So I have no way how to properly explain this. I have no idea how to solve this problem.
Given the following:
In a 3D world, we have a normalized vector $V(x_v, y_v, z_v)$, a starting point $S(x_s, y_s, z_s)$ and a radius r, as illustrated in the following image, and the vector $V$ is on the same plane as the entire orange circle with middle point $S$ and radius r, and given that the dotted line below is always parallel to the "floor", so the two points of our blue triangle on the orange circle have an equal y value.:
Now given that the angle of alpha is 60 degrees, s.t. the blue triangle becomes an equilateral triangle with sides of length r.
What I want, is the following, and I honestly have no clue how:
Given n concentric arcs between the two lines of the blue triangle, equally spaced s.t. $r = n \cdot d$.
Can I somehow get the function of each of those arcs somehow?
I know this question was asked so incredibly vague that it might not even make sense, but I tried my hardest to explain this the best i could.
The radius of each arc is $r_k=\frac kn r$ where $k$ ranges from $1$ to $n$. The points can be calculated by taking intervals of $\theta$ from $0$ to $\alpha$ as $x=r_k \cos (\frac {\pi-\alpha}2+\theta),y=r_k \sin (\frac {\pi-\alpha}2+\theta)$, the standard parameterization of a circle.
Added: the vertex of your angle is assumed to be the origin of the coordinates. For 3D you can restrict $\theta$ to $[0,\alpha /2]$, which will give you the point in the $xy$ plane with $x \ge 0$. Now you can revolve it around the $y$ axis. The $y$ values are all the same. Given the $x$ we calculated for a given $\theta$ the point is $(x',y,z)$ with $x'=x \cos \phi, z=x \sin \phi$ as $\phi$ ranges from $0$ to $2\pi$ The shape is no longer an arc, which is what led me to the 2D solution. It is now a spherical cap.