5515-Year Luni-Solar Cycle

[Irv Bromberg]

On 2009 Apr 28, at 20:59 , Helios wrote:

All the 5515 year cycle has to do is survive to within one day all of the
long-term physical effects mentioned by Irv B.. I think it's about right but
it should expire when it's finished and not be used multiple times.

Irv replies:  OK, I stand corrected.  For a lunisolar cycle it is often necessary to employ very long cycles to obtain a particular mean month or mean year.  For example the Gregorian Easter computus is a 5,700,000-year cycle, and the Dee Easter computus is a 131,670-year cycle.  There is no requirement for any significant portion of the cycle to elapse during the useful lifetime of the cycle.

On 2009 Apr 28, at 11:11 , Irv Bromberg wrote:

What is the "calendar season" that you are targeting?  (northward equinox?)

What is your target calendar mean year?  (mean northward equinoctial year presently = 365d 5h 49m)

Irv replies:  Helios has selected a cycle based on an assumed exact integer correspondence of the solar and lunar cycles.  Thus I can answer my own questions quoted above.  Some of this information appeared in previous messages about this cycle, but here I gather it in one place (please correct me if any arithmetic is wrong):

The 5515-year cycle has 39 repeats of a cycle of a 1749 lunar month subcycle = 68211 lunar months.

Each 1749-month subcycle contains exactly 51649 days so the mean month is exactly 29+928/1749 days = 29d 12h 44m 2+514/583s (the fraction is about 0.88s).

Thus it has 928 full months and 821 deficient months, with 35 repeats of the 3-yerm 49-month subcycle and also contains two stand-alone 17-month yerms, for a total of 107 yerms.

The mean year is exactly 51649*39/5515 = 365+1336/5515 days = 365d 5h 48m 50+290/1103s (the fraction is about 0.263s).

(There is not an integer number of weeks:  neither the 51649 days of the 1749 lunar month subcycle nor the full cycle 39*51649=2014311 days is divisible by 7.)

The calendar season calculated for 353-year intervals and middle year 1765 is presently stable at an ecliptic solar longitude of about 9.9° with the less stable calendar season at about 196°.  At the next interval, middle year 2118, it is stable at about 11.8°.  So it is slightly short compared to the mean northward equinoctial year, but that will allow it to "hold on" to the equinox longer.  My SOLEX-based Solar Calendar Drift spreadsheet shows that with epoch = Gregorian epoch the equinox will slowly drift later in the calendar year until it reaches a maximum of about 1/2 day late around year 5000, then will slowly drift back, returning to zero drift about year 8000, then will rapidly drift earlier in the calendar year, reaching 1/2 day early around year 9000 and 1 day early around year 9600.  This is quite respectable performance and rather long-lived for a cycle intended to approximate the northward equinox.

I wouldn't agree that 1 day of drift for the lunar component of a calendar cycle is at all acceptable, but anyhow the cycle's mean month is of good accuracy for the present era.  After manually entering its mean month excess of 2+514/583 seconds, my Lunar Calendar Drift spreadsheet shows that with a present-era lunar epoch it will reach about 2h late around year 3700, 4h late around 4500, 6h late in 5000, 8h late in 5600, 10h late around year 10000, 12h late around 6500, and 24h late around year 8400.  This performance is about as good as can be obtained with any fixed mean month that is reasonably accurate for the present era.

On 2009 Apr 28, at 11:28 , Palmen, KEV (Karl) wrote:

The only merit to the 5515-year cycle is that it is an exact multiple of a lunar 1749-month cycle as explained in earlier notes about this cycle.

Irv replies:  OK, so even though the much shorter 353-year cycle is an exact multiple of 4366 months, the 5515-year cycle has the merit of containing a shorter exact lunar subcycle of 1749 months.  I don't find this particularly compelling, but this made me I realize that I have never carried out a similar yerm-oriented breakdown of the 353-year / 4366-month cycle.

Helios wrote:  I will name the 5515 year cycle the "Meritorious Cycle" ( just kidding ).

Irv replies:  Works for me!

-- Irv Bromberg, Toronto, Canada

<http://www.sym454.org/>

[Helios]

Dear Karl P
I still think that 37 leap months must be dropped.
I get 2068 years that satisfy

( 3*Y + 3 )MOD( 8 ) < 3

in the entire 5515 year cycle.
------------------------------------------------
by grouping into threes ( regard the even number )

2, 5, 7
10, 13, 15
18, 21, 23
...
5498, 5501, 5503
5506, 5509, 5511
5514 ( end )
the amount must be as 3n + 1. Karl's 2067 is divisable by three
------------------------------------------------
However I'm now thinking the 5515 year cycle is not that good for this outré calendar, instead a luni-solar cycle whose number of years is divisible by 8 would be better. I have learned that a month is dropped nearly every 149 years, on average, from the octaeteris to keep pace with a year.

