Sphere Packings in $\mathbf{2}\frac{\mathbf{1}}{\mathbf{2}}$ Dimensions

Abstract

This paper investigates cylindrical sphere packings, that is, patterns of uniform spheres with mutually disjoint interiors which are all tangent to a common cylinder. The key unifying themes are the existence and uniqueness of hexagonal packings, in which each sphere is tangent to six others. Constructions are both intuitive and subtle, but result in the complete characterization in terms of integer parameter pairs $(m,n)$. Interesting questions in rigidity and density are encountered. Density questions arise because the packings, being of equal diameter, lie within the space between inner and outer cylinders. This density problem hovers between the 2D and 3D sphere packing cases, and though it is not solved here, it is conjectured that the hexagonal packings are densest for the countable number of cylinders which support them. Other geometric objects are along for the ride, including equilateral triangles and the packings’ dual graphs, which are associated with patterns of carbon atoms forming buckytubes. Interesting structural rigidity questions also arise.

1. Introduction

This paper concerns sphere packings on cylinders in three-dimensional space. That is, infinite configurations of congruent spheres with mutually disjoint interiors which are all tangent to the outside of a given cylinder. Figure 1 will set the scene with hexagonal packings, which play a key role in our work; the labels by pairs of integers will be explained shortly.

The objects involved here, spheres and cylinders, are so familiar that the reader is encouraged to imagine hands-on constructions—or actually do hands-on constructions, as the author has enjoyed with undergraduate students. Getting started is only slightly more difficult than it would appear: Figure 1 illustrates hexagonal packings, in which each sphere is tangent to six others. These play a central role in our story here, and despite their simple and pleasing visual nature, characterizing these packings presents engaging challenges. Combinatorics enters in a crucial way, with abstract triangulations representing patterns of sphere tangency. These can be realized concretely using yet another familiar, manipulable object, the flat equilateral triangle. Figure 2 illustrates the triangulations behind three of the hexagonal packings in Figure 1; they are realized as cylindrical polyhedra formed by equilateral triangles, each associated with a triple of mutually tangent spheres.

Figure 2 also illustrates what nature produces with its own manipulations: paired with the polyhedra are the familiar patterns of buckytubes, representing carbon atoms and their bonds. The combinatorics of buckytubes are in one-to-one correspondence with the duals of triangulations of hexagonal sphere packings. Our interests here are geometric, but there may well be concrete applications in the study of carbon structures, which are of great interest in physics and material science. The packings themselves are of scientific interest, as they are common in the modeling of many natural columnar structures, from plant cells to foams, crystals, and nanoparticles. The term “sphere packing” will, for many readers, raise the issue of packing density. In our case, the spheres have the same diameter, so they lie in the space between two cylinders. How much of the volume can they occupy? Our discussion of density questions shows that there is much work yet to be done with these cylindrical sphere packings.

Overview

The geometric objects involved here—cylinders, spheres, triangles—are so familiar that this paper will begin with a heuristic approach that might convey the sense of discovery which physicists enjoy but which all too often is obscured in mathematical writing. In Part A, we construct sphere packings with a hexagonal-themed approach—one that a reader with the right maker skills might even realize concretely. Applying thought experiments, visualization, and intuition, we reach a characterization of all cylindrical packings with hexagonal combinatorics. Part B continues the heuristic approach with a second construction of hexagonal packings using equilateral triangles. Although this does not provide a new proof of existence and uniqueness, it raises interesting questions in a major related topic, rigidity theory. In Part C, we then move to discuss packing density, a topic with a long mathematical history. That leaves Part D of the paper, which is mathematically rigorous; we fill in the formal, computational, and at times annoying details which justify our earlier heuristics.

2. Part A: Hexagonal Packings

Let us set the environment for our constructions. We work in ${\mathbb{R}}^3$ with cylindrical coordinates $(r,\theta ,z)$. Cyl$\left(r\right)$ will denote a cylinder of radius $r\ge 0$ whose central axis is the z-axis. This will be the inner cylinder in packings, and one attaches spheres of diameter 1 with mutually disjoint interiors tangent to this inner cylinder. The result is called a Cylindrical Sphere Packing (CSP) on Cyl$\left(r\right)$. Spheres tangent to the outside of Cyl$\left(r\right)$ will also be tangent to the inside of the outer cylinder Cyl$(r+1)$. The region between these is called the solid cylinder for r and is the setting for the density issue discussed in Part C. There is a crucial additional cylinder involved, the mid cylinder Cyl$(r+1/2)$. This contains the centers of the spheres of the packing; since it enters nearly all our computations, we denote its radius by $R=r+1/2$.

2.1. A.1. The Construction

There are myriad ways to build CSP’s, but we will concentrate on a systematic approach based on hexagonal flowers. For a given cylinder Cyl$\left(r\right)$, start by attaching the “base” sphere $S_0$ at the point $(r,0,0)$; in most images, $S_0$ will appear as pale red, while the other spheres will be light blue. Attach a sphere $S_1$ to Cyl$\left(r\right)$ so that it is tangent to $S_0$, and then another, $S_2$, which is tangent to both $S_0$ and $S_1$. (By convention, $S_2$ is chosen so the triple $\{S_0,S_1,S_2\}$ of mutually tangent spheres is counterclockwise oriented as viewed from outside the cylinder.) As we will see shortly, other spheres are added as the construction proceeds. Figure 3 shows the situation we have reached at this point. In the picture, the Euclidean triangle formed by the centers of these three spheres is obscured; it is equilateral with unit edge lengths.

Now, continue construction counterclockwise around $S_0$, sphere-by-sphere: whenever a tangent triple $\{S_0,S_j,S_{j+1}\}$ is already in place, compute $S_{j+2}$ to get the next tangent triple $\{S_0,S_{j+1},S_{j+2}\}$.

A Pleasant Surprise: When you work your way around and are ready to position $S_7$, you will find that it precisely matches $S_1$! You have completed a hex flower for $S_0$—a pattern of 6 successively tangent spheres with mutually disjoint interiors all tangent to and surrounding the central sphere $S_0$.

This closure property applies to the completion of hex flowers about any central sphere. We will prove closure in Part D, but accepting that for now, the final construction stage is a recursive process to be carried out ad infinitum: (1) Among the spheres already placed on Cyl$\left(r\right)$ at a given stage, find some tangent pair which is missing one of its shared neighbors. (2) Compute the location and put that new sphere in place. (3) Return to step (1) and repeat.

It may seem at first that one can complete a hex flower for every sphere, leading to a hexagonal CSP. But perhaps the reader has already anticipated an obstruction.

An Unpleasant Surprise? As the pattern of added spheres grows outward from $S_0$, those being added to the right around the cylinder might not be compatible with those added to the left. There might be spheres whose hex flowers cannot be completed because the next tangent sphere to be added would improperly overlap a sphere already in place.

Indeed, this is the typical situation: we will see that the process will generically fail to produce a hexagonal sphere packing. Figure 4 illustrates two failed attempts starting with the same parameters: everything looks good on the front, but when you peek around the back, you see that things do not fit. The ragged gap is what you might expect; the more pleasant gap along a helix is called a line-slip, and we will say more about it later.

We will have a lot to say about this construction, but let us start with issues that may have occurred to the reader. First is that the radius r might be too small for Cyl$\left(r\right)$ to support even a single hex flower, much less a hexagonal packing. That is the case, and we establish the positive lower bound $r_{\min }\sim 0.01963$ in §D.1. Second is that due to the choices in the reiterated step (1), the resulting configuration may not be unique (see Figure 4). In fact, it will be unique if and only if it is a hexagonal packing. Our task in the next section is to identify those isolated cases in which a hexagonal packing emerges.

