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

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. :p

[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.:o :blush:

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!

[Diogenes]

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.

DavidErm, 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:
http://gabe.stonehillnews.com/jpegs/sphere%20assy.jpg

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. :(

[69dodge]

In Jarom's dodecahedral arrangement, the seven spheres in the "middle plane" aren't actually all in the same plane; the six outside spheres alternate high and low. That leaves more room for the other spheres. The three top spheres are above the three lower middle ones, and the three bottom spheres are below the three higher middle ones. So, there's some wiggle room, and the spheres don't touch each other (except that they all touch the one in the very center).

On the other hand, if seven spheres are arranged hexagonally in a plane, and then three more are placed on top of them, each of the top three does touch the other two as well as touching three of the lower seven. (Besides reasoning it out mathematically, I just tried it with some marbles. Those who suggested that others try it should have tried it themselves first. :D)

[Ducky]

I'm glad we are working this out for circular pies.

The square ones suck. You cut them up and you never get the corner one, then there's a fight, then the cops get called and your party gets broken up for illegal drug possession.....

Or does that only happen to me?

[CaveDave]

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:
http://gabe.stonehillnews.com/jpegs/sphere%20assy.jpg

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.

I am thoroughly confused now: after looking at the page cited below by Voracity;

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.
I thought I saw a fit problem, and now I'm not so sure.

I tried to find error with your diagram and thought I had, then I looked again and it vanished. :confused:

I would like to find that you and I are correct, but I reserve judgement for the moment.

Thanks for your support, one way or the other.

Dave

[chipmunk stew]

I am thoroughly confused now: after looking at the page cited below by Voracity;


I thought I saw a fit problem, and now I'm not so sure.

I tried to find error with your diagram and thought I had, then I looked again and it vanished. :confused:

I would like to find that you and I are correct, but I reserve judgement for the moment.

Thanks for your support, one way or the other.

DaveThe confusion probably comes from the fact that the link provided partially answers ReFLeX's broader question about spheres in n-dimensions, but doesn't answer the question about close packing.

To be honest, I don't know what CurtC and Jarom were on about.

Try this link:
http://mathworld.wolfram.com/HexagonalClosePacking.html

[Ririon]

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.

Buy yourself a new set of pool balls if you're serious about the game. The ones you have are of differing size. (Or you're messing with people's heads.) As several people have figured out in various ways here, it IS a tight fit mathematically. Trust me. I deal with hexagonally close packed 3D crystal structures every day. No, not the woo kind of crystals. And actually they're "fcc", but that's another story.

[CurtC]

To be honest, I don't know what CurtC and Jarom were on about.Try this Wikipedia article (http://en.wikipedia.org/wiki/Kissing_number_problem):It is easy to arrange 12 spheres so that each touches a central sphere, but there is a lot of space left over, and it is not obvious that there is no way to pack in a 13th sphere. (In fact, there is so much extra space that any two of the 12 outer spheres can exchange places without any of the outer spheres losing contact with the center one.)

[DrMatt]

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?

It's 12, not 18.

[CurtC]

Another way to imagine this, if you don't have magnetic spheres available, would be to use Geomags (http://www.geomagsa.com/). By the way, my eight-year-old son got a set of these for Christmas last year, and I have spent hours playing with them.

Wouldn't you agree that doing this with spheres is equivalent to packing equilateral triangles, where the vertex of the triangles would correspond to the sphere centers? Imagine it in two dimensions first, six balls surrounding a center one, then replace that with six equilateral triangles sharing a center point. The task of packing spheres in 3D is equivalent to building a structure with only equilateral triangles.

Using Geomags, it's quickly obvious that this does not work. You can arrange four triangles to make a nice tetrahedron (which corresponds to three spheres on the bottom, one sitting in the gap on top) but you can't keep adding equilateral triangles to make a solid 3D structure. They don't fit, there will be some that can't touch each other, and it will be "loose."

[Jarom]

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.
I agree. I retract many of my earlier statements after further consideration and looking at chipmunk stew's helpful diagram. Although it is possible to arrange 12 spheres around a central sphere with wiggle room, this is an arrangement of 12 spheres without wiggle room.

Now I just have to borrow the cave CaveDave slunk off to. He won't be needing it any more.

Sorry, CaveDave!
(Besides reasoning it out mathematically, I just tried it with some marbles. Those who suggested that others try it should have tried it themselves first. )Yes, I deserved that.
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?Indeed they are. I saw the error and missed the point.

Jarom

[ReFLeX]

It's 12, not 18.
Glad you caught up with us...

So they do fit close together!? I'm still unsure.

