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So, the last time I dealt with augmentations, we were considering augmentations of duoprisms with n-prism pyramids and Stott-expanded n-prism pyramids (i.e., n-gonal magnabicupolic rings/wedges). While searching for CRFs containing J12, one thing that occurred to me is that certain polyhedral prisms can be augmented by n-prism pyramids or n-gonal magnabicupolic rings too.

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.

A similar augmentation of the great rhombicosidodecahedron x5x3x prism with pentagonal magnabicupolic rings (aka 10-prism||5-gon) should also be possible, though probably the 6-prisms and cubes can't be also augmented due to the dichoral angles in the 6-prism||3-gon being too big to remain convex.

In the same way, prisms of the other 3D uniforms with pentagonal/decagonal faces should also be similarly augmentable.

In any case, the interesting thing about the dodecahedral prism is that it can also be augmented with another kind of augment: the 5-pyramid prism (as opposed to the 5-prism pyramid). This augment can only be non-adjacent, and produces an (augmented dodecahedron)-prism, as opposed to an augmented (dodecahedral prism). So you see, there's some ambiguity going on here with our present naming scheme, as Klitzing has mentioned on one occasion. The Stott-expanded equivalents involving the x5o3x prism should also exist.

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.

A similar augmentation of the great rhombicosidodecahedron x5x3x prism with pentagonal magnabicupolic rings (aka 10-prism||5-gon) should also be possible, though probably the 6-prisms and cubes can't be also augmented due to the dichoral angles in the 6-prism||3-gon being too big to remain convex.

In the same way, prisms of the other 3D uniforms with pentagonal/decagonal faces should also be similarly augmentable.

In any case, the interesting thing about the dodecahedral prism is that it can also be augmented with another kind of augment: the 5-pyramid prism (as opposed to the 5-prism pyramid). This augment can only be non-adjacent, and produces an (augmented dodecahedron)-prism, as opposed to an augmented (dodecahedral prism). So you see, there's some ambiguity going on here with our present naming scheme, as Klitzing has mentioned on one occasion. The Stott-expanded equivalents involving the x5o3x prism should also exist.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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.

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.

So, first of all we calculate as the inradius rho of the unit-edged dodecahedron as rho = sqrt[(25+11 sqrt(5))/40] = 1.113516.

Then we calculate the circumradius r of the unit-edged icosahedron as r = sqrt[(5+sqrt(5))/8] = 0.951057.

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)) = tau

--- rk

PS: You surely would like to produce cute pics of that fellow. Don't you, quickfur?

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

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)) = tau^{2}/ sqrt(5) = 1.170820.

Unfortunately, that is wrong. The square pyramid cells of the augment are not coplanar, so the elements of the icosahedron do not coincide with any of the dodecahedron's elements in projection. Your calculated value of A produces points that lie inside the bulk of the dodecahedral prism, so the convex hull is still just the unaugmented prism.

I solved A via another method: we know that x5o3o lies on the hyperplanes w=±1 (for edge length 2), and o5o3A lies on w=0. Since o5o3A can be written as A*epacs<1,0,phi> (for edge length 2*A), we solve for edge length 2 betwen the points <0,1,phi^2,1> and <A, 0, A*phi, 0>, which, after some convoluted algebraic calculations, turns out to be A=3/√5. This gives an augmented prism with the requisite equal edge lengths, as checked by my polytope viewer.

[...]

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.

The coordinates are as follows:

- Code: Select all
`# x5o3o:`

apacs<phi, phi, phi> ~ <±1>

epacs<0, 1, phi^2> ~ <±1>

# o5o3A: (A=3/√5)

(3/√5)*epacs<1, 0, phi> ~ <0>

I'll upload the model files to the wiki.

EDIT: uploaded to the omniaugmented dodecahedral prism page.

Last edited by quickfur on Sat Apr 12, 2014 9:06 pm, edited 1 time in total.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

after some convoluted algebraic calculations, turns out to be A=3/√5.

