Dividing a sphere into twelve "identical" shapes.

[tar]

A few days ago I solved a challenge I had made myself, and think the division is neat, with a lot of nice symmetries and such, but several searches on the web and I have not found its name, or a breakdown of it, to study the angles and descriptions and such. So I am presenting it here, in the hopes that someone will recognize it, and point me to a link with its description.

 

Its made of clay and was done by eye with pen and steak knife, so its not exact, but it works.

 

What is this division called?

 

Regards, TAR

Edited March 26, 2014 by tar

[md65536]

A 12-sided polyhedron is a dodecahedron. This would be a "rhombic dodecahedron." As a sphere it is a spherical rhombic dodecahedron.

 

http://en.wikipedia.org/wiki/File:Rhombic_dodecahedron_spherical.png

[tar]

Thank you MD65536,

 

That's it!

 

Regards, TAR

 

Do the wedges have a name? A pyramid with a rounded base? Or is there some name for the curvature being 1/12th of a sphere?

Edited March 26, 2014 by tar

[imatfaal]

Thank you MD65536,

 

That's it!

 

Regards, TAR

 

Do the wedges have a name? A pyramid with a rounded base? Or is there some name for the curvature being 1/12th of a sphere?

 

 

Or is there some name for the curvature being 1/12th of a sphere?

 

The steradian is the SI unit of solid angle - http://en.wikipedia.org/wiki/Steradian

[tar]

imatfaal,

 

OK, thanks, but I am having a little trouble understanding the % difference between the surface area subtended by the steradian and the surface area subtended by my TARadian unit.

 

From Wiki:

"Because the surface area A of a sphere is 4πr², the definition implies that a sphere measures 4π (≈ 12.56637) steradians. By the same argument, the maximum solid angle that can be subtended at any point is 4π sr."

 

If it takes around 12.56637 sr to subtend the sphere, and exactly 12 tr to subtend the sphere, then the tr is 1.0742 of an sr, or the sr is.9549 of a tr.

 

I sort of like the tr better since they can be fitted together to cover the whole sphere exactly, whereas a circle on a sphere is hard to do anything with, visualizationwise. You can't readily see how many are going to fit around the sphere, or how to handle that funny shape inbetween you get when you touch 3 or 4 cones together.

 

In any case the division you get, the shape you get when you carve the lines (radii) of a spherical rhombic dodecahedron to the center, can not be called a steradian, as it is neither conical in shape, nor circular at the subtend point, nor does it subtend a unit area of the surface of the sphere. It must have another name.

 

Regards, TAR

you could cut up a globe into those diamond shapes and lay them all out where you can see them at once

you can't do that so easily with circles

[md65536]

If it takes around 12.56637 sr to subtend the sphere, and exactly 12 tr to subtend the sphere, then the tr is 1.0742 of an sr, or the sr is.9549 of a tr.

 

I sort of like the tr better since they can be fitted together to cover the whole sphere exactly, whereas a circle on a sphere is hard to do anything with, visualizationwise. You can't readily see how many are going to fit around the sphere, or how to handle that funny shape inbetween you get when you touch 3 or 4 cones together.

 

In any case the division you get, the shape you get when you carve the lines (radii) of a spherical rhombic dodecahedron to the center, can not be called a steradian, as it is neither conical in shape, nor circular at the subtend point, nor does it subtend a unit area of the surface of the sphere. It must have another name.

 

Regards, TAR

 

you could cut up a globe into those diamond shapes and lay them all out where you can see them at once

you can't do that so easily with circles

"tr is 1.0742 of an sr" means that for a unit sphere the rhombic shapes have a surface area 1.0742 times that of a unit area on the sphere. The area doesn't need to be a circle. For example, a square 1 unit by 1 unit has 1 unit of area. On a unit sphere it subtends 1 sr.

 

You may prefer your shapes to unit areas. Similarly, you might imagine cutting a unit circle into 6 equal slices, creating 6 equilateral triangles within the circle. Each triangle has a side length of 1, and its arc of the circle is slightly more. So you may say "I prefer to measure circumference using 6 special units instead of 2pi standard units," but probably no one will care.

 

 

If I had to describe the shape of the wedges, the best I can think of is the "intersection of a rhombic pyramid and a smaller ball centered at the pyramid's apex".

 

You might try to see if you can cut up a world map into these shapes. Do the continents fit nicely in them? Perhaps someone has done it before, or you could create a new map projection.

 

 

 

 

 

One thing I notice: On the vertices shared by 4 pieces, there are basically 2 straight lines cutting across each other. You can join 3 pieces together (those that share a 3-piece vertex) and end up with a triangle shaped section of the sphere's surface. Then you'd have a sphere cut into 4 identical pieces, which you would get if you cut a ball into a spherical tetrahedron, with each wedge being roughly a tetrahedron with a rounded face.

