Circles and Pi

[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. )

 

[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:

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.

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.

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

The 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.htmlhttp://mathworld.wolfram.com/CubicClosePacking.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:

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. 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]

Tight fit

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!?!

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.

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




12 Closest packed around 1

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.

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?