I second that. - Found my first try to be wrong myself. Meanwhile evaluated it similar to your way independently.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Yay, good to have confirmation.

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?

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
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

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.

"As you can see ..."

Well, if I'm honest, those are not too plastic, these renders. And you Need 2 to Show ist all-over structure.

(At least you managed to show the non-corealmity.)

But what about a close-up projective view without visuality clipping, showing all 3 layers concentrically?

(I.e. kind of similar to the known one of ico?

)

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

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

okay, a bit more elaborate:

- 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
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Would pentagonal prism || line also work as an augment?

- Marek14
- Pentonian
**Posts:**1190**Joined:**Sat Jul 16, 2005 6:40 pm

Marek14 wrote:Would pentagonal prism || line also work as an augment?

As the "line" has the same size as the "prism" height, it follows that the "line"-ends become co-realmic with the dodecahedral base. That is, the dodechadron becomes augmented itself. But this is Johnsonian only if augmentations will be non-adjacent. - This finally just results simply in J58-prism, J59-prism, J60-prism, and J61-prism.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Klitzing wrote:[...]

But what about a close-up projective view without visuality clipping, showing all 3 layers concentrically?

Is that better?

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:Is that better?

Yes

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

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.

Well, yes... but what about a construction that produces something new? Such as, one where a given polyhedral prism would let you scale the lacing length in such a way as to allow J12 cells to fill in the sides, perhaps?

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Okay, the same should work when using a cube with now 4-fold faces and 3-fold corners. Then you could adjoin 4-fold antiprisms to the squares and put 3-fold pyramids inbetween. Then you likewise would result in a bottom cuboctahedron. And sure you also can go on in the same way to get back again to the cube.

Because there is a square-antiprism included, the dihedral angles are harder to be calculated. Especially interesting then would that between the tetrahedra and the equatorial pseudo cell, the cuboctahedron. If those would come out to be 90 degrees, then we could join the tetrahedra pairs to J12 each (just as the 4-fold pyramids had been joinable for the icositetrachoron).

--- rk

Because there is a square-antiprism included, the dihedral angles are harder to be calculated. Especially interesting then would that between the tetrahedra and the equatorial pseudo cell, the cuboctahedron. If those would come out to be 90 degrees, then we could join the tetrahedra pairs to J12 each (just as the 4-fold pyramids had been joinable for the icositetrachoron).

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Isn't this the same as gluing two cube||cuboctahedron segmentochora back-to-back at the cuboctahedron cell? We already know the former is CRF. In fact, I have the model of cube||cuboctahedron already. A quick query of the model in my polyview viewer yields the dichoral angle between the cube and the square antiprism as 104.258°. So unfortunately, no J12 is possible here.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

It is not truely the angle at the squares between the cube and the antiprism, which matters. It rather would be the angle at the triangles between the tet and the co, which is of interest here.

But that one would not be 90 degrees, I fear now as well:

Consider the inradius squared of the co at the triangles, which is 2/3.

And the circumradius squared of the cube is 3/4.

Thus the axis of the trigonal pyramids (= tet), which is connecting these 2 points, won't be upright.

Rather those tets are slanted by sqrt(3/4)-sqrt(2/3) = 0.049529 to the outside! I.e. the formerly intended mirroring join of 2 cube||co segmentochora (at the co) not only would not produce J12s, but even would not be convex at those tet bases.

--- rk

But that one would not be 90 degrees, I fear now as well:

Consider the inradius squared of the co at the triangles, which is 2/3.

And the circumradius squared of the cube is 3/4.

Thus the axis of the trigonal pyramids (= tet), which is connecting these 2 points, won't be upright.

