MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There is a well known pattern that turns up in nature involving the golden ratio $\phi = \frac{\sqrt{5}-1}{2}$.

Sunflower Spiral

To get this "sunflower spiral" pattern, put the $k$th node at an angle of $2\pi \phi k$ and a radius of $\sqrt{k}$. The effect that this has is that each node is nicely spaced apart from its neighbours.

There are a couple of different ways to think about what this spiral is doing: it's something like a reasonably good packing algorithm in 2 dimensions that works for any number of points being packed. Another way to think about it, looking at the projection of the pattern onto the circle, is as a sequence of measures on the circle, where adding the $k$th puts a point weight at it's argument. This sequence of measures on the circle is near uniformly spaced at each step of the sequence.

The question is: Is there a similar spiral-like thing for 3 (or more) dimensions?

That is (for 3 dimensions), we want to put a sequence of nodes down in $\mathbb{R}^3$ so that the $k$th node is at radius $\sqrt[3]{k}$ from the origin, and so that the resulting pattern is reasonably well packed, in whichever sense analogous to those above you want. Is there a natural way to generate such a sequence on $S^2$ (perhaps by iterating a relatively simple function?), like there is for the sunflower spiral on $S^1$?

share|cite|improve this question
@Henry: Is that a curiosity question, or you have reasons to ask it? I wonder whether you have tried to construct 3D analogues yourself. Something interesting or nothing? – Wadim Zudilin May 16 '10 at 2:16
Is the sunflower spiral the densest packing in $\mathbb{R}^2$ constructed in this manner? That is, by continued scaling and rotation – Robby McKilliam May 16 '10 at 2:58
@Wadim: Some time ago I tried choosing a sequence of points on $S^2$ by moving a frame around: effectively the algorithm for a person walking around on the surface of the sphere to determine the next point would be to turn left by $\theta_1$ then walk forward by $\theta_2$. This would always seem to converge to some tour of a small number of points, which perhaps should be obvious in retrospect. – Henry Segerman May 16 '10 at 3:34
@Wadim: I also tried just getting a reasonable "packing sequence" on the sphere, by putting the next point of the sequence at the point furthest from all already chosen points of the sequence (using a Voronoi tile algorithm). This works to make a sequence, but it isn't some simple iteration and it looks random: no interesting patterns. – Henry Segerman May 16 '10 at 3:34
@Robby: J.N.Ridley, in "Packing efficiency in sunflower heads" shows that among patterns with $r(n) = \sqrt{n}$ and $\theta(n) = \lambda n$ for some $\lambda \in \mathbb{R}$, the sunflower spiral gives the most efficient packing pattern, meaning that the infimum of distances between nodes is maximised. The choice of $r(n) = \sqrt{n}$ is reasonable because it means that the ratio of the number of points in a circle centered at the origin to the area of that circle approaches a non zero finite limit as the radius goes to $\infty$. – Henry Segerman May 16 '10 at 5:55

This is just bold speculation: The spiral pattern you see arises from a homomorphism from the integers to the unit circle, and the irrationality of $\phi$ gives it a pleasant aperiodicity. In higher dimensions, there isn't an abelian group structure on spheres, so the naive generalization using $d$-parameter families of integers doesn't work. On the other hand, there are infinite discrete subgroups of orthogonal groups whose orbits have dense image in a sphere, so you can take the orbit of a point under the action, ordering by something like word length. An explicit example of such a faithful action of $F_2$ on $S^2$ is given in David Speyer's blog post, using the matrices $\begin{pmatrix} 3/5 & 4/5 & 0 \\ -4/5 & 3/5 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 3/5 & 4/5 \\ 0 & -4/5 & 3/5 \end{pmatrix}$.

I have no idea what the resulting "spiral" will look like, or how "well-packed" it is.

Addendum: If you choose suitably random orthogonal matrices, you will still get a faithful action of $F_2$. You might have fewer near-collisions with a "less hyperbolic" group, but I don't know much about the geometry.

share|cite|improve this answer
This looks promising: These show words up to length 5. The sphere has some transparency, so points on the far side are partially obscured by the sphere. In the version with labels, "a" is the inverse of "A", ditto for "b" and "B", and Mathematica draws all labels unobscured. It's pretty clear that there are some pairs of points that are very close to each other, so this choice of generating matrices seems like it could be improved upon in terms of "well-packing". – Henry Segerman May 16 '10 at 8:18
This seems to be an incomplete version of Lubotzky, Philips, Sarnak, "Hecke operators and distributing points on the sphere". – Victor Protsak May 16 '10 at 19:20

Bold speculation again, but I would guess the answer for $\mathbb{R}^3$ is 'no'. Heuristically, as r increases you want all of your spheres to 'move a bit' so that they can 'slot into the gaps between the previous spheres'. If this is going to happen in any natural way you're going to be able to project down onto a nowhere vanishing continuous vector field on $S^2$, which isn't going to happen.

