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Let $\cal{A}$ denote the family of all twice-differentiable simple open arcs $A$ in $\mathbb{R}^3$ satisfying the following properties:

(1) at each point of every arc $A\in\cal{A}$ the curvature of $A$ is at most $1$;

(2) no perpendicular projection of $A$ to any plane is an injection.

Question 1. What is the infimum of the length $|A|$ of $A\in\cal{A}$ ?

Question 2. Is there an arc $A_{min}$ of minimum length among all arcs in $\cal{A}$? If so, is it unique (up to isometry)?

Remark. This question is related to the notion of rope knots, see: Hans Stricker, Some questions about ideal knots

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In the absence of other ideas, I'll risk posting something that perhaps satisfies (2), and could be arranged to satisfy (1):

          OverUnder
If indeed this satisfies (2), then perhaps its length could approach $2 \pi$, by moving the over/under crossings close together, and closer to the top of the diagram.

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    $\begingroup$ This looks like a reasonable conjecture. $\endgroup$
    – Wlodek Kuperberg
    Commented Sep 6, 2018 at 12:56
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    $\begingroup$ The two "end parts" of the arc can be made very short, yet wrap around the "main part" many times, which would make the example even more convincing. The hard part would be proving that in $\mathcal{C}$ must be longer than $2\pi$. $\endgroup$
    – Wlodek Kuperberg
    Commented Sep 7, 2018 at 21:57
  • $\begingroup$ @WlodekKuperberg: Nice idea to wrap several times! Then (2) is clear. That $\cal{C}=\cal{A}$ must be longer than $2\pi$ almost seems obvious, but I don't see a clear proof. $\endgroup$
    – Joseph O'Rourke
    Commented Sep 7, 2018 at 22:25
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This doesn't seem close to tight, but there's an easy lower bound of $\pi$. If you have a curve shorter than $\pi$, choose any projection direction perpendicular to the curve at its midpoint. The resulting projected curve will be monotonic and therefore non-self-intersecting.

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    $\begingroup$ A wild guess: $\sqrt{4 \pi} = 2 \sqrt{\pi} \approx 3.5 > \pi$ might be a lowerbound, because one needs to block all $4 \pi$ surface of a unit sphere of projection directions, and each crossing in projection identifies two points on the curve. $\endgroup$
    – Joseph O'Rourke
    Commented Sep 15, 2018 at 1:56
  • $\begingroup$ What is a monotonic curve? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 15, 2018 at 5:43
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    $\begingroup$ One that can be projected onto a line with no self-intersections. In this case the projection direction is again the one perpendicular to the projected curve at the (former) midpoint. You can tell that the projection onto a line has no self-intersections because the tangent vector can't turn far enough to be projected to zero. $\endgroup$
    – David Eppstein
    Commented Sep 15, 2018 at 5:46

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