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'Twas the night before Christmas, and throughout the net
Not a question was stirring, at least---not yet.
The tree was all draped with garlands and lights,
With presents beneath for morning delights.

As I enjoyed this warm-hearted scene,
A question arose as if in a dream:
The tree is a cone, the garlands are O's.
I wonder how many can fit of those?

Has this been studied? Can you shed some light?
Merry Christmas to all! And to all a good night!


                Cone4View1


Q1. Have packings of congruent disks on (finite) cones been explored?

Notation. Let $C$ be a (finite) right circular cone. A generator of $C$ is a segment connnecting the apex $a$ to a point on the rim of the base. Let $\alpha$ be the angle of the cone at $a$ when cut open along a generator; so $\alpha$ is the total surface angle incident to $a$. Let $L$ be the length of a generator. The surface area of $C$ is $A=A(L,\alpha)=\pi L^2 (\alpha/(2 \pi))$.

Q2. For a fixed surface area $A$, which cone angle $\alpha$ permits the most unit-radius disks to be packed on $C$?

In the example above, $L= 3 \sqrt{2} \approx 4.14$ and $\alpha = \frac{2}{3} \pi = 120^\circ$; so $A=6 \pi$, which is the area of $6$ unit-radius disks, although I only packed $4$ disks (and so achieved a density of $\delta=\frac{2}{3}$).

As $\alpha \to 0$, no unit-radius disk can fit, and for $\alpha = 2 \pi$, we are packing disks in a circle of radius $L$, a heavily studied problem. In the absence of an answer to Q2, let me suggest:

Q3. Can the behavior of the packing density along the $A(L,\alpha) = \mathrm{const}$ hyperbola-like curves be described qualitatively? E.g., is the density $\delta$ unimodal along those curves, perhaps for sufficiently large $A$?


                SurfaceAreaContours


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    $\begingroup$ MathOverflow — the new Ladies' Diary. [‟All mathematical Questions in The Diary were stated in verse until 1729.”] $\endgroup$ – Marius Kempe Dec 24 '14 at 17:20
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    $\begingroup$ On Q1. I heard about many different packing problems, but not about that. May be Erich Friedman knows more. Merry Christmas and Happy New Year! :-) $\endgroup$ – Alex Ravsky Dec 24 '14 at 17:25
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    $\begingroup$ There's also covering n disks by the smallest wedge. That point of view may lead to quick optimal results. $\endgroup$ – The Masked Avenger Dec 24 '14 at 18:35
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Circle packings have been considered for $\alpha=2\pi$, which is the planar case; in that case the densest packing are the incircles of a hexagonal mesh.

While the planar case is trivial, two non-planar cases with an exact answer can be derived from it: place the apex at a point, that is the common corner of three of the meshes' hexagons and cut out a sector of either $\frac{2\pi}{3}$ or $\frac{4\pi}{3}$ and glue together the remains along the cuts.

For arbitrary $\alpha$ one could take inspiration from nature's way of packing seeds in circular disks, the most prominent example being the arrangement of the sunflower's seeds.
In the planar case, the $n$-th circle center would be placed at $\left(x(n),y(n)\right) := \left(c*n*\phi\cos(n*\phi),c*n*\phi*\sin(n*\phi)\right)$,
where $\phi:=2\pi\left(1-\frac{\sqrt{5}-1}{2}\right), n\in\mathbb{N}, c\in\mathbb{R}^+$.
In the case of a cone, the planar coordinate would have to be augmented by a z-coordinate, that lifts the planar point to the cone.
If the density of the circle-centers shall be proportional to surface area, then the radius of the planar spiral must rather be proportional to the square root of the turning angle and not be an archimedean spiral.

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    $\begingroup$ "in that case the densest packing are the incircles of a hexagonal mesh": For the infinite plane, Yes. But disks packing in a circle are more complicated. E.g., here is 20 disks. So I think your remarks concerning $\frac{2}{3}\pi$ and $\frac{4}{3}\pi$, only hold for infinite cones, or in the limit $L \to \infty$. $\endgroup$ – Joseph O'Rourke Dec 25 '14 at 12:53
  • $\begingroup$ You are right, I had the infinite plane in mind; I should have mentioned that. $\endgroup$ – Manfred Weis Dec 25 '14 at 13:08

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