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In a comment to my recent question about covering segments by a disk, 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?

In a comment to my recent question about covering segments by a disk, 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?

In a comment to my recent question about covering segments by a disk, 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 $n$?

In a comment to my recent question about covering segments by a disk, 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?

In a comment to my recent question about covering segments by a disk, 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 must be possible to do much better for large $n$, as the arcs to cover become tiny. In fact, it seems reasonable to expect $\lim\limits_{n\to\infty} r_{min}=0$.

If that is true, at what rate will $r_{min}$ decay? Or what are good upper bounds for large $n$?

In a comment to my recent question about covering segments by a disk, 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 $n$?