Palmen, KEV (Karl)-2 wrote:

Dear Helios, Irv and Calendar People

I think the most important property of the 5515-year cycle is that it is
a whole number (39) of an accurate lunar calendar cycles (of 1749
months). Helios does not seem to exploit this property enough, but
instead tries to exploit the property that the number of lunar months is
divisible by 99 and so it a whole number (689) of Octaeterides.

The whole number of lunar calendar cycles is divisible by 13. This
suggests to me using a solar calendar of 13 months in every year
alongside the lunar calendar with a 1749-month cycle.

I'm aware that if each solar year of the 5515-year cycle were divided
into 13 solar months, then 5515 solar months would have exactly three
cycles of 1749 lunar months. This is equal to 53 octaeterides and so 424
Octaeteris years. It is also equal to 424 3/13 solar years and so the
Octaeteris year falls one solar month behind the solar year exactly once
every lunar cycle of 1749 lunar months.


-----Original Message-----
From: East Carolina University Calendar discussion List
[mailto:[e-mail]] On Behalf Of Helios
Sent: 28 April 2009 06:15
To: [e-mail]
Subject: Re: 5515-Year Luni-Solar Cycle

The symmetrical distribution of 13-month years in the 5515 year cycle we
use

( 2031*Y + 2757 )MOD( 5515 ) < 2031

and a look at the very beginning and end years ( 13-month years )

2, 5, 7, 10, 13, 15, 18, 21

5495, 5498, 5501, 5503, 5506, 5509, 5511, 5514

evidently also obey the octaeteris accumulator

( 3*Y + 3 )MOD( 8 ) < 3

KARL SAYS: The first year that does not satisfy the Octaeteris
accumulator is year 24, whose 5515-year cycle accumulator is 1866 which
is less than 2031 so is a 5515-year cycle leap month year, but its
Octaeteris accumulator is 3 which is not less than 3. Exceptions will
occur more frequently later on as the octaeteris drifts out of place.
Towards the end of the 5515-year cycle the Octaeteris would have drifted
a three whole years late so will align itself again with the 5515-year
cycle years.
There'll be similar agreement about 1/3 and 2/3 through the 5515-year
cycle.

HELIOS CONTINUES:
If I am counting correctly, I get 2068 years that satisfy

( 3*Y + 3 )MOD( 8 ) < 3

in the entire 5515 year cycle.

KARL SAYS:
There are (3/8)*13*424 = 3*689 = 2067 not 2068 of them and
only a minority of these also satisfy ( 2031*Y + 2757 )MOD( 5515 ) <
2031 .

HELIOS CONTINUES:
We should like 2031 13-month years in the
5515 year cycle so 37 of these years we could remove the leap month to
make
an octaeteris calendar that is occasionally adjusted about every 149
years.
These 37 months equal the 3 years that is the difference between 5515
years
and 5512 octaeteris years.

KARL SAYS:
It's 36 years (2067=2031+36) rather than 37 to remove 13th month from.
Then you'd also need to insert three years of 12 months to bring the
5512 years (from the 689 octaeterides) up to 5515 years. Seems
pointless.
The property of the 5515-year cycle having a number of lunar months
divisible by 99 is no longer exploited.

I think it would be better to have a solar calendar with 13 solar months
in every year.
The solar months (normally of 28 days) can be arranged in a Helios cycle
of 5515 solar months, so a solar month M has an extra (29th) day if and
only if

( 527*M + 2757 ) MOD (5515) < 527

Each solar 5515-month cycle is exactly equal to three lunar 1749-month
cycles.
The resulting solar years would also form a Helios cycle of 5515 years,
such that year Y is has 366 days if and only if

( 1336*Y + 2757 ) MOD (5515) < 1336.

Now if you run an Octeateris calendar against this using the lunar
1749-month cycle, you find that the Octeateris year runs one solar month
late (on average) every lunar 1749-month cycle elapsed.


Karl

10(08(05

--
Scanned by iCritical.

[Karl Palmen]

Dear Helios and Calendar People

Comments below.

