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If a first period of 13 * 20 = 260 days is ending with *Ha6-24 (where 6 * 24 = 144), then it ought to begin with 4 * 29 = 116 dark nights. 144 + 116 = 260. Expressed in glyphs: 144 * 3 = 432 (which happens to be equal to (648 + 648) / 3 glyphs. In order to reach 648 also as the number of glyphs on side a I have counted the 'imaginary' glyphs in the burned area but not the 18 possible glyphs in the empty space at the end of line Ha12. 299 of these 432 glyphs are needed to count to *Ha6-24 from Ha1-1. 432 - 299 = 133 must therefore be at the end of side b, which is from Hb10-39 onwards:
Evidently this is close to an important event, because the parallel P text ends with *Pb11-60, parallel with Hb10-50. If we count the day number at Hb10-39 we can first divide by 3 and 516 / 3 = 172. Then we can add 64 and reach 236. Should we add also 63 the day number becomes 299 (i.e. the same day number as the glyph number of *Ha6-24). 299 can possibly be read as 29 * 9 = 261, or 'one more' than 260. As to the important event we presumably should think 16 * 29.5 = 472 = 300 + 172 (= 516 / 3), because 472 = 2 * 236 (where 236 = 172 + 64). Rona in Hb10-47 could illustrate the 'downfall' of the year. 104 * 7 = 728 and 728 / 2 = 364. |
