## Quasicrystals

### Quasicrystals

Hello everyone. I'm trying to figure out how to make 2 more polyhedrons in addition to the rhombic triacontahedron and rhombic icosahedron. These 4 different shapes are augmented together in various ways in a non-periodic fashion to form a model of a quasicrystal which has no gaps and has true 5-fold icosahedral symmetry. The other polyhedra are all made entirely from golden rhombi - like the RTC, they include the following:

1. Rhombic Icosahedron - this is the RTC minus one of the zones around the center. 20 faces with different radii. I have figured this model out by manipulating an OFF file of an RTC.

2. Rhombic Dodecahedron - take a rhombic icosahedron described above and remove another zone resulting in a polyhedron with 12 sides. Not to be confused with the other RD which has a different ratio of diagonal lengths across each face equal to the square root of 2. Has to have diagonal length ratio equal to 1.618...

3. Rhombohedron - take a rhombic dodecahedron as described above and remove a zone resulting in a 6-sided polyhedron. Rob explained to me earlier how to get a rhombic hexecontahedron as a stellation of the RTC. 20 rhombohedrons can be put together surrounding a central point to make a rhombic hexecontahedron.

I can easily make all of these polyhedra with my box-jointed wood tiles as pictured below, but I want to use Great Stella to determine how it all fits together using augmentation. I have the 2 largest models of the 4 I need for this project. Any help making the Rhombohedron and Rhombic Dodecahedron would be greatly appreciated!

Here's a video where Professor Paul Steinhardt of Princeton discusses quasicrystals and shows his 3D printed models. https://www.youtube.com/watch?v=a0wo_yAh0Ps

1. Rhombic Icosahedron - this is the RTC minus one of the zones around the center. 20 faces with different radii. I have figured this model out by manipulating an OFF file of an RTC.

2. Rhombic Dodecahedron - take a rhombic icosahedron described above and remove another zone resulting in a polyhedron with 12 sides. Not to be confused with the other RD which has a different ratio of diagonal lengths across each face equal to the square root of 2. Has to have diagonal length ratio equal to 1.618...

3. Rhombohedron - take a rhombic dodecahedron as described above and remove a zone resulting in a 6-sided polyhedron. Rob explained to me earlier how to get a rhombic hexecontahedron as a stellation of the RTC. 20 rhombohedrons can be put together surrounding a central point to make a rhombic hexecontahedron.

I can easily make all of these polyhedra with my box-jointed wood tiles as pictured below, but I want to use Great Stella to determine how it all fits together using augmentation. I have the 2 largest models of the 4 I need for this project. Any help making the Rhombohedron and Rhombic Dodecahedron would be greatly appreciated!

Here's a video where Professor Paul Steinhardt of Princeton discusses quasicrystals and shows his 3D printed models. https://www.youtube.com/watch?v=a0wo_yAh0Ps

I don't know what to do to get more precise than 15 decimal places? I've read in your documentation that it is preferable to have at least 17 decimal places of accuracy. What would you suggest?

I was also able to make the Golden Rhombic Dodecahedron by augmenting the Rhombohedron onto itself, and then added a third one as an excavation. The convex hull of this shape is the shape I need - or at least within a few atoms. I guess I just need more accuracy. Any help is appreciated.

Steve

There are two kinds of rhombohedron, depending on whether you stretch or squash the cube along a diagonal. Once the two rhombohedra are scaled to the same edge length, all your figures can be assembled from copies of just these two, and in this respect they bear a close parallel to the original rhombic Penrose tiles.

A very short introduction is given at http://www.steelpillow.com/polyhedra/qu ... uasicr.htm

I don't know if this helps.

A very short introduction is given at http://www.steelpillow.com/polyhedra/qu ... uasicr.htm

I don't know if this helps.

Cheers,

Guy. Guy's polyhedra pages

Guy. Guy's polyhedra pages

Thank you, Guy. I need the trigonal type of rhombohedron where all the faces are the same shape - which will be golden rhombi. So stretching a cube along its 3-fold axis (which passes through opposite corners of the cube) should keep all six faces the same shape. It's not difficult except for the fact that the required precision exceeds the precision of the measurements that the program shows. A model that looks correct may be off beyond a certain number of decimal places. I think this also prevents the program from indicating all the various kinds of symmetry that the model should have which might affect the possible orientations when you augment it?guy wrote:There are two kinds of rhombohedron, depending on whether you stretch or squash the cube along a diagonal. Once the two rhombohedra are scaled to the same edge length, all your figures can be assembled from copies of just these two, and in this respect they bear a close parallel to the original rhombic Penrose tiles.

A very short introduction is given at http://www.steelpillow.com/polyhedra/qu ... uasicr.htm

I don't know if this helps.

Has anyone made this model before who could send the .stel file? I'm still monkeying around with it, but haven't gotten anywhere.

Yes, both the rhombohedra I describe have this property - the one with the golden rhombs lengthways, the other with them sideways.

Are you any good at geometry and basic algebra? It is reasonably easy to develop formulas for the stretch/shrink ratios.

Then feed the formula into a high-precision calculator such as http://apfloat.appspot.com/

Are you any good at geometry and basic algebra? It is reasonably easy to develop formulas for the stretch/shrink ratios.

