The curious creature in the center of the world down below the flat surface on which the sun god stands presumably is equivalent to the dark mago:

The crack in the roof of the little chamber then will correspond to the 'crack in the carapace of the turtle'. 4 straight necks are holding the sun floor high. Then there are 6 fluid members in a more horizontal orientation, the world under the sun floor is down in the water.

Next page from the link 'elucidated':

Against the background of the calendar cycle of expanding creation a more precise map of the yearly path of the sun is presented. The dilemma of 365.25 days being shorter than 432 is solved in an elegant way by locating the end of the solar cycle to the same place as the end of the long cycle: Ha1-25 Ha1-26 Ha1-27 Ha1-28 Ha1-29 Ha1-30 ⅓ 1 Hb9-60 Hb9-61 Hb9-62 (1122) Hb9-63 Hb9-64 Hb9-65 (1125) (1122 - 26) / 3 = 365⅓ day 1 The 42 night long voyage of sun into the underworld (from winter solstice to Hb9-63) is only imaginary (similar to moving away from the line of real numbers into the other dimension of imaginary numbers). Hanau in Hb7-42 is not real, the outline of the figure is not closed: Hb7-38 Hb7-39 Hb7-40 (996) winter solstice (day 399 = 996 / 3 + 58) Hb7-41 Hb7-42 Hb7-43 day 1 in the underworld Real time continues from winter solstice in day 390 directly to the dark mago at Hb9-63 (where 63 possibly alludes to a reversed 36). With the calendar cycle beginning again at Hb9-63 there must be 13 * 30 = 390 days in real time (12 * 30 days is not enough for a year) according to the calendar. 432 - 42 = 390. This of course means there must be a time overlap of ca 25 days (= 390 - 365⅓), with the old year continuing into the beginning of the new year. Calendar day 1 with the dark mago is also day 366⅓ of the old year. 390 - 366⅓ = 24⅔. Counting 24 glyphs from the dark mago we arrive at glyph number 1122 + 24 = 1146 (from Ha1-1), and adding the fraction (⅔) we come to Hb10-22. The old year is tagata (complete) and vae says he is leaving, then follows (significantly) a niu glyph: 21 Hb9-63 Hb9-64 Hb9-65 (1125) Hb10-22 (1147) Hb10-23 Hb10-24 day 1 7 day 9 Counting with 1 glyph per day we can advance from the dark mago in Hb9-63 to Hb10-22 and find day number 365⅓ + 24⅔ = 390. The glyphs keep order in the text. In the 9th of the dark nights of the calendar cycle the old year is leaving. It is winter solstice according to the calendar of the solar cycle. But the new year goes according to another calendar. The above is only one example of how different calendar cycles are integrated in an intricate way. The 432 * 3 = 1296 glyphs in H express both the positions in the long calendar cycle of creation and the positions in the shorter cycles (of which the solar cycle above is just one of several).

If we should allow the old year to continue beyond day 390 also in the solar cycle calendar, we must add 3 * 42 = 126 glyphs, a number which seems to allude to 26 (at Ha1-26).

Day 9 according to the dark mago is day 365⅓ + 9 = 374⅓ according to the solar cycle. The solar cycle is also moving ahead with 3 glyphs per day. I have made a mistake above in my dictionary page.

Yet, the logic seems to work, it explains the glyphs (Hb10-22--23).

A calendar which works with 1 glyph per day indeed seems to begin with the dark mago as day 366 = 1:

			21
Hb9-63 (1123)	Hb9-64	Hb9-65		Hb10-22 (1147)	Hb10-23	Hb10-24
366	367	368		390	391	392

We must therefore do it right and see if the glyphs also agree with moving forward with 3 glyphs per day:

Hb9-63 at 1123 must be day 1123 / 3 = 374⅓ counted from Ha1-1. Adding 58 days we reach 432⅓.

First we renumber Hb9-63 from 374⅓ to 365⅔ (by reducing with 26 / 3 = 8⅔). Then we must add (390 - 365⅔) * 3 = 73 (= 3 * 24 + 1) glyphs (a number equal to 365 / 3).

The glyph which stands at the end of day 390 will be number 1123 + 73 = 1196 (= 1296 - 100):

	65
Hb9-63 (1123)		Hb10-64	Hb10-65	Hb10-66 (1191)
365⅔	21⅔	387⅔	388	388⅓

Hb10-67	Hb11-1 (1193)	Hb11-2	Hb11-3	Hb11-4 (1196)
388⅔	389	390 (= 1196 - 26) / 3

The disappearing strange glyph (probably a sun glyph) at Hb11-1 at the end of day 389 does indeed suggest we have done it right. Tagata at the end of day 390 has a tao elbow ornament at left, presumably a sign of the old cycle now being in the past.

