PETIN M.I.
PETIN - METON calendar
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Example XIII. The definition of a lunar and solar dates.
Initial data: Date 15, Month 6, 2005 AD (lunar calendar).
Date 01, Month 1, 1 AD (lunar calendar) - 1 721 436 JD
January 01, 1 AD (solar calendar) - 1 721 426 JD
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What is a solar date X, which according to
a date 15, Month 6, 2005 AD ?
I. The subtraction of dates:.
1. Date 15, Month 6, 2005 AD .. (the sum)
- Date 01, Month 1, 1 AD ... (the augend)
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There are 14 days and 5 lunar months and 2004 years between date 01, Month 1, 1 AD
and date 15 Month 6, 2005 AD (SimpleM), i.e.:
14 days , 5 months , 2004 lunar years ……….(the addend)
II. The definition of JD value for the above addend.
2. 2004 : R1 = 6,59, where: R1 = 304 from
the table 6 ( http://Petin22Mikhail.narod.ru/index.htm )
for interval of (304 – < 6 384 ) years.
We take 6 instead of 6,59 and
6 x 304 = 1824 years
6 x R2 = 666 210 days, where: R2 = 111 035 from the table 6
for interval of (304 – < 6 384 ) years.
3. 2004 – 1824 = 180 years
4. 180 : R1 = 3,16, where: R1 = 57 from the table 6
for interval of (>57) – 285 years.
We take 3 instead of 3,16 and
3 x R1 = 171 years
3 x R2 = 62 457 days, where: R2 = 20 819 from the table 6
for interval of (>57) – 285 years.
5. 180 – 171 = 9 years corresponds to 3249 days (table 5 - http://Petin21Mikhail.narod.ru/index.htm )
6. The sum of days
666 210
62 457
3 249
148 correspond to fife months (see point 1 and
table 4 – http://Petin21Mikhail.narod.ru/index.htm )
14 (see point 1 )
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732 078 days, i.e. the JD value between date 01, Month 1, 1 AD (or January 11, 1 AD,
JD = 1 721 436 ) and date 15, Month 6, 2005 AD.
III. The definition of JD for date 15, Month 6, 2005.
7. 732 078 i.e. addend, JD
1 721 436, i.e. augend JD for date 01, Month 1, 1 AD (or January 11, 1 AD)
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2 453 514 days,, i.e. the sum - JD for date 15, Month 6, 2005.
IV. The passing from date 15, Month 6, 2005 AD (JD = 2 453 514 ) to the Gregorian date X
8. 2 453 514 days , i. e. JD for date 15, Month 6, 2005 AD (the sum)
- 1 721 426, i.e. JD of January 01, 1 AD ( the augend)
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732 088 days …………………………............(the addend)
9. 732 088 : K2 =5, 011, where: K2 = 146 097 from
table 2 - ( http://Petin20Mikhail.narod.ru/index.htm )
for interval of 146 098 – 1 022 679 days.
We take 5 instead of 5,011
5 x K1 = 5 x 400 = 2000 years, where: K1 = 400 from table 2
for interval of 146 098 – 1 022 679 days.
5 x K2 = 730 485 days, where: K2 = 146 097 from the table 2.
10. 732 088
- 730 485
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1 603 days – interval between January 01, 2001 AD and a solar date X.
11. 1603 : K2 = 1,097, , where: K2 = 1461 from the table 2
for interval of 1462 – 35 064 days.
We take 1 instead of 1,097 and
1 x K1 = 1 x 4 = 4 years, where: K1 = 4 from the table 2
for interval of 1096 – 1 461 days.
1 x K2 = 1461 days, where: K2 = 1461 from the table 2.
12. 1603 days
- 1461
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142 days (the augend) - interval between January 01, 2005 AD and a solar date X.
13. It is necessary to define a numbers of full months (Gregorian).
A nearest mini modMg (table 1) is 120 days, i.e. the full months are the [1+2+3+4] months,
142 days
- 120
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22 days - interval between May 01, 2005 AD and a solar date X.
14. Date X is equal to:
May 01, 2005 AD
+ 22
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May 23, 2005 AD, Weekday - Lunaday (Full Moon Day).
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