mjhjmtu

_{[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 $U = 2pi R = L$ (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

What you want can be achieved using circle arcs, centered at $(0,r)$, of radius $r$ and central angle $2pi R/r$, 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:
$$
x=pi Rover 2thetasin2theta,quad y=pi Rover 2thetaleft(1-cos2thetaright),
$$
where $theta$ is the polar angle.
This also gives the simple polar equation:
$$
rho=pi Rsin thetaovertheta,quad 0lethetalepiover2.
$$

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 $f_t:[-pi, pi]tomathbbR^2$ where $tin[0,1]$ by
beginalign
f_t(s) &= left(fraccos(s(1-t))-11-t, fracsin(s(1-t))1-tright)&t<1\
f_1(s) &= (0, s)&
endalign

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).