There is no end to the complications. Why, for instance, should we not move forward to day 400?

We can do it in several different ways, by moving backwards from the dark mago at 433, by moving forward from the dark mago at 365⅔ counting 1 glyph per day, or by counting 3 glyphs per day:

433 = 400 + 33. We need to move backwards 3 * 33 = 99 glyphs, which means to glyph number 1123 - 99 = 1024 at Hb8-18. Jackpot!

Hb8-15	*Hb8-16	*Hb8-17	*Hb8-18 (1024)	*Hb8-19	*Hb8-20
399			400

Pb9-33 (1025)	Pb9-34	Pb9-35	Pb9-36	Pb9-37

8 * 18 = 12 * 12. The parallel P text has a pito at position 1025.

If we move forward with 1 glyph per day, we need to move only 10 glyphs further than Hb10-22

			21
Hb9-63 (366)	Hb9-64	Hb9-65		Hb10-22 (390)	Hb10-23	Hb10-24
day 1			7	day 9
Hb10-25	Hb10-26 (394)	Hb10-27		Hb10-28	Hb10-29	Hb10-30
day 10				day 11
Hb10-31	Hb10-32 (400)	Hb10-33		Hb10-34	Hb10-35	Hb10-36
day 12				day 13

Also this alternative is good. Day 12 from the dark mago coincides with the variant of tapa mea which means 'eating' (the sun). It confirms that the suggested 390 at Hb10-22 indeed could be correct.

We must also try with 3 glyphs per day counted from day 1. Here we need to add 3 * 10 = 30 glyphs to Hb11-1 in order to reach the first third of day 400:

Hb10-67	Hb11-1 (1193)	Hb11-2	Hb11-3	Hb11-4 (1196)
388⅔	389	390 (= 1196 - 26) / 3

Hb11-26	Hb11-27	Hb11-28	Hb11-29	Hb11-30	Hb11-31
398			399

Hb11-32	Hb11-33	Hb11-34 (1226)	Hb11-35	Hb11-36	Hb11-37
400 (= 1226 - 26) / 3			401

I am not equally convinced by this variant, although it may be correct (there are 11 nights from Hb11-1 to day 400 and reversed double tapa mea with 5 feathers occur in both places. Both 398 and 399 seem to account for a great cardinal point, maybe 2 nights are needed when we are encroaching on night number 400?

But I must change a little (redmarked below) in the glyph dictionary page:

... Calendar day 1 with the dark mago is also day 366⅓ of the old year. 390 - 366⅓ = 24⅔. Counting 24 glyphs from the dark mago we arrive at glyph number 1122 + 24 = 1146 (from Ha1-1), and adding the fraction (⅔) we come to Hb10-22. The old year is tagata (complete) and vae says he is leaving, then follows (significantly) a niu glyph:

			21
Hb9-63	Hb9-64	Hb9-65 (1125)		Hb10-22 (1147)	Hb10-23	Hb10-24
day 1			7	day 9

Counting with 1 glyph per day we have advanced from the dark mago in Hb9-63 to Hb10-22 and found day number 365⅓ + 24⅔ = 390. The glyphs keep order in the text. In the 9th of the dark nights of the calendar cycle the old year is leaving. It is winter solstice according to the calendar of the solar cycle. But the new year goes according to another calendar.

However, to simplify matters we have here counted with 1 glyph per day instead of 3 glyphs per day (which was used to move from Ha1-27 to Hb9-63), and yet the text clearly indicates the end of the year at Hb10-22. We could equally well have moved ahead with 3 glyphs per day and reached the end of the year (day 390) further on and described with other glyphs:

Hb10-67	Hb11-1 (1193)	Hb11-2	Hb11-3	Hb11-4 (1196)
388⅔	389	390 (= 1196 - 26) / 3

The curious Hb11-1 here probably depicts the vanishing sun, and tagata in Hb11-4 has an elbow ornament (a tao sign) at left which means the old cycle now is in the past.

The above is only one example of how different calendar cycles are integrated in an intricate way. The 432 * 3 = 1296 glyphs in H express both the positions in the long calendar cycle of creation and the positions in the shorter cycles (of which the solar cycle above is just one of several).