2.2. A.2. The Hexagonal Cases

When does this construction process succeed in building a hexagonal CSP (denoted as an hCSP)? We approach this question by observing success and discovering the key parameters. All our reasoning will be proven in Part D. So let us assume we encounter a configuration $\mathcal{S}$ of spheres tangent to Cyl$\left(r\right)$ in which every sphere is the center of a hex flower, as with the examples in Figure 1. There is an unmistakable visual uniformity within each hCSP. There are strings of spheres forming spirals around the cylinder—indeed, there are three families of spirals (two typically stand out visually while the third may be harder to see). All the spirals from any one family will fill out the full packing. Examples are illustrated in Figure 5.

Focus on any spiral and note its regularity: the centers $(R,\theta ,z)$ actually lie along a helical arc, hinting at an underlying isometry: combining a translation in the z direction with an associated rotation about the z-axis gives an isometry of ${\mathbb{R}}^3$ which maps each sphere of that spiral to the next. In other words, in a given spiral path, the heights z and angles $\theta$ of successive sphere centers change by constant amounts. This key observation leads us to a particular closed string of tangent spheres of special interest.

The base sphere $S_0$ of $\mathcal{S}$ has a flower with a counterclockwise list $\{S_1,S_2,\cdots ,S_6\}$ of petals. Necessarily one of these petals, say $S_1$, has height $z_1$ less than or equal to that of $S_0$, while its counterclockwise neighbor, $S_2$, has height $z_2$ greater than $S_0$. Start taking steps from $S_0$ along the spiral through $S_0$ and $S_1$. There are two possibilities:

(1)

After taking some number m of steps you have returned to $S_0$.

(2)

After taking m steps, you find that an additional n steps along another spiral (namely, one parallel to that through $S_0$ and $S_2$) will bring you back to $S_0$.

In either case, the result is a closed string of spheres with disjoint interiors which starts and ends at $S_0$. We will call this an $(m,n)$-necklace and we refer to the full packing $\mathcal{S}$ as an $(m,n)$-hCSP. (In case (1), we take $n=0$.)

As one might guess from studying Figure 6, $\mathcal{S}$ is just an infinite stack of copies of its necklace. We exploit that in the proof of the following theorem, which shows that the pairs $(m,n)$ serve to parameterize all hCSP’s.

Theorem 1.

For every pair $(m,n)$ of integers with $m\ge n\ge 0$ and $m+n\ge 3$ there is an essentially unique radius r whose cylinder Cyl$\left(r\right)$ supports an $(m,n)$-hCSP. This hCSP is unique up to isometries of ${\mathbb{R}}^3$. Moreover, every hCSP is an $(m,n)$-hCSP for some such pair $(m,n)$.

Our proof is based on observations and a dynamic thought experiment; formal justifications are in Part D, along with details on how to compute the relevant parameters.

Proof. 

From the given m and n, we first need to identify appropriate geometric parameters. Suppose $\mathcal{S}$ is a hCSP with $(m,n)$-necklace. Using our earlier observations, the positions of the first two spheres of the necklace, $S_0$ and $S_1$, are enough to reconstruct the entire packing. We can arrange that $S_0$ is centered on the mid cylinder at $(R,0,0)$ and $S_1$ is centered at $(R,{\theta }_1,z_1)$. Recall that $R=r+1/2$ and that $z_1\le 0$. The only geometric parameters we need, then, are $r=r(m,n)$ and $\zeta =\zeta (m,n)$, with $\zeta$ playing the role of $z_1$. We may assume henceforth that $m\ge n$; if not, one can reflect the sphere packing in the $(x,y)$-plane to get a CSP with the roles of m and n interchanged.

Two Special Cases: For cases $n=0$ and $n=m$ we compute explicit r and $\zeta$ values. When $n=0$, we have $m\ge 3$ and an $(m,0)$-hCSP will have a necklace of m spheres that stretches horizontally around Cyl$\left(r\right)$. The edges connecting necklace centers forms a regular unit-sided p-gon inscribed in a circle of radius R. An easy computation gives R and we conclude

$$\mathbf{For} \mathbf{an} (m,0)\mathbf{-}\mathbf{hCSP}\mathbf{:}r(m,0)={\displaystyle \frac{1}{2sin\left(\frac{\pi }{m}\right)}}-1/2,\zeta (m,0)=0,m\ge 3.$$

(1)

For an $(m,m)$-hCSP, we see from symmetry in the $(x,y)$-plane that the sphere $S_1$ and its tangent neighbor $S_2$ will have centers directly above one another, implying $\zeta =-1/2$. The sphere S neighboring $S_1$ and $S_2$ and opposite $S_0$ will have height 0. The line from the center of $S_0$ to the tangency point of $S_1$ and $S_2$ has length $\sqrt{3}/2$, as does the line from there to the center of S. Replicating around the cylinder, there will be $2m$ of these segments forming a regular $2m$-gon inscribed in a circle of radius R, and we conclude

$$\mathbf{For} \mathbf{an} (m,m)\mathbf{-}\mathbf{hCSP}\mathbf{:}r(m,m)={\displaystyle \frac{\sqrt{3}}{4sin\left(\frac{\pi }{2m}\right)}}-1/2,\zeta (m,m)=-1/2,m\ge 2.$$

(2)

The General Case: Assuming now that $m>n>0$, we consider a string of $m+n$ tangent spheres as though it were a string of pearls—a pearl necklace. We will be draping this necklace around a cylinder Cyl$\left(r\right)$. If r were equal to $r(m+n,0)$, as computed earlier, then the $m+n$ pearls would form a choker about Cyl$\left(r\right)$, a tight horizontal necklace. If r were any larger, the necklace could not reach around Cyl$\left(r\right)$. On the other hand, we know there is a smallest cylinder that supports a hexagonal flower. The r we need is somewhere between these extremes.

For our thought experiment, consider decreasing r from its maximum value and watching the behavior of the necklace draped on Cyl$\left(r\right)$. First, however, we must add some rigidity: Let S denote the $m^{\mathrm{th}}$ pearl along the necklace. For a given r, if you gently pull S downward, you will reach a point where it can physically go no lower. The tension in the necklace will then force the first m pearls to lie along a downward sloping geodesic—the shortest path along Cyl$\left(r\right)$ from $S_0$ to S. The remaining n pearls must lie along an upward-sloping geodesic, wrapping the rest of the way around from S to $S_0$. We call this particular necklace $(m,n)$-elegant. It is clearly unique for a given cylinder Cyl$\left(r\right)$. Since geodesic paths on cylinders are helical, this means that in an $(m,n)$-elegant necklace, the heights of the first m pearls drop in equal successive steps, which we will denote by $\zeta \le 0$, while the heights of the remaining n pearls rise in equal successive steps, denoted by $z_2$. It is elegant necklaces that we will work with from now on.

For each Cyl$\left(r\right)$, the lowest pearl S in its $(m,n)$-elegant necklace is at height $m·\zeta$. This lowest pearl will clearly drop in height as r decreases, so $\zeta$ becomes more negative as r decreases. But what are we watching for? If an elegant necklace were part of a global hexagonal sphere packing, then S would be the center of a hex flower of spheres in that packing. One of its petals would be the next sphere with height increment $\zeta$ (extending the spiral direction of the first m pearls), and its neighboring petal would be the pearl after S in the necklace, with height increment $z_2$. Look at the necklaces of Figure 7 to visualize the situation. For each of those $(7,4)$-elegant necklaces, we show a green sphere $S_{\mathrm{G}}$ tangent to S with height increment $\zeta$, while the blue sphere $S_{\mathrm{B}}$ is the ${(m+1)}^{\mathrm{st}}$ pearl of the necklace itself, tangent to S with height increment $z_2$. What you want to watch is the distance between the centers of $S_{\mathrm{G}}$ and $S_{\mathrm{B}}$. Denote this distance by $\delta =\delta \left(r\right)$. If this necklace were part of a hexagonal packing on Cyl$\left(r\right)$, then $S_{\mathrm{G}}$ and $S_{\mathrm{B}}$ would necessarily be tangent, so $\delta \left(r\right)=1$. This distance function $\delta$ provides our opening!

