Circles and Pi

[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:

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:

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) 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) 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 icosahedron

No 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 has twelve faces, whereas the regular icosahedron 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.

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 structure mentioned above has the same packing density as the Cubic Close Packing 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 structure, the shape is called a triangular orthobicupola:

In the Cubic Close Packing structure, the shape is a cuboctahedron:

 

[ReFLeX]

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