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A taut string can be unwound from base circle $r_{base}$ in a plane. The successive string positions are geodesics and the orthogonal trajectories themselves are involutes:

$$ r \sin \psi = r_{base};\,r \cos \psi = r_{base} ; $$

where $r$ is radius and $\psi$ the angle a geodesic or involute makes to the meridian.

The first is Clairaut's geodesic relation and second I propose now for general involutes.

It appears that a thread can be generalized to get unwound not only on a plane but also on all surfaces of revolution to form an orthogonal net of involutes/ geodesics.

Please help find the orthogonal net parameterization with relevant comments.

This can be verified for example on popcorn/ice-cream cup cones for physical realization. I had verified the net physically by unwinding a taut thread on some spheres along with their cutting great circles as well ... I have numerically calculated them separately for an easy Cone case on arc basis and its image is attached:

It is possible to choose any arbitrary surface of revolution desired. (But on a sufficiently convex surface to avoid telegraphing).

Involutes & Geodesics on a Cone

EDIT1:

Differentiating Clairaut constant $r \sin \psi$ we get geodesic curvature $ (\phi = $ slope of meridian).

$$ \kappa_{geodesic} = \psi^{'}+ \sin \phi \sin \psi/r = 0 $$

Similarly by differentiating constant $r \cos \psi$ I define involute curvature in $\mathbb R^3$ as

$$ \kappa_{involute} = \psi^{'}- \frac {\sin \phi \cos^{2}\psi}{r \sin\psi } =0 $$

and these definitions have been used in numerical integration forming the above image.

Both these constants $ r_{base} $ are isometric invariants seen on the top of image.

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