How to map a line into circle Is there any transformation mapping a line of length l into a circle of radius r such that $l = 2\pi*r$?
 A: My attempt:
Consider a circle on a sphere which is passing through a the North Pole of the sphere.
Use stereographic projection technique, to map every point on a circle (other than north pole) to a plane. Image of these points lying on the straight line. Also, it is a bijection.
(But there is one point on the circle that is North Pole does not map with any point).
Edit:
Assume the radius of sphere =$r$.
North pole= $(0,0,0)$
A: Yes, and here’s one way to do this. Assume the line is on the $x$ axis. First map the domain of the line to $[0,2\pi]$ by enforcing an affine map $f$. Then  map $[0,2\pi]$ to the circle of radius $r$ using the map $g: \theta \to (r \cos \theta, r \sin \theta)$. The composition $g \cdot f$ does the trick.
A: Simple geometric approach: do the projection from point on circle the farthest from line. E.g.,  let your line be x axis, and your circle be a unit circle with center at [0, 0.5]. Now draw the line from point [0, 1] of the circle through every other point of that circle until it intersects with x axis, and you "unfolded" circle into the line. Reverse the process and you "fold" it back into circle.
A: $$ s= r \theta , \theta =s/r $$
For a full closed circle radius R of perimeter length $ 2 \pi R $ at each arc length $s$..
$$ x=R \sin (s/R); y=  R (1-\cos (s/R)) $$
We start with a flat line, left end fixed at the origin. We change center angle as well as the radius of curvature preserving arc length product same. Done by increasing subtended angle at  arc / sector center and decreasing radius of curvature.
The map is a limited "path isometry," a mapping function could be defined..
The bent arc ( the tip locus traces out a cardoid like curve not shown here.)
Keeping the same arc-length through factor p, $ l =  2 \pi r, r=1/p $ is maintained constant at each bent configuration as shown:
EDIT1:
Mathematica code includes animation of bending phenomenon, forwards and backwards. (line curls up to circle and flattens back to a line).
$$x=(r/p) \sin (s/r); y=  (r/p) (1-\cos (s/r)); 0<p<1 ; $$
Btw,in Mechanics of materials it is a realization of Euler- Bernoulli bending Law $ EI \kappa = M $ where bending moment M increases curvature $ \kappa$, EI being the elastic constant/ bending rigidity.