When r takes its maximum value, the necklace is a horizontal choker, $S_{\mathrm{G}}$ and $S_{\mathrm{B}}$ would be identical, and $\delta \left(r\right)=0$. However, as suggested in Figure 7, as r decreases, $S_{\mathrm{G}}$ and $S_{\mathrm{B}}$ move apart, so $\delta \left(r\right)$ increases. Indeed, if you visualize the necklace on progressively smaller cylinders, you will see it drooping until it eventually collapses—the pearls cannot remain tangent to the cylinder without beginning to overlap. At some point, the pearl before S and the pearl after S in the necklace will be tangent to one another, meaning $S_{\mathrm{G}}$ and $S_{\mathrm{B}}$ must be far apart and $\delta \left(r\right)>1$. As $\delta \left(r\right)$ is clearly a continuous function, the intermediate value theorem ensures some radius r for which $\delta \left(r\right)=1$ and $S_{\mathrm{G}}$ and $S_{\mathrm{B}}$ are tangent. For r, this necklace is called the $(m,n)$-Goldilocks necklace. Once you have the $(m,n)$-Goldilocks necklace, our earlier construction method lets you fill out the rest of an hCSP. Indeed, stacking translated and rotated copies of the necklace will give you the full packing on Cyl$\left(r\right)$.

A closer study of $\delta \left(r\right)$ shows that the radius r for which $\delta \left(r\right)=1$ is unique, so we will be justified in writing $r=r(m,n)$. The argument is presented in Part D, but is based on another bit of geometry: connecting the centers of $S,S_{\mathrm{G}},$ and $S_{\mathrm{B}}$ forms a triangle in three-dimensional space with two sides of unit length. If $\alpha$ denotes the angle at S in this triangle, then $S_{\mathrm{G}}$ and $S_{\mathrm{B}}$ will be tangent if and only if this triangle is equilateral, if and only if $\alpha =\pi /3$, if and only if $cos\left(\alpha \right)=1/2$. As it happens, as r decreases from its maximal value $r(m+n,0)$, the function $cos\left(\alpha \right(r\left)\right)$ is strictly decreasing. Therefore, it can take the value $1/2$ only once, namely, when $r=r(m,n)$, proving uniqueness.

Summarizing, given $m>n>0$, there is a single radius r whose $(m,n)$-elegant necklace is $(m,n)$-Goldilocks, and hence part of a $(m,n)$-hCSP on Cyl$\left(r\right)$. Conversely, we have observed that every CSP contains an $(m,n)$-elegant necklace for some m and n, and this is clearly $(m,n)$-Goldilocks. Thus, hCSP’s are (up to isometries) in one-to-one correspondence with integer pairs $(m,n)$, where $m\ge n\ge 0$. □

One consequence of this theorem is confirmation of our earlier comments: for all but countably many radii r, our construction of hex flowers on Cyl$\left(r\right)$ starting from a tangent pair $S_0,S_1$ will encounter an unpleasant surprise—hex flowers that cannot be completed. In Part D, we present a table containing the exceptional values $r(m,n)$, those that support hCSP’s for $m=2,3,\cdots ,13$. The reader is encouraged to see what can be gleaned from this table. This open question in particular may come to mind:

Open Question 1.

Does there exist a cylinder Cyl$\left(r\right)$ which supports an $(m,n)$-hCSP for two distinct pairs $(m,n)$, $m\ge n\ge 0$?

In Part D, we present, for parameters m and n, a non-linear system of two equations for r and $\zeta$. While the radii $r(m,0)$ and $r(m,m)$ are explicit, other radii are only obtained by numerically solving this system. That means this question may be difficult to address.

3. Part B: Rigidity

Viewing a cylinder as simply a rolled up plane, the reader may see other ways to construct hCSP’s. We discuss an appealing but naive approach that does not quite work. However, a slightly trickier one succeeds while also leading to interesting rigidity questions.

It is well known that the plane is the universal cover of the cylinder. We will formalize this in Part D, but as an analogy, picture rolling a cardboard tube (the cylinder) so it picks up wrapping paper (the plane) to make it easier to store. One ends up with several sheets of paper over every point of the tube, but identifying sheets over the same points as a single layer leaves just a cylinder.

With this image in mind, the reader is offered this thought experiment: Start with a close-packed hexagonal lattice of uniform spheres resting on the plane. (This is just one of the planar sheets of spheres within a cannonball packing of space.) Consider Cyl$\left(\rho \right)$ as an infinite rolling pin that has glue on it. Start rolling across the tops of the spheres, each sphere adhering to the rolling pin as it passes by. In your mind’s eye, are you starting to see the emergence of a CSP?

Of course, there is a difficulty we have faced before: As you keep rolling along, the spheres already adhering to the rolling pin will start to encounter new spheres still in the lattice. If the direction in which we are rolling and the radius of the rolling pin are just right, each attached sphere that encounters a new sphere will match it identically, so they just merge into one, and we continue rolling along. In this instance, have we realized a hCSP? I thought so when I first imagined this process—but I was wrong! Due to the curvature of the rolling pin, spheres tangent in the plane will become separated as they are picked up. So we do not get a sphere packing! Still, the attempt makes clear how sensitive the construction is to the radius of the cylinder and its rolling direction, and one can see in a natural way how different helical patterns of spheres arise.

Disappointing? Perhaps, but also suggestive. Let us unroll an existing hCSP $\mathcal{S}$ and see what we learn. What we unroll, however, is the mid cylinder, since it contains the sphere centers. The centers for each triple of tangent spheres in $\mathcal{S}$ define a unit-sided equilateral triangle, and these attach to one another to form a cylindrical polyhedron. The unrolling process is depicted in Figure 8.

Six triangles meet at every vertex of this polyhedron. We have cut it open along one of its spiral edge paths, and it opens up to lie perfectly flat in the plane; it becomes part of the familiar equilateral tiling of the plane, which we denote by $\mathcal{H}$. Note that $\mathcal{H}$ is not covering the cylinder itself, but rather the cylindrical polyhedron. Our unrolling defines a fundamental region $\Omega$ in $\mathcal{H}$, whose opposite edges are to be identified. In Figure 8 the red stars are points associated with the center of $S_0$; the vector $\overline{F}$ from one to the next generates the fundamental group of the covering. Especially note the blue edge path connecting successive stars: it has $m=4$ steps along one axis and $n=2$ along another, so $\mathcal{S}$ is an $(4,2)$-hCSP.

Now you may see a real opportunity to construct hCSP’s by reverse engineering this unrolling—though there are some complications. The procedure starts with integers $(m,n)$, and a fundamental vector $\overline{F}$ formed, as in Figure 8, with m steps along one axis and n along another in $\mathcal{H}$. Place a cylinder on the plane at the initial end of $\overline{F}$ and with axis perpendicular to $\overline{F}$. Roll it towards the terminal end, picking up vertices as it rolls along. If you have chosen a cylinder of the proper radius r, the beginning and end of $\overline{F}$ will find themselves at the same point on the cylinder. You have now built the polyhedron for an $(m,n)$-hCSP, and all that is left is to place a sphere at every vertex.

This construction is quite subtle, so the reader is encouraged to spend some time visualizing the mechanics. As the vertices are picked up and triangles are added to the growing polyhedron, the structure has to crinkle up because the triangles remain connected, yet must take their various tilts. It is not easy to picture: the vertices end up on the cylinder, but the faces, to remain flat and equilateral, find themselves magically inside. You might think at first that $2\pi r=|\overline{F}|$, but the crinkled nature of the polyhedron forces r to be larger.

