PETIN M.I.

                         PETIN - METON calendar

 

    Contents - http://Petin1Mikhail.narod.ru/index.htm          

 

  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

      -------------

        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

                     --------------

                       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          

             ------------------

            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

                 -----------------------

        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 : 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)

                     --------------

                       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)    

        ------------------

        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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