Going Around In Circle

 The presence of $q$ does not guarantee that $q+2$ is prime.

where the bisector of arc $2(q+2)$ intersects arc $2b$ at a integer marking.  A line through the 1st marking on arc $2b$ intersect arc $2(q+2)$ at the $a^{th}$ mark.

$q+2=a*b$

if $a$ is not an integer neither $a$ nor $b$ is a factor of $q+2$.

$q$ is prime means none of the 1st intercept lines (lines through the first mark and the center of the arc) below it, cross it at an integer marking.

The first mark intercept lines approaches the vertical as the arc value approach $2q$.

The first mark intercept from arc $2q$ will be very close to the first mark of $2(q+2)$.

The widest 1st mark comes from $q=3$ or an arc of length $6$.  If this 1st intercept line crosses on $R_a$ on arc $2(q+2)$then the integer marks between the first mark on arc $2(q+2)$ and $R_a$ will have no intercepts from all first mark intercept from arcs below it, when $q+2$ is prime.

The arc $2q=6$ divides the right angle in $\cfrac{\pi}{12}$ this angle extends a width,

$\cfrac{\pi}{12}*\cfrac{2}{\pi}2(q+2)=\cfrac{1}{3}(q+2)$

on arc $2(q+2)$.

So, $q+2$ is prime only if there are no intercepts from its third mark to the $\cfrac{1}{3}(q+2)$ mark due to all 1st mark intercept lines below it.

This narrows the range from which a factor of $q+2$ can be found, to a number up to $\cfrac{1}{3}(q+2)$ after considering the number $3$.

All numbers above arc $b=3$ will mark, with a 1st intercept line, to the left of the 1st intercept line from $b=3$, and narrows the range from which a factor can be found.

Only $b$ that are odd are considered.

The 1st intercept steps in decreasing step angle given by $\cfrac{\pi}{4b}$.

The step angle size of $q$ is $\cfrac{\pi}{4q}$.  If, an integer marking at $n$, intercepts with the first intercept line from arc $b$.

$n\cfrac{\pi}{4q}=\cfrac{\pi}{4b}$

then     $q=nb$

then $q$ has a factor $n$, and $b$.

There is no further information to indicate $q+2$ to be prime.