quickfur wrote:The first example that comes to mind is the dodecahedral prism. The dichoral angle between the lacing prisms is the same as the dihedral angle of the dodecahedron, that is, 116.565°. Since the dichoral angle between the pentagonal prism and the square pyramid cells in the pentagonal prism pyramid is only 13.283°, and 2*13.283 + 116.565 = 143.131 < 180°, that means we can augment all of the pentagonal prisms with pentagonal prism pyramids to get a CRF omni-augmented dodecahedral prism with 2 dodecahedra, 24 pentagonal pyramids, and 60 square pyramids. The lace tower of this CRF would be x5o3o || o5o3A || x5o3o for some as-yet unknown value of A.
Klitzing wrote:quickfur wrote:[...] The lace tower of this CRF would be x5o3o || o5o3A || x5o3o for some as-yet unknown value of A.
Well, what is that A?
When projecting that tower along its axis, it becomes clear, that this o5o3A would be exactly that scaled icosahedron, which is face-centers inscribed into the unit-edged dodecahedron x5o3o.
[...]
We finally divide that rho by that r, in order to get the edge size A: rho/r = sqrt[(25+11 sqrt(5))/(25+5 sqrt(5))] = (3+sqrt(5))/(2 sqrt(5)) = tau2 / sqrt(5) = 1.170820.
[...]
PS: You surely would like to produce cute pics of that fellow. Don't you, quickfur?
# x5o3o:
apacs<phi, phi, phi> ~ <±1>
epacs<0, 1, phi^2> ~ <±1>
# o5o3A: (A=3/√5)
(3/√5)*epacs<1, 0, phi> ~ <0>
after some convoluted algebraic calculations, turns out to be A=3/√5.
quickfur wrote:This construction makes me wonder about another route to finding CRFs: suppose we have some CRF polyhedron X, and we make a 4D X-prism. Now, obviously, we can set the hyperplanes of the two X's to be exactly a unit edge length apart, then the result will be CRF. However, what if we allow the two hyperplanes to vary in relative heights to other values? Obviously, the edges will no longer be unit length. But what if we then augment the resulting non-CRF with tall augments (instead of shallow augments like the pentagonal prism pyramid)? The result will be non-convex, because these augments will stick out from the side of the prism with concave valleys between them. But suppose we can vary the heights of the hyperplanes, such that at some value of the height, we can close up the gaps between these augments with CRF cells?
In other words, this is sorta like a variation of the bridged-augment idea, except that the augments and the bridging pieces do not need to be CRF, as long as their surface cells in the result are CRF. By allowing the internal cells to be non-CRF, this should give us more flexibility to construct shapes that may otherwise not close up in a CRF way, maybe?
quickfur wrote:Klitzing wrote:PS: You surely would like to produce cute pics of that fellow. Don't you, quickfur?
Yes, so here's a render of it:
As you can see, the projection envelope is a pentakis dodecahedron. The pentagonal pyramids, obviously, overhang the dodecahedron; however, only ever so slightly. The square pyramids (not shown here) are not coplanar, so we don't get a rhombic triacontahedron envelope. Well, let's take a look at some of these square pyramids:
As you can see, these augments only protrude slightly from the base prism, so the square pyramids are not coplanar. I'm using parallel projection here, so you can see some of the pentagonal pyramids projected as pairs of scalene triangles hanging off the sides of the yellow dodecahedron.
Klitzing wrote:quickfur wrote:This construction makes me wonder about another route to finding CRFs: suppose we have some CRF polyhedron X, and we make a 4D X-prism. Now, obviously, we can set the hyperplanes of the two X's to be exactly a unit edge length apart, then the result will be CRF. However, what if we allow the two hyperplanes to vary in relative heights to other values? Obviously, the edges will no longer be unit length. But what if we then augment the resulting non-CRF with tall augments (instead of shallow augments like the pentagonal prism pyramid)? The result will be non-convex, because these augments will stick out from the side of the prism with concave valleys between them. But suppose we can vary the heights of the hyperplanes, such that at some value of the height, we can close up the gaps between these augments with CRF cells?
In other words, this is sorta like a variation of the bridged-augment idea, except that the augments and the bridging pieces do not need to be CRF, as long as their surface cells in the result are CRF. By allowing the internal cells to be non-CRF, this should give us more flexibility to construct shapes that may otherwise not close up in a CRF way, maybe?