Perhaps $\mathbb{R}^4=\mathbb{C}^2$ will provide more luck: something hopf bundle-ish, dropping a 2 spheres of points down at an angle of $2\pi\phi k$ (under the natural $S^1$ action) at a radius of $^4\sqrt{k}$? Maybe?

share|cite|improve this answer

One possible way to fill R3 is to divide up R3 into onion-layers, and use some sort of "generalized spiral points" to fill each onion-layer -- kind of like winding a string from N pole to S pole to N pole to S pole.

In each layer, the spiral starts at the North pole and spirals out something like the "sunflower spiral", eventually crosses the equator. The Southern equator is (more or less) identical to the Northern equator. After the path crosses the equator, it spirals in smaller and tighter curves until it reaches the South pole. The whole spiral is vaguely similar to Sphere Spirals by M.C. Escher. This spiral is one solution to the "place n equally-spaced points on a sphere" problem -- it may not be "the best way to pixelize a sphere", it may not be the best solution to the "Thomson problem", but there is no one "best way to pixelize a sphere". Then you move out an onion-layer and start spiralling back from South to north.

Going at it from another direction: perhaps somehow number the spheres used to form Waterman polyhedra? That gives a much more regular and close-packed set of points (sphere centers) -- but it's not obvious to me if there is any nice formula to convert from sphere-number to xyz coordinates.

share|cite|improve this answer

So, generalizing the "spiral" sense in which this could be defined (as in, the radial almost-spirals that radiate outwards from a region around the center, such as the sequence 3,11,19,40,61,82,103,124) could we define a 3d analog of this as sea-shell shaped almost-concentric shells spiraling out from a region?

In other words, is there a packing consisting of of sea-shells, that satisfies whatever properties we'd like in terms of distributions of integers put on each "leaf" of the sea-shells?

If what we were considering initially was an actual spiral, e.g., $r(\theta)=\theta$, I would say something like the map in spherical coordinates $r(\theta,\phi) = \phi$

enter image description here

which is a 2-dimensional "leaf" generalizing the spiral from before, which one could presumably tile similarly to the original spiral? Or tile subject to packing conditions?

So, could we look at a definition in terms of packing, and define a similar spiral in terms of "leafs" of shells that are "almost" like $r(\theta,\phi) = \phi$, given that our initial spiral was "almost" like $r(\theta) = \theta$?

This seems more in line with a good generalization of the "spiral" nature of this problem, as opposed to thinking or putting points on concentric spheres. But I don't see an obvious way to define this in a natural way that automatically makes the sea-shell shells "almost" shells in the same way as the spiral does this, nor a "natural" one that has the $r=k^{1/3}$ dependence you want. Presumably, the natural conditions would be more obvious if the "good" properties of the spiral you want to keep are more clearly stated? It seems 3d is not constrained enough by the definition in terms of only radii and angles to come up with a unique "obviously correct" solution.

share|cite|improve this answer
Apart from the first few in your example sequence, you're picking out one of the spirals for which consecutive terms differ by a particular Fibonacci number (in this case 21). In the 2D version, these spirals are apparent only for a small range of radii, before larger Fibonacci number differences show up in neighbouring points. So presumably in 3D there would have to be many different sea-shells with different "angles" (rates of growth, corresponding to the increasing Fibonacci number differences). – Henry Segerman May 22 '10 at 14:30
On $r=k^{1/3}$, this choice ensures that the density of points inside large spheres centered at the origin approaches a limit as the radii go to infinity. I guess it isn't necessary to the spirit of the problem to require this, but it seems sensible. – Henry Segerman May 22 '10 at 16:48

Think of the sunflower as a series of almost-triangles of increasing size centered on the central point, with integers at each of their points.

It's a couple steps more complicated, but try starting with a tetrahedral arrangement around the initial point. This arrangement has four natural gaps, which 5 through 8 can go into. I'm sure there's some rational way, from here, that consecutively larger almost-tetrahedrons can be put together around the central point. It might not have some of the more interesting properties of the sunflower sequence, and it's possible that it has arbitrary (or just wrong) elements, but it has the benefit of being scalable to n dimensions (if it works at all).

share|cite|improve this answer

I think a good reference for this is "The discrepancy method" by Bernard Chazelle, page 59 and following. It's available at to read. Chapter title is "An orbital construction for points on a spehere", and the method is compared to $\phi$ for the circle.

share|cite|improve this answer
Cruftless version of that URL: – J. M. Sep 27 '10 at 20:24
Thanks, I learned a new "word". – Paul-Olivier Dehaye Sep 28 '10 at 8:31

Have a look at this spiral here :

[scroll down bellow the video]

or to see it bigger go here:

maybe it will help

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.