In a [comment to my recent question about covering segments by a disk][1], Gerhard Paseman has suggested a generalisation: replacing the segments of the original $n$-gon by a simple closed (say, convex) curve, of which a certain number of points are removed. **How to pack the pieces into a minimal disk without rotation or overlap?**  

Of course this "curve puzzle problem" (as I'd like to call it) opens a bunch of possibilities. So I'd suggest to start with a unit circle, from which $n$ equidistant points are removed. What we want is thus to pack the $n$ remaining arcs into a minimal disk.  

For small $n$ at least, that seems tricky. For $n=4$, if the four arcs are assembled in the form of a windmill, we can pack it at best into a disk with $r=\sqrt{\dfrac{35-6\sqrt{10}}{20}}\approx0.895$. I think this construction is best possible for $n=4$.  

What happens as $n$ grows?
 
At first glance, the tractrix construction (*warning: nobody knows yet whether it is asymptotically best possible for covering the unit segments of all directions!*) will also approximately yield a packing for the arcs, but at second glance, it is possible to do much better for large $n$, as the arcs to cover become tiny. In fact, we have $\lim\limits_{n\to\infty} r_{min}=0$. E.g. for even $n\ge6$, the arcs can be arranged in a flower-like shape fitting in a disk of radius $2\sin\frac{\pi}n$. 
 
> Is $r_{min}\sim\dfrac{2\pi} n$? Or are there better upper bounds for large $n$?


  [1]: http://mathoverflow.net/questions/222536/the-smallest-disk-containing-all-sides-of-an-n-gon?noredirect=1#comment551578_222536