a1	50	50	b1	*51 (?)	51
a2	58	108	b2	48	99
a3	52	160	b3	47	146
a4	56	216	b4	51	197
a5	59	275	b5	57	254
a6	*69?	344	b6	54	308
a7	*51?	395	b7	50	358
a8	*54?	449	b8	*54 (?)	412
a9	*53?	502	b9	65	477
a10	*67?	569	b10	67	544
a11	*58?	627	b11	53	597
a12	*21?	648	b12	*51?	648
sum	*648?		sum	*648?

We will now count once more. There is no problem with the first 15 glyphs:

11
	Hb8-12	Hb8-13	Hb8-14	Hb8-15

Then comes a question mark, because there could, at least theoretically, once have been a glyph between Hb8-15 and Hb8-101:

...
	Hb8-101	Hb8-102	Hb8-103	Hb8-104	Hb8-105
Pb9-33		Pb9-34	Pb9-35	Pb9-36	Pb9-37

But when I earlier thought about this, I decided there was no glyph missing:

It is quite possible that the creator of H left an empty space to the right of Hb8-15 (where 8 * 15 = 120), a glyph which is drawn open to the right. This empty space (if it was designed to be there) should not be counted, I guess.

I see no reason to change my earlier judgment. Therefore we can assign the (probably) correct original ordinal numbers in line Hb8:

Ha5-18	Ha10-35	Hb2-35	Hb6-45	Hb7-32
*Hb8-29	*Hb8-30	Hb9-1	Hb11-7

Next, we should count from Ha1-1. If we can trust my table over number of glyphs per line (above), then the result will be:

Ha5-18 (234)	Ha10-35 (537)	Hb2-35 (734)	Hb6-45 (947)	Hb7-32 (988)
*Hb8-29 (1035)	*Hb8-30 (1036)	Hb9-1 (1061)	Hb11-7 (1199)

I have redmarked Ha5-18 (which we already know) and two glyphs which seem to be rather close in character. Hb8-30 and Hb9-1 are twins with wings are drawn to be different. The rest of the glyphs have other and more obvious signs. Now we count. A first suggestion which quickly emerged:

		540
Hb7-31	Hb7-32 (988)		Ha5-17	Ha5-18 (234)	Ha5-19	Ha5-20
546 = 3 * 182

709			39
	Hb6-45	Hb6-45 (947)
750 = 3 * 250

3 glyphs per day is maybe a basic trait in the H text. 540 = 3 * 180 could be half a year. Increasing 750 to 756 then gives 3 * 252. And 252 = 7 * 36, a good number. This solution could begin with:

Hb7-33 (989)	Hb7-34	Hb7-35	Hb7-36	Hb7-37 (993)
Hb7-38 (994)	Hb7-39	Hb7-40	Hb7-41	Hb7-42 (998)

We should notice that Hb7-38 is our special honui glyph (of the ariga erua kind). It comes 6 glyphs (2 days?) beyond Hb7-32:

329
	Hb7-30	Hb7-31 (987)	Hb7-32
330
	Hb7-33	Hb7-34 (990)	Hb7-35
331
	Hb7-36	Hb7-37 (993)	Hb7-38

But there are of course three different ways to structure 3 glyphs together into one day.

If there are 3 glyphs per day, then 648 glyphs = 216 days, and the whole text will be 2 * 216 = 432 days long (=  12 * 36 days).

Day 329 ends with Hb7-32 (according to my guess). 32 * 9 = 288 = 8 * 36. 7 * 32 = 224 = 8 * 28.

At Hb7-38 we can count 7 * 38 = 266, the number of Te Pou (= 9 * 29.5 = 265.5).

Advancing from day 330 (day 1), taking triplets of glyphs for each day, we will reach day 180 after 540 glyphs:

Ha4-56	Ha5-1	Ha5-2	Ha5-3	Ha5-4	Ha5-5
Ha5-6	Ha5-7			Ha5-8	Ha5-9
Ha5-10	Ha5-11	Ha5-12	Ha5-13	Ha5-14	Ha5-15	Ha5-16 (232)

232 (at Ha5-16) is glyph number 540 counted from Hb7-33 (989).  988 + 540 - 1296 = 232:

	539
Hb7-32 (988)		Ha5-16 (232)	Ha5-17	Ha5-18	Ha5-19	Ha5-20
	540

The suggested system of triplets of glyphs seems to break down because of such glyphs as Ha5-7 and Ha5-13 (redmarked above). Once again we find 5-7 at a special glyph. And neither is the quartet Ha5-17--20 congruent. But if we combine all the redmarked glyphs they can stand for 2 days (= 182 - 180).

And we had 2 glyphs (days) intercalated at the beginning of side b of G:

Ga8-26	Gb1-1	Gb1-2	Gb1-3

This solution has 180 + 2 days following Hb7-32, which - we can assume - is at winter solstice (the person has sunken down up to his neck).