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In extension of this phi research:
I have been trying to ascertain for some time what proportions a VBM torus should be. Randy and I have had many discussions, and tried a whole variety of ideas, making visualisations in 3d to see if resonances that accord with the simplicity of the maths can occur.
I have found one way of applying proportions to tori of different scale maps that accepts one rule as it's premise. It is a mathematical experiment which leads to a very specific conclusion. But I am not trying to promote this as THE solution to the proportions issue. But these Tori fit nicely together with totally exact phi ratio proportions, and this has a certain beauty to it which is worth checking out in any case. Its only in mod9 at the moment, so I will need to check mod25 and mod49 sometime. But I tried so so many calculations before landing on phi as the being the correct proportions.
The rule? This is that each diamond on the grid surface of a torus should be equal in surface area to every other diamond. And in every torus (all of which layer up with an onion skin/russian doll effect resembling the electromagnetic field of a planet) this diamond area will always remain the same.
check out some of the data here:
http://vortexspace.org/display/~tom+barnett/Phi+VBM+Tori+Discoveries
T
25,49,121,…..
the ones that will work will be prime squares, maybe even cubes?
here is a list of what might work, 25,49,121,169,289,361,529,625,841,961,1225,1369,1681,…...
1/4(6n-(-1)^n+3)^2 is the equation that will generate this sequence. 744711744711744711
Notice that each number you listed above (25, 49, 121, 169, etc...) is a multiple of 12, + 1. Your formula above is also centered around 6n. 6n (+ or -) 1 produces a set of numbers containing the set of all prime numbers. All primes in hexadecimal end in 1 or 5, and all primes in duodecimal end in 1, 5, 7, or (11). It is useful to look at this stuff in base 12. In addition (pardon the pun), the 24 digit repeating pattern in the Fibonacci series in base 12 (I am using "a" for 10 and "b" for 11) is 0,1,1,2,3,5,8,1,9,a,7,5, and the second half -- 0, 5, 5, a, 3, 1, 4, 5, 9, 2, b, 1. Notice that the only number with which no Fibonacci number ends in base 12 is 6, however superimposed and added together (which would be the same as writing the digits in a circle and adding together the numbers positioned opposite to each other), the first and second half form the pattern 0, 6, 6, 0, 6, 6, 0, 6, 6, etc...
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