Rather those tets are slanted by sqrt(3/4)-sqrt(2/3) = 0.049529 to the outside! I.e. the formerly intended mirroring join of 2 cube||co segmentochora (at the co) not only would not produce J12s, but even would not be convex at those tet bases.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

You're right, the angle that matters is not the cube / square antiprism angle, but the cube / tetrahedron angle, which my polyview viewer calculated to be 86.522° across the shared vertex. So you are right, cube || cuboctahedron || cube would be concave around at the common faces of the tets.

All hope isn't lost, however. Is it possible to augment the result at those concave areas to make it convex? A preliminary calculation suggests that the outer angle between the two tets in the concave area (i.e., the dichoral angle of an augmenting piece) should be 173.0445863006°. Do we know of any 4D CRF that has two tetrahedra with this dichoral angle? Such a CRF would be able to fit into these concave gaps and produce a (hopefully) convex result.

All hope isn't lost, however. Is it possible to augment the result at those concave areas to make it convex? A preliminary calculation suggests that the outer angle between the two tets in the concave area (i.e., the dichoral angle of an augmenting piece) should be 173.0445863006°. Do we know of any 4D CRF that has two tetrahedra with this dichoral angle? Such a CRF would be able to fit into these concave gaps and produce a (hopefully) convex result.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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).

--- rk

No, so far I have not encountered any angle with a value of 173.044(something).

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

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.

Correct.

No, so far I have not encountered any angle with a value of 173.044(something).

[...]

I'm thinking that if any such augmenting piece exists, it would probably not be a simple shape, because that is quite an unusual dichoral angle. And even if such a piece were found, it would likely introduce new non-convexity around the square antiprisms, which would in turn require more augments to make it convex again, and by then, we would have lost pretty much all of the original structure that we were interested in. So this may not be a fruitful direction of search.

But on a different note, one thing we can do is to determine which polyhedral prisms are augmentable, and perhaps find an algorithm that can enumerate and count these augmentations.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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.

- Marek14
- Pentonian
**Posts:**1190**Joined:**Sat Jul 16, 2005 6:40 pm

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

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

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

Yeah, so far I'm having fun researching the various possible vertices It looks like the more complex vertex type, the more constraints it has, leading to number that is actually manageable. I was really glad when the algorithm I use actually found a combination corresponding to the maximally augmented (5,20)-duoprism 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.

- Marek14
- Pentonian
**Posts:**1190**Joined:**Sat Jul 16, 2005 6:40 pm

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? )

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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? )

Okay, let's have a look into "possible" cases here.

The dihedral angles of trippy = pt||trip are provided here.

And those of tricuf = trip||{6} are provided here.

The tricuf (trigonal cupolifastigium) has 2 types of trips: 3 trips as lacing cells, and 1 trip for the non-degenerate base cell.

A) First consider now the point being atop of one of the lacing trips. Then this trip has the mirror symmetry of a tent only.

For its lacing squares we would have to add tricu-4-trip = 65.905 and trip-4-squippy = 65.905, getting 131.810.

For its lacing triangles we would have to add tet-3-trip = 127.761 and trip-3-tet = 52.239, getting 180 degrees exactly.

(I.e. here these 2 tets combine to a trigonal dipyramid (tedpy) indeed.)

But for its base square we have to add trip-4-trip = 131.810 and trip-4-squippy = 65.905, getting 197.715, which is concave!

B) Next consider the point being atop that single base trip. That one has full symmetry.

For the triangles we would have to add here tricu-3-trip = 52.239 and trip-3-tet = 52.239, getting 104.478.

For the squares we would have to add trip-4-trip = 131.810 and trip-4-squippy = 65.905, getting 197.715, which is concave again!

Thus both do ask for concave dihedral angles.

But only the former would combine tets into tedpies.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Maybe I made a mistake somewhere. Or it merely results in a locally convex vertex in otherwise nonconvex polychoron, that's possible as well. My logic is as follows:

I am researching vertex figures in shape of a quadragonal pyramid. This particular one is a pyramid whose base edges are all length 1 and lateral edges have lengths, in order, 1, sqrt(2), sqrt(3) and sqrt(2). This corresponds to a vertex figure with five triangles, two squares and a hexagon meeting at the vertex.