 

Likewise you could do the opposite and cut each of your 12 wedges into 2 equal triangular pieces, or 4 etc. Or you could take the spherical tetrahedron and cut each piece into 3 identical pieces in other ways (if you cut down the middle of an edge, you get your rhombus shapes, but if you cut from the corners you can make triangular shapes) and get other divisions of a sphere into 12 identical pieces.

[Janus]

Just for fun, I rendered a version of the sphere divided into 12 identical sections:

 

[tar]

Janus,

 

That's great? How'd you do that?

 

Is that actual or vitual?

 

Regards, TAR

 

Can we call that section a Janus if we find it does not yet have a name?

 

Regards, TAR

 

MD65336,

 

Your'e right, we already have an SI unit for solid angle, that is nicely based on Pi so everything works out nicely already, nobody has the need for, or would care about a special unit.

 

However, I still find it rather neat, and workable in visualizing how solid space fits together.

 

It evovled, from an earlier "discovery" of mine drawing on clay spheres, and trying to work out the closest pack possible of same size balls. I know these things already have names, have been discovered and studied and measured and such, but it gives me a little sense of ownership to have found it out for myself. Here is my ping pong ball version of how space is put together.

 

 

Twelve balls, fit exactly around a center ball, which results in some nice symmetries and intersection of hexagonal and square planes, when built out.

 

Regards, TAR

Edited March 28, 2014 by tar

[Janus]

Janus,

 

That's great? How'd you do that?

 

Is that actual or vitual?

 

Regards, TAR

 

It's a CGI image I created with a 3-D modeling program.

 

Here's an animated version showing a sphere being built piece by piece.

 

[tar]

Janus,

 

I received a "this video is private" attempting to view the build.

 

But you're still my hero. That is fantastic that you can do that.

 

 

Here are the 4 identical sections that MD noticed.

[Janus]

 

I received a "this video is private" attempting to view the build.

 

 

I reset the setting, try it now.

[tar]

Janus,

 

That worked. Thanks.

 

How much is that 3-D modeling program?

 

How much is the Janus brain to operate it?

 

I was wondering if you could do the ping pong ball figure.

 

Another long term challange I have is to understand the shape you get when you build out the ping ball ball lattice, one "spherical" layer at a time. If twelve are on the first layer, how many on on the second? And the third? and so on.

 

With this particular arrangement of "fitting" 90 and 60 degree angles, the intersecting hexagonal planes that establish square planes as well as you build out, I have a hard time visualizing what is going to happen. And have not determined what I should consider a complete "next layer". Seems its going to take a lot of ping pong balls and some correct decisions and insights...none of which I have obtained as of yet.

 

Regards, TAR

[Janus]

Janus,

 

That worked. Thanks.

 

How much is that 3-D modeling program?

 

How much is the Janus brain to operate it?

 

I was wondering if you could do the ping pong ball figure.

 

Another long term challange I have is to understand the shape you get when you build out the ping ball ball lattice, one "spherical" layer at a time. If twelve are on the first layer, how many on on the second? And the third? and so on.

 

With this particular arrangement of "fitting" 90 and 60 degree angles, the intersecting hexagonal planes that establish square planes as well as you build out, I have a hard time visualizing what is going to happen. And have not determined what I should consider a complete "next layer". Seems its going to take a lot of ping pong balls and some correct decisions and insights...none of which I have obtained as of yet.

 

Regards, TAR

Rendering software- free

Modeling software- free

Expertise in usage- priceless!

 

Here's the ping pong figure being built up glass ball by glass ball:

 

[Janus]

Here's a video of the building of the next "layer":

 

[tar]

Janus,

 

That is fantastic.

 

Except we need the top three flipped to get the intersecting hexagonal and square planes. That is, each ball on the first layer is in line with the center ball and a ball directly on the other side of the center ball. Where you see the two balls on the bottom three, you should see one ball on the top. In your build the top three were in the same holes as the bottom three. You have to rotate the whole third plane 60 degrees into the next set of homes.

 

Then make the next layer.

 

Regards, TAR

You are still my hero. Can you PM me the rendering and modeling links?

 

The priceless part I guess I will just have to borrow from you on occasion. (just for fun)

[Janus]

Janus,

 

That is fantastic.

 

Except we need the top three flipped to get the intersecting hexagonal and square planes. That is, each ball on the first layer is in line with the center ball and a ball directly on the other side of the center ball. Where you see the two balls on the bottom three, you should see one ball on the top. In your build the top three were in the same holes as the bottom three. You have to rotate the whole third plane 60 degrees into the next set of homes.