In the $(m,0)$ and $(m,m)$ cases, the mechanics are easier to visualize (and explicit radii are already known). However, as mechanically attractive as this approach to hCSP’s is, there is little hope that one can prove existence and uniqueness for general $(m,n)$ via this route.

Nevertheless, these experiments lead to another broad topic: rigidity. For a general reference on rigidity, the reader is directed to [1]. Constrained to lie in the solid cylinder about Cyl$\left(r\right)$, the spheres of an hCSP are jammed—they are packed together too tightly to move. This is rigidity in a 2D sense. The more interesting issue is rigidity in the 3D sense. Picture the skeleton of an hCSP, the edges between tangent spheres, as a physical structure in space (though infinite in extent). Treat the edges as unit-length bars, each joined at its ends to five other bars via flexible joints. Is that structure mechanically rigid? That is, is this the only shape it can take? Figure 9 illustrates that the answer is negative for $(m,m)$-hCSP’s. For other $(m,n)$ pairs, however, the author finds it difficult to imagine any such flexibility.

Open Question 2.

Is the skeleton for an $(m,n)$-hCSP rigid when $n\ne m$?

Another aspect of rigidity is suggested in the following conjecture. This is motivated by a foundational theorem in the development of circle packing, a topic initiated by Bill Thurston [2] and largely distinct from sphere packing; see [3,4]. That theorem is this: Suppose $\mathcal{P}$ is a collection of circles in the plane having mutually disjoint interiors and a tangency graph that forms a hexagonal triangulation of the plane. Then all circles of $\mathcal{P}$ must have the same radius. I.e., $\mathcal{P}$ is a penny packing. This suggests the following conjecture, which we will not pursue here:

Conjecture 1.

Suppose $\mathcal{S}$ is a configuration of spheres tangent to a common cylinder having mutually disjoint interiors and a tangency graph that forms a hexagonal triangulation of the cylinder. Then, all spheres of $\mathcal{S}$ have the same radius.

In other words, up to a scaling factor, $\mathcal{S}$ must be one of our hCSP’s. For additional work motivating this conjecture, see the paper [5] about Doyle spiral circle packings.

4. Part C: Packing Density

The topic of “sphere packing” is surprisingly broad and has a mathematical history reaching back two millennia: what is the densest arrangement of congruent spheres in a given region, that is, the arrangement that takes up the greatest proportion of the volume? The two- and three-dimensional cases are most familiar. In the plane, the “penny packing”, a hexagonal arrangement of congruent discs, is easily proven to maximize area density. However, the step to dimension 3 is far from easy. In fact, the famous Kepler Conjecture of 1611 proposed that in three-dimensional space, the “cannonball packing” would maximize volume density. This was only proven in 1998 by Thomas Hales [6]. Again, the pattern has a hexagonal character because the spheres are arranged in planar sheets, like the penny packing of discs, which are then stacked atop one another. Mathematicians have, of course, pushed to higher dimensions, where the results are even more challenging. The density problem in dimensions greater than 3 remains open, except for dimensions 8 and 24, where the solutions earned Maryna Viazovska the Fields Medal in 2022. These packings provide the basis for important error correcting codes (see [7,8]).

For cylindrical sphere packings, the density issue—like the rigidity issue—hovers somewhere between the two- and three-dimensional cases because the spheres live in 3D space but hug a 2D surface. The open question can be said to concern density in 2 and 1/2 dimensions:

Open Question 3.

What sphere packing in the solid cylinder about Cyl$\left(r\right)$ occupies the greatest proportion of the volume?

Spheres tangent to Cyl$\left(r\right)$ are also tangent to the inside of Cyl$(r+1)$, and there has been research in the physics literature about the density of packing within cylinders. These questions arise in various natural and physical situations, such as cylinders containing cells, crystals, soft spheres, or foams. See [9,10,11] and works cited there. These studies often model systems using solid spheres, and their critical parameter is $D/d$, the ratio of the diameter of the cylinder to that of the spheres. For $D/d$ less than roughly $2.715$, simulations suggest that the densest packing will have all spheres tangent to the cylinder wall, while for larger values of $D/d$ there may be spheres interior to the cylinder. With our conventions, this $D/d$ cutoff converts to $r<0.3574$, and from Table 1 one can see that the $(2,1)$-, $(2,2)$-, $(3,2)$-, $(4,1)$-, and $(5,0)$-hCSP’s fall within this setting. The work of physicists has been largely based on simulations, both physical and computer-based, as well as numerical approximations and analytic searches. Though this work consistently shows that nature prefers the hexagonal motif, purely mathematical claims about density remain open.

Suppose $\mathcal{S}$ is a CSP on Cyl$\left(r\right)$. Since the solid cylinder about Cyl$\left(r\right)$ has infinite volume, a natural way to define density takes the limit of densities in truncated cylinders. (Density is also called “average density” or “packing fraction”.).

For $h>0$, let n be the number of spheres with centers having height z in $[-h,h]$. We then define the pre-density and density as

$${\mathrm{den}}_h\left(\mathcal{S}\right)=12h(2r+1)/n,\hspace{1em}\mathrm{den}\left(\mathcal{S}\right)=\underset{h\to \infty }{lim}{\mathrm{den}}_h\left(\mathcal{S}\right).$$

Only those having a limit are competitors for maximal density. But the limiting process is not without its problems. Given a CSP on Cyl$\left(r\right)$ with maximal density, one can randomly remove any billion spheres, and what is left has equal density. So, in practice, one searches for packings that follow a motif that seems to optimize pre-densities. The search process is tricky to justify, computing pre-densities is difficult, and proving maximality is even harder. With the classical 2D/3D cases in mind, along with nature’s apparent preferences, we are led to this conjecture. An affirmative answer, at least for small r, is posited in the physics literature [10,12], but a mathematical justification remains to be found.

Conjecture 2.

If Cyl$\left(r\right)$ supports a hexagonal cylindrical sphere packing $\mathcal{S}$, then $\mathcal{S}$ has the maximal density among all sphere packings on Cyl$\left(r\right)$.

There is one case in which we can affirm this conjecture, if you are willing to treat the plane as a cylinder with infinite radius. The solid cylinder is now the slab of space between planes $z=0$ and $z=1$. Maximal volume density for the hexagonal packing of spheres within this slab follows easily from the well-known maximal area density for the penny packing in the plane. The slab is the union of the Voronoi cells of the sphere centers, the Voronoi cell of a sphere center consisting of all points closer to that center than to any other. The global density is identical to the density of a sphere within its cell.

$${\displaystyle \frac{\mathrm{Sphere} \mathrm{volume}}{\mathrm{Cell} \mathrm{volume}}}={\displaystyle \frac{6/\left(4\sqrt{3}\right)}{\pi /6}}\sim 0.6045997881.$$

As $m+n$ grows and the cylinder Cyl$\left(r\right)$ associated with the $(m,n)$-hCSP $\mathcal{S}$ grows, the density of $\mathcal{S}$ will converge to this value, roughly $60\%$. I ask the reader to test their intuition: Is the density of an $(m,n)$-hCSP larger or smaller than $60\%$? The intuition of the author failed on this question, but the reader can settle the issue with the data provided in §D.5. As to proofs, §D.5. also presents some evidence for the $(m,m)$-hCSP cases of the Conjecture. The approach is limited, however, and even for these limited cases, the arguments remain incomplete.

