I'm stuck with my 'Balls in a vacuum'

[ch@rlatan]

Hey,

Can anyone help me with my imaginary experiment?

In a large glass tank I place air-filled balls of thin rubber (like a balloon but perfectly spherical). I fill the tank to the top with the rubber balls, seal the lid and evacuate the tank of air. The internal pressure difference causes the balls to completely fill the vacuum. But what shape do the balls take? Ignoring any effects caused by the balls distorted against the wall of the tank and any gravitational effects.

Trying to imagine this shape without doing the experiment has proved torturous for me. The structure is clearly going to be crystalline in nature but I need to determine exactly how. Can anybody help? It's a helluva lot tougher than it sounds.

 

[tiny-tim]

Hey ch@rlatan!

If you don't interfere with them, I think it'll depend on how you originally packed the spheres …

eg, if you packed them in a cubic lattice, I think they'll become cubes.

But if you shake them about while they're expanding, I expect they'll try to find the lowest-energy configuration.

Perhaps an easier question, with the same result, is what is the lowest-energy configuration if they remain spheres?

 

[xlines]

Hey,
In a large glass tank I place air-filled balls of thin rubber (like a balloon but perfectly spherical). I fill the tank to the top with the rubber balls...

Well, it will for certain matter how do you fill them up. Also, I am not sure you can achieve a perfectly symmetric initial condition with glass ball. You probably can, but I am not sure. But you can do so for sure, with tetrahedral with HCP packing of rubber balls. What happens on evacuation, which should be painfully slow, is an interesting problem.

I would go with primitive cell of HCP if there are many balls and that's obvious. Not sure if it is correct.

A personal note: When I saw title of the thread, I really thought you would need assistance of some other nature ;)

 

[ch@rlatan]

Hey tiny-tim


Yes they would be cubes if I packed them square.
And No your question is not valid because it is impossible for them to remain spheres - think about how the spaces between the spheres have to be taken up by the membrane of the ball. As the pressure builds and the balls' membranes start seeking out those gaps, points on the balls that were tangent prior to the evacuation will start to press against one another to become facets of what I think is a crystalline structure.

But what does it look like?

 

[ch@rlatan]

OK let's make these balls very slippery by adding a very fine oil so no internal friction occurs. It doesn't matter how they are packed as long as none are forced. The vacuum will arrange them naturally.

 

[Jobrag]

Assuming that the natural shape will be one that stresses the material that the ballons are made of least. What is the shape that will allow complete packing of the tank while giving the maximum volume to surface area ratio?

 

[skeptic2]

I believe that the space between the spherical balls with tetrahedral packing would be a tetrahedron but with concave sides. As the air is evacuated, the surface of the balls would expand toward the center of the tetrahedral spaces. It seems likely that given uniform spheres and interior pressures that the balls would transform into tetrahedrons.

 

[DaveC426913]

I believe that the space between the spherical balls with tetrahedral packing would be a tetrahedron but with concave sides.

No. Count the number of contact points with other spheres in a tetrahedral arrangement. Each of those contact points will become a face.

Note that each sphere is in contact with exactly 12 other spheres (3,6,3), regularly spaced around it. Those contact points will become faces of the final ball shape.

A solid shape with 12 faces is a dodecahedron.

 

[ch@rlatan]

Hey Dave

Glad somebody got their balls out.

Here's the problem. Look at the face of those steel balls - particularly the three in the middle. Each of those is surrounded by 6 others that are showing in the same plane. The dodecahedron is not consistent with this observation. Plus every time I work it out I arrive at fourteen contact points (4/6/4). What you are deriving is taken specifically from the tetrahedron. While there may be some tetrahedronic gaps to fill, I don't believe that the overall shape is swayed by this.
But good stuff and great pictures.

 

[Q_Goest]

Hey Dave

Glad somebody got their balls out.

Here's the problem. Look at the face of those steel balls - particularly the three in the middle. Each of those is surrounded by 6 others that are showing in the same plane. The dodecahedron is not consistent with this observation. Plus every time I work it out I arrive at fourteen contact points (4/6/4). What you are deriving is taken specifically from the tetrahedron. While there may be some tetrahedronic gaps to fill, I don't believe that the overall shape is swayed by this.
But good stuff and great pictures.

How about a "rhombic dodecahedron"

Ref: http://en.wikipedia.org/wiki/Rhombic_dodecahedron

 

[DrGreg]

How about a "rhombic dodecahedron"
Ref: http://en.wikipedia.org/wiki/Rhombic_dodecahedron

That looks like the answer to me.

http://en.wikipedia.org/wiki/Rhombic_dodecahedron]The[/PLAIN] rhombic dodecahedron can be used to tessellate 3-dimensional space. It can be stacked to fill a space much like hexagons fill a plane.