 

Then make the next layer.

 

Regards, TAR

You are still my hero. Can you PM me the rendering and modeling links?

 

The priceless part I guess I will just have to borrow from you on occasion. (just for fun)

I get what you wanted, but both ways are valid solutions, in that the next "layer" fits snug to the layer inside. The alternating square and triangle panels is more of a esthetics issue. I'd post a image of the other solution, as I have it rendered, but my image hosting site is giving me problems right now.

 

POV-ray is the renderand can be found at: http://www.povray.org

 

Moray, which is the modeling program, can be found here: http://www.stmuc.com/moray/

[tar]

Janus,

 

Wonderful.

 

The fliped third hexagonal plane though is more valid here, because it renders the exact arrangement of the pen holes in my clay sphere. And the pen holes are exactly in the center of each of the diamonds. It's THAT arrangement I desire to see the "next" layer of. I have been unable to achieve it in actual balls, but you can do it in virtual ones. So this is very exciting for me, as I have had it on my mind for well over a decade.

 

The retention of the intersecting hexagonal and square planes is crucial.

 

MD65536,

 

"(if you cut down the middle of an edge, you get your rhombus shapes, but if you cut from the corners you can make triangular shapes)"

 

I cut them in half the way you said, and then again. The rhombus shape is retained to the same proportions, and I think its great.

I noticed that the long axis of the diamond appears to be 1/4 of a sphere circumfrence long, and I am guessing/estimating that the short axis of the diamond is 1/6 of a sphere. These are nice figures, considering the 60 degree matrices and the 90 degree matrices that this particular arrangement of pen holes around a sphere, sets up.

 

Also gives some credence to entertaining the TARadian as a useful measure of solid angles as that the figures cut are 1 tr, 1/16 tr and 1/64 tr, respectively, and they designate 1/12, 1/48, and 1/192 of a sphere respectively. In a rather solid and sure and regular way that is easy to visualize and compute.

 

Regards, TAR

[Janus]

Here's the version you wanted:

 

 

Upon looking a little further, what this shape represents is a Hexahedron called a triangular dipyramid (basically two tetrahedrons joined at the base) with the "corners" lopped off.

 

Here it is with the corners added back on in a different color to highlight.

 

Edited March 30, 2014 by Janus

[tar]

Thankyou Janus.

 

That's the most beatiful thing I have ever seen. Its the figure I have been flirting with, but could never get a hold of. Thanks a million. (actually I have seen more beautiful things...but how the square planes and the hexagonal planes conspire at those angles to make each other, is really, really nice.) Symmetry to the max.

 

Thanks a million. (by the way, my computer is way to underpowered for the POV and I couldn't get the modeling program to run, so your putting that together for me is PRICELESS.)

 

You remain my superhero.

 

Best Regards, TAR

[md65536]

Rendering software- free

Modeling software- free

Expertise in usage- priceless!

How are you modelling these? I would try calculating locations using a pov-ray script, but it sounds like Moray is easier. Do you have to figure out where to put everything yourself, or does it automatically fit the pieces so they're touching?

Thanks a million. (by the way, my computer is way to underpowered for the POV and I couldn't get the modeling program to run, so your putting that together for me is PRICELESS.)

You should be able to run POV on even very old computers. You can render in low resolution without antialiasing to speed it up. With some basic scripting and some math I think you could replicate some of these images.

 

 

If you like fitting spheres physically, you might try magnetic buckyballs

https://www.google.ca/search?q=buckyballs&tbm=isch

(Do they still make them? There are several other similar brands, but last I heard they stopped selling them due to swallowing hazards.)

It can be frustrating when the magnets don't want to go where you want them to, but many interesting shapes can be put together.

 

 

 

 

So I was thinking about this a little more, wondering why you care about the surface area of a sphere. In your construction, if you expand your spheres uniformly to take up all space, you should end up with dodecahedrons. Basically, each of the 12 points where your spheres touch should form a face.

 

On the other hand, if you shrunk your spheres to the smallest convex polyhedron that still shares all of the same (12) contacts (Ah, it looks like this notion is a "dual" http://en.wikipedia.org/wiki/Dual_polyhedron), you would get an icosahedron, with 12 vertexes but 20 triangular faces. However it's not a regular icosahedron, because not all of the triangles are equal. Edit: I may have this wrong. Perhaps you get an icosahedron with 12 vertexes and 14 faces (6 of them rectangles).

