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TRANSLATIONS
There remains, however, another possible way out of the dilemma. We can try to work backwards:
After Hb1-226 the parallel text in Q has an inserted text sequence (magenta below), which is found neither in H nor in P:
Then follows:
I believe we can ignore Hb2-14 etc, although the glyphs probably are connected in some way with the earlier glyphs (cfr Hb2-16 with Hb1-228) the theme has changed. Next, I suggest the beginning of line Hb2-1 is a natural place at which to begin a new sequence. If so, then we will have 13 glyphs, maybe arranged like this:
4 + 4 + 5 = 13. It does not seem possible to use the pattern 7 + 6 etc. If Hb2-1 does not mark a new unit, we should add the glyphs at the end of line b1:
Hb1-227--228 have features common with Hb2-6 and Hb2-11. The pattern is now 3 + 4 + 4 + 5 = 16, a rather good sum. Hb2-7--8 connects to even earlier glyphs, however, and we then will reach 2 + 3 + 4 + 4 + 5 = 18, an acceptable sum (because 36 / 2 = 18):
Yet earlier we have 8 + 6 = 14 glyphs with a possible partial parallel in B:
Even further backwards we at last can connect with our number problem:
We now recognize the pattern at the end of each sequence (Hb1-206--210), and we can count to 10 in the first of the similar sequences. It is just to read the number labels. In the first sequence of this kind we have 10 glyphs, after that arrives a special sequence with 8 glyphs:
Then comes the 2nd of the sequences which ends with GD21 (hua poporo):
We have now arrived at the sum 10 + 8 + 8 = 26 glyphs, counted from Hb1-201. If we add 3 more we reach 29 and at the same time to the end of line Hb1:
I think we could add one more glyph before Hb1-201, because then we will reach 30. 1 + 10 + 8 + 8 + 3 = 30. Though, of course it would be nice to add two, having 12 glyphs before the special section:
We need not bother, I think, about what comes after Hb1-218.
Are we defeated? Is there any way to determine whether there were 1 or 2 glyphs at the ?-mark? There are 30 glyphs from the ?-mark to the end of line b1 if we assume there was just 1 glyph at that place. Dividing by 2 we get 15 and counting from the questionmark 15 glyphs ahead we reach up to and including Hb1-214:
Due to the parallel glyphs in B I guess there was just 1 glyph at the question mark. 15 glyphs from the question mark up to and including Hb1-214, then followed by another 15 to the end of the line. 51 = 14 + 7 + 15 + 15. Furthermore, Hb1-214 is loking backwards. Furthermore, otherwise we would reach 315 (an odd and evil number) before the cosmic 6 + 20 at the center of side b. Reflecting further: Bb6-13 (kua motu te pito o te fenua) corresponds to Qb4-7, which is no. 21 counting from Qb3-105 (corresponding to Hb1-1):
In Pb3-21 and Qb4-7 I imagine we can see henua with breast and nipples, perhaps meaning the pregnant 3 wives of the sun (120 days each). Beyond 20, a fundamental unit, arrives te pito, that is natural. If we then add 28 more glyphs, we reach 7 (as in Qb4-7) + 28 = 35 (i.e. half of 70 which we have discovered measures the 'moon month', i.e. 420/12):
Beyond this exclusive piece of text (only present in Q) we have in H (the Q text is destroyed here):
The inserted part in Q (before Hb1-227) should tell us that a new phase begins with Hb1-227. Therefore, the number pattern in line Hb1 should be: 51 = 14 + 7 + 15 + 12 + 3 = 20 + te pito + 30 = 20 + 28 + 3. I think we now can change from 51? to 51 (?) in our table of how many glyphs there are on side b in H. To be more exact we should put an asterisk in front to indicate a reconstructed number: *51(?). |
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