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calculating the moon

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  • mnempi
    Here is a new method to calculate the full moon for years CE mentally, and without using any mnemonics, and with a precision of +/- 1 day over a period of many
    Message 1 of 2 , Aug 1, 2007
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      Here is a new method to calculate the full moon for years CE mentally,
      and without using any mnemonics, and with a precision of +/- 1 day
      over a period of many millennia.

      I do everything in the Julian calendar. There is no problem in
      switching to the Gregorian date. For every century there is a number
      d#, which, if added to the Julian date, gives the Gregorian date. E.g.
      d#(H=20) = d#(H=19) = 13, so that 2007 august 1 (Julian) = 2007 August
      14 (Gregorian).

      You need a start value. Let us take the 7th april 0 CE.
      Let us (as is common) express the dates as difference to march 21 (I
      write march in the Julian, March in the Gregorian system, so that 0 CE
      march 21 = 0 CE March 19).

      For the first 200 years or so I just calculate the usual way
      ("Dionysian style"), but with april 7 instead of april 5 as Easter
      full moon for 0 CE.

      Example J = 97 CE

      The usual calculation is g (J=97) = (15 + 19 x 97 mod 19) mod 30 = 23,
      but instead I calculate g* (J=97) = (17 + 19 x 97 mod 19) mod 30 = 25,
      or rather I use 2 (= 17 - 15) as start value for 0 CE and calculate
      g* (J=97) = (2 + g(J=97)) mod 30 = 25

      Now, given a year J, I analyze J into multiples of 448 and 160, like
      our present year 2007 = 4 x 448 + 160 + 55

      Two rules to be observed:
      For every 448 year subtract 3 (or add 27)
      For every 160 year add 1 (or subtract 29)

      So we arrive at
      g* (J=2007) = (2 - 4 x 3 + 1 + g (55)) mod 30 = (2 - 12 + 1 + 8) mod
      30 = -1 (or 29, but we should take the smaller number in order to make
      comparison with the corresponding Gregorian date more easy).
      g* = -1 means that we should expect the Easter full moon on march 20 =
      April 2, which is exactly the date of the observed full moon and the
      date of the Gregorian Easter computation.

      The same result can be obtained by dropping 160:
      2007 = 4 x 448 + 215
      g* (2007) = (2 - 4 x 3 + g(215)) mod 30 = 2 - 12 + 9 = -1

      Another example: J = 2301 CE

      Here the astronomical tables say Full moon on March 25, and the same
      says the Gregorian calculation. Let us see, what our mental
      calculation says to that.
      As d#(H=22) = 16, you have March 25 = march 9, so we should expect g*
      = -12 or rather g* = 18.

      solution:
      2301 = 5 x 448 + 61
      g* (2007) = (2 - 5 x 3 + g (61)) mod 30 = (2 - 15 + 1) mod 30 = 2 + 15
      + 1 = 18 bingo

      Yet another example: J = 4449

      The Gregorian Easter moon is on March 22, one day earlier than the
      astronomical calculation which expects the full moon on March 23 very
      early 0 : 55 Greenwich time.
      As d# (H=44) = 31 you have March 22 = february 19 = march (21 - 30).

      So our question is, if we can calculate something like g* (4449) = -30
      (or rather g* = 0).

      solution
      4449 = 9 x 448 + 2 x 160 + 97
      g* (2007) = (2 - 9 x 3 + 2 + g (97)) mod 30 = (2 - 27 + 2 + 23) mod 30
      = 0 bingo

      A last example: J = 9000 CE

      9000 = 20 x 448 + 40
      g* (9000) = ( 2 - 20 x 3 + g (40)) mod 30 = (2 - 60 + 23) mod 30 = 25,
      so we expect the full moon on april 15 = June 20, as d# (H=90) = 66.
      Well, as this is somewhat too far away from March 21, we have just to
      subtract 90 (= 3 x 30) days and reach March 22.

      But the astronomical computation says: "June 19",and "March 21" so we
      are in both cases one day late.

      The Gregorian Easter calculation, on the other hand, offers the same
      date as our mental calculation, viz. March 22.

      I hope that the examples convince you that the number 448 plays a
      powerful role in the matter.

      The trick using 448 (= 4 x 112) and 160 (= 10 x 16) occured to me, as
      I am studying pre Dionysian Easter computation in general and the
      112-year Pascha table of Hippolyte in particular. There is a passage
      in Geminos ELEMENTA ASTRONOMIAE (dating from the first cent. BC) which
      gives the hint.

      For years BC I use a somewhat different period, namely 1120 (= 10 x
      112) which subtracts 7. With this tool you can calculate mentally the
      full moon of 10 millennia BC to the day.
    • mnempi
      To continue: It is even better not to take the astronomically exact date 0 CE april 7 but one day later april 8 as starting value (though I cannot explain the
      Message 2 of 2 , Nov 24, 2007
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        To continue:

        It is even better not to take the astronomically exact date 0 CE april
        7 but one day later april 8 as starting value (though I cannot explain
        the reason why).

        Over the Christian centuries up to now, there is not one single bad
        result (difference to the astronomical value > 2 days) 99,4 % of the
        results are good (difference < 2), 65 % excellent (difference 0).

        And all this can be calculated without any use of mnemonics, just
        multiples of 448, 160, 16. No need of the number 19 at all. And
        everything deduced froma a pre-christian source! I call it in honour
        of this source the Geminos-System.