As for the general situation, most cylinders Cyl$\left(r\right)$ do not support hCSP’s, and Open Question 3 becomes much more difficult. In the physics literature, the parameter $D/d$ is varied continuously in simulations, and hCSP’s emerge only for isolated values. For other values, hexagonal close packing still seems to be nature’s preference, but with spiral patterns of gaps—what are called line-slip packings. This Figure 10 illustrates cylinders of three different radii, so we are not comparing their densities. Rather, note that while the packing on the left would appear to be rigid, it has some spheres which are tangent to just five neighbors. The other two packings have open gaps and may be somewhat flexible. However, it appears that the middle packing, with its small gap, could not accommodate any additional spheres. On the other hand, one can conceive of adjustments to the packing on the right that would allow additional spheres and thus perhaps a larger density.

5. Part D: Computations

The developments and claims in Parts A, B, and C depended on the reader’s familiarity with spheres, cylinders, and triangles and on their ability to visualize and mentally carry out manipulations. This section provides mathematical rigor as we confirm various steps in these earlier heuristic arguments.

First, some general and standard background facts. Isometries of ${\mathbb{R}}^3$ are homeomorphisms which preserve Euclidean distance, hence they preserve cylinders, spheres, and flat triangles. In the sequel, we restrict attention to cylinders with central axis the z-axis and to isometries that map the z-axis to itself, and hence map CSP’s to CSP’s. They are generated by combinations of translations in the z-direction, rotations about the z-axis, and reflections in the coordinate planes. We can apply an isometry to move any sphere in a CSP so that its center lies on the positive x-axis, as we have done with the base sphere $S_0$. A useful observation for later is that given any two points on a common cylinder, there is an isometry which interchanges those two points.

5.1. D.1. Smallest Cylinder

Our definition of CSP allows the cylinder Cyl$\left(r\right)$ to have any radius $r\ge 0$. However, we are interested in hexagonal flowers. When $r=0$, the attached spheres must all touch the z-axis, and the best one can hope for is shown in Figure 11a. This may look promising until you note that flowers have only 5 petals, one exactly opposite the central sphere. Problems persist briefly as r gets larger. One might be led to the CSP of Figure 11b, where $r\sim 0.0303300857$ and $\zeta =-1/2$. Base $S_0$ and the spheres directly above and below do have hex flowers, but as the view from the back shows in Figure 11c, the other spheres have room for only five neighbors.

A close look at Figure 11c will note gaps that are ringed by four spheres. These suggest room for improvement. Decreasing $\zeta$ from the initial value $-1/2$ introduces a twist, and as you know if you have ever played with a rope—twisting makes it smaller around. Here, the twisting ends with the hex flower of Figure 11d, having $\zeta \sim -0.31619807367$. At this point, the opposite petals $S_1$ and $S_4$ are tangent to one another, so it is clear that we can not decrease r any further.

Lemma 1.

The minimum radius r for which Cyl$\left(r\right)$ supports a hex flower is $r\sim 0.0196290419$. Moreover, this hex flower extends to an hCSP, namely, the $(2,1)$-hCSP shown in Figure 1.

In the notation we have adopted, this minimal radius is denoted $r(2,1)$. Henceforth we assume our cylinders have radius $r\ge r(2,1)$.

5.2. D.2. Hex Flower Closure

It is evident physically when working with a real-life cylinder and real-life (uniform) spheres, that if you position a sphere, place a tangent petal sphere, and then successively add additional tangent petal spheres, a hex flower will result—that is, as you place the sixth sphere, it will be precisely tangent to the first sphere. We now prove this.

Lemma 2.

Suppose that $r\ge r(2,1)$, that S is a sphere attached to Cyl$\left(r\right)$, and that $S_1$ is a sphere tangent to S. If spheres $\{S_2,S_3,S_4,S_5,S_6,S_7\}$ are successively tangent spheres placed about S, then $S_7$ is identical to $S_1$.

Proof. 

Without loss of generality, we may assume that S is centered at $(R,0,0)$, while $S_j$ is centered at $(R,{\theta }_j,z_j),j=1,2,\dots ,7$. We will show that ${\theta }_4=-{\theta }_1$ and $z_4=-z_1$. Repeating that argument will show that ${\theta }_7=-{\theta }_4$ and $z_7=-z_4$, and hence $S_7$ is identical to $S_1$.

The argument depends on a study of equilateral triangles (all unit-sided in the following). Denote by $t_j$ the triangle formed by centers of the tangent triple $\{S,S_j,S_{j+1}\}$. Let $e_j$ be the line segment between (the centers of) S and $S_{j+1}$. This is an edge shared by $t_j$ and $t_{j+1}$, and it is evident that these are the only equilateral triangles with vertices on Cyl$\left(R\right)$ which have $e_j$ as an edge.

We refer to $t_{j+1}$ as the reflection of $t_j$ in $e_j$. The justification lies with an observation we made earlier: there is an isometry T that maps Cyl$\left(R\right)$ to itself and interchanges S and $S_{j+1}$. This clearly interchanges $t_j$ and $t_{j+1}$, and therefore $S_{j+2}=T\left(S_j\right)$. The change in angle (the $\theta$ coordinate) from $S_{j+1}$ to $S_{j+2}$ is the negative of the change in angle from S to $S_j$, which is ${\theta }_j$. We conclude that ${\theta }_{j+2}={\theta }_{j+1}-{\theta }_j$. Likewise with heights, $z_{j+2}=z_{j+1}-z_j$. Reflecting at times $t_1$ and $t_2$, we have the successive expressions

$${\theta }_3={\theta }_2-{\theta }_1,z_3=z_2-z_1,{\theta }_4={\theta }_3-{\theta }_2,z_4=z_3-z_2.$$

In consequence,

$${\theta }_4={\theta }_2-{\theta }_1-{\theta }_2=-{\theta }_1\mathrm{and}{z}_4=z_2-z_1-z_2=-z_1.$$

Repeating by reflecting in $t_4$ and then in $t_5$, we have ${\theta }_7=-{\theta }_4={\theta }_1$ and $z_7=-z_4=z_1$. In other words, $S_7$ equals $S_1$. □

This result confirms our observations about the three spiral paths within an hCSP; see Figure 5. In each path, one enters a flower via (the center of) one petal, passes through the central sphere, and exits via the opposite petal. Now we see that the increments in angle and height in each step are identical, implying that those centers lie on a common helix. (Of course, in an $(m,0)$-hCSP, one path degenerates to a closed ring, while in an $(m,m)$-hCSP, one path degenerates to an infinite vertical path.)

5.3. D.3. Universal Covers

It is well known that the plane is the universal cover for a cylinder. In studying sphere packings on a given cylinder Cyl$\left(r\right)$, the centers are given in cylindrical coordinates by $(R,\theta ,z)$, where $R=r+1/2$. Since only the coords $\theta$ and z vary, it is convenient to define a universal covering map ${\Phi }_R:(\theta ,z)\mapsto (R,\theta ,z)$ from the $(\theta ,z)$-plane onto the mid cylinder Cyl$\left(R\right)$. This is $2\pi$ periodic in the first coordinate, so ${\Phi }_R(\theta +2\pi ,z)={\Phi }_R(\theta ,z)$. A convenient fundamental domain is the strip $\left\{\right(\theta ,z):0\le \theta <2\pi \}$.

Using ${\Phi }_R$, spheres attached to Cyl$\left(r\right)$ may be identified with their “centers” as points in the $(\theta ,z)$-plane. Given a hex flower on Cyl$\left(r\right)$, geodesics between centers on Cyl$\left(R\right)$ lift to straight lines, so on the plane that hex flower defines three straight line segments. We are now in position to formalize the general hex construction undertaken in §A.2. by working in the $(\theta ,z)$-plane.