Ah, eg. ico = oct || pseudo co || oct ...
--- rk
Marek14 wrote:Would pentagonal prism || line also work as an augment?
Klitzing wrote:[...]
But what about a close-up projective view without visuality clipping, showing all 3 layers concentrically?
quickfur wrote:Is that better?
Klitzing wrote:Klitzing wrote:quickfur wrote:This construction makes me wonder about another route to finding CRFs: [...]
In other words, this is sorta like a variation of the bridged-augment idea, except that the augments and the bridging pieces do not need to be CRF, as long as their surface cells in the result are CRF.[...]
Ah, eg. ico = oct || pseudo co || oct ...
--- rk
okay, a bit more elaborate:--- rk
- Take oct||oct;
- attach to all the 8 prisms a trip||dual {3} each (thereby blending out right those prisms);
- then stretch the former prisms by x:q=1:sqrt(2),
- and you could introduce 6 further octs = pt||pseudo{4}||pt into the gaps:
- this is what ico would be.
Klitzing wrote:Thus that angle at {3} between co and tet of the cube||co then would be 180 - 86.522 = 93.478, I suppose.
No, so far I have not encountered any angle with a value of 173.044(something).
[...]
Marek14 wrote:I found that truncated cuboctahedral prism can have its octagonal prisms augmented with square || octagonal prisms. This will transform the cubes around it in elongated square pyramids/dipyramids.
Klitzing wrote:Marek14 wrote:I found that truncated cuboctahedral prism can have its octagonal prisms augmented with square || octagonal prisms. This will transform the cubes around it in elongated square pyramids/dipyramids.
Haha, well that's right. And perhaps indeed so far not being considered on its own. But, you know, gircope is nothing but the equatorial monostratic segment of prit. And prit in turn is known to be augmentable by oct||sirco, cf. that post of quickfur: resulting in owau prit. - Thus your shape is nothing but the equatorial bistratic segment of that. And the {4}||op augmentation, you use, in turn will be nothing but bidiminished oct||sirco segmentochora (oct = squippy + {4} + squippy, sirco = squacu + op + squacu).
--- rk
Marek14 wrote:[...] And I found a finished CRF that actually has triangular dipyramid cells, though in retrospective it was kind of obvious ( hexagon || triangular prism augmented with triangular prismatic pyramid) -- both of these segmentochora have the same height.
quickfur wrote:Marek14 wrote:[...] And I found a finished CRF that actually has triangular dipyramid cells, though in retrospective it was kind of obvious ( hexagon || triangular prism augmented with triangular prismatic pyramid) -- both of these segmentochora have the same height.
Are you sure this is correct? I tried to construct a model of it, but it appears that some of the dichoral angles at the joining hyperplane are concave, so the result is non-convex. (Or did I make a calculation error, as usual? )
Of course! In fact, this idea is the basis of a school-project of mine.JMBR wrote:Hey. I'm looking this page for some time but no idea came until now. Good work. I was thinking about two ideas, one inside this topic. For each of the 63 combinations of faces of the dodecahedron it's possible to generate a CRF, right?
Except for the 63, the calculations in my school-project give the same results as you have got here.The tetrahedral prism admits just one lateral augmentation, and the octahedral admits four types of this. The cubical prism is just the tesseract, already counted. But the biggest surprise is the rhombicosidodecahedral prism. It's dichoral angle is 148° between a cube and a pentagonal prism, so it's also possible to make more 63 CRFs augmenting with 13° pentagonal prism pyramids.
I don't know if my calculations are wrong, neither I can render these shapes. I'm very curious about the result.
I don't yet quite understand what you mean here. When placing three trangular prisms (trips) around an edge gives (line||triangular prism). This polytope has quite a small circumradius, so it will be challenging making it into a big sphenocorona. three cubes inevitably need to be closed up by (some derivative of) a thesseract.And an equivalent of sphenocorona may be possible. Starting with three cubes or triangular prisms around an edge and putting pyramids on the other faces creates a luna-like shape, which I don't know if it can be completed with tetrahedral coronae.
student91 wrote:P.S. If you haven't seen it before, I suggest you should take a look at the aritcle about segmentachora, it is quite well-written and segmentochora are quite simpoe to understand, just a little bit more complicated than prisms.
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