The short question is: how exactly is SU(3) realized with ropes?

The long question: There is this idea that deformations of a configuration of three infinitely long, flexible ropes that cross each other can be mapped to the eight Gell-Matrices, the generators of SU(3).

SU(3) appears in the quantum harmonic oscillator and in qutrits. So a visualization of SU(3) with ropes is useful to quantum information theory.

The SU(3) idea is mentioned in http://arxiv.org/abs/0905.3905 on page 35. It seems that the idea started in this way: deformations of configurations of TWO ropes reproduce Dirac's string trick and behave like the Pauli matrices of SU(2). Deformations of configurations of ONE rope reproduce U(1).

THREE ropes apparently yield a relation between eight different versions of the third Reidemeister move and the eight Gell-Mann matrices. But the paper is too terse for me to see the relation in detail. A literature search does not bring up anything related to this idea. And I got no answer to my email.

Can anybody help to understand the details?

Added points:

Peter's answer below mentions a relation between the braid group $B_3$ and SU(3). A Google search does not yield anything about this topic. Can anybody provide a reference?

A graph similar to that of Joseph's answer below is also part of the paper. But I am not interested in QCD or unification: I'd like to understand how the deformations in that graph yield or correspond to the Gell-Mann matrices $\lambda_1$ to $\lambda_8$. I can see that the deformations of the graph correspond to $F_1$, $F_2$ and $F_3$, where $F_i=e^{i \pi \lambda_i / 2}$. This gives $F_1^4=F_2^4=F_3^4=1$, the unit matrix, as it should. Also the SU(2) subgroup is generated as it should. Next, the $\lambda_8$ deformation behaves as expected. But I cannot see (so far) that $\lambda_1 \lambda_4= \lambda_6/2 + i \lambda_7/2$. Can anybody provide a hint?