# Curves of constant curvature on an ellipsoid

It is not difficult to see that the curves of constant geodesic curvature on a geometric sphere are all circles: simple, closed curves that are geometric circles lying in a plane:

My question is:

What are the curves of constant (positive) geodesic curvature on an ellipsoid?

Earlier Dmitri Panov asked a more general MO question, "Curves of constant curvature on $S^2$." My question is more specific. I would be interested to know how large is the class of simple (non-self-intersecting), closed, constant-curvature curves on an ellipsoid, whether there are nonsimple closed curves, whether there are infinitely long curves, etc. Dmitri's question revealed that many general questions are open, but perhaps there has been a special study made of the ellipsoid? I have so far not found any literature specifically on this. Thanks for pointers!

Addendum. I've made a few experiments which suggest that simple, closed curves of constant curvature might not be uncommon. My calculations were incorrect---Sorry to mislead!

Further edit. I've now rewritten the calculations, which (I think!) are now correct within numerical accuracy. Here is one section of a constant-curvature curve on the ellipsoid $\frac{x^2}{2^2}+y^2+z^2=1$:

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do you want a geometric picture or a mathematical formula? – Will Sawin Jul 20 '12 at 14:58
@Will: A geometric picture! – Joseph O'Rourke Jul 20 '12 at 17:59
In the degenerate case of a cylinder, you can clearly get non-simple closed curves curves. Draw a circle on a piece of paper and roll the paper up into a tube so that it self-intersects. For a prolate ellipsoid whose long axis is long compared to the scale set by the curve's curvature, you should get the same behavior. – Ben Crowell Jul 22 '12 at 20:19
It's strongly counterintuitive to me that randomly chosen initial conditions would lead with nonzero probability to a simple closed curve. For the curves you showed, did you choose initial conditions that had special symmetry? Did you verify with high precision, or only visually, that they returned to the same point with the same tangent vector? – Ben Crowell Jul 23 '12 at 5:07
@Ben: You were right to be suspicious! My method of computing the angular turn at each point of the curve was incorrect. I agree that closure should not be so prevalent. – Joseph O'Rourke Jul 23 '12 at 11:57