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SteveG
Joined: 24 Jul 2011 Posts: 9 Location: Arkansas

Posted: Fri Aug 05, 2016 7:51 pm Post subject: 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 nonperiodic fashion to form a model of a quasicrystal which has no gaps and has true 5fold 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 6sided 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 boxjointed 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


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SteveG
Joined: 24 Jul 2011 Posts: 9 Location: Arkansas

Posted: Fri Aug 05, 2016 10:53 pm Post subject: 


Update. I realized I could use nonuniform scaling to stretch a cube along a 3fold axis by a factor of 1.851229520935533 to make the square faces into golden rhombi for a Golden Rhombohedron. I arrived at this scale factor by process of elimination (my 7th grade math teacher would have called this the "Houdini method"  but it works for me) comparing the long diagonal length to the same diagonal on an RTC with the same edge length. This is the most digits I can get on my iphone calculator. I keep getting errors that suggest my model is not accurate enough when augmenting with other golden rhombus polyhedra, and the program crashes.
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 

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guy
Joined: 11 Feb 2008 Posts: 75 Location: England

Posted: Sat Aug 06, 2016 6:47 pm Post subject: 


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/quasicr/quasicr.htm
I don't know if this helps. _________________ Cheers,
Guy. Guy's polyhedra pages 

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SteveG
Joined: 24 Jul 2011 Posts: 9 Location: Arkansas

Posted: Sat Aug 06, 2016 10:36 pm Post subject: 


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/quasicr/quasicr.htm
I don't know if this helps. 
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 3fold 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?
Has anyone made this model before who could send the .stel file? I'm still monkeying around with it, but haven't gotten anywhere. 

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guy
Joined: 11 Feb 2008 Posts: 75 Location: England

Posted: Sun Aug 07, 2016 7:03 pm Post subject: 


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 highprecision calculator such as http://apfloat.appspot.com/ _________________ Cheers,
Guy. Guy's polyhedra pages 

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SteveG
Joined: 24 Jul 2011 Posts: 9 Location: Arkansas

Posted: Mon Aug 08, 2016 2:26 am Post subject: 


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 highprecision calculator such as http://apfloat.appspot.com/ 
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 3D 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.
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/systemdesktoptools/hipercalcscientific2323330204/
Has anyone else here made a model of a quasicrystal?
Steve 

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guy
Joined: 11 Feb 2008 Posts: 75 Location: England

Posted: Mon Aug 08, 2016 9:04 am Post subject: 


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. onethird of a cube diagonal.
The problem is now reduced to one of calculating the ratio between the pyramid risingedge 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 risingedge length 1 in all cases, correct scaling is assured. _________________ Cheers,
Guy. Guy's polyhedra pages 

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SteveG
Joined: 24 Jul 2011 Posts: 9 Location: Arkansas

Posted: Tue Aug 09, 2016 4:50 am Post subject: 


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. onethird of a cube diagonal.
The problem is now reduced to one of calculating the ratio between the pyramid risingedge 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 risingedge 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 3fold 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 

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guy
Joined: 11 Feb 2008 Posts: 75 Location: England

Posted: Tue Aug 09, 2016 9:01 am Post subject: 


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 3fold axis? 
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.
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 crossratios 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 

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SteveG
Joined: 24 Jul 2011 Posts: 9 Location: Arkansas

Posted: Tue Aug 09, 2016 7:23 pm Post subject: 


With all four polyhedra in memories, I started with a rhombohedron and augmented a Bilinski dodecahedron to 3 faces...
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...


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SteveG
Joined: 24 Jul 2011 Posts: 9 Location: Arkansas

Posted: Wed Aug 10, 2016 5:26 am Post subject: 


This may or may not be assembled in the proper sequence. There are many ways to assemble the 4 types of polyhedra without gaps. Apparently there is a sequence where the crystal can be assembled endlessly without voids or overlapping polyhedra boundaries. This is a lot of fun to play with! 

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guy
Joined: 11 Feb 2008 Posts: 75 Location: England

Posted: Wed Aug 10, 2016 8:49 am Post subject: 


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. _________________ Cheers,
Guy. Guy's polyhedra pages 

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