I should note that I don't require the quadrangle to be planar -- the only condition is that all five vertices must lie on the unit hypersphere.

Now, there are two necessary conditions a vertex figure to actually exist in a CRF polychoron: combinatoric and computational.

The combinatoric condition means there must exist CRF polyhedra that could theoretically fit into the faces of the vertex figure.

The computational condition means that a set of equations describing the vertex figure must have a solution.

If you have combinatoric solution, but not computational, you can't fit the polyhedra together (a triangular dipyramid pyramid would be an example -- it's impossible to erect tetrahedra on triangular dipyramid to have them meet in a single point).

If you have computational solution, but not combinatoric, you could fit the polyhedra together, but you'd have to use nonexistent ones (for example, a triangular face with lengths 1, sqrt(2) and (1 + sqrt(5))/2 is forbidden because no CRF polyhedron has a (3,4,5) vertex.

So, this vertex figure (3333-3463 in my notation) is notable by the fact that the computational condition is relaxed there. It seems that no matter how I skew the base quadrangle, the polyhedra still fit. Using this on a triangular dipyramid leads to two solutions.

One of those solutions is flat (all vertices turn out to be in the same hyperplane) and corresponds to cutting triangular dipyramid out of triangular orthobicupola. The other solution would have vertex figure with these coordinates (all endpoints of edges from the origin):

1.00000 0.00000 0.00000 0.00000

0.50000 0.86603 0.00000 0.00000

0.50000 0.28868 0.81650 0.00000

0.50000 -0.67358 0.54433 0.00000

0.00000 0.57735 -0.20412 0.79057

First four, with w coordinate 0, are the vertices of the quadrangle, while the fifth one is the apex of the pyramid.

Now, quadrangle pyramid can be cut in two tetrahedra. If a diagonal of the quadrangle corresponds to a shortchord of a polygon, we can think of the vertex figure as of a blend of two tetrahedral vertex figures where two cells merge. In this case, one of the diagonals is 1 and we can cut the vertex figure in this diagonal and along the two side edges of length sqrt(2) in two tetrahedra with base edges all equal to 1. Side edges will be sqrt(2), sqrt(2) and sqrt(3) for one, and 1, sqrt(2) and sqrt(2) for the other.

The first vertex figure, consisting of tetrahedron, triangular prism and two triangular cupolas, appears in hexagon || triangular prism at the vertex of the hexagon. The second one appears in triangular prismatic pyramid at the vertex of the triangular prism.

Now I see the problem. My result only says that the tetrahedra will blend together if I augment it in this way, but it's not a sufficient condition as the remaining cells of both won't combine in a convex way.

The thing is that while you're trying to build CRF polychora globally, as wholes, I started to look into the details to create the basic "building blocks" -- and the results are pretty interesting once I got the computational algorithm right. So I have a whole bunch of convex CRF vertices, but I don't yet know much about how to connect them properly. The analysis of tetrahedral vertices is complete and I have over 40 that are theoretically possible to exist but I don't know of any specific example of an "acrochoron" that would fit there. And it looks like the more complex vertex figures will often be blends of simpler ones...

I am researching vertex figures in shape of a quadragonal pyramid. This particular one is a pyramid whose base edges are all length 1 and lateral edges have lengths, in order, 1, sqrt(2), sqrt(3) and sqrt(2). This corresponds to a vertex figure with five triangles, two squares and a hexagon meeting at the vertex.

I should note that I don't require the quadrangle to be planar -- the only condition is that all five vertices must lie on the unit hypersphere.

Now, there are two necessary conditions a vertex figure to actually exist in a CRF polychoron: combinatoric and computational.

The combinatoric condition means there must exist CRF polyhedra that could theoretically fit into the faces of the vertex figure.

The computational condition means that a set of equations describing the vertex figure must have a solution.

