Cubes, squares and Fibonacci shifts

Why is there a 1/2 in the Golden Ratio?  Mainly because the recurrence relation is a shift once and add, and if you look at the derivation, you will see the 1/2 is needed to avoid a coefficient in the relationship: R^2 -R-1 = 0.

So, the canonical digits (Phi)^N look like :

2^(-N) * (1+root(5))^N, and can be put further into:

2^(-N) * 2^(N*log2(1+root5)) or 2^[N(log2(1+root(5))-1)], if you like twos binary.

So, the 2 results from sampling the shift, once for each shift. This Nyquist sampling is sampling to be orthogonal to  phase error introduced by the shift, which has phase noise falling at twice the quant rate.

Anyway, I bring this up because there is a cubic being solved, in the proton to generate the quarks.  It is mainly related to curvature of the unit circle, it is too curved. So phase which cannot meet the curvature fly off, going Gamma, as we say.  Call it quantization noise.  The unit circles that survive meet some minimum curvature requirement, mainkly satisfyin a relationship in R^3 that keeps curvature optimum. So the system has this perfect number with solutions that are interleaved base three and base two. Somewhere, in getting this 2/3 charge we ended up with a shift three times in the Fibonacci series.

Physicists call a unit circle which is too curved a unit circle which curves faster than light travels. I call it a curve that curves faster than light samples.

This ruler, it gets expanded and accommodates the base three curvature management. Its maximum entropy when the available bits finite