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Re: IVY MIKE CALENDAR
by Karl Palmen
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Dear Helios and Calendar People
The calendar structure is not obvious from the formulae given and is considerably more simple than these formulae.
This arises from the facts that 709=21+16*(22+21) and (33 + 692) mod 709 = 16 = (33-1)/2.
The calendar is an example of a calendar Helios suggested several years ago.
The terms start with day C=1 and alternate between 21 and 22 days for a 33-term cycle (a term-yerm) of 709 days.
The first day C of each term has ( ( 33*C + 692 ) mod 709 ) < 33. For a 22-day term the value of this expression is less than 17 so that adding 33*21=693 makes it less than 709. For a 21-day term adding 33*21=693 makes it more than 709. For each 33-term cycle (term-yerm), the value of this expression for the first day of each term is
16, 00, 17, 01, 18, 02, 19, 03, ... 29, 13, 30, 14, 31, 15, 32.
Seventeen of these 709-day cycles have 12053 days, which is very close to 33 years and gives a mean year of 365.2424242424... days. So Helios defines a year to be equal the 17 terms. A year has 365 days if it either begins with a 21-day term or contains the first term of the 709-day cycle (term-yerm) else it has 366 days (by having a 22-day term at both the beginning and end).
The 33-term cycle is a Helios cycle, therefore these years form a Helios Cycle, and so the 366-day years are 3, 7, 11, 15, 19, 23, 27 and 31 of each 33-year cycle.
I take it that JD is the JD at the start of the day. Calendar conversion algorithms usually use the JD at noon, which is an integer. Lance Latham did this, so does Calendrica (explicitly)
http://emr.cs.iit.edu/home/reingold/calendar-book/Calendrica.htm .
This leads to day C=1, being the day that begins JD 2434316.5 and so be JD 2434317, which I make to be 31 October 1952.
See http://isotropic.org/cgi-bin/date.pl?date=31+Oct+1952 .
Helios also needs to work out formulae to convert T and D back to C or JD.
Also needed are the simple formulae to convert T to Y (year) and S (term of year) and back.
If each day of the term is replaced by a 17-day period (which I call a dove). Each term expands to form a year and the 709-day term-yerm becomes a 33-year cycle. This idea I put forward as the Dove calendar
http://www.the-light.com/cal/kp_Dove.html
Helios's formulae can be modified to find the Dove calendar date for a given day.
I set the Dove calendar epoch so that the last Dove of each year nearly always has the northward equinox (especially in US EST).
Today (26 June 2009) is Ff 2009 in the Dove calendar (6th day of 6th Dove).
Karl
10(10(04
-----Original Message-----
From: East Carolina University Calendar discussion List [mailto:CALNDR-L@...] On Behalf Of Helios
Sent: 26 June 2009 03:48
To: CALNDR-L@...
Subject: IVY MIKE CALENDAR
In previous post, I have favored a 17-fold-division calendar. Now I have the
equations that do the job. The Ivy Mike calendar is based on the 33-year
cycle. It's very simple.
-------------------------------------------------------------------------------
Here are the formulas;
C = JD - 2434315.5
T = FLOOR[ ( 33*C + 692 ) / 709 ]
D = C - 21*( T - 1 ) - FLOOR[ 16*T / 33 ]
Where JD = Julian date, C = Day Count, T = Term or 17th-of-year, D =
day-of-term
-------------------------------------------------------------------------------
I use Mayan ( Haab ) names for convenience. The similarity to the Haab is in
name only. There is the ommision of "Uo" and "Uayeb" from a Haab set of 19
names.
01) Pop ( 9th = Autumn Mid-quarter* )
02) Zip
03) Zotz ( 9th = Winter Solstice )
04) Tzec
05) Xul ( 9th = Winter Mid-quarter )
06) Yaxkin
07) Mol ( 13th = Spring Equinox )
08) Chen
09) Yax ( 16th = Spring Mid-quarter )
10) Zac
11) Ceh ( 19th Summer Solstice )
12) Mac
13) Kankin
14) Muan ( 1st = Summer Mid-quarter )
15) Pax
16) Kayab ( 6th = Autumn Equinox )
17) Cumku
*Mid-quarters are based on a solar declination = ( + or - ) 16.694°
-------------------------------------------------------------------------------
Here are 17 "solar glyphs" I am fond of.
http://www.nabble.com/file/p24214208/17glyph.jpg
-------------------------------------------------------------------------------
So let's input JD = 2455008.5 ( JUN 26, 2009 )
C = 20693
T = 964
D = 3
that is, Mac 3 of year 56
--
View this message in context: http://www.nabble.com/IVY-MIKE-CALENDAR-tp24214208p24214208.html
Sent from the Calndr-L mailing list archive at Nabble.com.
