Gnomonical arrangement


If those seventy stems really reflect the secret about this old enigma, if they so were the key and the first sure entry to the Minoan secret, you would expect some kind of correspondence between the stem-signs inside and outside of the stems; and yes, this was exatly, what was found


Comparing 'the unfolded arrangement' with a picture-lottery seems significant, in which the pair of stems, subsidiary all stems, are playing-squares (point 1, 2), and all other signs are pieces (skin colour). It is a familiar situation, that one or more pieces are missing. Let us imagine conversely, that we have had a mingling together with some outside pieces. We therefore charge ourselves with the task to sort out the pieces, that are in excess.

The remaindersigns (point 5) have no correspondence with the playing squares, and are so to be sorted out without further ado. Putting down the pieces (point 3x), where they fit into the squares (sky blue), will finaly leave us with nine pieces (point 4), which were represented in the squares, but they are now in excess, because the playing squares are already occupied. The dolphin (sign K) for instance is present in two stems, but it occurs four times independently, there are so to say two uncovering dolphins in excess. Those nine stemsigns are to be sorted out too, like the remaindersigns. - So are the thorns (point 6).


Of cause, this is not an argue in favour of the inscription as a picture-lottery, I just want to indicate, that a principle of a similar kind can be applied with success; that some complicated, but symmetrical proportions, between the signs on the basis of their functions, are unveiled.

By observing the numbers 18, 22 and 26, it struck me, that square-sides on respectively 10, 12 and 14 hold circumferences, which are related to these numbers. This compels an almost unambiguous and most expressive way, in which 'the unfolded situation' is to be arranged, as the signs of the 33 pair of stems are to be set up as circumferences in a quadratic framework 'the folded arrangement'. The upper half of the framework is uncovered (characters in) stems, and the lowest part is those stems, which have cover from the detached stemsigns (point 3x).

As it does show, the covering signs do not only make up a gemination of the lowest part of the framework, they keep themselves within the areas in their half, which are marked out by the three stemgroups. I have divided the frame into six zones, each containing 22 signs. The six zones are symbolized by Aa, Ab, Ba, Bb and Ca, Cb.

Bb for instance is those 22 signs in second position, which are gathered in the bottom left-hand corner of the frame. Together with Ba, Bb is able to establish 22 covering stems, of which 11 are dissimilar. As regards positions, (the right sign in a stem holds first position and vice versa)

the covering stemsigns (Bb, Cb, Ba) consist of 33 signs in first position and 33 signs in second position. The upper half (Aa, Ca, Ab) then get the same bisection of the positions, obviously.The zones Ca and Cb hold each 11 signs in both first and second position. Together they compose 22 stems (11 dissimilar) crosswise of the median line. It is seen, that multiples of eleven are reflected in a lot of new facets, although there are some limited ways to castle the signs, within those by positions and stemgroups restricted areas. This gnomonic arrangement, I believe, is the most ideal way to illustrate those symmetrical pro portions, which are unquestionably available in the inscription. The arrangement especially substantiate the legitimacy of the three stemgroups. There are however irregularities to be mentioned: Stemgroup II holds 18 covering and 18 covered, but 22 uncovered signs; While stemgroup III has 22 and 22, but 18 signs. If you consider the inscription as a numerical system, it would probably had made a more convincing impression, if the equal conditions had been respectively: 18, 18 and 18 plus 22, 22 and 22, together with the 26, 26 and 26 signs of signgroup I; though the very irregularities may be promising for a more complex application, than a mere ornamental play with some imprints and their quantities, on the part of the designer of the Phaistos disc.



With this I have reverted to the calendar possibility. Each of the four halves, the fourth half being the absent units in the reduced stems, is systems of 29 well-organized pair of signs (58 components) and 8 immediately unpaired characters. Although, there is one condition for this complete symmetrical arrangement: That the obliterated reduced element in A08 unambiguously is set to be the sign "P". Thus the Gnomonic arrangement compels the identity of the missing final character in A08.

The left and the right part of the covering reduced stems (Bb,Ba) constitute 22 stems, 11 stem-forms. The bottom (Cb) with the lid (Ca): 22 stems,and 11 forms.

You'll find 70 stems in the inscription as a whole. As much as 66 of these stems are entering into stem-pairs. Giving 4 unpaired stems in excess. Those 66 stems fit into a frame, that I call the Gnomonical arrangement. *Stemgroup I is aqua, stemgroup II is yellow, stemgroup III is pink. In the same way. You will find 75 stem signs in total, which are disconnected from stems. 66 of these isolated stem signs suit into pairs with the signs inside of the 70 stems. Giving 9 signs in excess. In other words: They will be covering exactly one half part of the components in the 66 stem pairs (as a duplicate of the lower part of the arrangement above). All in all comparable to a picture-lottery. Harmony everywhere! manifesting itself as multiples of eleven.

To end up the calculation. The inscription moreover contains 29 'remainder signs', which is my designation to the left over signs different from the 215 stem signs. Giving: 132+8 +66+9 +29 =244 signs in total. If you add the 17 thorns, then the total is 261 units.

Doctor of Philology Professor, Emeritus Ernst Doblhofer,

Die Entzifferung vershollener Schriften und Sprache (1961). About the Phaistos disk: "Possibly a professional investigator will sooner or later win the laurels promised to the one who solves the riddle of this clay plaque, which can be seen today in the Heraclion Museum. Or perhaps a brilliant amateur will solve the mystery of the spiral images and, like a modern Theseus, find the way out of this new labyrinth of the island of Minos".


Skriv en ny kommentar: (Klik her)
Tegn tilbage: 160
OK Sender...
Se alle kommentarer

| Svar

Nyeste kommentarer

30.07 | 22:18


26.01 | 11:29


20.01 | 18:12

Vi er kristne organisation dannet for at hjælpe folk med behov for økonomisk hjælp. Så hvis du går igennem økonomiske vanskeligheder, eller hvis du har økonomisVi er kristne organisation dannet for at hjælpe folk med behov for økonomisk hjælp. Så hvis du

15.12 | 16:29

3% lånetilbud gælder nu for mere detaljer, vi tilbyder alle former for lån her blot ansøg nu:

Du kan lide denne side