PETIN M.I.
PETIN - METON calendar
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Examples VII - VIII. The examples of calculations of
the PETIN-METON calendar (lunar part)
VII. Example. The passing from 01 Lunation 2 1996
(coinciding with January 21, 1996, 2 450 104 JD) to 01 Lunation 2, 1 AD
1. 1996 lunar year
-1, i.e. the correction between BC/AD
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1995 lunar year
2. 1995 : R1 = 1,54, where: R1 = 1292 from the table 6 - http://Petin22Mikhail.narod.ru/index.htm
for interval of (>1292) – 3 876 years.
We take 1 instead of 1,54 and
1 x 1292 = 1292 years
1 x R2 = 471 900 days, where: R2 = 471 900 from the table 6
for interval of (>1292) – 3876 years.
3. 1995 – 1292 = 703 years
4. 703 : R1 = 2.,31 where: R1 = 304 from the table 6
for interval of (>304) – 1216 years.
We take 2 instead of 2,31 and
2 x R1 = 608 years
2 x R2 = 222 070 days, where: R2 = 111 035 from the table 6
for interval of (>304) – 1216 years.
5. 703 – 608 = 95 years
6. 95: R1 = 1,67 where: R1 = 57 from the table 6
for interval of (57) – 171 years.
We take 1 instead of 1,67 and
1 x R1 = 57 years
1 x R2 = 20 819 days, where: R2 = 20 919 from the table 6
for interval of (>304) – 1216 years.
7. 95 - 57 = 38 years
8. 38 : R1 = 1 where: R1 = 38 from the table 6
for interval of (38) – 57 years.
We take 1 and
1 x R1 = 38 years
1 x R2 = 13 879 days, where: R2 = 13 879 from the table 6
for interval of (>38) – 57 years.
9. The quantity of days
471 900
222 070
20 819
13 879
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728 668 days, i.e. the quantity JD from 01 Lunation 2, 1 AD, Monday
10. Finally JD for 01 Lunation 2, 1 AD
2 450 104 days, i.e. the sum
- 728 668 days, i.e. the addend
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1 721 436 days, (the augend), i.e. JD for 01 Lunation 2, 1 AD.
11. Using the mentioned above method III. Return account , we shall receive that
01 Lunation 2, 1 AD corresponds to January 11, 1 AD (Gregorian calendar).
VIII. Example. The subtraction of dates of the PETIN-METON calendar (Meton part).
1. Date 22, Month 3, 2004 AD, i.e. Easter (April 11, 2004).. (the sum)….........2 453 107 JD
- Date 01, Lunation 2, 1 AD …………………….... … (the augend)……..1 721 436 JD
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There are 3 lunar months and 21 days and 2003 years between 01 Month, 1 AD and 22 Month 3, 2004 AD (SimpleM), i.e.:
21 days , 3 months , 2003 lunar years ………..………………(the addend)
2. 2003 : R1 = 6,59, where: R1 = 304 from the table 6
for interval of (>304) – 11 856 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) – 11 856 years.
3. 2003 – 1824 = 179 years
4. 179 : R1 = 3,14, where: R1 = 57 from the table 6
for interval of (>57) – 285 years.
We take 3 instead of 3,14 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. 179 – 171 = 8 years corresponds to 2894 days (table 5 - http://Petin21Mikhail.narod.ru/index.htm )
6. The sum of days
666 210
62 457
2 894
89 correspond to three Months (see point 1 of VIII. Example and table 4 )
21 (see point 1 of VIII. Example)
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731 671 days, i.e. the quantity JD between 01Month 1, 1 AD (or January 11, 1 AD,
JD = 1 721 436 ) and 22 Month 3, 2004.
7. Finally JD for 01 Month 1, 1996
731 671 i.e. addend,
1 721 436, i.e. augend JD for 01 Month 1, 1 AD (or January 11, 1 AD)
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2 453 107 days,, i.e. the sum - JD for 22 Month 3, 2004.
8. Using the mentioned above method III. Return account , we shall receive that
22 Month 3, 2004 corresponds to April 11, 2004
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