# Circle-arc number of a knot

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:

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:

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:

(The black triangle vertices indicate the circle centers on the plane before their arcs are twisted into 3D.)

• 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

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 $1.$
• [As you note, her $C^2$ condition is different from my $C^1$ condition.] – Joseph O'Rourke Aug 10 '12 at 21:51
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 $C(K)=O(cr(K))$.