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I would like to build knots in $\mathbb{R}^3$ from arcs of unit-radius (planar) circles, joined together at points where the tangents match. Thus the knot will have curvature $1$ at all but the joints. Here is an example of how two arcs might join:
                Joined Arcs

Define the circle-arc number $C(K)$ of a knot $K$ as the fewest number of such arcs from which one can build a nonselfinterecting curve in space representing $K$. This number is analogous to the stick number of a knot, except that the pieces are arcs, and there is a tangent-joining condition.

I would be interested to learn of bounds on $C(K)$ in terms of other knot quantities, for example, the stick number, or the crossing number cr$(K)$.

Here is an example of what I have in mind. It appears that one might be able to build a trefoil from six arcs, something like this:
                Trefoil Circle Arcs
However, the above picture is actually planar, and I have not verified carefully that this is achievable in $\mathbb{R}^3$!

Has this concept been studied before? If so, pointers would be welcomed. Thanks!

Addendum. The trefoil can be realized with six arcs:
      Trefoil: 6 Arcs
(The black triangle vertices indicate the circle centers on the plane before their arcs are twisted into 3D.)

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In your planar diagram above, I suspect you might acheive 9 arcs by splitting each big arc directly (perhaps approximately) in half. Gerhard "Think Of Those Magic Rings" Paseman, 2012.08.10 –  Gerhard Paseman Aug 11 '12 at 0:03
    
How do you make such wonderful pictures, Joe? Is there a resource for this? (is it Mathematica?) –  Jon Bannon Aug 23 '12 at 15:49
    
@Jon: Thanks! :-) Yes, that one was produced via Mathematica, with a little post-Photoshop (because what I can get out of Mathematica directly is lower quality). –  Joseph O'Rourke Aug 23 '12 at 15:54
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3 Answers 3

Not exactly your question, but in this paper, (knots of constant curvature, Jenelle McAtee, 2004), the author shows that every knot can be representated as a $C^2$ curve of constant curvature, so if you don't insists on planar curvature arcs, the answer is that your number equals $1.$

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@Igor: Thanks for this reference, of which I was unaware! But the number is not 1, unless I misunderstand (which is quite possible!). She shows the number is finite, if helices are included. The number I defined is the number of distinct arcs. A number of 1 means just the trivial knot represented as a geometric circle. –  Joseph O'Rourke Aug 10 '12 at 21:46
    
[As you note, her $C^2$ condition is different from my $C^1$ condition.] –  Joseph O'Rourke Aug 10 '12 at 21:51
    
@Joseph: you do misunderstand what I am saying (my fault). You are asking for PLANAR arcs of constant curvature one. What I am saying is that you remove the planarity condition, you only need a single arc of constant curvature 1. But in fact, if you read her construction (she uses the braid representation of the knot, you can tweak her argument to get your planar arcs...) –  Igor Rivin Aug 10 '12 at 23:59
    
@Igor: Ah, I see your point! Apologies for misunderstanding! –  Joseph O'Rourke Aug 11 '12 at 0:28
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You can bound this number from below by the crossing number. The projection of the arcs of two unit circles can cross in at most two points, so $cr(K)\leq C(K)(C(K)-1)$. Also, you ought to be able to bound it quadratically from above by the grid number. If you have a knot presented by a grid diagram, you can represent the knot by a linear number of segments of linear length. Each one of these can be made into a linear number of arcs of unit circles by putting in wiggles. Since grid number is bounded above linearly by crossing number, one obtains inequalities of the form $$ \sqrt{cr(K)}\leq C(K)\leq A\cdot cr(K)^2$$ for some constant $A$. Notice that the grid number can sometimes be like $O(\sqrt{cr(K)})$, so I don't expect the upper bound to be sharp. For example, for certain torus knots you'll have $C(K)=O(cr(K))$.

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Brilliant! Thanks so much! –  Joseph O'Rourke Aug 16 '12 at 23:12
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       McAtee
This figure is from the paper by Jenelle McAtee to which Igor refers. Her results are impressive! Her goals are worthy but rather different from mine.

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