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Which closed space curves of constant curvature are there?

Two families spring to mind: circles and closed helices around a torus (both having - in addition - constant torsion).

What other families of curves of constant curvature are there?

Are there especially knots of constant curvature? Can - for example - the trefoil knot be realized by a curve of constant curvature?

Added: If it were possible to wind a curve around a cylinder, and to bend the cylinder back to itself - while keeping the curvature of the helix constant by appropriate adjustments -, thus forming an eventually arbitrarily knotted torus with a closed helix on it, one could have every knot as a curve of constant curvature. The paper Knots of Constant Curvature might be related to this point of view. It is interesting, but it is not exactly what I am looking for.

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    $\begingroup$ There are some formulas (which I don't remember) ... given funtions "curvature" and "torsion" of variable "arc length", to compute (up to orthogonal transformation) the curve. Then we need to ask when this result is periodic (with constant curvature). $\endgroup$
    – GEdgar
    Feb 20 '13 at 22:23
  • $\begingroup$ The curvature and torsion determine a curve up to rigid motion. Therefore, the curves of constant curvature and torsion are helices, circles and lines. A helix-type curve on a torus is not one of these. $\endgroup$
    – user53153
    Feb 20 '13 at 22:46
  • $\begingroup$ @5pm. It maybe was a misguided "guess by intuition". I removed the remark, anyway. But I still cannot "see" immediately why the curvature and torsion of a well-adjusted helix on a torus can not be constant. $\endgroup$ Feb 20 '13 at 22:59
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    $\begingroup$ It would be interesting to ask this question for curves in $\mathbb{S}^3$ as well. There, for example, you can realize torus knots as geodesics (by embedding $t\mapsto \frac{1}{\sqrt{2}}(e^{ipt},e^{iqt})$). $\endgroup$
    – Neal
    Feb 20 '13 at 23:00
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    $\begingroup$ @Neal: It certainly would be interesting. But I am not so familiar with $S^3$ (having almost no geometric intuitions). Anyway: I appreciate your hint on realizations of torus knots as geodesics! (Watching some torus knots in $R^3$, cycling around the torus, I just wondered whether they could be geodesics...) $\endgroup$ Feb 20 '13 at 23:02
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Can - for example - the trefoil knot be realized by a curve of constant curvature?

Here is a numerically computed solution in the affirmative.

constant curvature trefoil knot

It is a $C^2$ curve with constant curvature $\kappa=1$ and piecewise constant torsion, made by gluing together six pieces of helices. Three of the pieces have torsion $\tau_1=-0.3$ and arc length $2s_1\approx4.13159$, and the other three have torsion $\tau_2=2$ and arc length $2s_2\approx0.67712$. The image below shows the six pieces, along with the $xy$ plane illustrating the three-fold symmetry about $z$. More details follow.

the six pieces

Starting from $x_0=(0,0,0)$, we can define a helix $\gamma_1$ with tangent $(0,\cos\theta_0,\sin\theta_0)$, normal $(-1,0,0)$, curvature $\kappa$, and torsion $\tau_1=-0.3$ (chosen somewhat arbitrarily). After arc length $s_1$, we terminate $\gamma_1$ and continue another helix $\gamma_2$ from that point, with matching tangent, normal, and curvature, but with different torsion $\tau_2=2$. Say this meets the $xy$ plane after an arc length $s_2$, at which point we terminate it. This is one "monomer" which we will repeat six times to construct the whole curve. At each end, the curve can be extended by rotating it a half turn about the normal vector at the end point.

monomer

Suppose the position, tangent, and normal at the end of $\gamma_2$ are $x_2$, $t_2$, and $n_2$ respectively, which are functions of $\theta_0$, $s_1$, and $s_2$. For the whole curve to be $C^2$ continuous and to be closed after six repetitions, we just require that (i) $x_2$ lies on the $xy$ plane, (ii) the projection of $t_2$ to the $xy$ plane has rotated $2\pi/3$ radians from its original direction, and (iii) $n_2$ has no component in $z$. That is, $$x_2\cdot(0,0,1)=0,\\t_2\cdot(\cos\tfrac{2\pi}3,\sin\tfrac{2\pi}3,0)=0,\\n_2\cdot(0,0,1)=0.$$ These are three equations in the three variables $\theta_0$, $s_1$, and $s_2$, and can be solved numerically. For the chosen values of $\tau_1$ and $\tau_2$, I got $\theta_0 \approx 0.605278$, $s_1 \approx 2.0658$, $s2 \approx 0.33856$. I picked the torsions fairly large so that the knot is easy to see; with still larger values, the numerical method failed to find a solution.

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  • $\begingroup$ (+100) but I wish I could have the Mathematica codes.;-) $\endgroup$
    – Mikasa
    Jun 4 '13 at 12:41

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