Mathematics and the art of linearizing the circle

[I edited the question and put stronger emphasis on "constant curvature" than on "naturalness".]

One of the most prominent problems of ancient mathematics was the squaring of the circle: to construct the square with the same area as a given circle.

A related problem is linearizing the circle: to find a natural transition between a given line segment of length $$L$$ (having constant curvature $$0 = 1/\infty$$) and the circle with circumference $$U = 2\pi R = L$$ (with constant curvature $$1/R$$) going through circle segments of length $$L$$ (with intermediate constant curvatures $$1/R'$$, $$\infty > R' > R$$).

The question is: Along which paths do the points of the line segment move to finally yield the circle?

This is how the transition looks like:

The points of the line segments follow these paths:

as can be seen here:

To be honest: Even though these paths look very much like circle segments, I'm not quite sure and I didn't define them by an explicit formula (which I didn't have at hand) but heuristically using some support points and splining.

My questions are:

• Are these paths really circle segments?

• If so: How to parametrize them?

• If not so: What kind of curves are they otherwise?

Please allow me – freely associating – to compare the pictures above with this (artificially symmetrized) picture of The Great Wave off Kanagawa

• Do we have an actual definition of what it means for this transition to "appear natural"? – Morgan Rodgers Feb 27 at 17:56
• I think it would be "natural" to think of a family of curves $\gamma_t$, $t\in[0,1]$, of constant curvature such that $\gamma_0$ is the flat line (curvature 0), $\gamma_1$ is the circle (curvature 1/R) and each $\gamma_t$ has curvature $t\cdot1/R$. – Mars Plastic Feb 27 at 18:06
• I think you could probably make a map that "appears natural" using circles for the paths, and you could also probably make a different map without using circles that also "appears natural". So without a definition of what that term means, I don't know what answer you are looking for. – Morgan Rodgers Feb 27 at 18:08
• Maybe I'm missing something, but this 'transition' seems quite arbitrary to me. – rafa11111 Feb 27 at 18:20
• You ask about making it "appear natural", but it's not clear what this means. You have things that look like circle segments, and you ask if they "really are circle segments", but we just have a picture, so how can we tell? – Morgan Rodgers Feb 27 at 18:34

What you want can be achieved using circle arcs, centered at $$(0,r)$$, of radius $$r$$ and central angle $$2\pi R/r$$, with $$r$$ varying between $$1$$ and $$+\infty$$. But I don't know if that is "natural" or not. Here's how it looks:

EDIT.

Each endpoint of the arc describes a spiral curve, given by: $$x={\pi R\over 2\theta}\sin{2\theta},\quad y={\pi R\over 2\theta}\left(1-\cos{2\theta}\right),$$ where $$\theta$$ is the polar angle. This also gives the simple polar equation: $$\rho={\pi R}{\sin \theta\over\theta},\quad 0\le\theta\le{\pi\over2}.$$

• That's essentially my approach. Good to know, that an expert like you came to the same conclusion. Concerning "natural": which approach could be more "natural"? – Hans-Peter Stricker Feb 27 at 19:23
• How (and with which tools) did you create your elegant animated gif? (I had a hard time with mine.) – Hans-Peter Stricker Feb 27 at 19:45
• I used GeoGebra, which is free: www.geogebra.org – Aretino Feb 27 at 20:20
• Can you share this plot with me, i.e. save the plot and share the link? – Hans-Peter Stricker Feb 27 at 20:49
• The path of the endpoints is some kind of spiral: $$\left(\pi{\sin t\over t},\pi{1-\cos t\over t}\right).$$ – Aretino Feb 27 at 21:20

If we want this transition to have all the points on the boundary of a circle at all times, then it makes most sense to parameterize by the radius of this circle (and apply a transformation to get it in terms of finite time later). For simplicity, I will also have the transition be to a vertical line.

We shall have the radius of the circle $$C_r$$ be $$r$$ and centre be $$(-r,0)$$, such that $$(0,0)$$ is on $$C_r$$ for all $$r$$. The coordinates of the point at arclength $$s$$ from $$(0,0)$$ are then given by $$(r (\cos(s/r) - 1), r \sin(s/r))$$.

The most natural way to transition would likely be varying the curvature at a constant rate; thus we create a family of curves $$f_t:[-\pi, \pi]\to\mathbb{R}^2$$ where $$t\in[0,1]$$ by \begin{align} f_t(s) &= \left(\frac{\cos(s(1-t))-1}{1-t}, \frac{\sin(s(1-t))}{1-t}\right)&t<1\\ f_1(s) &= (0, s)& \end{align}

Since the goals here seem to be rather subjective, I would attempt this and see how it looks to you (beyond making substitutions as needed to result in a horizontal line).

• Note that the paths of the endpoints are given by setting $s = \pm \pi$ in your parametric curve. The resulting paths are not circle segments (as the OP guessed they might be); this can be seen by calculating their curvature. – Michael Seifert Feb 27 at 21:57