The angle in the mystery number, corrected for 1/f

Take three point charges in free space. One has 1/3 charge, the other two have1/3, opposite charge. They have an L1 spot, where all forces balance. They never meet but they never escape.  Move them Cos^2 theta apart, and they have a new L1 spot on space.  The angle between the small charge is theta.  That particle has also spit the triangle, and the angle is 15, or pi/12.
2* Cos(PI/12)^2 -1,  which is, 15 degrees, for the single splits his his degrees of separation. Cos(15 degrees) + 1/Cos(15 d) = , 2.268, 1/2 + sqrt(3) = 1/(1/2+sqrt(3)) . So they say 22.7, the rotten rounders.

Any way, I love it and it explains a lot, whether you like bubbles of rubber bands. It makes the quarks work, so naturaly I plugged it in has a wave quant ratio, and sure enough it went quark wild, opening up six wave slots across the peak in my spectrum.  So, yes, let me repeat, it dont matter if you believe in rubber bands of bubbles, the only different is management of the grid, so the barbarion bubbles can bash from quant in space

Now my prediction for 1+sqrt(2) fell short, it sorta did what I expected, but nothing special came out, except it likes quants in half units sometime. But the sqrt two quant will get stuck at the electron level, its just too redundant, and traps at the electron. So, Markov five, the sqrt(221)/5 made it to the peak with one great number.  The bigger the Lagrange, the farther they travel, but the match probability drops.

Ok, puzzle. Which came first the 221 or the sqrt(3)?

So Mr Guy with the 22.7, nice work, great puzzle, and I did not buy the article, sorry, the puzzle was too fun.

Anyway, those roots get a z = e^ (i  * theta), in the Reimann discrete spiral system. i is complex numbers, z is two dimensional, having: D and d, d = D+1. i is cyclic over two Pi. D + i*k] is a valid location, in the map.. We can make the complex log work as well as we want, multiply, add subtract, all within their bounds. It is discrete, quantized.

 He looks too serious, dishevel that hair, be nutty.