PETIN M.I.
PETIN - METON calendar
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Examples IX - X. The examples of calculations of the PETIN-METON
calendar (lunar part)
IX. Example. The passing from 01 Month 1, 1 AD (January 11, 1 AD, 1 721 436 JD) to 01 Month 11 2083
1. 2083 lunar year
-1, i.e. the correction between BC/AD
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2082 lunar years
2. 2082 : R1 = 6,85, where: R1 = 304 from the table 6 - http://Petin22Mikhail.narod.ru/index.htm
for interval of (>304) – 11 856 years.
We take 6 instead of 6,85 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. 2082 – 1824 = 258 years
4. 258 : R1 = 4,53 where: R1 = 57 from the table 6
for interval of (>57) – 285 years.
We take 4 instead of 4,53 and
4 x 57 = 228 years
4 x R2 = 83 276 days, where: R2 = 20 819 from the table 6
for interval of (>57) – 285 years.
5. 258 – 228 = 30 years.
6. 30 : R1 = 1, 58 where: R1 = 19 from the table 6
for interval of (>19) – 38 years.
We take 1 instead of 1,58 and
1 x R1 = 19 years
1 x R2 = 6 940 days, where: R2 = 6 940 from the table 6
for interval of (>19) – 38 years.
7. 30 – 19 = 11 years corresponds to 3988 days
(table 5 , column Simple - http://Petin21Mikhail.narod.ru/index.htm )
8. The quantity of days
666 210
83 276
6 940
3 988
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760 414 days, i.e. the quantity JD from 01 Month 1, 1 AD, Monday to
01 Month 1, 2083 AD Monday.
9. There are 10 lunar months between 01 Month, 2083 and 01 Month 11, 2083, i.e.
10 months = 295 days (table 5)
10. Finally JD for 01 Month 11, 2083 AD
295 days correspond to 10 months (see table 4)
760 414 days, i.e. the addend
1 721 436 days, (the augend), i.e. JD for 01 Month 1, 1AD.
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2 482 145 days, i.e. the sum
8. Using the mentioned above method III. Return account , we shall receive that
01 Month 11, 2083 AD corresponds to October 12, 2083 AD(Gregorian calendar).
X. Example. The subtraction of dates of the PETIN-METON calendar (Meton part).
1. Date 01, Month 11, 2083 AD
- Date 01, Month 1, 1 AD ……………………..… … (the augend)…..1 721 436 JD
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There are 10 lunar months plus 2082 years between 01 Month, 1 AD and 01 Month 11, 2083 AD, i.e.:
10 months plus 2082 lunar years ………..………………(the addend).
2. 2082 : R1 = 6,85, where: R1 = 304 from the table 6
for interval of (>304) – 11 856 years.
We take 6 instead of 6,85 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. 2082 – 1824 = 258 years
4. 258 : R1 = 4,53, where: R1 = 57 from the table 6
for interval of (>57) – 285 years.
We take 4 instead of 4,53 and
4 x R1 = 228 years
4 x R2 = 83 276 days, where: R2 = 20 819 from the table 6
for interval of (>57) – 285 years.
5. 258 – 228 = 30 years.
30 : R1 =1,11, where: R1 = 19 from the table 6
for interval of (>19) – 38 years.
We take 1 instead of 1,11 and
1 x R1 = 19 years
1 x R2 = 6 940 days, where: R2 = 6 940 from the table 6
for interval of (>19) – 38 years.
6. 30 – 19 =11 years corresponds to 3988 days (table 5 , column Simple)
7. The sum of days
666 210
83 276
6 940
3 988
295 correspond to ten months (see table 4)
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760 709 days, i.e. the quantity JD between 01Month 1, 1 AD
(or January 11, 1 AD, JD = 1 721 436 ) and 01 Month 11, 2083.
8. Finally JD for 01 Month 11, 2083
760 709 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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2 482 145 days,, i.e. the sum - JD for 01 Month 11, 2083 AD.
9. Using the mentioned above method III. Return account , we shall receive that
01 Month 11, 2083 AD corresponds to October 12, 2083 AD (Gregorian calendar)
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