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.)
 A: 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 $1.$
A: 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 $C(K)=O(cr(K))$. 
A:        
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.