The calendar structure is not obvious from the formulae given and is considerably more simple than these formulae.
This arises from the facts that 709=21+16*(22+21) and (33 + 692) mod 709 = 16 = (33-1)/2.
The calendar is an example of a calendar Helios suggested several years ago.
The terms start with day C=1 and alternate between 21 and 22 days for a 33-term cycle (a term-yerm) of 709 days.
The first day C of each term has ( ( 33*C + 692 ) mod 709 ) < 33. For a 22-day term the value of this expression is less than 17 so that adding 33*21=693 makes it less than 709. For a 21-day term adding 33*21=693 makes it more than 709. For each 33-term cycle (term-yerm), the value of this expression for the first day of each term is
16, 00, 17, 01, 18, 02, 19, 03, ... 29, 13, 30, 14, 31, 15, 32.
Seventeen of these 709-day cycles have 12053 days, which is very close to 33 years and gives a mean year of 365.2424242424... days. So Helios defines a year to be equal the 17 terms. A year has 365 days if it either begins with a 21-day term or contains the first term of the 709-day cycle (term-yerm) else it has 366 days (by having a 22-day term at both the beginning and end).
The 33-term cycle is a Helios cycle, therefore these years form a Helios Cycle, and so the 366-day years are 3, 7, 11, 15, 19, 23, 27 and 31 of each 33-year cycle.
I take it that JD is the JD at the start of the day. Calendar conversion algorithms usually use the JD at noon, which is an integer. Lance Latham did this, so does Calendrica (explicitly)
http://emr.cs.iit.edu/home/reingold/calendar-book/Calendrica.htm .
This leads to day C=1, being the day that begins JD 2434316.5 and so be JD 2434317, which I make to be 31 October 1952.
See http://isotropic.org/cgi-bin/date.pl?date=31+Oct+1952 .
Helios also needs to work out formulae to convert T and D back to C or JD.
Also needed are the simple formulae to convert T to Y (year) and S (term of year) and back.
If each day of the term is replaced by a 17-day period (which I call a dove). Each term expands to form a year and the 709-day term-yerm becomes a 33-year cycle. This idea I put forward as the Dove calendar
http://www.the-light.com/cal/kp_Dove.html
Helios's formulae can be modified to find the Dove calendar date for a given day.
I set the Dove calendar epoch so that the last Dove of each year nearly always has the northward equinox (especially in US EST).
Today (26 June 2009) is Ff 2009 in the Dove calendar (6th day of 6th Dove).
Karl
10(10(04
-----Original Message-----
From: East Carolina University Calendar discussion List [mailto:CALNDR-L@...] On Behalf Of Helios
Sent: 26 June 2009 03:48
To: CALNDR-L@...
Subject: IVY MIKE CALENDAR
In previous post, I have favored a 17-fold-division calendar. Now I have the
equations that do the job. The Ivy Mike calendar is based on the 33-year
cycle. It's very simple.
-------------------------------------------------------------------------------
Here are the formulas;
C = JD - 2434315.5
T = FLOOR[ ( 33*C + 692 ) / 709 ]
D = C - 21*( T - 1 ) - FLOOR[ 16*T / 33 ]
Where JD = Julian date, C = Day Count, T = Term or 17th-of-year, D =
day-of-term
-------------------------------------------------------------------------------
I use Mayan ( Haab ) names for convenience. The similarity to the Haab is in
name only. There is the ommision of "Uo" and "Uayeb" from a Haab set of 19
names.
01) Pop ( 9th = Autumn Mid-quarter* )
02) Zip
03) Zotz ( 9th = Winter Solstice )
04) Tzec
05) Xul ( 9th = Winter Mid-quarter )
06) Yaxkin
07) Mol ( 13th = Spring Equinox )
08) Chen
09) Yax ( 16th = Spring Mid-quarter )
10) Zac
11) Ceh ( 19th Summer Solstice )
12) Mac
13) Kankin
14) Muan ( 1st = Summer Mid-quarter )
15) Pax
16) Kayab ( 6th = Autumn Equinox )
17) Cumku
*Mid-quarters are based on a solar declination = ( + or - ) 16.694°
-------------------------------------------------------------------------------
Here are 17 "solar glyphs" I am fond of.
http://www.nabble.com/file/p24214208/17glyph.jpg
-------------------------------------------------------------------------------
So let's input JD = 2455008.5 ( JUN 26, 2009 )
C = 20693
T = 964
D = 3
that is, Mac 3 of year 56
--
View this message in context: http://www.nabble.com/IVY-MIKE-CALENDAR-tp24214208p24214208.html
Sent from the Calndr-L mailing list archive at Nabble.com.
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