-----Original Message-----
From: East Carolina University Calendar discussion List [mailto:CALNDR-L@...] On Behalf Of Helios
Sent: 30 April 2009 01:08
To: CALNDR-L@...
Subject: Re: 5515-Year Lunisolar Cycle

Dear Karl P
I still think that 37 leap months must be dropped.
I get 2068 years that satisfy

( 3*Y + 3 )MOD( 8 ) < 3

in the entire 5515 year cycle.
------------------------------------------------
by grouping into threes ( regard the even number )

2, 5, 7
10, 13, 15
18, 21, 23
...
5498, 5501, 5503
5506, 5509, 5511
5514 ( end )
the amount must be as 3n + 1. Karl's 2067 is divisable by three
------------------------------------------------
However I'm now thinking the 5515 year cycle is not that good for this outré
calendar, instead a luni-solar cycle whose number of years is divisible by 8
would be better. I have learned that a month is dropped nearly every 149
years, on average, from the octaeteris to keep pace with a year.

KARL SAYS:
Yes. Three years need adding to the 5512 resulting from the 689 octaeterides being corrected. If one of those three years has a leap month, then 37 would need to be removed. However I see no point in this at all, because the 8-year cycle is broken by those 3 years (as Helios seems to have realised). I reckoned the 36 for the 689 whole octaeterides.

An alternative that crossed my mind is to have a leap month once every 33 months to form an octaeteris cycle, but correct it by turning a leap month into an ordinary month. This would preserve the 99-month cycle, but force the leap months to drift through the 8-year cycle. After each correction, the leap months would move one month later in the 8-year cycle. 36 corrections would be needed after which the leap months would have drifted over 3 years bringing 5512 to 5515. I think this can be made into a Helios cycle, if the 17th, 50th and 83rd month of each Octeateris is made into a leap month, except corrections. Correction would occur once every 57*33 or 58*33 months in a ratio of 7 to 5.

I have considered the idea of correcting an Octeateris with 2923.5 days on average, by dropping a month of 29 days. This also corrects the short lunar month (29.53030303 days) arising from the 2923.5 day Octeateris. This I have sent to the list. If correction is done once every 149 years, one gets a 2384-year cycle with 878 leap months and 870,739 days with a mean year of 365.24287 days and a mean month of 29.5305908 days.

However, both these suggestions, like all other schemes of correcting the Octaeteris by removing a month from a year suffers from a two month jitter (or worse).


Another idea is my Annuary Calendar http://www.hermetic.ch/cal_stud/palmen/anry.htm .


Karl

10(08(06



Palmen, KEV (Karl)-2 wrote:

>
> Dear Helios, Irv and Calendar People
>
> I think the most important property of the 5515-year cycle is that it is
> a whole number (39) of an accurate lunar calendar cycles (of 1749
> months). Helios does not seem to exploit this property enough, but
> instead tries to exploit the property that the number of lunar months is
> divisible by 99 and so it a whole number (689) of Octaeterides.
>
> The whole number of lunar calendar cycles is divisible by 13. This
> suggests to me using a solar calendar of 13 months in every year
> alongside the lunar calendar with a 1749-month cycle.
>
> I'm aware that if each solar year of the 5515-year cycle were divided
> into 13 solar months, then 5515 solar months would have exactly three
> cycles of 1749 lunar months. This is equal to 53 octaeterides and so 424
> Octaeteris years. It is also equal to 424 3/13 solar years and so the
> Octaeteris year falls one solar month behind the solar year exactly once
> every lunar cycle of 1749 lunar months.
>
>
> -----Original Message-----
> From: East Carolina University Calendar discussion List
> [mailto:CALNDR-L@...] On Behalf Of Helios
> Sent: 28 April 2009 06:15
> To: CALNDR-L@...
> Subject: Re: 5515-Year Luni-Solar Cycle
>
> The symmetrical distribution of 13-month years in the 5515 year cycle we
> use
>
> ( 2031*Y + 2757 )MOD( 5515 ) < 2031
>
> and a look at the very beginning and end years ( 13-month years )
>
> 2, 5, 7, 10, 13, 15, 18, 21
>
> 5495, 5498, 5501, 5503, 5506, 5509, 5511, 5514
>
> evidently also obey the octaeteris accumulator
>
> ( 3*Y + 3 )MOD( 8 ) < 3
>
> KARL SAYS: The first year that does not satisfy the Octaeteris
> accumulator is year 24, whose 5515-year cycle accumulator is 1866 which
> is less than 2031 so is a 5515-year cycle leap month year, but its
> Octaeteris accumulator is 3 which is not less than 3. Exceptions will
> occur more frequently later on as the octaeteris drifts out of place.
> Towards the end of the 5515-year cycle the Octaeteris would have drifted
> a three whole years late so will align itself again with the 5515-year
> cycle years.
> There'll be similar agreement about 1/3 and 2/3 through the 5515-year
> cycle.
>
> HELIOS CONTINUES:
> If I am counting correctly, I get 2068 years that satisfy
>
> ( 3*Y + 3 )MOD( 8 ) < 3
>
> in the entire 5515 year cycle.
>
> KARL SAYS:
> There are (3/8)*13*424 = 3*689 = 2067 not 2068 of them and
> only a minority of these also satisfy ( 2031*Y + 2757 )MOD( 5515 ) <
> 2031 .
>
> HELIOS CONTINUES:
> We should like 2031 13-month years in the
> 5515 year cycle so 37 of these years we could remove the leap month to
> make
> an octaeteris calendar that is occasionally adjusted about every 149
> years.
> These 37 months equal the 3 years that is the difference between 5515
> years
> and 5512 octaeteris years.
>
> KARL SAYS:
> It's 36 years (2067=2031+36) rather than 37 to remove 13th month from.
> Then you'd also need to insert three years of 12 months to bring the
> 5512 years (from the 689 octaeterides) up to 5515 years. Seems
> pointless.
> The property of the 5515-year cycle having a number of lunar months
> divisible by 99 is no longer exploited.
>
> I think it would be better to have a solar calendar with 13 solar months
> in every year.
> The solar months (normally of 28 days) can be arranged in a Helios cycle
> of 5515 solar months, so a solar month M has an extra (29th) day if and
> only if
>
> ( 527*M + 2757 ) MOD (5515) < 527
>
> Each solar 5515-month cycle is exactly equal to three lunar 1749-month
> cycles.
> The resulting solar years would also form a Helios cycle of 5515 years,
> such that year Y is has 366 days if and only if
>
> ( 1336*Y + 2757 ) MOD (5515) < 1336.
>
> Now if you run an Octeateris calendar against this using the lunar
> 1749-month cycle, you find that the Octeateris year runs one solar month
> late (on average) every lunar 1749-month cycle elapsed.
>
>
> Karl
>
> 10(08(05
>
> --
> Scanned by iCritical.
>
>
>

