Making a quark matrix

That list of numbers from my spread sheet of the 1/2 + sqrt(3) factors in the spectral chart is the quark matrix. Quarks are finite complex polynomial in discrete Reimann space with b = log(1+Q) being the scalar exponent of the generator. Then  q = base^b-1. generates the matrix.  So, q is the set of indices, {0,1,..k} such that q^i/baud, for i in k, is the ith value of the complex polynomial in discrete space.  Baud is the set of all ordered indices in the finite number line.


So, a quick review. The Afimov state completes out picture, more or less. We have a ring of yardstick, Y, comprising the characteristic roots:  
1/2 + sqrt(5)/2, 1+sqrt(2), and 1/2 + sqrt(3), over the finite number line baud, baud = Log(Y+1), are indices of the finite number line. The  number line is of dimension 3, having three spectral modes. The fourth spectral mode, curvature of the unit sphere, is 3/2 + 3*sqrt(221)/10, which makes motion.

We can use 0 = Null, and we are orthogonal, or we can say 0 > small e, and we have coupling constants. Thus the sample rate of multiply, c, is 1/e.
A conic on the number line baud, obeys the rules of hyperbolics.  So, sinh(q)+cosh(q) is the band width of q, and sinh(q)^2+cosh(q)^2 is the power spectra, I think (check me on this).  The quarks are pi/6, at 1/2 sqrt(3). How that works as a function, I am not sure, but it counts three times, n*pi/3 *k + root(1/2+sqrt(3)), I have not worked all this out. The professional mathematicians will finish up the theorems. Many thanks to Weinberg, Higgs, Yang-Mills, and the whole team of the standard model.

The counting pattern of the quarks factors, without the 1+sqrt(2). Leaving out the spinners, makes this chart a bit distorted, but these are the spectral paints created at the peak of the proton by the root 1/2 +sqrt(3). Putting the spinners back in removes about half those peaks, but leaves more room for the remaining. The number after 30 are is the Gluon spectral space, I am sure.



This is how we do with a the ring of yardsticks.