TRANSLATIONS

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The trail we are following is to '... compare the distribution of the internal parallel glyph sequences (all of which are found on the opposite side of the tablet, none on the same side) ...' in order to find out why there are 6 extra hau tea after Ab4-79. To recapitulate:

... There is a better alternative to what I have just suggested, viz. that the 6 extra hau tea glyphs at the end of side b correspond to waxing moon. The Spring Sun time has no corresponding hau tea glyphs on side b (according to my reading of the text), therefore a sense of balance proposes that the 'spring' moon time (presumably not documented on side a), i.e. Waxing Moon, is worthy of a place on side b.

Spring Sun:
Aa1-5 Aa1-6 Aa1-7 Aa1-8

Aa1-19

Aa1-21

Aa1-23

Aa1-27

Aa4-43

Aa4-46

Aa4-54

Aa4-70

Waxing Moon:
Ab5-8 Ab6-23 Ab6-71 Ab7-64

The first internal parallel text is, as we have seen, Ab1-14--20 (to be compared with Aa4-12--18):

b
13 14 15 16 17 18 19
14 15 16 17 18 19 20
a
263 264 265 266 267 268 269
264 265 266 267 268 269 270

The double rows of ordinal numbers in the table are there in order to see whether the counting should begin with the 1st glyph on the side or with the 2nd glyph.

Beginning with the 2nd glyph on side a implies that the first three lines (a1-a3) have 250 (= 52 * 10) glyphs together (90 - 1 + 85 + 76).

Beginning with the 1st glyph on side b implies that the first three lines (b1-b3) have 244 (= 122 + 100) glyphs together (82 + 85 + 77).

I regard this as evidence for counting that way, i.e. the π glyph (no. 314) on side b should be Ab4-69 (e mea):

On side a we have found the π glyph to be Aa4-64 (kua hura te ragi):

In the table above I have painted with red the ordinal numbers according to what so far seems to be the correct counting methods, one method for side b and another for side a.

The discovery that 250 glyphs constitute lines a1-a3 makes it possible to compare the parallel sequence above in a new way, which possibly was intended by the creator:

b  
14 15 16 17 18 19 20
a  
263 264 265 266 267 268 269
  14 15 16 17 19 20

Red-marked numbers correspond to the ordinal numbers on side a minus 250. No. 263 is curious, it seems to be located at the wrong place. The parallel glyph on side b (which is the first of the two sides to be read) instead has no. 18.

No. 263 has two parts (left and right) and if we count each part separately, and if we in our minds relocate no. 263 to place no. 18 + 250 = 268, then the two following glyphs (i.e. those numbered 268 and 269) will become 19 (= 268 + 1 - 250) respectively 20 (= 269 + 1 - 250). Is this some kind of message implying that we should increase with 1 unit the ordinal numbers on side a from glyph no. 268 onwards?

It seems as if no. 263 should have two implied numbers. No. 268 will then be number 269 and that may be the reason behind all this, because a henua marks a time period (here, it seems, qualified by the following hau tea) and a period should end with 20. Was this the sole reason we could of course instead make a lesser adjustment into:

b  
14 15 16 17 18 19 20
a  
263 264 265 266 267 268 269
13-14 15 16 17 18 19 20

But then 264 - 250 = 14 will not agree with the red 15 above.

Anyhow, the counting to π should perhaps not be disturbed, because Aa4-64 certainly has both a design and ordinal number (4-64 - i.e. 4 and 8*8) which marks importance:

Aa4-63 Aa4-64 Aa4-65 Aa4-66 Aa4-67
Aa4-68 Aa4-69 Aa4-70 Aa4-71 Aa4-72

On the other hand, the renumbering process occuring due to the suggested splitting up of no. 263 into two parts to be counted separately results in a kind of harmony with the parallel glyphs on side b:

Ab5-1 Ab5-2 Ab5-3 Ab5-4 Ab5-5
Ab5-6 Ab5-7 Ab5-8 Ab5-9 Ab5-10

The 'dangling ball' in Ab5-1 implies that it is possible that also the 'dangling ball' in Aa4-64 should be regarded as connected to the Rei at its left. After renumbering (by adding 1 unit from Aa4-13 onwards) Aa4-63 could be the proper π glyph on side a.

We must now recall our earlier table over glyphs which could be regarded as split up in a left and a right part. Are there other such glyphs (than Aa4-12) somewhere in the sequence from Aa1-1 up to the π glyph? Yes, indeed:

3

10

13

Aa1-15

Aa1-66

Aa1-70

3

Aa2-45

Aa2-46

Aa2-60

3

Aa4-6

Aa4-12

Aa4-72

1

Aa5-69

3

Ab2-45

Ab4-12

Ab8-63

Most of the discussion above now evaporates. Let us therefore summarize:

Beginning to count with the 1st glyph on side b implies that the first three lines (b1-b3) have 244 (= 122 + 100) glyphs together.

Beginning to count with the 2nd glyph on side a implies that the first three lines (a1-a3) will have 250 (= 52 * 10) glyphs together.

In the first internal parallel sequence of glyphs we find a sequence of glyphs similar to  Ab1-14--20 on the other side (a) and 3 lines later (a4) in the text.

The ordinal numbers of the glyphs (counted in line a4) is though not 14--20 but 12--18.

At this point it occurs to me that perhaps the explanation for the dissimilar glyph numbers may be that we should disregard 2 of the glyphs before Ab1-14. Can I find them?

Ab1-1 Ab1-2 Ab1-3 Ab1-4 Ab1-5 Ab1-6
Ab1-7 Ab1-8 Ab1-9 Ab1-10 Ab1-11 Ab1-12 Ab1-13

No, I cannot. If forced to bet on a pair I would hesitate between Ab1-1--2 and Ab1-12--13, but would have difficulties deciding. Maybe the glyphs instead should be regrouped like this:

Ab1-1 Ab1-2 Ab1-3
Ab1-4 Ab1-5 Ab1-6 Ab1-7
Ab1-8 Ab1-9 Ab1-10 Ab1-11
Ab1-12 Ab1-13 Ab1-14

Suddenly a solution offers itself by analogy: We have grouped the glyphs on side a wrongly. Aa4-6 and Aa4-12 must belong together, because they are exactly alike.

We remember the patterns in the henua glyphs, which patterns probably were used to mark glyphs belonging together (with a triplet redmarked as an example):

The shapes are either straight or bent with concave side to the left.

Aa4-12 belongs to a grouping immediately before the internal parallel we are discussing:

Aa4-1 Aa4-2 Aa4-3 Aa4-4 Aa4-5 Aa4-6
Aa4-7 Aa4-8 Aa4-9 Aa4-10 Aa4-11 Aa4-12

We no longer need to press Sb2-2 into the pattern of the internal parallel in Tahua:

-
Ab1-14 Ab1-15 Ab1-16 Ab1-17 Ab1-18 Ab1-19 Ab1-20
14 15 16 17
...
Sb2-2 Sb2-32 Sb2-33 Sb2-34
-
Aa4-12 Aa4-13 Aa4-14 Aa4-15 Aa4-16 Aa4-17 Aa4-18
  264 265 266 267
Counting from Aa1-1 gives the redmarked numbers immediately above. Subtracting 251 for the first three lines (a1--a3) results in the ordinal numbers -13, -14, -15 and -16. Subtracting 250 instead (i.e. disregarding Aa1-1 in the counting) results in 14, 15, 16 and 17 in harmony with the ordinal numbers in line b1.

The results confirm that on side a counting should start with Aa1-2.