TRANSLATIONS

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I decided not to add another 'Mercury' page here, although the resemblance between the Turkestan cosmos and what we can read in the rongorongo texts is intriguing:

Much more information can certainly be gained from what lies inside the double perimeter of the 'wheel of time'. For instance is the inside band broader close to where there are 10 circles in the outside band, the same type of sign as we can see in the band outside the hare of the moon.

The toppling mountain combines with the bottom one to define ⅓ of the wheel. This seems to be the beginning, in the dark sector of the year. Here we can be certain that it is the year and not the month or some other cycle.

The hare inside the moon is at the beginning of the year. This beginning comes beyond high summer (the bottom mountain). But is lies before autumn equinox, winter solstice is cutting the cycle and with this as a point of reference we can conclude that the hare moon lies towards the end of the summer half of the astronomical year. The winter half is full with 'cubes' and other strange objects, the summer half contains only Moon, Sun, Venus, Mercury and the summer mountain.

The summer mountain (high summer) comes beyond midsummer. On its plateau old Sun will be sacrificed so that a new son can be generated. Counting 9 months ahead we indeed find the newborn giant baby, Tama Nui. He is born around spring equinox. The sky winter is ending with him.

Spring sun together with moon hare apparently define a sky summer which is ⅓ of the year. This explains why the moon hare is not located at autumn equinox.

The winter mountain has 3 * 4 + 1 * 5 = 17 'fingers'. This number refers to the beginning we have seen in the rongorongo texts. With the toppling mountain the old year (another one than we just have been discussing) vanishes. 17 * 24 = 408 = the day number before Rogo at Gb6-26.

4 * 5 + 3 * 4 + 1 * 5 = 20 + 17 = 37 = 36 + 1. It is the time of 'one more'. Let us stop here.

Next page is the last from 'shown to be true':

 

Calendar III is the shortest of them:

winter 4 * 25 = 100
Ye1-1 Ye1-2 Ye1-3 Ye1-4
summer
Ye2-1 (120) Ye2-2 Ye2-3 Ye2-4 (180) Ye2-5 Ye2-6
10 * 20 = 200
Yf1-1 (240) Yf1-2 Yf1-3 Yf1-4 (300)
winter 4 * 25 = 100
Yf2-1 (325) Yf2-2 Yf2-3 Yf2-4 (400)

Following my intuition (based on much experience of course) I can read that there are 10 glyphs with hidden tops. I think these indicate the lightest sun part of the year. Giving 20 days per glyph it becomes 200 days. The rest of the year is then distributed evenly between 4 + 4 glyphs (signifying the night time of the moon).

I have then tried with 25 (the number of Saturn - the opposite male character compared with the Sun) to define a total of 400 days. Had I instead chosen to give these 8 glyphs 20 days each, the result would have been:

winter 4 * 20 = 80
Ye1-1 Ye1-2 Ye1-3 Ye1-4
summer
Ye2-1 (100) Ye2-2 Ye2-3 Ye2-4 (160) Ye2-5 Ye2-6
10 * 20 = 200
Yf1-1 (220) Yf1-2 Yf1-3 Yf1-4 (280)
winter 4 * 20 = 80
Yf2-1 (300) Yf2-2 Yf2-3 Yf2-4 (360)

I think this alternative is more reasonable. The simplest solution should be chosen as the basic reading, and to have 20 for all glyphs is to prefer.

Moreover, summer is still living at Yf1-4. After the death of summer sun it will look as in Yf2-2. Therefore Yf1-4 cannot be at 300, that is too high a number.