ETA: After rethinking, I'm going to have to side with chipmunk stew and CaveDave. I looked closer at the Kissing Number diagram and the gaps must have been successfully suggested to me because I no longer see them...

[69dodge]

The task of packing spheres in 3D is equivalent to building a structure with only equilateral triangles.

Using Geomags, it's quickly obvious that this does not work. You can arrange four triangles to make a nice tetrahedron (which corresponds to three spheres on the bottom, one sitting in the gap on top) but you can't keep adding equilateral triangles to make a solid 3D structure. They don't fit, there will be some that can't touch each other, and it will be "loose."
Have you looked at chipmuck stew's link: http://mathworld.wolfram.com/HexagonalClosePacking.html? Especially the picture on the right.

There are some squares too, not just triangles. But that doesn't mean that the three spheres on top don't touch each other. They do.

I'm not sure exactly what you mean by "packing spheres in 3D". I guess you could come up with a reasonable definition for which it's impossible.

[GodMark2]

Buy yourself a new set of pool balls if you're serious about the game. The ones you have are of differing size. (Or you're messing with people's heads.) As several people have figured out in various ways here, it IS a tight fit mathematically. Trust me. I deal with hexagonally close packed 3D crystal structures every day. No, not the woo kind of crystals. And actually they're "fcc", but that's another story.

Ririon is correct, if you packed the balls that way, and there were spaces, you need a new set of balls.

Ririon is also incorrect. This does NOT show that it is a “tight fit” mathematically. For it to be a tight fit, it is not sufficient to show that an arraignment CAN be bade with no spaces, but that, for the same number of balls, there does NOT exist a packing with spaces. For spheres in Euclidean space, there is certainly one such arraignment; in fact, it can be made such that all twelve balls touch the center sphere, but none of each other.

Place the centers of the spheres along the center normal of a dodecahedron with it’s faces tangent to the center sphere. As shown in an earlier post (by Jarom) these spheres will be at too great a spacing to touch any of its neighbors. This shows that 12 spheres can NOT be packed tightly.

[CaveDave]

I agree. I retract many of my earlier statements after further consideration and looking at chipmunk stew's helpful diagram. Although it is possible to arrange 12 spheres around a central sphere with wiggle room, this is an arrangement of 12 spheres without wiggle room.

Now I just have to borrow the cave CaveDave slunk off to. He won't be needing it any more.

Sorry, CaveDave!
Yes, I deserved that.
Indeed they are. I saw the error and missed the point.

Jarom
Thanks, but I think I'll sit in the doorway so I don't have far to move either way.;) Everyone is welcome, I picked one with lots of room.:)

I would expect there would be an obvious solution to the question, but it looks like all these fine minds and excellent references STILL have trouble agreeing!?! :boggled:

Thanks chipmunk stew, Ririon and others for the support and references. Thanks to those on the other side for making this all interesting.

My old brain is gettin' fatigued from all the flip-flopping caused by this thread.:D

Dave

[CurtC]

I just got out the Geomags and the digital camera to demonstrate what I'm talking about. This first picture is equivalent to six balls plus the center ball, in a plane, all touching each other:

http://www.ccdominoes.com/pics/pi1.jpg

Instead of real spheres, which are hard to hold, the situation is perfectly modeled by the Geomags, I hope we all agree.

Now add the three balls sitting on top of those, each touching its neighbor:

http://www.ccdominoes.com/pics/pi2.jpg

Now add three more just like it underneath, and you have twelve balls all around the center ball, and all making contact with their neighbors:

http://www.ccdominoes.com/pics/pi3.jpg

In this arrangement, all the balls are touching, but they are *not* tightly packed. This structure is not solid, it's squishy. In the last picture, all the outer balls are still touching the center ball, but they're not all touching each other.

http://www.ccdominoes.com/pics/pi4.jpg

If the first seven balls are constrained to lie in an exact plane, then the structure is rigid, but in free space, there is no such constraint. What's happening is the square that you see in the second picture becomes a rhombus when it's squished, and then some of the balls will no longer be touching. Twelve balls cannot be tightly packed around a center one.

[Atlas]

12 Closely packed with no central ball

http://www.earth360.com/math_spheres_Al__12_spheres.jpg (http://www.earth360.com/math_spheres.html)


12 Closest packed around 1

http://www.earth360.com/math_spheres_B__12_around_1.jpg

Interesting pictures of closely packed spheres here ...
http://www.earth360.com/math_spheres.html

[DrMatt]

I'd like to see how two exchange places without losing contact with the central sphere. Last I checked, each outer sphere is just touching the central sphere, and is surrounded by other spheres which will push it away from the central sphere if you try to move it in any direction. Plus, it looks to me like 60 degree packing is the most compact available, and "switching place" involves extending at least one of the angles to not be 60 degrees and thus not be most compact--which will force that sphere to lose contact with the center sphere... No?