Constructions started with base sphere $S_0$ centered at $(0,0)$ and a tangent sphere $S_1$ centered at $({\theta }_1,z_1)$. We then computed $S_2$ tangent to $S_0$ and $S_1$, with center $({\theta }_2,z_2)$. Define the vectors $\overline{u}=\langle {\theta }_1,z_1\rangle$ from (the center of) $S_0$ to $S_1$, $\overline{v}=\langle {\theta }_2,z_2\rangle$ from $S_0$ to $S_2$, and $\overline{w}=\overline{v}-\overline{u}=\langle {\theta }_2-{\theta }_1,z_2-z_1\rangle$ from $S_1$ to $S_2$. The six centers about $(0,0)$ are in the directions $\{\overline{u},\overline{v},\overline{w},-\overline{u},-\overline{v},-\overline{w}\}$. The construction continues by successively adding neighbors to complete the hex flower of any existing sphere. Proceeding ad infinitum in the plane, we generate the hexagonal lattice

$$\Lambda =\{a\overline{u}+b\overline{v}:(a,b)\in {\mathbb{Z}}^2\}.$$

In the generic situation, projecting these points to centers on a mid cylinder Cyl$\left(R\right)$ will lead to distinct centers that are too close—their spheres would not have disjoint interiors. This can only be avoided if $\Lambda$ is $2\pi$-periodic.

Lemma 3.

Given Cyl$\left(r\right)$, tangent spheres $S_0,S_1$ generate an hCSP on Cyl$\left(r\right)$ if and only if the associated lattice Λ is $2\pi$-periodic, if and only if Λ contains $(2\pi ,0)$, if and only if there exist integers $m,n$ so that $m\overline{u}+n\overline{v}=(2\pi ,0)$.

Figure 12 illustrates the situation for the $(7,3)$-hCSP: • the lattice $\Lambda$ with edges for tangencies; • the shaded fundamental domain; • the spiral through $S_0,S_1$ in red; • the spiral through $S_0,S_2$ in green; • the spiral through $S_0,S_3$ in blue; • the covering map; • the projection to the mid cylinder Cyl$\left(R\right)$. In this figure, one can see a concrete meaning for m and n: the full lattice is formed by $m=7$ parallel copies of the green spiral or $n=3$ copies of the red spiral. Though harder to see, it is formed as well by $l=m+n=10$ copies of the blue spiral. In biological studies of plant growth, 3-tuples $(l,m,n)$ are known as parastichy numbers and arise in the context of phyllotaxis, the growth of spiral patterns in plants, such as sunflowers, pine cones, or corn (See [12,13]). We will stick with our $(m,n)$ notation.

A cautionary note about covering maps: ${\Phi }_R$ is not the classical textbook covering map—the covering map used, e.g., in the related physics literature (see [12]). It differs by a scaling by factor R in the first variable, so infinitesimal distances in the plane and the mid cylinder are not equal (except for the $(6,0)$-hCSP, in which case $R=1$). Neither is ${\Phi }_R$ the covering map discussed in relation to Figure 8, which is actually covering an underlying polyhedral cylinder.

Lemma 3 confirms the second statement in Theorem 1. However, this approach is not helpful in proving the first statement, so in the next subsection, we prove the mathematical details to support that part of our heuristic proof.

5.4. D.4. Monotonicity Results

We confirm various claims made about necklaces on cylinders Cyl$\left(r\right)$ in the proof of Theorem 1. Integers m and n are given with $m>n>0$. We search for appropriate geometric parameters r and $\zeta$ by considering $(m,n)$-elegant necklaces in the covering lattice described above: the first m sphere centers are incremented by the lattice vector $\overline{u}=\langle {\theta }_1,\zeta \rangle$ while the remaining n are incremented by $\overline{v}=\langle {\theta }_2,z_2\rangle$, with $\zeta \le 0$ and $z_2>0$. Two key equations reflect the fact that a necklace must be closed:

(a)

Height closure: $m\zeta +n{z}_2=0$.

(b)

Angle closure: $m{\theta }_1+n{\theta }_2=2\pi$.

Height closure implies $z_2=-(m/n)\zeta$, meaning we can eliminate $z_2$ in subsequent expressions. The radius r enters because when moving between tangent spheres on Cyl$\left(r\right)$ the increments in angle $\theta$ and in height z are related. Specifically

$$\theta =arccos(1-{\displaystyle \frac{(1-z^2)}{2R^2}}),R=r+1/2.$$

Computation gives

$$\partial \theta /\partial z={\displaystyle \frac{-z}{R^{2}sin\left(\theta \right)}},\partial \theta /\partial R=\partial \theta /\partial r={\displaystyle \frac{-(1-z^2)}{R^{3}sin\left(\theta \right)}},$$

meaning that the angles are decreasing as functions of $\left|z\right|$ and of r.

Consider an $(m,n)$-elegant necklace on Cyl$\left(r\right)$. We claimed in our proof of Theorem 1 that this necklace is unique. If $\zeta$ were to become more negative, then both $\left|\zeta \right|$ and $|z_2|=(m/n)|\zeta |$ would increase, implying both ${\theta }_1$ and ${\theta }_2$ would decrease, contradicting the angle closure condition. Likewise, if $\zeta$ were to become less negative, both $\left|\zeta \right|$ and $|z_2|$ would decrease, ${\theta }_1$ and ${\theta }_2$ would increase, again contradicting the angle closure condition. Thus, there is only one $(m,n)$-elegant necklace on Cyl$\left(r\right)$ and hence only one associated height increment $\zeta$.

Next, applying implicit differentiation to the angle closure condition (b) gives

$${\displaystyle \frac{d\zeta }{dR}}=-{\displaystyle \frac{m(1-{\zeta }^2)sin\left({\theta }_2\right)+n(1-{(m/n)}^2{\zeta }^2)sin\left({\theta }_1\right)}{Rm\zeta [sin\left({\theta }_2\right)+(m/n)sin\left({\theta }_1\right)]}}.$$

Since $\zeta <0$, the expression for ${\displaystyle \frac{d\zeta }{dR}}$ is positive. Therefore, as we have claimed, when the radius r of the cylinder decreases, then the height $m\zeta$ of the lowest sphere S of the $(m,n)$-elegant necklace decreases—i.e., the necklace droops lower.

We showed by the intermediate value theorem that there is some radius r so that Cyl$\left(r\right)$ supports an $(m,n)$-Goldilocks necklace. The final claim to verify is that this r is unique. Recall that this involved the lowest sphere, S, in $(m,n)$-elegant necklaces along with two tangent spheres (see Figure 7). The first of these, $S_{\mathrm{G}}$, is displaced from S at $\langle 0,0\rangle$ by the vector $\overline{u}=\langle {\theta }_1,\zeta \rangle$ and the second by vector $\overline{v}=\langle {\theta }_2,z_2\rangle$. Let $\alpha \left(r\right)$ be the angle at S in the triangle formed by these three, and let $C\left(R\right)=cos\left(\alpha \right(r\left)\right)$, a function of $R=r+1/2$. It suffices to show that $dC/dR>0$.

