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 : R= 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)ģ.........2 453 107  JD

-  Date 01, Lunation 2,      1 AD       ģģģģģģģģ....  ģ (the augend)ģģ..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 ģģģ..ģģģģģģ(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 : R= 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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