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I've looked around and maybe I'm just searching for the right terms to find the answer. I know that the impedance of ladder line has to do with the distance between wires:
What is the generalized form to calculate feedline impedance between two parallel wires separated by some distance?
Does velocity factor or wire radius affect that impedance?
Yes, wire diameter matters. Assuming the two wires are round and of equal diameter, the impedance is $\frac{120}{\sqrt{\epsilon_r}} \mathrm{acosh}\,\frac{D}{d}$, where $\epsilon_r$ is the relative permittivity (dielectric constant) of the material between the wires, $D$ is the distance between wire centers, and $d$ is the diameter of the wires.
In cases where the spacing is significantly greater than the diameter, you can use the simpler formula $\frac{120}{\sqrt{\epsilon_r}} \ln \frac{2D}{d}$. This has less than 1% error when $\frac{D}{d}$ is more than 3.6.
The velocity factor is simply $\frac{1}{\sqrt{\epsilon_r}}$, as user10489 points out.
Estimating $\epsilon_r$ can be tricky; for ladder line you need to figure something based on the material and the "fill factor" (how much of the space between the wires is occupied by plastic and how much of it is air). The goal of "open wire" feedline is to reduce this fill factor to nearly 0, which makes $\epsilon_r$ decrease to nearly 1, and VF increase to nearly 1.
The dielectric constant of polyethylene is around 2.3; the $\epsilon_r$ of ladder lines made with PE webbing is between around 1.2 and 1.55, based on published velocity factors between 0.8 and 0.91.
The ARRL Handbook and the ARRL Antenna Book both have an entire chapter on transmission lines. However, the chapter in the Antenna book is more detailed.
Velocity factor is $$VF = \frac{1}{\sqrt\epsilon}$$ where $\epsilon$ is the dielectric constant of the material between the wires.
