PETIN M.I.

       The World PETIN - METON calendar

 

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Example  XIII. The definition of a lunar and solar dates.

 

                    Initial data:   Date 15, Month 6, 2005 AD (lunar calendar).

                                          Date 01, Month 1,       1 AD (lunar calendar)  -  1 721 436 JD

                                          January 01,                  1 AD (solar calendar)  -  1 721 426 JD

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

                                          What is a solar date X, which according to

                                           a date 15,  Month 6, 2005 AD ?

 

I.  The subtraction of dates:.

 

     1.   Date 15, Month 6,   2005 AD   .. (the sum)

        -  Date 01, Month 1,         1 AD  ... (the augend)

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

        There are 14 days and 5 lunar months and 2004 years between date 01, Month 1, 1 AD

and date 15 Month 6, 2005 AD  (SimpleM),   i.e.:           

          14 days , 5 months , 2004 lunar years ……….(the addend)        

 

II.  The definition of  JD value  for the above addend.

 

     2.   2004  : R₁ = 6,59,  where:  R₁ = 304 from

                                                      the table 6  ( http://Petin22Mikhail.narod.ru/index.htm  )

                                                      for interval of (>304) – 7 296 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) – 7 296 years.

     3.   2004 – 1824 = 180 years

     4.   180 : R₁  = 3,16,  where:  R₁ = 57 from the table 6

                                                       for interval  of  (>57) –  285 years.

       We take 3 instead of  3,16 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.        180 – 171 = 9 years  corresponds to 3249 days (table 5 - http://Petin21Mikhail.narod.ru/index.htm  )

                                                        

     6.    The sum of days

                        666 210         

                          62 457

                            3 249

                               148     correspond to  five  months (see point 1 and

                                          table 4 – http://Petin21Mikhail.narod.ru/index.htm  )

                                14       (see point 1 )

                     --------------

                       732 078 days, i.e. the  JD  value  between date 01, Month 1, 1 AD (or January 11, 1 AD, 

          JD = 1 721 436 ) and  date 15, Month 6,  2005 AD.

 

III.  The definition of  JD for date 15, Month 6,  2005.    

 

      7.   732 078        i.e. addend,          JD

         1 721 436,       i.e. augend           JD for date 01, Month 1, 1 AD (or January 11, 1 AD)    

        ------------------

        2 453 514  days,,  i.e. the sum  -  JD for date 15, Month 6,  2005.      

 

          IV.  The passing  from date 15, Month 6,  2005 AD (JD = 2 453 514 )  to  the Gregorian date X

 

       8.     2 453 514  days ,  i. e.  JD  for date 15,  Month 6,  2005 AD  (the sum)     

    - 1 721 426,  i.e.  JD  of   January  01, 1 AD                       ( the augend)

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

         732 088  days              …………………………............(the addend)     

 

 

                         

9.   732 088 : K₂ =5, 011,     where:  K₂ = 146 097 from

                                                              table 2  -  ( http://Petin20Mikhail.narod.ru/index.htm )

                                                              for interval  of 146 098 – 1 022 679 days.

 

 

We take 5 instead of   5,011

                    5 x K₁ = 5 x 400 = 2000 years,  where:  K₁ = 400 from table 2 

                                                                           for interval  of 146 098 – 1 022 679 days.

                    5 x K₂  = 730 485  days,   where:  K₂ = 146 097 from the table 2.     

 

10.        732 088

           - 730 485 

          ----------------

                 1 603 days – interval between January 01, 2001 AD and a solar date X.

11.   1603 : K₂ = 1,097,  ,     where:  K₂ = 1461 from the table 2

                                                         for interval  of 1462 – 35 064 days.

           We take  1 instead of   1,097 and

                    1 x K₁ = 1 x 4 = 4 years,   where:  K₁ = 4  from the table 2 

                                                                           for interval  of 1096 – 1 461 days.

                    1 x K₂  = 1461  days,   where:  K₂ = 1461  from the table 2.

12.      1603 days

         - 1461

       --------------

             142 days   (the augend)   -  interval between January 01, 2005 AD and a solar date X

 

13.    It is necessary to define a numbers of full months (Gregorian).

 

       A nearest mini modM_g (table 1) is 120 days, i.e.  the full months are the [1+2+3+4] months,

   

  142 days

- 120

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    22  days  - interval between May 01, 2005 AD and a solar date X.

 

14.    Date X  is equal to:

               May 01, 2005 AD

            +         22

               ----------------

               May 23, 2005 AD,   Weekday  -  Lunaday   (Full Moon Day). 

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