PETIN M.I.

       The World PETIN - METON calendar

 

   Contents – http://NewWorldCalendar.narod.ru/index.htm               

 

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

          -------------

             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 – < 6 384 ) 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 – < 6 384 ) 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

                     --------------

                      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.  

             ------------------

            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 World PETIN-METON calendar   (Meton part).

     1.   Date 01, Month 11,   2083 AD 

        -  Date 01, Month 1,           1 AD       ……………………..…  … (the augend)…..1 721 436  JD

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

  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 – < 6 384 )  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 – < 6 384 ) 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)

                     --------------

                       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)    

        ------------------

        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) 

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

     Contents – http://NewWorldCalendar.narod.ru/index.htm                

 



Hosted by uCoz