 

Looking at a regular icosahedron, I wonder if that is a better shape? Of course the symmetry is nice, but is the goal here to pack the spheres as densely as possible? I made one with buckyballs and it looks pretty dense. Perhaps Janus could model one??? Here's how you would do it: Have one layer of 5 balls in a ring. Fit another 5-ball ring on top of that (rotated 2pi/10). Each ring leaves a "hole" in the middle into which you fit 1 ball as a top and bottom "layer". So that's 12 balls again, in 4 layers however it looks very similar length between opposite vertexes as does your irregular icosahedron. What I don't know is:

- Is this shape actually smaller? (Should be a reasonable challenge to calculate)

- Do these shapes fit together as well as yours?

I wonder if the best way to pack a single layer (which you have) is not necessarily the best way to pack a volume? It seems so, but perhaps shifting the balls slightly out of a flat layer might give up some space to be exploited???

 

Edit: The icosahedron shapes you end up with (a ring of 3, then one of 6, then one of 3) has 12 balls on the outside but it can fit another one in the middle, so I'm pretty sure it stacks balls more efficiently than the regular icosahedron can.

 

Definitely interesting... I might try to produce a povray scene sometime if curiosity eventually beats laziness. I'm sure there are scripted ball-stacking povray scenes that can be found on the web, too.

Edited March 31, 2014 by md65536

[Janus]

Thankyou Janus.

 

That's the most beatiful thing I have ever seen. Its the figure I have been flirting with, but could never get a hold of. Thanks a million. (actually I have seen more beautiful things...but how the square planes and the hexagonal planes conspire at those angles to make each other, is really, really nice.) Symmetry to the max.

 

Thanks a million. (by the way, my computer is way to underpowered for the POV and I couldn't get the modeling program to run, so your putting that together for me is PRICELESS.)

 

You remain my superhero.

 

Best Regards, TAR

Really? Just how old a system do you have? Even if you can't run version 3.7, They still offer 3.6 and that ran well on my last computer, which was 6+ years old when I replaced it a year ago. And if that won't run, they have archived even the older versions.

 

When you say that you can't get the Modeler to run, do you mean that it won't even start-up? Again, it ran fine on my old computer. You need POV-Ray to use it anyway, because Moray uses its rendering engine to actually generate the image.

How are you modelling these? I would try calculating locations using a pov-ray script, but it sounds like Moray is easier. Do you have to figure out where to put everything yourself, or does it automatically fit the pieces so they're touching?

Yes you do have to figure it out for yourself, but it can be simpler when you can see the model visually (top, front, side and camera views) There are also some short cuts you can take by creating groups or unions and duplicating them, etc.

 

Moray is an older modeling program, and since the Pov-ray team took over its licensing hasn't been developed further (its last version was mated to POV-Ray 3.5). There are likely better modelers out there, but I've become comfortable with it and its free.

 

It does have its glitches. Sometimes if you try to copy and paste in a scene that is already complex, it can crash. The nice thing is that since it exports directly as a Pov-ray scene file, you can work around this. I basically can model the different objects in the scene and then assemble the scene through the Pov-ray editor.

For example, this is how I assembled this image. Each of the figures were modeled separately in Moray and then assembled in a single scene by editing a Pov-ray scene file.

 

 

This ability also helps when you want to incorporate some of the features in the newer versions of Pov-ray that Moray doesn't support. (like the new pavement and tile patterns in 3.7)

 

It also had some nice features, like a mesh editor, which allows you to generate a mesh from any object and then manipulate the mesh by points, lines or faces. You can also pan, track, dolly and orbit the camera with the mouse.

 

As far as fitting the spheres more tightly, here is a comparison. On the right is Tar's original method, on the left, a different method both use 13 total spheres. The encompassing boundary spheres around both are of equal size. As can be seen, the one on the left fits entirely inside the boundary, while the one on the right doesn't.

 

Edited March 31, 2014 by Janus

[tar]

Janus,

 

I am working with an old laptop, without its original keyboard, that my daughter used in highschool. She is now a doctoral candidate at VT. I have put so many things on it and taken them off, the registry is probably a tangled mess.

 

Anyway, I will try an older version of the Moray. Do you have to link the POV to the Moray, or the Moray to the POV.

 

And thanks a lot for chrushing my hopes with the pentagonal pack. Looks like MD wins the close pack challenge.

 

Still I think the intersecting planes in mine are sweet. As I recall, I had built the pentagonal one and discarded it, for some reason.

The multiple axis in the one I settled on, looked more promising. But non-the-less, it looks like the triangle wins over the diamond...when it comes to a close packed sphere.

 

You have any problems building out the one on the left? There was some reason I didn't like it. Something didn't work out. Can't remember what it was.

