Marek14 wrote:Why is face 39 not closed...?
I don't know, ask quickfur... all I'm doing is passing your input to his program after all.
Marek14 wrote:Why is face 39 not closed...?
quickfur wrote:[...]
I see, so basically it's the same procedure that we applied to o5x3o3o to delete a bunch of equatorial vertices and collapse the two halves together to form the bilbiroes. That's certainly an interesting idea. Maybe I can try deleting some vertices from my o5x3o3x model to see what comes out...
student91 wrote:quickfur wrote:[...]
I see, so basically it's the same procedure that we applied to o5x3o3o to delete a bunch of equatorial vertices and collapse the two halves together to form the bilbiroes. That's certainly an interesting idea. Maybe I can try deleting some vertices from my o5x3o3x model to see what comes out...
I just discovered that if you delete all vertices of which abs( x-coordinate ) <phi/2 (or just phi if the edge-length is two ), and then add the vector (-phi/2,0,0,0) to the vertices with x-coordinate>=phi/2 and add the vector (phi/2,0,0,0) to the vertices with x-coordinate <=-phi/2, you should get the bilbiro'd o5x3o3x. you only have to insert the pentagonal prisms to make it CRF then.
quickfur wrote:student91 wrote:quickfur wrote:[...]
I see, so basically it's the same procedure that we applied to o5x3o3o to delete a bunch of equatorial vertices and collapse the two halves together to form the bilbiroes. That's certainly an interesting idea. Maybe I can try deleting some vertices from my o5x3o3x model to see what comes out...
I just discovered that if you delete all vertices of which abs( x-coordinate ) <phi/2 (or just phi if the edge-length is two ), and then add the vector (-phi/2,0,0,0) to the vertices with x-coordinate>=phi/2 and add the vector (phi/2,0,0,0) to the vertices with x-coordinate <=-phi/2, you should get the bilbiro'd o5x3o3x. you only have to insert the pentagonal prisms to make it CRF then.
Yes, that's what I did last night, but it turns out that 12 of the o5x3o's have their pentagons removed, resulting in non-CRF cells with phi-scaled pentagons. These pentagons moreover are too close to each other in the northern/southern hemispheres, resulting in sub-unit edges, and also introduce a whole bunch of non-CRF pyramids of the triangular cupola fragments. However, deleting these pentagons from these non-CRF diminished o5x3o's turns them into CRF pentagonal rotunda, and introduces some double-length edges. Probably adding a few more vertices (your pentagonal prisms?) will make it CRF. We shall see. I'm currently trying to fix an error I made in the model somewhere else, but the lattice enumeration algorithm is taking a long time (too many cells in o5x3o3x!).
Edit: the problematic pentagons have vertices at x=±1, so subtracting phi/2 causes them to introduce sub-unit edges.
student91 wrote:[...]
Those f-pentagon vertices should be removed. To make it CRF again, you should insert one pentagonal prism per "gap", as shown in my lace city's. After that's done, everything should be CRF. I don't know a fast way to calculate the coordinates of the pentagonal prisms, but they should be located at two x5o3A's, with A some number. I know it's not the easiest CRF to calculate coordinates for .
quickfur wrote:Hmm. Using the trick of "completing the hexagon", that is, by adding the edge vectors of the incomplete hexagons in the bisected hexagonal prisms to find the missing points, I managed to obtain two sample points of the pentagonal prisms on either side of a J91, and measuring the distance between them yielded A=6*phi. I'm going to try to compute x5o3A now and see if that works...
Klitzing wrote:quickfur wrote:Hmm. Using the trick of "completing the hexagon", that is, by adding the edge vectors of the incomplete hexagons in the bisected hexagonal prisms to find the missing points, I managed to obtain two sample points of the pentagonal prisms on either side of a J91, and measuring the distance between them yielded A=6*phi. I'm going to try to compute x5o3A now and see if that works...
If phi = 1.618, then it cannot be true, at least as long you use unit edges.
Consider the following:
those pentagon vertices will have a distance which allows for a (non-straight) path, being a shortchord of a hexagon (at hip), then a shortchord of a pentagon (at bilbiro), and then again of a hexagon (at hip). That is, according to the triangle inequality (or rather: tetragon inequality, hehe) you'd have A ≤ h + f + h = 1.732 + 1.618 + 1.732 = 5.082.
Whereas your value would be 9.708.
Still, your idea by completing the hexagons should work - if done correctly...
--- rk
Marek14 wrote:BTW, does anyone have off files for all the ursachora and 600-cell lunes?
quickfur wrote:Marek14 wrote:BTW, does anyone have off files for all the ursachora and 600-cell lunes?
I haven't built all of the ursachora, but everything that I made renders of, I have the .def files, so I can convert them to .off easily. So whatever has images posted here or on the wiki, I can give you the .off files. I'll see if I can post some of them tomorrow, but I might be busy, so maybe Saturday.
Marek14 wrote:BTW, does anyone have off files for all the ursachora and 600-cell lunes?
quickfur wrote:Sorry, I forgot to mention that I customarily work with edge length 2, so the value of A for me is 6*phi, but if you're using unit edge it should be 3*phi. And I just built the model, and all edge lengths are equal, so looks like it's CRF after all! I'll post images later -- I just tweaked something and running the convex hull / lattice enumeration again, so by tomorrow I'll be able to post images.