--
View this message in context: http://www.nabble.com/5515-Year-Luni-Solar-Cycle-tp20632776p23302623.html
Sent from the Calndr-L mailing list archive at Nabble.com.

[Karl Palmen]

Dear Helios and Calendar People

-----Original Message-----
From: East Carolina University Calendar discussion List
[mailto:CALNDR-L@...] On Behalf Of Helios
Sent: 28 April 2009 06:15
To: CALNDR-L@...
Subject: Re: 5515-Year Luni-Solar Cycle

The symmetrical distribution of 13-month years in the 5515 year cycle we
use

( 2031*Y + 2757 )MOD( 5515 ) < 2031

and a look at the very beginning and end years ( 13-month years )

2, 5, 7, 10, 13, 15, 18, 21

5495, 5498, 5501, 5503, 5506, 5509, 5511, 5514

evidently also obey the octaeteris accumulator

( 3*Y + 3 )MOD( 8 ) < 3

KARL SAYS:
You can get an exact match by using the Helios 19-year cycle ( 2, 5, 7,
10, 13 15, 18 ) and its truncation to 11 years ( 2, 5, 7, 10 ) and
arranging them as a Helios cycle. You'll need 16 truncated to 11 years
out of a total of 297.

NB: 5515 = 297*19 - 16*8 and 2031 = 297*7 - 16*3.

Karl

10(08(06
--
Scanned by iCritical.

[Irv Bromberg]

On 2009 Apr 29, at 11:35 , Brij Bhushan Vij wrote:

A 353-year cycle has 4365.998 lunation.

An important aspect of the 5515-year and 1556-year cycles being discussed is that by definition they have an integer number of solar years, lunar months, and calendar days (and 7-day weeks in the case of the 1556-year cycle).  These integer relationships exactly fix the calendar mean year and calendar mean month.  How good this works out astronomically is another matter.

This is that of 334-years/4131 lunation (4131.00098121 lunation); and another 19-years shorter 315-years/3896 lunation (3896.0039194 lunation).

With reasonably accurate calendar mean months the 334- and 315- year cycles will end up having excessively short calendar mean years, with respect to the northward equinox, although the 334-year cycle will have a mean year very close to the so-called mean tropical year, around 365+31/128 days, one of your favorites.  The 315-year cycle will be shorter still.  The 296-year cycle is of course even shorter, but is nevertheless well suited to the present-era mean north solstitial year -- but I was unable to find such a short cycle having an integer number of years, months and days.

-- Irv Bromberg, Toronto, Canada

<http://www.sym454.org/>