A table of 5.5 : {6, 11, 17, 22, 28, 33, 39, 44, 50, 55, 61, 66}. Or if rounded off to the lower figure {5, 11, 16, 22, 27, 33, 38, 44, 49, 55, 60, 66}.

A decisive reckoner according to the Phaistos disc is an altering table of { 5, 7, 5, 7,...}, or its prolongation to {12 ,17, 12, 17, ...}, which has a significant geometrical expression. .

For you who admit, you conceive my gnomonical arrangement of the stem-elements, I shall try to disentangle a (29,5 x12) +11 calendar.

The upper half part of the frame consist of the 59 stem-elements, deducted for 7 unsuitable stems (2z), and 4 unpaired stems (1). The 29 small bluish grape-coulered squares constitute part A's stem-elements (2x), whereas the olive squares are part B's do (2y).

The lower red-heat part of the frame contains a reduced copy of the above 59 stems added up with 7 suitable short-form stems (0), which are among the 9 uncovering reduced stems outside the frame (4). Now the reduced stems count 29 elements, which are recognisable of category I(3x) Those are the blood-orange-coulered squares of the lower frame. Furthermore we have 30 Jaffa-oranges of category II (see calendar 261), which are abbreviated by two choices (3y). Besides of the stem-signs, the inscription holds 29 unrelated elements (elements are of two units) designated remainder-signs (5).

The remaining two lunations is simply the material that is left over. Say 16 overriped apples of category III (3z) together with the 11 stem-elements in excess (2z) (1) attached with 6 of the 17 slashes. Finally 11 intercalary slashes.

Did you get it? This will tot up the numbers of sign-units in defined portions to: 29+30+29+30+29+30+29+30+29+30+29+30 =354 +11. In full accordance with a moon-calendar.

By the aid of these two tables, it becomes easy to synchronize the yearly movements of moon and sun.

7 +5 +7 +5 +7 = **31**

5 +7 +5 +7 +5 = **29**