Applying an isometry, we may assume S is centered at $(R,0,0)$, $S_{\mathrm{G}}$ at $(R,{\theta }_1,\zeta )$, and $S_{\mathrm{B}}$ at $(R,{\theta }_2,z_2)$. In Euclidean coordinates, the 3D vectors from S to $S_{\mathrm{G}}$ and to $S_{\mathrm{B}}$ are, respectively, the unit vectors

$$\overline{U}=(R-Rcos\left({\theta }_1\right),Rsin\left({\theta }_1\right),\zeta ),\mathrm{and}\overline{V}=(R-Rcos\left({\theta }_2\right),Rsin\left({\theta }_2\right),z_2).$$

Let us agree to some abbreviations: $c_j=cos\left({\theta }_j\right)$ and $s_j=sin\left({\theta }_j\right)$, for $j=1,2$, and also $c_{12}=cos({\theta }_1-{\theta }_2)$ and $s_{12}=sin({\theta }_1-{\theta }_2)$. These will greatly facilitate the computations involving $C\left(R\right)$.

$$\begin{array}{cc}\hfill C\left(R\right)& =\overline{U}·\overline{V}=R^2(c_1-1)(c_2-1)+R^2{s}_1{s}_2+\zeta z_2\hfill \\ \hfill & =R^2(1-c_1-c_2)+R^2{c}_{12}-(m/n){\zeta }^2\hfill \\ \hfill & =1-R^2-({\displaystyle \frac{m^2+n^2}{2n^2}}){\zeta }^2+R^2{c}_{12}-(m/n){\zeta }^2\hfill \\ \hfill & =1-R^2-({\displaystyle \frac{{(m+n)}^2}{2n^2}}){\zeta }^2+R^{2}c12.\hfill \end{array}$$

Computing the derivative, we see

$$\begin{array}{cc}\hfill {\displaystyle \frac{dC}{dR}}=& -2R-({\displaystyle \frac{{(m+n)}^2}{n^2}})\zeta {\displaystyle \frac{d\zeta }{dR}}+R^2{s}_{12}({\displaystyle \frac{d{\theta }_2}{dR}}-{\displaystyle \frac{d{\theta }_1}{dR}})+2R{c}_{12}\hfill \\ \hfill =& -2R+({\displaystyle \frac{{(m+n)}^2}{mn}}){\displaystyle \frac{(m(1-{\zeta }^2)s_2+n(1-{(m/n)}^2{\zeta }^2)s_1+(m-n)s_{12})}{R(n{s}_2+m{s}_1)}}\hfill \\ \hfill & +2R{c}_{12}.\hfill \end{array}$$

Note that ${\theta }_2\le {\theta }_1\le 2{\theta }_2$, so $({\theta }_2-{\theta }_1)<{\theta }_2$. We get the inequalities $s_2\le s_1\le 2s_2$ and $c_{12}\ge c_2$, and, in turn,

$$2R{c}_{12}\ge 2R{c}_2=2R(1-{\displaystyle \frac{1-{(m/n)}^2{\zeta }^2}{2R^2}})=2R-{\displaystyle \frac{(1-{(m/n)}^2{\zeta }^2)}{R}}.$$

Now we note a succession of inequalities:

$$\begin{array}{cc}\hfill {\displaystyle \frac{dC}{dR}}& \ge ({\displaystyle \frac{{(m+n)}^2}{mn}}){\displaystyle \frac{(m(1-{\zeta }^2)s_2+n(1-{(m/n)}^2{\zeta }^2)s_1)}{R(n{s}_2+m{s}_1)}}-{\displaystyle \frac{(1-{(m/n)}^2{\zeta }^2)}{R}}\hfill \\ \hfill & >({\displaystyle \frac{{(m+n)}^2}{mn}}){\displaystyle \frac{(m(1-{(m/n)}^2{\zeta }^2)s_2+n(1-{(m/n)}^2{\zeta }^2)s_2)}{R(n{s}_1+m{s}_1)}}-{\displaystyle \frac{(1-{(m/n)}^2{\zeta }^2)}{R}}\hfill \\ \hfill & >({\displaystyle \frac{{(m+n)}^2}{mn}}){\displaystyle \frac{(m+n)(1-{(m/n)}^2{\zeta }^2)s_2}{R(m+n)s_1}}-{\displaystyle \frac{(1-{(m/n)}^2{\zeta }^2)}{R}}\hfill \\ \hfill & =({\displaystyle \frac{{(m+n)}^2}{mn}}){\displaystyle \frac{(1-{(m/n)}^2{\zeta }^2)}{2R}}-{\displaystyle \frac{(1-{(m/n)}^2{\zeta }^2)}{R}}\hfill \\ \hfill & >0.\hfill \end{array}$$

This is what we wanted to show: the fact that $dC/dR>0$ implies that there will be only one cylinder Cyl$\left(r\right)$ supporting an $(m,n)$-Goldilocks necklace. This completes the details for the claims we made in the proof of Theorem 1.

That brings us to the practical problem of finding the values of r and $\zeta$ for a given pair $(m,n)$. We noted earlier the explicit solutions for cases $(m,0)$ and $(m,m)$, but to the author’s knowledge, there are no other explicit solutions. The approach to the proof of Theorem 1, wherein the value of r is decreased until an $(m,n)$-Goldilocks necklace is found, can be implemented numerically. This table provides approximate values of r for m in the range 2 to 13. There are some patterns in the data that confirm your intuition. As you might expect, the values are increasing across rows and down columns. As m grows, r grows, so Cyl$\left(r\right)$ has less and less curvature, and the packing looks increasingly like the hexagonal close packing of spheres attached to a plane, Cyl$(\infty )$.

Intuition about twisted ropes leads to a more subtle pattern: As you twist a rope more tightly, we expect its diameter to decrease. Consider elegant pearl necklaces with the same number of pearls, say, for example, 12 pearls, as in Figure 13. As n grows at the expense of m, the height increment $\zeta$ becomes more negative, the twist is more visually evident, and the cylinder radius decreases.

Table 1. Cylinder radii for $(m,n)$-hCSP’s, $m=2,\cdots ,13$.

$m\setminus n$	0	1	2	3	4	5	6
2	n/a	0.0196290419	0.1123724357
3	0.0773502692	0.1452616461	0.2431283644	0.3660254038
4	0.2071067812	0.2856098969	0.3846240416	0.5018804956	0.6315167192
5	0.3506508084	0.4325849755	0.5311569265	0.6444141170	0.7688700797	0.9012585384
6	0.5	0.5831845450	0.6807569165	0.7908960515	0.9111728813	1.0392003480	1.1730326075
7	0.6523824355	0.7359998206	0.8324219685	0.9399926944	1.0568342905	1.1811092662	1.3112354268
8	0.8065629649	0.8902678585	0.9855566370	1.0909457796	1.2049023274	1.3259498048	1.4527672542
9	0.9619022001	1.0455414663	1.1397724649	1.2432804142	1.3547712360	1.4730137330	1.5969015931
10	1.1180339887	1.2015408950	1.2948084394	1.3966743006	1.5060242195	1.6218126006	1.7430843299
11	1.2747327664	1.3580813817	1.4504786749	1.5508976867	1.6583736322	1.7719955302	1.8909327960
12	1.4318516526	1.5150326938	1.6066469807	1.7057899444	1.8116080132	1.9233014648	2.0401412351
13	1.5892907344	1.6723076354	1.7632209762	1.8612236940	1.9655605867	2.0755248045	2.1904720432
$m\setminus n$	7	8	9	10	11	12	13
7	1.4459414186
8	1.5842617208	1.7195498855
9	1.7254702745	1.8579272334	1.9936207664
10	1.8690058043	1.9988486482	2.1320179590	2.2680134412
11	2.0144434561	2.1418868738	2.2726988110	2.4064189762	2.5426391734
12	2.1614633356	2.2866935874	2.4153065939	2.5468806365	2.6810370913	2.8174391632
13	2.3098049512	2.4329930196	2.5595707414	2.6891317304	2.8213209281	2.9558226383	3.0923728858

Table 1. Cylinder radii for $(m,n)$-hCSP’s, $m=2,\cdots ,13$.

