Circles and Pi

[ReFLeX]

As one might guess, I'm a pool player. One question that I've been curious about for a while is, in a rack of billiard balls, why is it that six balls fit around one?

Like so:

.

.

When you assume that all balls are the same size, six seems a mathematically arbitrary number. What I wonder is, do six circles actually fit tightly around one or is that a deception and the number is actually 2∏?
(That's the best pi I could find).

 

[RandFan]

As one might guess, I'm a pool player. One question that I've been curious about for a while is, in a rack of billiard balls, why is it that six balls fit around one?

Like so:

.

.

When you assume that all balls are the same size, six seems a mathematically arbitrary number. What I wonder is, do six circles actually fit tightly around one or is that a deception and the number is actually 2∏?
(That's the best pi I could find).

I think this is basically the kepler problem only the Kepler problem dealt with stacking spheres.

Try this,

http://www.findarticles.com/p/articles/mi_qa3773/is_199903/ai_n8846815

Mathematics "Proves" What the Grocer Always Knew

This article is an exacting account of Thomas Hales recently announced solution to the Kepler problem. Almost four hundred years ago, Kepler conjectured that the "face-centered cubic lattice" arrangement was the most efficient packing of space by stacked spheres. Starting with an array of unit spheres covering the plane, with each sphere surrounded by six others, place atop of this array a second such array that is translated so each new sphere contacts three in the original planar pattern. Now translate these two layers vertically to fill space with layers having the following properties: within each layer every sphere contacts six others, and each sphere in one of any pair of adjacent layers contacts three in the other. The proportion of space occupied by this lattice was long known to be optimal among "lattice-packings" and now has been proved maximal among all possible packings. Kepler's insight resisted proof for centuries, and his conjecture appeared in Hilbert's classic list of problems. As the author of this Times article recently wrote elsewhere, "Of all the problems likely to replace Fermat's Last Theorem as the greatest unsolved problem in mathematics the best candidate is Kepler's."

The article does a good job of explaining the gap between belief and mathematical proof. Regarding Kepler's conjecture, belief is evidenced by grocers the world over, whose piles of oranges conform to the Hypothesized arrangement. Moreover, as packing theorist C. A. Rogers has put it: " . . . many mathematicians believe, and all physicists know" Kepler's conjectured packing is optimal. Hales' proof makes extensive use of machine computation, and so may cause uneasiness among a few skeptics. In this respect, it is similar to the somewhat controversial method of setting another great, recently solved problem in mathematics, the four-color conjecture (a beautiful outline of which is given in this article). Hales acknowledges the significant contribution of his doctoral student Samuel Ferguson, and gives his own internet site address, enabling readers to tap into a cornucopia of information about the Kepler conjecture and its solution (wv-arw.math.Isa.umich.edu/-hales).

 

[Marquis de Carabas]

A three-ball cluster creates an equilateral triangle imagining segments from the centres of the spheres through the points where the balls meet (assuming your balls are the same size). Each angle in an equilateral triangle is 60 degrees, 6 of those is 360.

 

[ReFLeX]

Ok, so. In 2 dimensions, 6 circles of equal size can touch the centre one. In 3 dimensions, how many is it? It seems to me to be 18 although I'm not certain. But is there then a formula into which we could insert the number of dimensions (2 or 3) and come up with those corresponding numbers?

 

[CurtC]

I'm also a pool player - I'm in a tournament tonight!

In three dimensions, spheres can't be packed tightly. You can imagine your seven-ball rack arrangement, placed atop a triangle of three balls (each touching your center one), then three more on the top layer, each of those also touching the center. That's twelve balls around the center one, or 13 total. The trouble is, that's not a tight fit, and if you imagine that they were stuck onto the center one, they could wiggle around some.

This seems like a bummer, but it's the reason that golf balls don't get stuck in the vending machine at the driving range!

 

[Schneibster]

As one might guess, I'm a pool player. One question that I've been curious about for a while is, in a rack of billiard balls, why is it that six balls fit around one?

Like so:

.

.

When you assume that all balls are the same size, six seems a mathematically arbitrary number. What I wonder is, do six circles actually fit tightly around one or is that a deception and the number is actually 2∏?
(That's the best pi I could find).

OK, here's the deal: there's 360 degrees in a full circle. For three balls to all be in contact with one another, they must be in an equilateral triangle. That's three 60 degree angles, right? Well, there's six of those in a full circle. So there you go.

 

[Vorticity]

Ok, so. In 2 dimensions, 6 circles of equal size can touch the centre one. In 3 dimensions, how many is it? It seems to me to be 18 although I'm not certain. But is there then a formula into which we could insert the number of dimensions (2 or 3) and come up with those corresponding numbers?