[Ririon]

Ririon is correct, if you packed the balls that way, and there were spaces, you need a new set of balls.

:)


Ririon is also incorrect. This does NOT show that it is a “tight fit” mathematically. For it to be a tight fit, it is not sufficient to show that an arraignment CAN be bade with no spaces, but that, for the same number of balls, there does NOT exist a packing with spaces. For spheres in Euclidean space, there is certainly one such arraignment; in fact, it can be made such that all twelve balls touch the center sphere, but none of each other.

Place the centers of the spheres along the center normal of a dodecahedron with it’s faces tangent to the center sphere. As shown in an earlier post (by Jarom) these spheres will be at too great a spacing to touch any of its neighbors. This shows that 12 spheres can NOT be packed tightly.

:jaw-dropp I WAS wrong!

I was only thinking of structures that can can be repeated as a lattice. five-fold symmetry is no good there. Some exceptions (quazicrystals) exist, just so that noone says I was wrong on that one, too.

http://en.wikipedia.org/wiki/Dodecahedron

SO:
1: If you want to have 6 spheres in a plane around the central one, you get a maximum of 12 spheres and no wiggle room.

2: If you don't care about keeping some of the spheres in a plane, you still get a maximum of 12 spheres and some wiggle-room to spare.

If you want to stack spherical fruit in your cornershop, use option 1. :)

Thanks, GodMark2!

[CurtC]

I'd like to see how two exchange places without losing contact with the central sphere. Last I checked, each outer sphere is just touching the central sphere, and is surrounded by other spheres which will push it away from the central sphere if you try to move it in any direction.Please refer to my post #34 above. The 12 balls can all exchange places without any losing touch with the center sphere, because there is a lot of wiggle room. There is almost enough room to fit a 13th ball, but not quite.

[luvhumility]

very good discussion here on the "kissing number"... Vorticity! I like your name! Good observation and conjecture here!

[luvhumility]

what if they all move together at the same time? are they really "touching as they all move?? This is an important assumption we all may make? Also why does nature seem to prefer odd numbers or harmonics in these types of problems? phi?

[chipmunk stew]

I just got out the Geomags and the digital camera to demonstrate what I'm talking about. This first picture is equivalent to six balls plus the center ball, in a plane, all touching each other:

http://www.ccdominoes.com/pics/pi1.jpg

Instead of real spheres, which are hard to hold, the situation is perfectly modeled by the Geomags, I hope we all agree.

Now add the three balls sitting on top of those, each touching its neighbor:

http://www.ccdominoes.com/pics/pi2.jpg

Now add three more just like it underneath, and you have twelve balls all around the center ball, and all making contact with their neighbors:

http://www.ccdominoes.com/pics/pi3.jpg

In this arrangement, all the balls are touching, but they are *not* tightly packed. This structure is not solid, it's squishy. In the last picture, all the outer balls are still touching the center ball, but they're not all touching each other.

http://www.ccdominoes.com/pics/pi4.jpg

If the first seven balls are constrained to lie in an exact plane, then the structure is rigid, but in free space, there is no such constraint. What's happening is the square that you see in the second picture becomes a rhombus when it's squished, and then some of the balls will no longer be touching. Twelve balls cannot be tightly packed around a center one.So there are two separate things going on here.

The first is, if you have seven spheres arranged as a "rack of billiard balls", you can, in fact, create a stable structure by adding three on top and those three will, in fact, be touching (as shown above a couple different ways). However, "close packing" is not the same as "tight packing". In the hexagonal close packing structure (which is called a "triangular orthobicupola") the center ball has twelve balls packed closely around it, but while some "sides" of the structure are "tight" three-ball configurations, others are "loose" four-ball configurations with gaps:

http://gabe.stonehillnews.com/jpegs/sphere%20assy%20gap.jpg

The other thing going on takes place in space, rather than in a flat rack of billiards, and that is, if you pack twelve spheres of equal radius and equally spaced with one another around a central sphere, they will not, in fact touch one another. The way to do this is to place them on the vertices of an icosahedron:

http://gabe.stonehillnews.com/jpegs/sphere%20assy%20icosahedron.jpg

The confusion came from incorrectly generalizing the implications from the second example to conclude that the top three balls in the first would not touch one another.

[chipmunk stew]

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. :(No, you're quite right. Kepler's problem is relevant to the first case above (the hexagonal close packing one).