$m\setminus n$	0	1	2	3	4	5	6
2	n/a	0.0196290419	0.1123724357
3	0.0773502692	0.1452616461	0.2431283644	0.3660254038
4	0.2071067812	0.2856098969	0.3846240416	0.5018804956	0.6315167192
5	0.3506508084	0.4325849755	0.5311569265	0.6444141170	0.7688700797	0.9012585384
6	0.5	0.5831845450	0.6807569165	0.7908960515	0.9111728813	1.0392003480	1.1730326075
7	0.6523824355	0.7359998206	0.8324219685	0.9399926944	1.0568342905	1.1811092662	1.3112354268
8	0.8065629649	0.8902678585	0.9855566370	1.0909457796	1.2049023274	1.3259498048	1.4527672542
9	0.9619022001	1.0455414663	1.1397724649	1.2432804142	1.3547712360	1.4730137330	1.5969015931
10	1.1180339887	1.2015408950	1.2948084394	1.3966743006	1.5060242195	1.6218126006	1.7430843299
11	1.2747327664	1.3580813817	1.4504786749	1.5508976867	1.6583736322	1.7719955302	1.8909327960
12	1.4318516526	1.5150326938	1.6066469807	1.7057899444	1.8116080132	1.9233014648	2.0401412351
13	1.5892907344	1.6723076354	1.7632209762	1.8612236940	1.9655605867	2.0755248045	2.1904720432
$m\setminus n$	7	8	9	10	11	12	13
7	1.4459414186
8	1.5842617208	1.7195498855
9	1.7254702745	1.8579272334	1.9936207664
10	1.8690058043	1.9988486482	2.1320179590	2.2680134412
11	2.0144434561	2.1418868738	2.2726988110	2.4064189762	2.5426391734
12	2.1614633356	2.2866935874	2.4153065939	2.5468806365	2.6810370913	2.8174391632
13	2.3098049512	2.4329930196	2.5595707414	2.6891317304	2.8213209281	2.9558226383	3.0923728858

The table entries suggest that $r(m-1,n+1)$ is less than $r(m,n)$—in other words, you see r increasing as you move up and to the right in the table. This is an observation, not a proof, but it is supported by intuition.

5.5. D.5. About Density

Has the reader made their best guess about packing density for hCSPs? Is the density greater or smaller than the roughly $60\%$ achieved by the hexagonal sphere packing on a plane?

Table 2. Representative densities for hCSPs.

$(m,n)$	$(2,1)$	(3,0)	(4,0)	(5,0)	(10,0)	(50,0)	(100,0)	(300,0)	(∞,0)
r	0.01963	0.07735	0.20711	0.35065	1.11803	7.46299	15.418112	47.247355	∞
density	0.50714	0.53033	0.56060	0.57582	0.59721	0.60430	0.60453	0.60459	0.60460

provides a sampling of approximate densities; aside from the $(2,1)$ case, I have chosen $(m,0)$-hCSP’s for consistency.

Turning now to the density Conjecture 2, it seems plausible that this is true in the limited case of $(m,m)$-hCSP’s. Here is a potential line of reasoning. Suppose $\mathcal{S}$ is an $(m,m)$-hCSP on Cyl$\left(r\right)$, some $m\ge 2$ and we wish to compare its density to that of ${\mathcal{S}}^{\prime }$, a CSP on the same cylinder. The direct approach would compare the numbers V and $V^{\prime }$ of spheres in truncated cylinders. Such counting may be straightforward for $\mathcal{S}$, but it is likely more involved for ${\mathcal{S}}^{\prime }$. We suggest, instead, moving to consideration of triangulations T and $T^{\prime }$ formed by the centers.

How do triangulations help? Truncations of T and $T^{\prime }$ to heights $[-h,h]$ will be triangulations of an annulus, hence with Euler characteristic $V-E+F=0$. The number $E_{\partial }$ of boundary edges is clearly bounded above independent of h, while V, E, and F grow without bound. Counting edges by faces gives

$$E=3F/2+E_{\partial }⟹V=F/2+E_{\partial }⟹V\sim F/2.$$

(3)

In light of this, we can compare the numbers of faces rather than the numbers of vertices in the truncated triangulations. And we might do this indirectly by looking at the face areas.

We lift T and $T^{\prime }$ under the covering map ${\Phi }_R$ so we can work in the plane, that is, in the $(\theta ,z)$-plane as described in §D.3. The triangles in the plane correspond to geodesic triangles on Cyl$\left(R\right)$, and their Euclidean areas are proportional to surface areas, with scaling factor $1/R$. We may be able to avoid counting by comparing areas: if the areas of triangles of T are no bigger than those of $T^{\prime }$, then the number of faces in a truncated portion of the triangulations will be no smaller, and we would conclude that den$\left(\mathcal{S}\right)\ge$ den$\left({\mathcal{S}}^{\prime }\right)$, as conjectured.

A triangle in the plane will be called “special for R” if the lifts of its three vertices under ${\Phi }_R$ determine a unit-sided equilateral triangle inscribed in Cyl$\left(R\right)$ (i.e., so the lifted vertices are centers for a tangent triple of spheres; see, e.g., Figure 2). Among the special triangles for R, there is only one (up to translations and horizontal flips) which has a side that is vertical. (Its projection to Cyl$\left(R\right)$ is also vertical and has geodesic length 1). We will denote this triangle by ${\Delta }_R$. One can show that for any R, ${\Delta }_R$ has an area strictly smaller than any other special triangle for R.

We return to our triangulations T and $T^{\prime }$. As $\mathcal{S}$ is an $(m,m)$-hCSP, all its faces are copies of ${\Delta }_R$. If we assume for a moment that the competing packing ${\mathcal{S}}^{\prime }$ is also hexagonal, the faces in $T^{\prime }$ would all have areas strictly larger than those of T. Without knowing anything further about the triangulations, we can conclude that the number of triangles in T will outrun the number in $T^{\prime }$, implying via (3) that den$\left(\mathcal{S}\right)\ge$ den$\left({\mathcal{S}}^{\prime }\right)$. Of course, the existence of such an ${\mathcal{S}}^{\prime }$ would provide a positive answer to Open Question 1. So what of more general competitors ${\mathcal{S}}^{\prime }$? One is tempted to posit that the area for faces of T is smallest in other situations.

Open Question 4.

If t is a triangle in the $(\theta ,z)$-plane whose corners lift to points on Cyl$\left(R\right)$ at least euclidean distance 1 apart from one another, is the area of t greater than or equal to the area of the special triangle ${\Delta }_R$?

A positive answer would ensure that $\mathcal{S}$ has maximal density among all CSP’s on Cyl$\left(r\right)$. It is tempting to exploit some extremal property of equilateral triangles, but note that ${\Delta }_R$ is not itself an equilateral triangle, nor is the corresponding geodesic triangle on Cyl$\left(R\right)$ equilateral—at least two of its geodesic sides are of length greater than 1.

Finally, note that this particular approach only applies when $\mathcal{S}$ is an $(m,m)$-hCSP. Other situations seem to require much more detailed analysis, perhaps like that employed by Hales [6], to understand the excluded volume. One might start by breaking the solid cylinder into Voronoi cells associated with the sphere centers: These cells are identical and would be bounded by planes and parts of the inner and outer cylinders. In any case, there seem to be interesting new questions here, and perhaps some of them will capture the reader’s attention.

6. Conclusions

This paper concerns cylindrical sphere packings, configurations of uniform spheres all attached to a common cylinder. As spheres attached to a surface, these packings hover between two and three dimensions. The familiarity of the geometric objects involved leads to heuristic arguments for the creation of hexagonal packings, in which each sphere is tangent to six others, and then to the complete characterization of these packings in terms of integer pairs $(m,n)$. Their visual appeal suggests natural questions and conjectures concerning the density of these sphere packings and the structural rigidity of associated polyhedra. In particular, in analogy to Kepler’s classical conjecture on the densest packing of spheres in three-dimensional space, the hexagonal sphere packings are conjectured to be the densest sphere packings on their respective cylinders. The final section of the paper provides rigorous mathematical proofs of earlier heuristic arguments.