles. This
> arises from the fact that 1749 is divisible by 33=99/3 so makes up
> 1749/99 = 17 2/3 octaeterides = 141 1/3 years, which is one third of 424
> Octeateris years.
>
>
> I realise now that a lunar calendar using the 1749-month cycle could
> have a 13-month solar calendar run alongside it so that three 1749-month
> cycles of 5247 lunar months are exactly equal to 5515 solar months. The
> 5155 solar months (424 years and three months) would have 527 days in
> addition to 28 days per month. In this period, there would be
> 5515-5247=268 solar months that do not contain the first day of a lunar
> month. A 527/5515 calendar like Victor's 43/450 or 28/293 calendar would
> do this. However there may be a way of using the lunar calendar to
> determine which solar month gets an extra day.
>
> One possibility is to have two long (29-day) solar months begin in each
> 17-month yerm and one long solar month begin in each short 15-month
> yerm. There'd be ten exceptional 17-month yerms every three 1749-month
> (107-yerm) cycles with just one long solar month beginning in it. This
> leads to 2*321 - 10 + 105 = 527 long solar months as required. Also
> between any two consecutive exceptions, there'd be exactly 103 solar
> months every 6 yerms of 98 lunar months.
> Another possibility is to have a long solar month after each solar month
> that misses either the 1st or 15th day of a lunar month, with three
> exceptions every 1749-month cycle. This leads to 268*2 - 3*3 = 527 long
> months as required.
>
>
>
> Karl
>
> 10(03(04
>
>
> -----Original Message-----
> From: East Carolina University Calendar discussion List
> [mailto:CALNDR-L@...] On Behalf Of Helios
> Sent: 01 December 2008 00:17
> To: CALNDR-L@...
> Subject: Re: Lunisolar Cycles made up of Short Lunar Cycles RE:
> 5515-Year Luni-Solar Cycle
>
> 5515-Year Luni-Solar Cycle
>
> Here's one more observation of the 5k515 ( or V_DXV ) lunisolar and
> solar
> cycle
>
> The cycle can now include the
> octaeteris year = 365 & 187 / 424 days
>
> The cycle of entirety
> = 4173 yerms
> = 5512 octaeteris years
> = 5515 years
> = 68211 months
> = 154947 trecena
> = 2014311 days
>
> This proportionality constant K = 5512 / 5515 could just serve to define
> this lunisolar year,
> Y = K*( octaeteris year )
>
> I don't know yet if the octaeteris leap-moon function
> ( 3*Y + 3 + ?? )MOD( 8 ) < 3
> 2, 5, 7, 10, 13, 15, 18, 21, ??, ??
> can be tweaked and functionally related to K to yield the same output as
> ( 2031*Y + 2757 )MOD( 5515 ) < 2031
>
> --
> View this message in context:
> http://www.nabble.com/5515-Year-Luni-Solar-Cycle-tp20632776p20764152.htm
> l
> Sent from the Calndr-L mailing list archive at Nabble.com.
>
> --
> Scanned by iCritical.
rectangular months because next month contains a special day. There won't be
another rectangular month until 7 more special days.
I define a rectangular month as a month where day 0 is on Sunday and the
last day of the month is on Saturday. Weeks are arranged Sunday through
Saturday as is the custom here in America.
The next special day is on -4-0-19, 09 Jan 2009 Gregorian. This is three
days after the troll in my yard solves a puzzle.
This isn't relevant to anything, but I find it interesting nonetheless. The
day number of the special day contains the same digits as the month and day
in the Gregorian calendar. Furthermore, the special day occurs on day 9 of
the Gregorian calendar and 9 days from the end of the month in the 28/293
calendar (the accumulator of the month is 9).
Victor
> -----Original Message-----
> From: East Carolina University Calendar discussion List [mailto:CALNDR-
> L@...] On Behalf Of Palmen, KEV (Karl)
> Sent: Monday, December 01, 2008 6:49 AM
> To: CALNDR-L@...