Then feed the formula into a high-precision calculator such as http://apfloat.appspot.com/

Cheers,

Guy. Guy's polyhedra pages

Guy. Guy's polyhedra pages

I'm a bit rusty and slow on my trig, but can hammer it out if I set my mind to it. Going from plane to 3-D can also be confusing. Guess that's from relying on Autocad to do my drawing with for a long time - which apparently isn't accurate enough.guy wrote:Yes, both the rhombohedra I describe have this property - the one with the golden rhombs lengthways, the other with them sideways.

Are you any good at geometry and basic algebra? It is reasonably easy to develop formulas for the stretch/shrink ratios.

Then feed the formula into a high-precision calculator such as http://apfloat.appspot.com/

I found a nice little calculator program for my pc called "HiPER Calc" that works like the kind I am used to except it has more decimal places. Hopefully that will do the trick. http://www.techworld.com/download/syste ... 2-3330204/

Has anyone else here made a model of a quasicrystal?

Steve

The trick is to choose the right slices through the 3D object. It is easiest to explain for the cube.

Pick one vertex and identify the three adjacent ones. Cut round these three to remove a triangular pyramid. Do the same on the opposite side so you now have two pyramids and a square antiprism.

These three solids are all the same height, viz. one-third of a cube diagonal.

The problem is now reduced to one of calculating the ratio between the pyramid rising-edge length and its height.

The same can be done with the stretched/squashed rhomb, by starting from the vertices along the stretch axis.

By making the pyramid rising-edge length 1 in all cases, correct scaling is assured.

Pick one vertex and identify the three adjacent ones. Cut round these three to remove a triangular pyramid. Do the same on the opposite side so you now have two pyramids and a square antiprism.

These three solids are all the same height, viz. one-third of a cube diagonal.

The problem is now reduced to one of calculating the ratio between the pyramid rising-edge length and its height.

The same can be done with the stretched/squashed rhomb, by starting from the vertices along the stretch axis.

By making the pyramid rising-edge length 1 in all cases, correct scaling is assured.

Cheers,

Guy. Guy's polyhedra pages

Guy. Guy's polyhedra pages

guy wrote:The trick is to choose the right slices through the 3D object. It is easiest to explain for the cube.

Pick one vertex and identify the three adjacent ones. Cut round these three to remove a triangular pyramid. Do the same on the opposite side so you now have two pyramids and a square antiprism.

These three solids are all the same height, viz. one-third of a cube diagonal.

The problem is now reduced to one of calculating the ratio between the pyramid rising-edge length and its height.

The same can be done with the stretched/squashed rhomb, by starting from the vertices along the stretch axis.

By making the pyramid rising-edge length 1 in all cases, correct scaling is assured.

I think I finally got it precise enough now, ended up calculating coordinates of each of the 8 vertices and making it into an off file. I understand what you were talking about, but for whatever reason the numbers were still not right. By the way, the solid in the middle between the two pyramids would be a triangular antiprism instead of square antiprism Would the length of the space diagonal always be equally divided by the 3 solids when stretching a cube along the 3-fold axis?

I was able to construct an accurate Bilinski Dodecahedron by augmenting with the new rhombohedron, and the symmetry indicators show all the appropriate symmetry, so I am happy now having a precise model of each of the 4 golden polyhedra needed for my Quasicrystal model. I'll post some pictures soon of what it looks like as I build it up. Thanks again for the help!

Steve

I usually have to take quite a few goes before I get the same result often enough to believe it. Also key here are the ratios in the golden rhomb between the diagonals and the side.SteveG wrote:I understand what you were talking about, but for whatever reason the numbers were still not right. By the way, the solid in the middle between the two pyramids would be a triangular antiprism instead of square antiprism Would the length of the space diagonal always be equally divided by the 3 solids when stretching a cube along the 3-fold axis?

Triangular indeed, I was just checking to see if you were awake (ahem).

Yes, the proportions or ratios along any linear stretch axis are always fixed. More interesting is the perspective view, a projective transformation in which cross-ratios between the various distances are preserved although simple ratios are not. But I doubt you will want to go there.

Looking forward to the screen shots...

Cheers,

Guy. Guy's polyhedra pages

Guy. Guy's polyhedra pages

Continued augmenting rhombohedra, and Bilinski dodecahedra forming a cavity for the center of the quasicrystal which is a rhombic triacontahedron...

Now working along a plane building model which is a copy of Paul Steinhardt model...

Edge view of model showing location of central RTC in relation to other polyhedra...

Very nice, congratulations and thank you for sharing.

How about removing a few building blocks so that a central cavity connects to the outside? Besides looking cool, such "holey" crystals have found various applications as catalysts, molecular or atomic filters, etc. but I have never seen a quasicrystal variant. Might trigger a brain cell in a crystallographer or a physical chemist somewhere, who knows.

How about removing a few building blocks so that a central cavity connects to the outside? Besides looking cool, such "holey" crystals have found various applications as catalysts, molecular or atomic filters, etc. but I have never seen a quasicrystal variant. Might trigger a brain cell in a crystallographer or a physical chemist somewhere, who knows.

Cheers,

Guy. Guy's polyhedra pages

Guy. Guy's polyhedra pages