This tessellation can be seen as the Voronoi tessellation of the face-centred cubic lattice. Some minerals such as garnet form a rhombic dodecahedral crystal habit. Honeybees use the geometry of rhombic dodecahedra to form honeycomb from a tessellation of cells each of which is a hexagonal prism capped with half a rhombic dodecahedron. The rhombic dodecahedron also appears in the unit cells of diamond and diamondoids. In these cases four vertices are absent, but the chemical bonds lie on the remaining edges.

 

[ch@rlatan]

EUREKA

Well done you guys, one and all.
Sweet little cherry on the top from Q_Goest.
Been wrestling with that puppy for some time now.
So Thanks

Ch@rlatan

 

[skeptic2]

ch@rlatan, I'm afraid I don't see the 14 contact points (4/6/4). I see only 12 (3/6/3) as Dave mentioned. But can dodecahedrons be packed without any leftover space?

 

[DaveC426913]

ch@rlatan, I'm afraid I don't see the 14 contact points (4/6/4). I see only 12 (3/6/3) as Dave mentioned.

Thanks. I doubted myself.

See attached graphic.

But can dodecahedrons be packed without any leftover space?

Yes.

 

[harriswillys]

This is not a new question - or perhaps its a novel restatement of a question that's intrigued physicists for decades. Soap bubbles provide the classic example of this maximum volume and minimum surface area problem.

Shapes will vary according to number and size of balloons (bubbles) and the shape of the container - there is no "one" default shape.

The determined experimenter could build a clear plastic box, obtain a child's bubble wand and blow uniform sized bubbles one by one to reach more specific conclusions.

 

[DaveC426913]

This is not a new question - or perhaps its a novel restatement of a question that's intrigued physicists for decades. Soap bubbles provide the classic example of this maximum volume and minimum surface area problem.

Shapes will vary according to number and size of balloons (bubbles) and the shape of the container - there is no "one" default shape.

The determined experimenter could build a clear plastic box, obtain a child's bubble wand and blow uniform sized bubbles one by one to reach more specific conclusions.

It certainly borrows a lot from the soap bubble question but really it's not the same.

For one, maximum volume, minimum surface area really has nothing to do with this problem.
For two, the spheres are stacked manually in a specific configuration, not allowed to find their own configuration.

It's really just a simple sphere packing problem.

(See https://www.physicsforums.com/showpost.php?p=2643180&postcount=14" where I illustrate the tetrahedral packing. You have a sphere that touches 12 adjacent spheres, and thus expands into a 12-sided dodecahedron. )

 

[harriswillys]

It certainly borrows a lot from the soap bubble question but really it's not the same.

For one, maximum volume, minimum surface area really has nothing to do with this problem.
For two, the spheres are stacked manually in a specific configuration, not allowed to find their own configuration.

It's really just a simple sphere packing problem.

(See https://www.physicsforums.com/showpost.php?p=2643180&postcount=14" where I illustrate the tetrahedral packing. You have a sphere that touches 12 adjacent spheres, and thus expands into a 12-sided dodecahedron. )

Well, sure, if you assume stacked spheres then yes, the shapes can indeed be predicted. If random sized balloons are placed randomly in a box and the air evacuated, you have the soap bubble problem.

I am not disagreeing with you, simply pointing out that different assumptions on the starting state of the thought-experiment will yield different results.

 

[ch@rlatan]

Hey skeptic2

"ch@rlatan, I'm afraid I don't see the 14 contact points (4/6/4). I see only 12 (3/6/3) as Dave mentioned" Originally posted by skeptic2

Yeah, in my original thinking I was confusing my vertices and facets, which is why I put the question to the forum. A rhombic dodecahedron has 12 facets and 14 vertices. 14 was the number I was hoping to find. I thought I was getting it in facets instead I got it in vertices.

Hey DaveC429613

"It's really just a simple sphere packing problem" Originally posted by DaveC429613

Yeah, I thought so at first, but Harriswillys has made me look at the experiment in a little more detail.

In this experiment let me state that; every balloon is identical to every other; the balloon membrane is elastic and frictionless; the size of the balls are very small relative to the tank so that the shape of the tank becomes insignificant; the experiment should be examined without gravitational effects, slight though they might be; it is not important that the balls are laid in any order nor that the tank is full.
All the energy from the vacuum is going to end up in the tension of the membrane. It's clear then that the membrane will seek minimum surface area and minimum volume to maintain a minimum tension. As all other balloon membranes seek to do the same, the tensions and therefore volumes will be the same.
The soap bubble consideration put forward by harriswillys got me thinking about fluidity. Could it be that the stacking analogy is just a little too rigid. Is it really analogous to the more frictionless and fluid alignment that the experiment above would attain. The beauty of the rhombic dodecahedron is that it fills space perfectly. Any proposed shape will have to fit that criteria. Some more investigation is needed.