Involutary Spirals

Suppose you have a chain with infinitely many flat (i.e., one-dimensional) links. The first link has length 1, and the length of each successive link is a fraction $f$ of the previous link’s length. As you might expect, $f$ is less than $1.$ You place the chain flat on a table and some ink at the very end of the chain (i.e., the end with the infinitesimal links).

Initially, the chain forms a straight line segment, and the longest link is fixed in place. From there, the links are constrained to move in a very specific way: The angle between each chain and the next, smaller link is always the same throughout the chain. For example, if the $N$th link and the $N+1$st link form a $40$ degree clockwise angle, then so do the $N+1$st link and the $N+2$nd link.

After you move the chain around as much as you can, what shape is drawn by the ink that was at the tail end of the chain?

While the problem already provides for flexibility in terms of the value of $f \in (0,1),$ we will generalize the fixed angle between successive links to $\theta \in (0, 2\pi).$ Let's first figure out the points on the one-dimensional chain where the ends of each link are depending on the values of $f$ and $\theta.$

Assume that $X_0 = (0,0)$ and that $X_1 = (1,0)$ is the end of the first chain link of unit length. The end of the next link will be at $X_2 = (1+f\cos\theta, f\sin\theta),$ or in complex notation $X_2 = 1 + fe^{i\theta}.$ Similarly, since the end of the third link and the second link will have an angle of $\theta$ between them, it will make an angle of 2\theta with the positive real axis. Also, since the length of the third link is $f^2,$ we have $X_3 = 1 + fe^{i\theta} + f^2e^{i2\theta}.$ So on and so forth, we eventually get $$X_n = 1 + fe^{i\theta} + \cdots + f^{n-1}e^{i(n-1)\theta} = \sum_{k=0}^{n-1} (fe^{i\theta})^k = \frac{1 - f^n e^{in\theta}}{1- fe^{i\theta}}.$$

Depending on the values of $f$ and $\theta,$ the sequence $X(f,\theta) = \{X_n(f,\theta)\}_{n \in \mathbb{N}}$ has slightly different properties but is generally a sequence that spirals around its limit of $X_\infty(f,\theta) = \frac{1}{1-fe^{i\theta}} \in \mathbb{C}.$ The figure below shows the final position of the chain for several values of $f$ and $\theta$ in orange.

Now by construction, we are constructing involutes of these piecewise linear tracings of these complex sequences. Denote $Y_n = Y_n(f,\theta) \in \mathbb{C}$ denote the end of the chain after the completion of $n$ rotations/chain links. We start with $Y_0 = \frac{1}{1-f} \in (1,\infty).$ When we fix the point $X_1,$ the remaining link length of the remaining chaing to be rotated is $\frac{1}{1-f} -1 = \frac{f}{1-f},$ so after rotating through an angle of $\theta$ with the positive real axis, we have $$Y_1 = X_1 + \frac{f}{1-f} e^{i\theta}.$$ After fixing the point $X_2 = 1 + fe^{i\theta}$ the remaining chain to be rotated is $\frac{1}{1-f} - 1 -f = \frac{f^2}{1-f}$, so after rotation through another angle of $\theta,$ we have $$Y_2 = X_2 + \frac{f^2}{1-f} e^{i2\theta}.$$ Continuing the process we have $$Y_n(f,\theta) = X_n(f,\theta) + \frac{f^n}{1-f} e^{in\theta},$$ for each $n \in \mathbb{N}.$

We can obviously extend the process to fill in the gaps continuously between the sequence of points $Y(f,\theta) = \{Y_n(f,\theta)\}_{n\in \mathbb{N}}.$ Namely, let $\tilde{Y}(\cdot \mid f,\theta) : [0, \infty) \to \mathbb{C}$ denote the point where the ink is (at the end of the chain) at time $t \geq 0,$ as you are wrapping it around the sequence $X(f,\theta),$ such that $\tilde{Y}(t \mid f, \theta) = Y_t(f,\theta)$ for each $t \in \mathbb{N}.$ In particular, we have $$\tilde{Y}(t \mid f, \theta) = \frac{1 - (fe^{i\theta})^{\lceil t\rceil}}{1-fe^{i\theta}} + \frac{f^{\lceil t\rceil}}{1-f} e^{it\theta}$$ which is shown in the figure above shows the mapping $\tilde{Y}$ in blue for various values of $f$ and $\theta$.