Let's look for some structure.
We have three triangles let's think about the three instincts,. Good now let's think about the six instinctual variant heads,. Good.
Now let's look at the image of the three triangles,. Good.
Now we live on a Planet that has a Moon that travels 12 times around the Earth every time the Earth completes one orbit around the Sun,. Good.
Now we will think about the three triangles and the Nine Points,. Ok
1-4-7 , 3+12+21 = 36 , 3+6 = 9
2-5-8 , 6+15+24 = 45 , 4+5 = 9
3-6-9 , 9+18+27 = 54 , 5+4 = 9
9+9+9 = 27 , 2+7 = 9 , (27 provides me a thought of braking down the two tens's "20" in to 10 and 10 and placing each, one ten at Point:3 and one ten at Point:6. The Severn units remaining provides me with the simple idea of placing one of each of the "7" units at all remaining Points: 4-5-2-1-7-8-9 , 1+2+4+5+7+8+9 = 36 , 3+6 = 9. This reminds me of Point:9
36+9 = 45 , utilising the principle of multiplying via 3 the number value of Point:3 provides the sum number value of 9. So now we have (1+2+4+5+7+8+9 = 36) plus three multiplied by three for one of the two remaining Points, Point:3 and now we have a number value of 45 (36+9 = 45).
36+18=54 , utilising the same process of multiplying via 3 the number value of Point:6 provides the sum number value of 18. So now we have (1+2+4+5+7+8+9 = 36) plus six multiplied by three for the last of the remaining Points, Point:6 and now we have a number value of 54 (36+18 = 54).
Ok so now we call in vision a diagram like this:
..................... 1
....... 1 ....................... 1
1 ....................................... 1
10 .................................. 10
............1 ............... 1
At this stage we will see this as a clue. And we will take this clue with another clue's (12) & (3) & the instinctual triangles.
Ok start at Point:1 envision a triangle with Point:1 at the apex of the triangle and the remaining two corners of the triangle are found at Point:7 and Point:4.
Ok start at Point:4 envision a triangle with Point:4 at the apex of the triangle and the remaining two corners of the triangle are found at Point:1 and Point:7.
Ok start at Point:7 envision a triangle with Point:7 at the apex of the triangle and the remaining two corners of the triangle are found at Point:4 and Point:1.
Or one of the other two remaking triangles.
Ok start at Point:9 envision a triangle with Point:9 at the apex of the triangle and the remaining two corners of the triangle are found at Point:6 and Point:3.
Ok start at Point:6 envision a triangle with Point:6 at the apex of the triangle and the remaining two corners of the triangle are found at Point:3 and Point:9.
Ok start at Point:3 envision a triangle with Point:3 at the apex of the triangle and the remaining two corners of the triangle are found at Point:9 and Point:6.
So we may as well do the final triangle since we have viewed the two of the three.
Ok start at Point:8 envision a triangle with Point:8 at the apex of the triangle and the remaining two corners of the triangle are found at Point:2 and Point:5.
Ok start at Point:5 envision a triangle with Point:5 at the apex of the triangle and the remaining two corners of the triangle are found at Point:8 and Point:2.
Ok start at Point:2 envision a triangle with Point:2 at the apex of the triangle and the remaining two corners of the triangle are found at Point:5 and Point:8.
Ok the next step is quite simple we go three steps each way from what ever Point we may first chose. Or we do the same action we find the other two parts of the equilateral triangle the two part's one left and one right of the apex. We will choose any number,.let's choose "5".
Point:5 envision a triangle with Point:5 at the apex of the triangle and the remaining two corners of the triangle are found at Point:8 and Point:2 so now let's take twelve baby steps in each direction from each of the two remaining corner's 8 and 2 towards the apex Point Point:5.
So let's start counting (2) 3 4 5 6 7 8 9 1 2 3 ((4)) four is the twelfth and if we do the same from Point:8 it goes (8) 7 6 5 4 3 2 1 9 8 7 ((6)).
Exactly why is it so,. I don't know other that the pattern is sound and the pattern is there.
Twelve is "1" from the ten's column and "2" from the unit's column in the decimal point system. Two ten's equals twenty ( cause-effect -- reconciliation ). Three parts, cause (1), effect (1), reconciliation (1).
___ Reconciliation
Effect _________ Cause
The primary triangle at the reactive level "personality" has a "multiplicity" ("6----3") (-3-6-9-) and it seems at the instinctual level there maybe a further six. I suspect that the hole man may utilising as many as "10" or as many as "9" ,9 plus one mechanical orientated multiplicity to adequately respond to cause and effect via a personality in a response to entropy In order to respond creatively with an awareness toward cultivating understands of the nature of the universe perceived or the cosmos that act as host.
We are made via (Greater Nature from Grater Nature) to view and understand the Cosmos, Greater Nature.
So 12 is like a paradox or a conundrum it's ten and two and two and one...., "15" the provides the counter the other wing. 10+2+2+1= 15
- For Point:5 ,(8) 9 1 2 3 4 5 6 7 8 9 1 2 3 ((4)). Fifteen Step's.
- For Point:5 ,(8) 7 6 5 4 3 2 1 9 8 7 ((6)). Twelve Step's
- For Point:5 ,(2) 1 9 8 7 6 5 4 3 2 1 9 8 7 ((6)). Fifteen Step's.
- For Point:5 ,(2) 3 4 5 6 7 8 9 1 2 3 ((4)). Twelve Step's.
Point5 has two subtypes (5w4w6) and (5w6w4) so in a sense at some level, Point5 has two types four wings.. or one energy point:5 two types 5w4w6 and 5w6w4, four wings. So now 54 becomes a number of interest. Fifty four step's inclusive. Or one hundred and eight and if we include a value of one for the Point number in this case "5" we can have a number value of 109 and 109 is the smallest prime number over 100.
Below is a set of diagrams and if we take two of the three as like a multiplicity, two of three parts we could think about the special prime number "11".
..................... 11
....... 11 ....................... 11
11 ....................................... 11
. 11 .................................... 11
............ 11 ............. 11
,
..................... 1
....... 1 ......................... 1
1 .........................................1
10 .................................. 10
.......... 1 ................ 1
..................... 10
....... 10 ....................... 10
10 ....................................... 10
.. 1 ..................................... 1
............ 10 ............. 10
Three multiplied by five equals fifteen.
Three multiplied by four eques twelve.
5+4= 9
15+12= 27
..................... 0
........ 0 ...................... 0
0 ....................................... 0
10 ................................. 10
........... 0 ............... 0
..................... 0
........ 0 ....................... 0
0 ........................................ 0
6 .......... Multiplicity ..........3
............ 0 .............. 0
10+6+10+3= 29
..................... 1
......... 1 .................... 1
1 ...................................... 1
10 ................................... 10
............ 1 ............. 1
29+7= 36 , 3+6= 9.
27+6+3+9+8+7+5+4+2+1= 72
..................... 10
....... 10 ....................... 10
10 ....................................... 10
. 1 ..................................... 1
............ 10 ............. 10