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Stem-determination

STEMDEFINITION : TWO CONNECTED GLYPHS THAT REPEAT ELSEWHERE, I CALL STEM-ELEMENTS.


Let me by means of an example introduce the reader into how a deduction of a stem is carried out in practice. I’ll choose a signgroup at random "B13", and then I intend to investigate its contents of stems, if any. A stem is a construction of two and only two signs. By this means, there are given three compositions for a signgroup with five signs as B13: 1) ØS RR F. 2) ØS R RF. 3) Ø SR RF. Taking the potential stems in the first composition, beginning with RR, this combination of signs is to be searched for in all 61 signgroups. - It is not found. The search proceeds with ØS - the result is still negative. Only the combination SR in the third composition has a repetition; namely in the signgroup B05. B05 consist of four signs, and by that it has two compositions: B SR K and BS RK. The stem SR is yet not guaranteed before both compositions of B05 have been tested. The first SR was tried with a positive result. The potential stems in the second composition cannot be recovered. The stem SR is by now determined. On the other hand, if the possible second compositions BS or RK were recovered, but only at one location, just as SR, then the contents of stems in B13 and B05 would have been evasive, cf. B22, B29 and A04. Thus, a majority in favour of the deduced stem is demanded before certainty is attained. In A17, A29 and A23, the combination WB emerges, but it is overlapping the stem BA, which is present thirteen times in all, by means of which WB is not to be a stem. The two identical signgroups A17 and A29 must, besides BA, of necessity hold two more pairs of stems, but a search determines that none of the compositions available are found in other signgroups, without which the stems cannot be deduced with certainty. At this point the outcome must be based on an estimate: That the position for a sign in a stem is to be respected, when it is known from a definable stem, due to the definable stems observing these position rules among themselves.


As a consequence of the stems, at a point you will arrive to this exactly same 'survey of all 261 visible units'. Here the 70 stems are arranged in the upper half part of the scheme, and the reduced elements below, as imperfect reflections of the stems as in water. Their 104 disregarded units are indicated by pale blue. Congratulation! when you at a late hour arrive to the gnomonical tabulation.



TRANSFORMATION OF THE GNOMONICAL ARRANGEMENT INTO CALENDRICAL 29 x 10


1). With the above statistic tabulation as point of departure starts a search for ground, that will explain why the number eleven is so significant. Examples: Basically the inscription consist of 242 regular glyphs, solved as 22 x11, those of the glyphs appears in 45 species. The stems are in 22 stem-forms made of 30 different glyphs. -In that the 70 stems are defined as pairs, the number of pairs must be of special interest, these are 33 pair of stems plus 4 unpaired . The 75 stem-signs outside of the stems are build of 22 unique glyphs, eight glyphs less than the stems, that had 30 different.. Yet another primary count: From 61 sign groups, 50 contain stems, complementary there are 11 non-stem sign groups. Divisibility by eleven possess priority. - In my gnomonical arrangement eleven is the basic module. For instance the six corners each are build of 22 units of which 11 are unique. This gnomonical frame is the essens of the overall multipla of eleven, that the inscription reveals due to the stems. It explains the question of "why eleven"?.

2). Another simultaneous and concurrent result of the stems are the self-given subgroupings of all signs, that they immediately secure. The 70 stem-elements, the 75 reduced stems, the 29 non-stem-elements, and the 17 thorns. Those four fundamental amounts show to hide a calendrical divisibility by twenty-nine. To be traced in 'the frame' too.








This is my linear display of 243 characters, which are connectible into a grand circle, if B30 is followed by A01 (A31 is placed in the center). A30 is then continued by B01.
To obtain the ideal of a calendar with eight months of 29 days, I have to isolate the 4 unpaired stems, as medial intercalary days.
The hierarchy of glyphs
DARE NOT TO SEEK AND YOU SHALL FIND NOTHING
In confidence with my definition of 70 stems, which leads to the gnomonical arrangement, ( no doubt  a perfect structure, almost of celestial symmetry), I continue 'relaxed' through inspired initiatives (without to fly the original colour of my investigation)  in search for what should be, a more comprehensive and  simpler understanding..
; Unravelling even more
According to myth, Thoth earned the extra five days of the 365-day calendar by gambling with Khonsu in a game of senet.
He won a portion of light from the moon (1/72) which equated to five new days.


CONTACT
This arrangement is called the husk and kernel calendar. It is using the sign-group structure as a replacement for abbreviated elements
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