        Compare the results of the Gregorian calendar based on Lilian epacts.
        Here you also have not a single bad result, but you have only 95,7 %
        good, and 55 % excellent results.

        The Geminos-system proves to be superior. If it had still been
        understood in the Middle Ages, the Gregorian reform might have been
        much simpler as in fact it was.

        With very little mnemonics you can obtain 100 % good results, this
        goes without saying.

        The Geminos-system continues with the same quality for many millennia
        into the future, surpassing the Gregorian system by far. Indeed it
        makes it look somewhat old-fashioned.

        Ulrich Voigt
        www.likanas.de

        --- In MentalCalculation@yahoogroups.com, "mnempi" <voigt@...> wrote:
        >
        > Here is a new method to calculate the full moon for years CE mentally,
        > and without using any mnemonics, and with a precision of +/- 1 day
        > over a period of many millennia.
        >
        > I do everything in the Julian calendar. There is no problem in
        > switching to the Gregorian date. For every century there is a number
        > d#, which, if added to the Julian date, gives the Gregorian date. E.g.
        > d#(H=20) = d#(H=19) = 13, so that 2007 august 1 (Julian) = 2007 August
        > 14 (Gregorian).
        >
        > You need a start value. Let us take the 7th april 0 CE.
        > Let us (as is common) express the dates as difference to march 21 (I
        > write march in the Julian, March in the Gregorian system, so that 0 CE
        > march 21 = 0 CE March 19).
        >
        > For the first 200 years or so I just calculate the usual way
        > ("Dionysian style"), but with april 7 instead of april 5 as Easter
        > full moon for 0 CE.
        >
        > Example J = 97 CE
        >
        > The usual calculation is g (J=97) = (15 + 19 x 97 mod 19) mod 30 = 23,
        > but instead I calculate g* (J=97) = (17 + 19 x 97 mod 19) mod 30 = 25,
        > or rather I use 2 (= 17 - 15) as start value for 0 CE and calculate
        > g* (J=97) = (2 + g(J=97)) mod 30 = 25
        >
        > Now, given a year J, I analyze J into multiples of 448 and 160, like
        > our present year 2007 = 4 x 448 + 160 + 55
        >
        > Two rules to be observed:
        > For every 448 year subtract 3 (or add 27)
        > For every 160 year add 1 (or subtract 29)
        >
        > So we arrive at
        > g* (J=2007) = (2 - 4 x 3 + 1 + g (55)) mod 30 = (2 - 12 + 1 + 8) mod
        > 30 = -1 (or 29, but we should take the smaller number in order to make
        > comparison with the corresponding Gregorian date more easy).
        > g* = -1 means that we should expect the Easter full moon on march 20 =
        > April 2, which is exactly the date of the observed full moon and the
        > date of the Gregorian Easter computation.
        >
        > The same result can be obtained by dropping 160:
        > 2007 = 4 x 448 + 215
        > g* (2007) = (2 - 4 x 3 + g(215)) mod 30 = 2 - 12 + 9 = -1
        >
        > Another example: J = 2301 CE
        >
        > Here the astronomical tables say Full moon on March 25, and the same
        > says the Gregorian calculation. Let us see, what our mental
        > calculation says to that.
        > As d#(H=22) = 16, you have March 25 = march 9, so we should expect g*
        > = -12 or rather g* = 18.
        >
        > solution:
        > 2301 = 5 x 448 + 61
        > g* (2007) = (2 - 5 x 3 + g (61)) mod 30 = (2 - 15 + 1) mod 30 = 2 + 15
        > + 1 = 18 bingo
        >
        > Yet another example: J = 4449
        >
        > The Gregorian Easter moon is on March 22, one day earlier than the
        > astronomical calculation which expects the full moon on March 23 very
        > early 0 : 55 Greenwich time.
        > As d# (H=44) = 31 you have March 22 = february 19 = march (21 - 30).
        >
        > So our question is, if we can calculate something like g* (4449) = -30
        > (or rather g* = 0).
        >
        > solution
        > 4449 = 9 x 448 + 2 x 160 + 97
        > g* (2007) = (2 - 9 x 3 + 2 + g (97)) mod 30 = (2 - 27 + 2 + 23) mod 30
        > = 0 bingo
        >
        > A last example: J = 9000 CE
        >
        > 9000 = 20 x 448 + 40
        > g* (9000) = ( 2 - 20 x 3 + g (40)) mod 30 = (2 - 60 + 23) mod 30 = 25,
        > so we expect the full moon on april 15 = June 20, as d# (H=90) = 66.
        > Well, as this is somewhat too far away from March 21, we have just to
        > subtract 90 (= 3 x 30) days and reach March 22.
        >
        > But the astronomical computation says: "June 19",and "March 21" so we
        > are in both cases one day late.
        >
        > The Gregorian Easter calculation, on the other hand, offers the same
        > date as our mental calculation, viz. March 22.
        >
        > I hope that the examples convince you that the number 448 plays a
        > powerful role in the matter.
        >
        > The trick using 448 (= 4 x 112) and 160 (= 10 x 16) occured to me, as
        > I am studying pre Dionysian Easter computation in general and the
        > 112-year Pascha table of Hippolyte in particular. There is a passage
        > in Geminos ELEMENTA ASTRONOMIAE (dating from the first cent. BC) which
        > gives the hint.
        >
        > For years BC I use a somewhat different period, namely 1120 (= 10 x
        > 112) which subtracts 7. With this tool you can calculate mentally the
        > full moon of 10 millennia BC to the day.
        >
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