Next page:

Probably each glyph means 1 day ('night'), and the 192 glyphs in the text can then be understood as somewhat more than half a year. The key measure is 22 glyphs, which can be explained as a way to express π (ca 22 / 7). As a kind of confirmation the first 60 glyphs, which form a fundamental unit, end with Ka3-14 (where 3-14 alludes to 3.14): 42 65 22 Ka3-14 (60) Ka3-15 Kb1-7 Kb4-17 (170) 2 * 22 = 44 3 * 22 = 66 60 + 6 * 22 = 192 The first 60 glyphs can be subdivided so that 1 * 22 appears in a highly meaningful position: 22 20 12 Ka1-1 Ka1-24 Ka2-1 Ka2-22 Ka3-1 Ka3-14 (60) 24 22 14 24 = 192 / 8 36 The intention was probably to introduce the fundamental numerical base of the text. The presumtive reader should first count the number of glyphs in the text, and then - of course - he would try to divide by 8. The number of glyphs in line a1 are 24 in order to confirm the result of his action. The reader would then go on and count the number of glyphs in line a2. The result would surely stay in his mind when he then continued. And he would discover that also glyph line b1 has 22 glyphs. And he would discover that beyond Ka3-14 there are 6 * 22 glyphs. 22 in line a2 certainly is meant to be seen together with those 14 which end with Ka3-14, thereby creating a suggestion that 22 and 14 together in some way will measure up to 36 (meaning 360). Addition it is not, because the glyphs in line a2 are not of the same kind as those in the interval Ka3-1--14. Presumably, instead, the sun cycle (22) is to be measured by the fortnights of the moon. 192 - 60 - 20 (b5) = 112 = 8 * 14. 112 = 8 * 14 for the 'day' is counted by a 'night' measure (14), while 80 = 60 + 20 = 8 * 10 for the 'night' is measured by a 'day' measure (10). 192 = 8 * (14 + 10).

The juxtapositioning of Ka2-22 and Ka3-1 illustrates how the spiritual fish-mountain wrapped in feathers is regenerated as a rising sun fish. At the end of the sea journey (cfr the canoe in Ka2-1, like a tapa mea without feathers) light will dawn.

Once again we have an example of a glyph playing a double role, here both as the reincarnation of Ka2-20 and as the beginning of the fortnight before sun reaches the island:

 

Ka3-1	Ka3-2	Ka3-3	Ka3-4	Ka3-5	Ka3-6	Ka3-7
Ka3-8	Ka3-9	Ka3-10	Ka3-11	Ka3-12	Ka3-13	Ka3-14

Ka3-1 is like the forward looking face of Janus. The steps of 7 seem to continue with 3 * 7:

 

Ka3-15	Ka3-16	Ka3-17	Ka3-18	Ka3-19	Ka3-20	Ka3-21
Ka4-1	Ka4-2	Ka4-3	Ka4-4	Ka4-5	Ka4-6	Ka4-7
Ka4-8	Ka4-9	Ka4-10	Ka4-11	Ka4-12	Ka4-13	Ka4-14

In Ka4-14 the tail of mago says the journey takes a new turn. If we then look in the other direction, we find Rei in Ka2-10 initiating a the 12-glyph sequence, or is it not rather a 14-glyph sequence:

 

Ka2-10	Ka2-11	Ka2-12	Ka2-13	Ka2-14	Ka2-15	Ka2-16
Ka2-17	Ka2-18	Ka2-19	Ka2-20	Ka2-21	Ka2-22	Ka3-1

Still sun has not arrived to the island, though, which happens beyond 60. Dawn (fire) comes first to the sky, then to the island.

With 60 days before sun rises over the island and 20 beyond the island, it is easy to imagine 360 - 80 = 280 = 10 * 28. But only half a year is (presumbly) described. There is nothing more to describe (according to the Romans, the creator of Q and the space of the K tablet).

192 - 12 = 180 is a better point of departure than 360. Maybe the first 12 glyphs belong to the previous year:

 

Ka1-1	Ka1-2	Ka1-3	Ka1-4	Ka1-5	Ka1-6
Ka1-7	Ka1-8	Ka1-9	Ka1-10	Ka1-11	Ka1-12

Ka1-12 could once have shown a variant of kiore+henua. Half of line a1 would then belong to the previous year and the other half to the new year.

Ka1-11 maybe shows a straight world tree combined with a mouth sign. Ka1-7--8 is one pair, Ka1-9--10 another (cfr the little thread instead of neck and head in Ka1-7 compared with the cut of 'band' in Ka1-9). The body in Ka1-7 is turned into a moon sign, the elbow carries a cycle of time (completed). A slightly bent henua has two line straight across, while in Ka1-9 a shape like a canoe is at right. The horizonal double lines look like those in Ka4-10 and in Aa1-11:

 

Ka1-7	Ka4-10	Aa1-11

 Presumably it characterizes the 2nd dark part of the year.

If we can calibrate the text of K to 180 glyphs (days), then we should find some recognizable sign at position 90 (i.e. 90 + 12) I think:

 

15
	Kb1-4	Kb1-5 (102)	Kb1-6	Kb1-7 (104)

14
	Ga4-20 (104)	Ga4-21 (105)	Ga4-22

Maybe 14 glyphs should be added instead of 12? Gb4-20--21 could together correspond to Kb1-7.

Another approach is to increase 20 at the end of the text to 24:

 

Kb4-15	Kb4-16 (169)	Kb4-17	Kb4-18	Kb4-19
168	4

We would then have twice 24 and twice 22:

 

a1	*24	67	b1	*22
a2	22		b2	*18	44
a3	14		b3	16
	7
a4	16	30	b4	15
				4	*24
a5	14		b5	*20
sum	*97		sum	*95

Aa4-1 (beginning) should be added to 67 to reach 68, and then 100 glyphs later comes Kb4-15 (end).

The method could have been used in several places. We have noted Aa3-1 as having a Janus function too.

Maybe we should add 97 + 2 = 99, because of counting Aa3-1 and Aa4-1 twice?