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 circle, from which $n$ equidistant points are removed. What we want is thus to pack the $n$ remaining arcs into a minimal disk.  
(EDIT) To keep the scaling similar to the one used before, we'll require the endpoints of each arc to have distance $1$.  
That doesn't seem very different from what Gerhard suggested as variant to start with: removing the midpoints of each segment of an $n$-gon, which amounts to 'bending' each segment in the middle slightly by $\frac{2\pi}n$ (so it is replaced by the legs of an angle $\pi-\frac{2\pi}n$). Note that in either case, we cannot cover all pieces by a half unit circle anymore (a nice fact allowing the general constructions in the original thread).

For $n=4$, if the four arcs are assembled in the shape of a windmill (shutter), we can pack it at best into a disk with $r=\dfrac{\sqrt{5}-1}{2}$. I think this construction is best possible for $n=4$.   

What happens as $n$ grows? I'm still working on the trig for the general $n$-fold shutter construction, thinking it should tend towards $r=1$, but there might as well be a better construction, like in the case of straight segments. So:
 
> What are good constructions for packing those $n$ arcs into a disk?
 
 

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