PETIN M.I.
The World PETIN
- METON calendar
Contents – http://NewWorldCalendar.narod.ru/index.htm
VII. Example. The passing
from 01 Month 1 1996
(coinciding with
1.
1996 lunar year
-1, i.e. the correction between BC/AD
-------------
1995 lunar year
2.
1995 : R1 = 6,56, 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
6,56 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. 1995 – 1824 = 171 years
4. 171 : R1
= 3, where: R1
= 57 from the
table 6
for interval of (>57) – 285
years.
We take 3 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. 171 – 171 = 0 years
6. The quantity of days
666 210
62 457
--------------
728 667 days, i.e. the quantity JD
from 01 Month 1, 1 AD, Monday
7. Finally JD
for 01 Month 1, 1 AD
2 450 103 days, i.e. the
sum
- 728 667 days, i.e. the addend
------------------
1 721 436 days, (the augend), i.e. JD for
01 Month 1, 1AD.
8. Using the mentioned above method III. Return account
, we shall receive that
01 Month 1,
1 AD corresponds to
VIII.
Example. The subtraction of dates of the World
PETIN-METON calendar (Meton part).
1. Date 22, Month
4, 2004 AD, i.e. Easter (
-
Date 01, Month 1, 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 4, 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 – < 6 384 )
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 – < 6 384 ) years.
3. 2003 – 1824 =
179 years
4. 179 : R1 = 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 of JD
between 01Month 1, 1 AD (or
JD = 1 721 436 ) and date 22, Month 4, 2004.
7. Finally JD for date 22, Month 4, 2004:
731 671 i.e. addend,
1 721 436, i.e. augend JD
for 01 Month 1, 1 AD (or
------------------
2 453 107 days, i.e. the sum - JD for date
22, Month 4, 2004.
8. Using the mentioned
above method III. Return account , we shall receive that
date 22, Month 4, 2004 corresponds
to
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