3. Next we have to get some order in the jumble inside, and we can make use also of the grid of help-lines. Each of the 3 rim crossings of the upper bowl are located at a horizontal such line:
First, however, we can see the L string is making a wave-like pattern and it is reaching above the 2nd coordinate line both where it is crossed over by the R string and at its end:
Our earlier sum 72 beads above the 2nd coordinate line is thereby increased to 81 = 3 * 27. Inside the bowl we had better move horizontally rather than vertically. Therefore we can continue with the L string. Outside the bowl we counted 22 beads, and inside we then added first 2 in order to reach the 2nd coordinate line and then 2 + 7 more which also were above this line, and 22 + 2 + 2 + 7 = 33 (= 3 * 11 = 81 - 3 * 16). Making the most of all the lines inside the bowl:
The result is 7 + 8 = 15 groups of numbers, with 24 + 17 = 41 beads. I have divided the total where the R string is crossing over the L string. If we should count also the hidden bead at this place the total will rise to 42. Thus 22 beads outside the bowl + 41 inside = 63 in all for the L string. Now to the shortest string, the C string:
I have divided it where the L string is crossing over the C string, and if we count also with the hidden bead at this place the total will rise to 30. The horizontal double help line partly hides one of the beads in the 2nd group, and if we should consider this to mean a fraction of a bead, then the total could be 30 - ½ = 29½. The number of beads on the C string (presumably the most important of them) is therefore somewhat uncertain. However, outside the bowl there are 20 beads and the basic number evidently could be 20 + 29 = 49. The longest string is the R string, and proceeding from above:
I have divided the string beads into 5 + 15 = 20 subgroups, which maybe has no relevance but which helps when securing the numbers. Once again, we could consider to increase the number from 59 to 60 by inserting the hidden bead between the first and the second part of the string:
On the R string we have only 16 beads outside, and 16 + 59 = 75. It is tempting to add 16 + 14 and see the total number of beads as 16 + 14 + 45 (where 45 = 360 / 8). We can now summarize the upper bowl:
But I think we should rearrange the table because the crossings are probably of major importance:
Number 106 could be a Sign which Flamsteed has inserted for us to deduce from the upper bowl. It is the distance from January 1 to Polaris, and such a number should be connected with the upper bowl rather than the lower:
Furthermore, 360 + 106 = 466, which could point at the heliacal rising of the Pleiades, 30 days later:
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