We ought to try to count the glyphs in Q, in order to see if winter solstice really could be located where the eyes disappear on ariga erua.

I have made a prelimary table based on, among other things, an assumption of 3 glyphs per day and a guess of 400 days described in the text (whereas H has 432 and P presumably 386):

a1		b1	51	51
a2		b2	46	97
a3		b3	39	136
a4		b4	39?	175?
a5		b5	42?	217?
a6		b6	37?	254?
a7		b7	39?	293?
a8		b8	32	325?
a9	45	b9	12??	337??
sum		sum	336???

Let us now begin not with line Qa1 because the text at the beginning of side a is damaged. Instead we can verify that there are 45 glyphs in line Qa9, no question about it (if we use Barthel and Fischer together).

But that is not enough information, we must compare with H and P too, in order to reach a reasonable estimate - at the end of our work - of the length of the Q text:

Then it must be noted that beyond the point where Qa9 has ended, and before glyph line Qb1 is beginning, the texts of H and P have more glyphs:

To reconstruct the number of H glyphs (parallel with line Qa8) is complicated. Qa8-1 is parallel with Ha8-34 while toward the end of line Qa8 the parallel text in H seems to have been destroyed by fire.

We should take it step by step. At the beginning, having ascertainted Ha8-34 as parallel with Qa8-1, the destroyed text arrives already where Ha8-35 should have been.

Earlier I have determined, as far as that is possible, that the next visible glyph,  Ha8-43, is parallel with Qa8-11--12. Between Qa8-1 and Qa8-11 there are 9 glyphs, where the H text presumably may have had 8.

Then H is intact for the rest of line Ha8, and the last glyph in the line is Ha8-54, at which point the parallel glyph is Qa8-23. So far Q has 23 glyphs while H has an estimated 21:

I have redmarked the glyphs between Qa8-38 and Qa8-38, because here Q has an inserted 7 glyph long sequence, not found in H or P:

Qa8-31	Qa8-32	Qa8-33	Qa8-34	Qa8-35	Qa8-36	Qa8-37

We can enter 38 with a question mark at H for line a8 in Q, the same number as in P:

Line a7 in Q probably has no more glyphs than what Fischer has depicted, because according to Barthel's drawing of the line there is hardly room for more than those two mutilated glyphs which Fischer has added.

The texts in H and P are parallel and intact, so we can fill in the numbers:

However, after the end of these glyphs, more glyphs come in H and P which have no corresponding glyph in Q:

The end of line Qa6 is destroyed, but from Barthel we can estimate there once were 4 à 5 glyphs there, let us guess at 5 which will result in 49 - an odd number, in agreement with the previous numbers (45, 47, and 41):

Line Qa5 can be determined to have 55 glyphs (another odd number). Barthel leaves room for ca 4 glyphs at the beginning of the line, but Fischer has documented 5 (although only partially seen). The parallel texts in H and P support those 5 glyphs.

Once again, however, we must documet that beyond the end of these 55 glyphs the texts of H and P continue  longer:

Qa4 is also damaged at the beginning. Barthel has left an open space for ca 3 glyphs, and Fischer has filled in only 1 of them. To the 41 documented by Fischer we can add 2 in order to reach 43 (an odd number). Also, the text of P support 2 lost glyphs, and line Pa4 is beginning where line Qa4 is beginning:

Line Qa3 is beginning all right, but the end is somewhat unclear. Looking at the drawing of the line in Barthel we can conclude that he has left open space for only 1 à 2 glyphs at the end. But Fischer has managed to capture a bit more and we can be fairly sure that there are 41 glyphs in the line, I will not enter any question mark for this number.

Beyond the end of Qa3 there are more glyphs in H and P:

Barthel leaves no empty space for missing glyphs in line Qa2, though many glyphs are drawn only partially visible. Fischer has added one more than Barthel as the first glyph in the line. At the end of the line he has 3 more than Barthel (where Barthel has left no room for additional destroyed glyphs), and the parallel texts in H and P support the existence of these glyphs. Counting according to Fischer's glyphs we have 43 for line Qa2. There is no reason to doubt the correctness.

Now we have reached the problematic line Qa1. According to Barthel's picture of side a the line is only ca half as long as the rest. From that we can expect ca 20-25 glyphs, but Barthel has only ca 13 in his drawing of the line.

From Fischers computer-generated glyphs it is easy to take a first step in identifying 16 glyphs (Qa1-201--216) which are in parallel with the H and P texts

Then there are 4 additional (Qa1-101--104) which are so destroyed that we cannot have any use for them (other than counting them). 4 + 16 = 20, which is at the lower end of the expected ca 20-25 glyphs.

But from the beginning of side a in H and P there are 14 respectively 13 glyphs, not 4. And beyond Qa1-216 there are 47 respectively 45 which cannot have any parallel in Q. Except perhaps the very first of them:

Ha1-25	Ha1-26		Ha1-27	Ha1-28	Ha1-29	Ha1-30	Ha1-31
Pa1-24	Pa1-25	Pa1-26	Pa1-27	Pa1-28		Pa1-29	Pa1-30
							...
Qa1-211	Qa1-212		Qa1-213	Qa1-214	Qa1-215	Qa1-216	Qa1-217

Ha1-32	Ha1-33	Ha1-34	Ha1-35	Ha1-36	Ha1-37	Ha1-38	Ha1-39
Pa1-31	Pa1-32	Pa1-33	Pa1-34			Pa1-35	Pa1-36
...
text destroyed

I think Ha1-32--37 (with parallel in P) is a separate 'sentence', not to be read together with the previous glyphs. Accordingly I guess the number of glyphs in Qa1 could be stated as 21 (with a great question mark). And I must delete the statement 'text destroyed' - there never were glyphs here in Qa1.

Moreover, I guess the 4 remnants of glyphs at the beginning of the Q text belong in parallel with Ha1-11--14 rather than being located earlier:

Ha1-9	Ha1-10	Ha1-11	Ha1-12	Ha1-13	Ha1-14	Ha1-15	Ha1-16
Pa1-8	Pa1-9	Pa1-10	Pa1-11	Pa1-12	Pa1-13	Pa1-14	Pa1-15
...	...
		Qa1-101	Qa1-102	Qa1-103	Qa1-104	Qa1-201	Qa1-202

But Ha1-9--10 (respectively Pa1-8--9) ought to have parallel glyphs in Q. I.e., we can raise 21 to 23, smack in the middle of the expected interval ca 20-25 glyphs.

And now we can add:

387 = 9 * 43 seems to be rather reasonable (as a sum where every term is an odd number).

The odd numbers invite to adding the pairwise, in which case 45 in Qa9 must be put together with the number of glyphs in line Qb1. 387 - 45 = 342 = 18 * 19.