> Subject: Re: 5515-Year Luni-Solar Cycle
>
> Dear Helios, Victor and Calendar People
>
> So the Octaeteris new year would drift through all the seasons
> (spring->summer->fall->winter) three times in the course of one
> 5515-year cycle.
>
> The Octaeteris has eight years (an even-number) so cannot be made into a
> Helios cycle. This may be why Helios cannot find a leap-moon function
> for it. However the three leap months can be spread perfectly evenly
> through the 99 months of Octeateris exactly one every 33 months. To form
> a Helios cycle of months, select the 17th, 50th and 83rd month as leap
> months in the 99-month cycle.
>
> The 5515-year cycle has 2014311 = 3 * (13^2) * 29 * 137 days and 5512 =
> (2^3) * 13 * 53. So there is a common divisor of 13 for the 5512
> Octeateris years. So the octaeteris cycle would repeat once every 424
> octaeteris years = 53 octaeterides = three 1749-month cycles. This
> arises from the fact that 1749 is divisible by 33=99/3 so makes up
> 1749/99 = 17 2/3 octaeterides = 141 1/3 years, which is one third of 424
> Octeateris years.
>
>
> I realise now that a lunar calendar using the 1749-month cycle could
> have a 13-month solar calendar run alongside it so that three 1749-month
> cycles of 5247 lunar months are exactly equal to 5515 solar months. The
> 5155 solar months (424 years and three months) would have 527 days in
> addition to 28 days per month. In this period, there would be
> 5515-5247=268 solar months that do not contain the first day of a lunar
> month. A 527/5515 calendar like Victor's 43/450 or 28/293 calendar would
> do this. However there may be a way of using the lunar calendar to
> determine which solar month gets an extra day.
>
> One possibility is to have two long (29-day) solar months begin in each
> 17-month yerm and one long solar month begin in each short 15-month
> yerm. There'd be ten exceptional 17-month yerms every three 1749-month
> (107-yerm) cycles with just one long solar month beginning in it. This
> leads to 2*321 - 10 + 105 = 527 long solar months as required. Also
> between any two consecutive exceptions, there'd be exactly 103 solar
> months every 6 yerms of 98 lunar months.
> Another possibility is to have a long solar month after each solar month
> that misses either the 1st or 15th day of a lunar month, with three
> exceptions every 1749-month cycle. This leads to 268*2 - 3*3 = 527 long
> months as required.
>
>
>
> Karl
>
> 10(03(04
>
>
> -----Original Message-----
> From: East Carolina University Calendar discussion List
> [mailto:CALNDR-L@...] On Behalf Of Helios
> Sent: 01 December 2008 00:17
> To: CALNDR-L@...
> Subject: Re: Lunisolar Cycles made up of Short Lunar Cycles RE:
> 5515-Year Luni-Solar Cycle
>
> 5515-Year Luni-Solar Cycle
>
> Here's one more observation of the 5k515 ( or V_DXV ) lunisolar and
> solar
> cycle
>
> The cycle can now include the
> octaeteris year = 365 & 187 / 424 days
>
> The cycle of entirety
> = 4173 yerms
> = 5512 octaeteris years
> = 5515 years
> = 68211 months
> = 154947 trecena
> = 2014311 days
>
> This proportionality constant K = 5512 / 5515 could just serve to define
> this lunisolar year,
> Y = K*( octaeteris year )
>
> I don't know yet if the octaeteris leap-moon function
> ( 3*Y + 3 + ?? )MOD( 8 ) < 3
> 2, 5, 7, 10, 13, 15, 18, 21, ??, ??
> can be tweaked and functionally related to K to yield the same output as
> ( 2031*Y + 2757 )MOD( 5515 ) < 2031
>
> --
> View this message in context:
> http://www.nabble.com/5515-Year-Luni-Solar-Cycle-tp20632776p20764152.htm
> l
> Sent from the Calndr-L mailing list archive at Nabble.com.
>
> --
> Scanned by iCritical.
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