 

Regards, TAR

[md65536]

As far as fitting the spheres more tightly, here is a comparison. On the right is Tar's original method, on the left, a different method both use 13 total spheres. The encompassing boundary spheres around both are of equal size. As can be seen, the one on the left fits entirely inside the boundary, while the one on the right doesn't.

How did you manage 13 on the left? Is there one in the center? The one I built with buckyballs wouldn't fit that. Can you easily render the intersections of pairs of spheres to show they're not overlapping?

 

You have any problems building out the one on the left? There was some reason I didn't like it. Something didn't work out. Can't remember what it was.

Yes, I don't see how they'll fit together as nicely as the hexagon-based ones.

 

Anyway while searching for a ball-stacking POV script I came across a page that claims, "Mathematicians have not yet reached consensus on a proof that a Barlow packing, including the face-centered cubic (fcc) and hexagonal (hcp) is actually the densest possible, although Gauss proved the fcc’s density of approximately 0.74 optimal for a lattice (any denser arrangement would have to be more random)." -- http://grunch.net/archives/56 I'm not sure how reliable this is.

 

Also: http://mathworld.wolfram.com/KeplerConjecture.html so it seems reliable.

Edited March 31, 2014 by md65536

[Janus]

Janus,

 

I am working with an old laptop, without its original keyboard, that my daughter used in highschool. She is now a doctoral candidate at VT. I have put so many things on it and taken them off, the registry is probably a tangled mess.

 

Anyway, I will try an older version of the Moray. Do you have to link the POV to the Moray, or the Moray to the POV.

You shouldn't have to do either. If you load in Pov-ray (I'd try ver. 3.6) and then load Moray, Moray will recognize POV-ray and make the connection for you. Then if you tell Moray to render, it will open POV-ray and start the render. If for some reason, it doesn't, you can do the following: once you've named and saved your Model file, go to the menu bar and hit "render" and then "export". This will create a POV-ray scene file that you can open and render directly in POV-ray.

And thanks a lot for chrushing my hopes with the pentagonal pack. Looks like MD wins the close pack challenge.

 

Still I think the intersecting planes in mine are sweet. As I recall, I had built the pentagonal one and discarded it, for some reason.

The multiple axis in the one I settled on, looked more promising. But non-the-less, it looks like the triangle wins over the diamond...when it comes to a close packed sphere.

 

You have any problems building out the one on the left? There was some reason I didn't like it. Something didn't work out. Can't remember what it was.

 

Regards, TAR

 

 

Don't hang your head quite yet. After stepping away for a while and then coming back to it, something started bothering about the whole thing. Then it hit me, neither method could fit into smaller sphere than the other could. Each method involves placing 12 balls surrounding and touching a single middle ball. But that automatically means that the smallest radius sphere that can enclose either group of balls has a radius 1 1/2 that of the balls. So I went back and looked at the my second model and found a small scaling error. It wasn't obvious, but it was enough to throw things off. It looks like its a tie.

 

Here's how the exterior vertices work for both: Your "square and triangle" version and the all triangles version:

 

Edited April 1, 2014 by Janus

[tar]

Janus,

 

As I recall, the tetrahedra on the left do not complete the sphere. That is, if you make 12 regular ones and fit them together, they "look" like they are going to complete a "sphere", but they are a "little" off. Resulting in the reality, that if you are to join all the edges on the outside...you can't. The radii going toward the center are just a little too long...and you wind up with spaces. This would mean that twelve balls would not exactly fit around a center sphere of the same radius, in this arrangement. However, it appears that twelve balls do fit exactly around the figure on the right, and the edges shown are exactly one radii long. As are the internal edges of the six pyramids and eight tetrahedra, that point toward the center (not shown in the wire frame). This results in the fact that placing a ball at each vertices, each with a diameter equal to the length of an edge, and one in the center of the figure will result in the ping pong ball arrangement, and each ball will use exactly half its diameter to reach the outside edge of each of the twelve balls that surround it.

 

Being that this positioning of balls is exactly the same as taking a regular cube and placing the appropriate size ball exactly at the midpoint of each of the twelve edges of the cube, and one of the same size in the exact center of the cube, the division of the cube into twelve equal solid angles, from its center point, is assured by this arrangement. As well, the division of the sphere in the center, into twelve equal solid angles is likewise insured. The diamond shapes in my original figure, are the result of forming the flat perpendiculars exactly half way between the adjacent pen holes(vertices).

 

So in this regard, my close packing, wins, and space can be discribed, from any one point, as either 6 square areas and 8 triangular ones, in this arrangement, or the twelve identical diamond shapes that result from this arrangement.

 

Regards, TAR