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 : R= 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 : R= 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 : R= 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 : R=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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