This page states "The logarithmic spiral can be constructed from equally spaced rays by starting at a point along one ray, and drawing the perpendicular to a neighboring ray... As the number of rays approaches infinity, the sequence of segments approaches the smooth logarithmic spiral"
My question is, if you change the angle you draw to the neighboring ray, does this still create a logarithmic spiral? If it does, what is the relationship between the angle you choose and the equation for a logarithmic spiral $r= ae^{b\theta}$. I assume changing the angle would change the a and b arbitrary constants?
Edit 2024-03-24 Found this online tool that demonstrates what my question is about. https://www.geogebra.org/m/DbQxZyN7#material/PRRbry9T
Yes, in some sense it still creates segments which look like a logarithmic spiral.
I'm too lazy to draw a picture, but I'll try to describe things
First, draw $n$ equispaced rays from the origin (denoted $O$), and draw a point $P_0$ on the $0$-th ray that is a distance $r_0$ away from the origin.
Now, for every integer $k = 0,1,\ldots$ draw a line segment from $P_k$ at an angle $\phi$ relative to the ray that $P_k$ is on and stop when you hit a point $P_{k+1}$ on the next ray. This means that $\angle OP_kP_{k+1} = \phi$. Also, since $P_k$ and $P_{k+1}$ are on adjacent rays, $\angle P_kOP_{k+1} = \tfrac{2\pi}{n}$. Since angles of a triangle add up to $\pi$, we have $\angle P_kP_{k+1}O = \pi-\tfrac{2\pi}{n}-\phi$.
Then, by using the law of sines on $\Delta P_kOP_{k+1}$, we have $$OP_{k+1} = OP_k \dfrac{\sin(\angle OP_kP_{k+1})}{\sin(\angle P_kP_{k+1}O)} = OP_k \dfrac{\sin(\pi-\tfrac{2\pi}{n}-\phi)}{\sin(\phi)} = OP_k \dfrac{\sin(\phi+\tfrac{2\pi}{n})}{\sin(\phi)}$$
Since $OP_0 = r_0$, we can use induction to show that the distance from the origin to $P_k$ is $$OP_k = r_0\left(\dfrac{\sin(\phi+\tfrac{2\pi}{n})}{\sin(\phi)}\right)^k$$
This shows that the points are getting exponentially further away from the origin as $k$ increases, which is like a logarithmic spiral.
In fact, if we orient the points such that first point $P_0$ is at polar coordinates $(r_0,0)$ and the direction of the spiral is counterclockwise, then the polar coordinates $(r_k,\theta_k)$ of the $k$-th point satisfy $$r_k = r_0\left(\dfrac{\sin(\phi+\tfrac{2\pi}{n})}{\sin(\phi)}\right)^k \quad \text{and} \quad \theta_k = \dfrac{2\pi k}{n}$$ which means that $$r_k = r_0\left(\dfrac{\sin(\phi+\tfrac{2\pi}{n})}{\sin(\phi)}\right)^{\dfrac{n\theta_k}{2\pi}}$$ Hence, all the points lie on the logarithmic spiral $r = r_0e^{b\theta}$ where $b = \dfrac{n}{2\pi}\log\left(\dfrac{\sin(\phi+\tfrac{2\pi}{n})}{\sin(\phi)}\right)$