This is known as the d-dimensional "kissing number". See this article for more info:

http://mathworld.wolfram.com/KissingNumber.html

In d=3, the number is 12.

 

[CaveDave]

In three dimensions, spheres can't be packed tightly.

WRONG!

You can imagine your seven-ball rack arrangement, placed atop a triangle of three balls (each touching your center one), then three more on the top layer, each of those also touching the center. That's twelve balls around the center one, or 13 total. The trouble is, that's not a tight fit, and if you imagine that they were stuck onto the center one, they could wiggle around some.

WRONG!

This seems like a bummer, but it's the reason that golf balls don't get stuck in the vending machine at the driving range!

???

All the center-to-center distances of touching equal-sized spheres are equal. They form equalateral triangles in 2 dimentions, and equalateral tetrehedrons (3 sided pyramids) in 3 dim. All the planar angles are equal @ 60 degrees. Any "looseness" would be caused by imperfections.

Dave

 

[Hawk one]

The solution is that you start playing snooker instead.

 

[Jarom]

WRONG!

All the center-to-center distances of touching equal-sized spheres are equal. They form equalateral triangles in 2 dimentions, and equalateral tetrehedrons (3 sided pyramids) in 3 dim. All the planar angles are equal @ 60 degrees. Any "looseness" would be caused by imperfections.

No, CurtC is quite correct. The easiest way to see this is to try it. Get yourself some marbles, golf balls, what have you. Lay out a filled hexagon of 7 spheres on a table, brace them so they won't roll, and set three more spheres on top, settling them in the indentations thus formed. You''ll see that these three spheres do not contact each other, and there will be a large enough gap that looseness can be ruled out.

If you don't want to try it, we can easily argue geometrically. We can pack 12 spheres equally spaced around a central sphere by circumscribing a dodecahedron (12 pentagonal faces) to the sphere, then attaching one sphere in the center of each face of this dodecahedron.

In order for the spheres to be touching each other, the angle formed at the center of the sphere by rays going through the points of tangency with two nearby spheres must be pi/3 (60 degrees, if you prefer; I don't.). This angle is going to be pi minus the dihedral angle of the dodecahedron. To see this, imagine a quadrilateral formed from the two points of tangency, the midpoint of the dodecahdron edge immediately between them, and the center of the sphere. Two of the angles of this quadrilateral are right angles.

Therefore, the spheres will touch if and only if the dihedral angle of a dodecahedron is 2pi/3. But in fact the dihedral angle of a dodecahedron is between 116 and 117 degrees (too long to derive here, see http://kjmaclean.com/Geometry/dodecahedron.html)

Therefore, the spheres don't touch. In a large clump of spheres, you'll have a lot of them forming tetrahedra, but there won't be a regular lattice of tetrahedra.

- Jarom

 

[ReFLeX]

This is known as the d-dimensional "kissing number". See this article for more info:

http://mathworld.wolfram.com/KissingNumber.html

In d=3, the number is 12.

24 dimensions? That is just a little bit insane. And indeed, in the picture of 12 spheres, they don't appear to be touching. Plus I tried it (on a snooker table that is!) and realized, yes, three rest on top while not quite fitting. Well, I learned something.

 

[CaveDave]

No, CurtC is quite correct. The easiest way to see this is to try it. Get yourself some marbles, golf balls, what have you. Lay out a filled hexagon of 7 spheres on a table, brace them so they won't roll, and set three more spheres on top, settling them in the indentations thus formed. You''ll see that these three spheres do not contact each other, and there will be a large enough gap that looseness can be ruled out.

If you don't want to try it, we can easily argue geometrically. We can pack 12 spheres equally spaced around a central sphere by circumscribing a dodecahedron (12 pentagonal faces) to the sphere, then attaching one sphere in the center of each face of this dodecahedron.

In order for the spheres to be touching each other, the angle formed at the center of the sphere by rays going through the points of tangency with two nearby spheres must be pi/3 (60 degrees, if you prefer; I don't.). This angle is going to be pi minus the dihedral angle of the dodecahedron. To see this, imagine a quadrilateral formed from the two points of tangency, the midpoint of the dodecahdron edge immediately between them, and the center of the sphere. Two of the angles of this quadrilateral are right angles.

Therefore, the spheres will touch if and only if the dihedral angle of a dodecahedron is 2pi/3. But in fact the dihedral angle of a dodecahedron is between 116 and 117 degrees (too long to derive here, see http://kjmaclean.com/Geometry/dodecahedron.html)

Therefore, the spheres don't touch. In a large clump of spheres, you'll have a lot of them forming tetrahedra, but there won't be a regular lattice of tetrahedra.

- Jarom

OK. It was I who was wrong.

To my eternal shame, I had overlooked some important properties of the third dimension in the ways the spheres fit together. I was wiewing them as groups of three only and didn't spot the collision.

