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I urge you to visit Remy Landau's website Hebrew Calendar Science and Myths for a thorough explanation of the calculations used in this calendar.
The measurements of time used to calculate this calendar are:
To find the beginning of a year, it is necessary to find out how many lunations have occurred since the Prota Selenogenesis, the beginning of the calendar. This has been calculated to have occurred on Monday, September 7th, 3760 BCE (see Year Zero) — its value is 2 days, 5 hours, 204 parts.
To do so, we use the formula:
We then multiply the value of the lunation by this number — keeping the three values of day, hour and part separate — and add to each part the appropriate value of the Prota Selenogenesis.
For example, the formula for the year 5758 produces the integer of (235 × 5758 - 234) ÷ 19, which is 71205.
Multiplying this by the value of the lunation, and then adding the value of the Prota Selenogenesis yields: 2064947 days, 854465 hours, 56465565 parts. This reduces to 2102728 days, 4 hours, 129 parts.
Now, how on earth are we supposed to work with measurements of millions of days? The answer is to divide it into manageable chunks of time, which we can:
The Gregorian calendar has a perfectly repeating cycle of 400 years, which works out to 146097 days — 365 × 400 + 97 (remembering that, in the Gregorian calendar, century years, those divisible by 100, are not counted as leap years unless they are also divisible by 4; e.g. 2000 CE will be a leap year, but 1900 CE was not).
Since the Prota Selenogenesis there have been 14 cycles of 400 Gregorian years, taking us up to Monday, September 7th, 1840.
14 × 146097 = 2045358, to which we add the 2 days of the Prota Selenogenesis to get a final value of 2045360, which we then subtract from our calculated value of 2102728 days to get 57370 days since Monday, September 7th, 1840 to the beginning of Hellene year 5758.
From that we subtract the value for 100 years — 365 × 100 + 24 = 36524 (again, remembering the century year rule) — to get 20846 days since September 7th, 1940 to the beginning of Hellene year 5758.
Then we begin repeatedly subtracting the value of 4 years — 365 × 3 + 366 = 1461 (remembering that there will be one leap year every four years) — from that date. Eventually we get to September 7, 1996, and find that there are 392 days left over. From that we subtract 365, and find that Hellene year 5758 starts 29 days after September 7, 1997 — that is to say, on October 2, 1997.
Now that we have the start date for our year, it becomes necessary to find out how long the year will be: whether a "regular" common year, a "full" common year, a "deficient" leap year, or a "regular" leap year (see Year Length).
To do this, we calculate the start date of the following year, in this case 5759. This works out to be September 21, 1998. Now we take the difference, which works out to 354 days.
Therefore Hellene year 5758, which begins on October 2, 1997, is a "regular" common year of 354 days, which means there is no leap month and the month Metageitnion has 29 days. (See Months for an explanation of the months and their lengths.)
Notice that we do not need to know what year in the Metonic cycle we are in — the initial formula's inclusion of "÷ 19" takes care of that. For example, the next leap year should be 5760, since 5760 ÷ 19 leaves a remainder of 3 and therefore Hellene year 5760 is the third year of the Metonic cycle. We can calculate that year 5760 will begin on September 10, 1999. The next year, 5761, begins on September 28, 2000 — 384 days later. Therefore Hellene year 5760 is a "regular" leap year of 384 days, which means that it includes the leap month Poseideon II and that the month Metageitnion has 30 days.
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