Good job, student91, looks like your idea worked!
quickfur wrote:[...] And I just built the model, and all edge lengths are equal, so looks like it's CRF after all! I'll post images later -- I just tweaked something and running the convex hull / lattice enumeration again, so by tomorrow I'll be able to post images.
Good job, student91, looks like your idea worked!
student91 wrote:[...]quickfur wrote:Good job, student91, looks like your idea worked!
Hurray!! I think we've now found most of the bilbiro'd polytopes.
It might just be possible that things with id-cuts will allow a relatively simple bilbiro-ing. The only things with id-cuts are as far as I know the o5o3x3o and the o5o3o3x. The latter probably won't be bilbiro-able. This means from now on I'll be looking for a bilbiro-ing of the o5o3x3o. I think that when I've investigated that one, I'll be able to say something about the other ways of bilbiro-ing you proposed.
Marek14 wrote:Here's a question:
I had a look at the hemi-600cell now. Of course, the dihedral angles at its icosidodecahedron won't allow gluing two together (which is also clear from its construction).
But -- what if the bilbiro process was used on this icosidodecahedron? Could that allow for duplicating the shape? Basically, it would be bilbiroing the middle layer of x3o3o5o, despite the fact that it usually doesn't contain any icosidodecahedra.
well, I thought bilbiro's are only derivable from id's, and also only from the .2.-view of the id. Now we've bilbiro'd all things with o5x3o3. So that means any similar bilbiro-ing (we might call it "obvious" bilbiro-ing) doesn't exist. Any less-obvious bilbiro-ing of course is not excluded, but I suspect it to be not very productive.quickfur wrote:student91 wrote:[...]quickfur wrote:Good job, student91, looks like your idea worked!
Hurray!! I think we've now found most of the bilbiro'd polytopes.
Oh? How do you know that? Did you discover something we don't know, that limits the number of bilbiro-able polytopes?
It might just be possible that things with id-cuts will allow a relatively simple bilbiro-ing. The only things with id-cuts are as far as I know the o5o3x3o and the o5o3o3x. The latter probably won't be bilbiro-able. This means from now on I'll be looking for a bilbiro-ing of the o5o3x3o. I think that when I've investigated that one, I'll be able to say something about the other ways of bilbiro-ing you proposed.
I'd love to hear about that!
Marek14 wrote:quickfur: Right, I guess it wouldn't work...
quickfur wrote:Marek14 wrote:quickfur: Right, I guess it wouldn't work...
I didn't say it wouldn't work, I just wasn't clear what you meant. Perhaps there is a way to insert bilbiro's between the two hemi-600cells that would produce a CRF. Might be worth looking into?
quickfur wrote:Marek14 wrote:Here's a question:
I had a look at the hemi-600cell now. Of course, the dihedral angles at its icosidodecahedron won't allow gluing two together (which is also clear from its construction).
But -- what if the bilbiro process was used on this icosidodecahedron? Could that allow for duplicating the shape? Basically, it would be bilbiroing the middle layer of x3o3o5o, despite the fact that it usually doesn't contain any icosidodecahedra.
Not sure what you mean, ...
student91 wrote:quickfur wrote:It might just be possible that things with id-cuts will allow a relatively simple bilbiro-ing. The only things with id-cuts are as far as I know the o5o3x3o and the o5o3o3x. The latter probably won't be bilbiro-able. This means from now on I'll be looking for a bilbiro-ing of the o5o3x3o. I think that when I've investigated that one, I'll be able to say something about the other ways of bilbiro-ing you proposed.
I'd love to hear about that!
I don't have anything yet, I just was saying what I was going to innvestigate next, and when I'm done investigating (or stuck) I'll of course share my results with this forum.
quickfur wrote:I just updated the crown jewel page and added Keiji's proposed CJx.y.z numbering system for them. These include all of the confirmed CRF crown jewels we found so far.
There are still 4 as-yet unnamed crown jewels, CJ4.5.1, CJ4.5.2, CJ4.5.3, and CJ4.6. Any suggestions?
I gotta run now, I'll create stub wiki pages for these crown jewels later, and link the images and .off files to them.
Keiji wrote:Thanks for the hard work quickfur!
I don't suppose we have coordinates/images/.def/.off for CJ4.5.2 anywhere? That's now the only 4D crown jewel without its own page.
quickfur wrote:Keiji wrote:Thanks for the hard work quickfur!
I don't suppose we have coordinates/images/.def/.off for CJ4.5.2 anywhere? That's now the only 4D crown jewel without its own page.
I didn't build it because basically it's CJ4.5.1 with the top part of the rectified 120-cell glued on both ends. It's just extra vertices (=slower convex hull times, slower rendering, etc.) away from the exciting action (i.e. where the J92's are). You could still create a page for it, and mention the cell counts, etc., which are all already known. It's just that it's not that much more interesting than CJ4.5.1 itself.
quickfur wrote:Hmm. Today I was reading up on the snub disphenoid on Wolfram, and it claims that the analytic solution for its coordinates requires solving a cubic equation?? How did Marek get a solution in terms of square roots??
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