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 (
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 : R1
= 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 : R1 = 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
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 : R1 = 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 : R1 =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
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
------------------
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
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