Didn't mean to snub you. In my case, I was just too lazy to read the article you linked to (Mathematics "Proves" What the Grocer Always Knew (http://www.findarticles.com/p/articles/mi_qa3773/is_199903/ai_n8846815)) until now, and I find visual explanations easier to digest for this kind of thing.

[RandFan]

No, you're quite right. Kepler's problem is relevant to the first case above (the hexagonal close packing one).

Didn't mean to snub you. In my case, I was just too lazy to read the article you linked to (Mathematics "Proves" What the Grocer Always Knew (http://www.findarticles.com/p/articles/mi_qa3773/is_199903/ai_n8846815)) until now, and I find visual explanations easier to digest for this kind of thing. Thanks, sometimes I feel like the 6 year old in a group of adults who are discussing world events and I'm interjecting what I saw on Sesame Street. Damn I wish I was smarter.

I agree about visualizing the problem and I love your graphics. :)

[69dodge]

The way to do this is to place them on the vertices of an icosahedronNo argument here, just a clarification: "the vertices of an icosahedron" is the same as "the centers of the faces of a dodecahedron".

[chipmunk stew]

No argument here, just a clarification: "the vertices of an icosahedron" is the same as "the centers of the faces of a dodecahedron".Yes, exactly right. I should have said "A way to do this..." The regular dodecahedron (http://mathworld.wolfram.com/Dodecahedron.html) has twelve faces, whereas the regular icosahedron (http://mathworld.wolfram.com/Icosahedron.html) has twelve vertices (and twenty sides which are equilateral triangles).

[CaveDave]

Thanks, sometimes I feel like the 6 year old in a group of adults who are discussing world events and I'm interjecting what I saw on Sesame Street. Damn I wish I was smarter.
Me too.

I agree about visualizing the problem and I love your graphics. :)
Same here.

Now, do I go back in, come back out, or stay sittin' in the cavemouth?

I've lost track.:)

Dave

[chipmunk stew]

Now, do I go back in, come back out, or stay sittin' in the cavemouth?
I think most of us are crowded in the cave's mouth at this point. :crowded:

I, for one, do understand now what CurtC and Jarom were on about, and humbly apologize for being, well, curt about the point they were making rather than working through what they were saying and trying to verify or refute it first.

[69dodge]

I'd like to see how two exchange places without losing contact with the central sphere. Last I checked, each outer sphere is just touching the central sphere, and is surrounded by other spheres which will push it away from the central sphere if you try to move it in any direction. Plus, it looks to me like 60 degree packing is the most compact available, and "switching place" involves extending at least one of the angles to not be 60 degrees and thus not be most compact--which will force that sphere to lose contact with the center sphere... No?I can't visualize exactly how to exchange two spheres, but I don't have any great difficulty believing that it's possible. Not all the angles are 60 degrees. Some are 90.

Starting from the hexagonal close packing arrangement, I don't think it's possible to move just one sphere at a time without some sphere losing contact with the center. But if you move a bunch simultaneously, you can change continuously to the dodecahedral/icosadedral arrangement---where none of the twelve surrounding spheres touches any other---while keeping them all in contact with the center throughout the change. Basically, rotate the top three spheres as a unit and the bottom three spheres as another unit, in opposite directions, about a vertical axis, while allowing the six outer spheres of the middle layer to be pushed alternately up and down, until each top sphere is directly above the contact point between two bottom spheres, instead of being directly above a bottom sphere as in the beginning. Then, there's room to separate all the twelve spheres from each other.

[chipmunk stew]

Concerning the close packing scenario, here's something I found interesting in reading. The Hexagonal Close Packing (http://mathworld.wolfram.com/HexagonalClosePacking.html) structure mentioned above has the same packing density as the Cubic Close Packing (http://mathworld.wolfram.com/CubicClosePacking.html) structure. The only difference is the relative orientations of the top three and bottom three spheres and the resulting shape of the structure.

In the case of the Hexagonal Close Packing (http://mathworld.wolfram.com/HexagonalClosePacking.html) structure, the shape is called a triangular orthobicupola (http://mathworld.wolfram.com/TriangularOrthobicupola.html):

http://gabe.stonehillnews.com/jpegs/sphere%20assy%20triangular%20orthobicupola.jpg

In the Cubic Close Packing (http://mathworld.wolfram.com/CubicClosePacking.html) structure, the shape is a cuboctahedron (http://mathworld.wolfram.com/Cuboctahedron.html):

http://gabe.stonehillnews.com/jpegs/sphere%20assy%20cuboctahedron.jpg

[ReFLeX]

No wonder the Politics forum is what it is: there is rarely a clear answer that satisfies everyone. I love it.