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Finally we ought to revise our old table:

1st calendar:

A

Ka2-1--6

6

9 10

B

Ka2-7--9

3

B

Ka2-10

1

C

Ka2-11--16

6

12 26

D

Ka2-17--22

6

E

Ka3-1--4

4

12 14

F

Ka3-5--8

4

G

Ka3-9--12

4

H

Ka3-13--14

2

2nd calendar:
1

Ka3-15

1

*73
1-6 Ka3-16--Ka4-14 spring 20
6-16 Ka4-15--Kb1-10 summer 26 *52
16-20 Kb1-11--*Kb2-14 *26
20-23 *Kb2-15--Kb3-8 humpback 12 26 27
24-28 Kb3-9--Kb4-6 rest of autumn 14
28

Kb4-7

winter solstice 1
28-29 Kb4-8--14 new year 7

From Ka4-15 up to Kb4-7 there are 3 * 26 = 78 = 6 * 13 glyphs. The rest (blackmarked numbers) sum up to 29, possibly indicating the winter half of the year.

Probably ua (Kb3-5) should be counted in the same way as when 26 was reached in the 1st calendar, i.e. by counting beyond (not from) the 1st Rei? Then ua will be number 108 = 3 * 36, a way to indicate autumn equinox (1 * 36 left of the year).

There are 9 + 1 = 10 blackmarked (excluded from the calendar) glyphs at the beginning of line Ka2. The total number of glyphs on the tablet was probably 192. The first line (Ka1) probably had 24 glyphs. 192 - 24 (Ka1) - 10 (Ka2-1--10) - 108 (Ka2-11--Kb3-5) = 50 (Kb3-6--*Kb5-20).

24 (Ka1) + 10 (Ka2-1--10) + 50 (Kb3-6--*Kb5-20) = 84. QED.

The missing 1 * 36 (last quarter of the year) is not the number of glyphs in the 1st calendar, we cannot find it. Instead we find 84:

Ka1-1--Ka1-24 24
Ka2-1--Ka2-10 10
Ka2-11--Kb3-5 (ua) 108
Kb3-6--*Kb5-20 50
sum 192

According to this pattern ua is the last glyph of the sun (6 * 18 = 108). Moon rules the rest of the year, 6 * 14 = 84.