TRANSLATIONS

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We should complete the excursion into how many glyphs there are in H. We left the question here:

a1 50 318 (?) b1 51? 314?
a2 58 b2 48
a3 52 b3 47
a4 56 b4 51
a5 59 b5 57
a6-1--119 43 (?) b6 54
a6-120--139 20 b7-1--6 6
a6-140--145 6 310? b7-7--26 20
a7 51? b7-27--50 24 314?
a8 54? b8 54?
a9 53? b9 65
a10 67? b10 67
a11 58? b11 53
a12 21? b12 51 (?)
sum 648? sum 648?

By using 7 + 6 + 7 (+ 6) as a pattern to look for we may be able to eliminate ?-marks. We begin with b1:

from up to and including number of glyphs
Hb1-1 Hb1-6 6
P has 15 glyphs, Q has 17. To reach the sum 51, H must have had 16 glyphs.
Hb1-201 Hb1-229 29

There are however remains of ca 6 more glyphs in the destroyed area:

... ... ... ... ...
Hb1-7 Hb1-8
... ... ...
Hb1-101 Hb1-102 Hb1-103 Hb1-104
... ... Ca 10 glyphs are missing totally, whereof about ½ before Hb1-101 and about ½ after Hb1-104.

That is what may be concluded from Barthel's picture of this part of H.

The difference in number of glyphs between P and Q (in the parallel area) is due to Pb3-14:

Pb3-14
Qb3-116 Qb3-117 Qb3-118


Otherwise the number of glyphs in P and Q are the same (in this area). If we follow Q we have:

Qb3-105 Qb3-106 Qb3-107 Qb3-108 Qb3-109 Qb3-110 Qb3-111
Qb3-112 Qb3-113 Qb3-114 Qb3-115 Qb3-116 Qb3-117 Qb3-118
Qb4-1 Qb4-2 Qb4-3 Qb4-4 Qb4-5 Qb4-6 Qb4-7
Qb4-8 Qb4-9 Qb4-10 Qb4-11 Qb4-12 Qb4-13 Qb4-14
The 24 magenta-marked glyphs probably constitute a unit (Pb3-24 is parallel with Qb4-10), because from Qb4-11 sequences seem to start which repeat (more or less).
Qb4-15 Qb4-16 Qb4-17 Qb4-18

Working with the 24 glyphs we must observe a 7-glyph parallel in B, from the beginning of line Qb4:

Pb3-15 Pb3-16 Pb3-17 Pb3-18 Pb3-19 Pb3-20 Pb3-21
Qb4-1 Qb4-2 Qb4-3 Qb4-4 Qb4-5 Qb4-6 Qb4-7
Bb6-7 Bb6-8 Bb6-9 Bb6-10 Bb6-11 Bb6-12 Bb6-13

We have, then, 6 glyphs from the beginning of line Hb1 which are parallel in appearance and number with what we see in P and Q:

Hb1-1 Hb1-2 Hb1-3 Hb1-4 Hb1-5 Hb1-6
Pb3-3 Pb3-4 Pb3-5 Pb3-6 Pb3-7 Pb3-8
Qb3-105 Qb3-106 Qb3-107 Qb3-108 Qb3-109 Qb3-110

 Then follows destroyed glyphs in H, where P has 6 and Q 8 glyphs:

Pb3-9 Pb3-10 Pb3-11 Pb3-12 Pb3-13 Pb3-14
Qb3-111 Qb3-112 Qb3-113 Qb3-114 Qb3-115 Qb3-116 Qb3-117 Qb3-118

In the next phase there are 7 glyphs in parallel (B, P, Q) while H still is unreadable:

Pb3-15 Pb3-16 Pb3-17 Pb3-18 Pb3-19 Pb3-20 Pb3-21
Qb4-1 Qb4-2 Qb4-3 Qb4-4 Qb4-5 Qb4-6 Qb4-7
Bb6-7 Bb6-8 Bb6-9 Bb6-10 Bb6-11 Bb6-12 Bb6-13

And at the end of the destroyed area in H we have a sequence with 3 glyphs in the parallel P and Q texts:

...
Hb1-201
Pb3-22 Pb3-23 Pb3-24
Qb4-8 Qb4-9 Qb4-10

Summary:

→ 6 14 (12) 24 (22)
Hb1-1 Hb1-6
→ 8 (6)
Qb3-111 Qb3-118
→ 7 10
Qb4-1 Qb4-7
3
Qb4-8 Qb4-9 Hb1-201

The difference between 24 (a cosmic number) in Q and 22 (unacceptable) in P is explained if we look at the beginning of line Pb3, where the missing 2 glyphs are found:

Pb2-35 Pb2-36 Pb3-1 Pb3-2 Pb3-3 Pb3-4 Pb3-5 Pb3-6
Qb3-102 Qb3-103 Qb3-104 Qb3-105 Qb3-106 Qb3-107 Qb3-108

H has no text in parallel with Pb2-35--Pb3-2. The text which in P and Q precedes Hb1-1 seems to be connected by way of Pb3-1--2 to the following text (also is found in H).

To reach 24 also in H, we must assume a parallel number of glyph as in Q. Therefore, there should be 52 and not 51 glyphs in line Hb1.

Sorting according to the pattern 7 + 6 etc we get:

Hb1-1 Hb1-2 Hb1-3 Hb1-4 Hb1-5 Hb1-6 Qb3-111
Qb3-112 Qb3-113 Qb3-114 Qb3-115 Qb3-116 Qb3-117
Qb3-118 Qb4-1 Qb4-2 Qb4-3 Qb4-4 Qb4-5 Qb4-6
7 + 6 + 7 + 4 = 24
Qb4-7 Qb4-8 Qb4-9 Hb1-201

However, considering the 7-glyph sequence with parallel in B (red-marked below) another pattern must rule:

Hb1-1 Hb1-2 Hb1-3 Hb1-4 Hb1-5 Hb1-6 Qb3-111
Qb3-112 Qb3-113 Qb3-114 Qb3-115 Qb3-116 Qb3-117 Qb3-118
Qb4-1 Qb4-2 Qb4-3 Qb4-4 Qb4-5 Qb4-6 Bb6-13
14 +  6 = 20, followed by te pito (borrowed from B), and then the start of something else
Qb4-8 Qb4-9 Hb1-201

The goal was not 24, it was - you could say - 14 (= 7 + 7), followed by 6 + te pito. Therefore, depending on how many glyphs there were in parallel with Qb4-8--9 (respectively with Pb3-22--23), there could have been just one glyph in H - i.e. the number for Hb1 may have been 51 (or 52). The question mark is not eliminated.