My most humble apologies to CurtC and all others.

Dave <slinks away to hide in cave... repeatedly falling on face from kicking self>

 

[RandFan]

Please forgive my naivete, is Thomas Hales solution to Kepler's problem not relevant in this instance?

 

[davidhorman]

To my eternal shame, I had overlooked some important properties of the third dimension in the ways the spheres fit together. I was wiewing them as groups of three only and didn't spot the collision.

I still can't understand why they don't fit perfectly (if you start with 7 spheres laid flat and add three more on top).

With spheres of radius r, the distance between the centres of neighbouring spheres will be 2r. And the distance between the centres of each group of 3 spheres (where the spheres of the next layer sit) will also be 2r, won't it?

Or, to put it another way, what's wrong in my poorly drawn diagram?

David

 

[Jarom]

I still can't understand why they don't fit perfectly (if you start with 7 spheres laid flat and add three more on top).

With spheres of radius r, the distance between the centres of neighbouring spheres will be 2r. And the distance between the centres of each group of 3 spheres (where the spheres of the next layer sit) will also be 2r, won't it?

No. Why would it?

Or, to put it another way, what's wrong in my poorly drawn diagram?

The distance that you've labeled as 'r' clearly isn't 'r', since it's the distance from the center of the sphere to a point outside the sphere.

In your diagram, if the top spheres are touching each other, then they aren't touching the lower outside spheres. In order for the top spheres to touch, they have to be raised enough that the only sphere in the bottom layer they can touch is the central sphere. This is another perfectly fine way of packing 12 spheres around the central sphere.

-Jarom

 

[davidhorman]

The distance that you've labeled as 'r' clearly isn't 'r', since it's the distance from the center of the sphere to a point outside the sphere.

Blimey, that's embarrassing.

Okay, those r aren't r, but, those two lines are parallel to each other, so the two lines i've labelled as 2r are correct, aren't they?

In order for the top spheres to touch, they have to be raised enough that the only sphere in the bottom layer they can touch is the central sphere.

But they're still directly over the gap between the three lower spheres, aren't they? In which case they can only touch all or none of the spheres below, not just the middle one... can't they?

David

PS I'm not disbelieving anyone - I just haven't grasped the explanations yet!

 

[Skeptical Greg]

Blimey, that's embarrassing.

Okay, those r aren't r, but, those two lines are parallel to each other, so the two lines i've labelled as 2r are correct, aren't they?



But they're still directly over the gap between the three lower spheres, aren't they? In which case they can only touch all or none of the spheres below, not just the middle one... can't they?

David

PS I'm not disbelieving anyone - I just haven't grasped the explanations yet!

Why don't you get some of those magnetic spheres and check it out..

I'ts really hard to illustrate cause you need a CAD program to really get it across.

 

[davidhorman]

Why don't you get some of those magnetic spheres and check it out..

I managed to rustle up a few ping pong balls, and there is a small gap between the upper spheres if I arrange them on the carpet, but it's not a very stable way of looking at the problem, and it doesn't help me see where my diagram is wrong:

a) The centres of alternate "wells" between the 6 spheres surrounding the central one, where each upper sphere will sit, form an equilateral triangle with sides of length 2r
b) therefore the centres of the spheres that sit in these wells will be at distances of 2r from each other
c) the spheres are of radius r, so they'll touch.

David

 

[chipmunk stew]

I managed to rustle up a few ping pong balls, and there is a small gap between the upper spheres if I arrange them on the carpet, but it's not a very stable way of looking at the problem, and it doesn't help me see where my diagram is wrong:

a) The centres of alternate "wells" between the 6 spheres surrounding the central one, where each upper sphere will sit, form an equilateral triangle with sides of length 2r
b) therefore the centres of the spheres that sit in these wells will be at distances of 2r from each other
c) the spheres are of radius r, so they'll touch.

David

Erm, I'm going to have to go with David here. The "gaps in the top three" explanations bothered me intuitively, so I modelled it in SolidWorks:

The ten spheres are all of equal size. The spheres were assembled by first laying the bottom seven tangential to one another with their centers coplanar. The top three were then laid in one at a time, each tangential to three bottom spheres, as though they were dropped in and allowed to settle. Lo and behold, the top three are touching one another. (You can infer this from the image above, and I confirmed it a couple different ways with the CAD tools.)

Maybe real-world spheres don't actually touch, but platonic ones sure do.

edit: "The ten spheres..." was "The twelve spheres..."
and also eta: CaveDave, don't slink away in shame. You were right.

 

[RandFan]

Please forgive my naivete, is Thomas Hales solution to Kepler's problem not relevant in this instance?

Dude, go away. You don't have a clue.