Counting 'Mother Earth' leads to the result that we have to count 96º twice because she is found also on the other side of the earth, also there with 36º below the equator and 60º north of the equator. 2 * 96 = 192 or equal to the number of glyphs I have reconstructed for the London Tablet (K). This number, in turn, is only part of the cycle, because half the cycle should be 'in the water'. 2 * 192 = 384 (which happens to be only 0.5 more than 13 * 29.5). And 400 - 384 = 16. Therefore the total old cycle should have been 12 * 16 + 13 * 16 = 192 + 208 = 400 = 16 * 25. If we contemplate the curious number of days in Q (368) it is rather unavoidable to think '36 times 8'. Multiplying 36 * 8 results in 288, the measure from day 1 up to and including the vanishing Rogo in Qb2-10:
We can add twice 32 (i.e. 64) days to the measure for twice 'Mother Earth' (36 + 24 + 36). 32 + 36 + 24 + 36 = 68 + 60 = 128. 2 * 68 + 2 * 60 = 136 + 120 = 256 = 4 * 64 = 8 * 32 = 16 * 16. 192 = 12 * 16 and 256 = 16 * 16, the difference is 64. 192 = 3 * 64 and 256 = 4 * 64. Therefore 448 = 7 * 64. In other words, there are 32 weeks from Qa1-1 up to and including Qb2-10. The 'coinage' seems to be 16. 288 = 18 * 16 and 448 = 28 * 16. Furthermore, twice 368 = 736 (the number of glyphs, as reconstructed by me, for Q) is equal to 46 * 16. That is, 288 + 448 = 736. Also in G and K is 16 used as the measure (16 * 29.5 = 472 respectively 16 * 12 = 192). There are 16 fortnigths from Qa1-1 up to and including Qb2-10. |
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