If you have combinatoric solution, but not computational, you can't fit the polyhedra together (a triangular dipyramid pyramid would be an example -- it's impossible to erect tetrahedra on triangular dipyramid to have them meet in a single point).

If you have computational solution, but not combinatoric, you could fit the polyhedra together, but you'd have to use nonexistent ones (for example, a triangular face with lengths 1, sqrt(2) and (1 + sqrt(5))/2 is forbidden because no CRF polyhedron has a (3,4,5) vertex.

So, this vertex figure (3333-3463 in my notation) is notable by the fact that the computational condition is relaxed there. It seems that no matter how I skew the base quadrangle, the polyhedra still fit. Using this on a triangular dipyramid leads to two solutions.

One of those solutions is flat (all vertices turn out to be in the same hyperplane) and corresponds to cutting triangular dipyramid out of triangular orthobicupola. The other solution would have vertex figure with these coordinates (all endpoints of edges from the origin):

1.00000 0.00000 0.00000 0.00000

0.50000 0.86603 0.00000 0.00000

0.50000 0.28868 0.81650 0.00000

0.50000 -0.67358 0.54433 0.00000

0.00000 0.57735 -0.20412 0.79057

First four, with w coordinate 0, are the vertices of the quadrangle, while the fifth one is the apex of the pyramid.

Now, quadrangle pyramid can be cut in two tetrahedra. If a diagonal of the quadrangle corresponds to a shortchord of a polygon, we can think of the vertex figure as of a blend of two tetrahedral vertex figures where two cells merge. In this case, one of the diagonals is 1 and we can cut the vertex figure in this diagonal and along the two side edges of length sqrt(2) in two tetrahedra with base edges all equal to 1. Side edges will be sqrt(2), sqrt(2) and sqrt(3) for one, and 1, sqrt(2) and sqrt(2) for the other.

The first vertex figure, consisting of tetrahedron, triangular prism and two triangular cupolas, appears in hexagon || triangular prism at the vertex of the hexagon. The second one appears in triangular prismatic pyramid at the vertex of the triangular prism.

Now I see the problem. My result only says that the tetrahedra will blend together if I augment it in this way, but it's not a sufficient condition as the remaining cells of both won't combine in a convex way.

The thing is that while you're trying to build CRF polychora globally, as wholes, I started to look into the details to create the basic "building blocks" -- and the results are pretty interesting once I got the computational algorithm right. So I have a whole bunch of convex CRF vertices, but I don't yet know much about how to connect them properly. The analysis of tetrahedral vertices is complete and I have over 40 that are theoretically possible to exist but I don't know of any specific example of an "acrochoron" that would fit there. And it looks like the more complex vertex figures will often be blends of simpler ones...

- Marek14
- Pentonian
**Posts:**1190**Joined:**Sat Jul 16, 2005 6:40 pm

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? 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. 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.

I don't know if my calculations are wrong, neither I can render these shapes. I'm very curious about the result. 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.

- JMBR
- Mononian
**Posts:**8**Joined:**Wed Mar 30, 2016 12:58 pm

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?

However, this school-project is in dutch, when writing it I had time-issues, and over all it could have been done better. But yeah, indeed any way of painting a dodecahedron's faces gives a new polytope in this case.

Back to the point, in my project I calculated there to be 82 distinct colorings of the dodecahedron instead of 63. I'm curious to know what could have gone wrong (considering the time-issues, it could very well be that I am wrong). How did you go on calculating 63?

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.

But could you explaiin what you mean by tetrahedral corona? I'm curious .

Kind regards, student91

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.

- student91
- Tetronian
**Posts:**328**Joined:**Tue Dec 10, 2013 3:41 pm

The idea behind augmenting prisms of the more unusual Johnson solids seems like an interesting one, though. I don't know if you could count those as crown jewels, but at least they would be a little more interesting than just boring old prisms.

The augmented snub disphenoid prism seems like it might be possible, due to the sharp angles between the top and bottom pairs of triangles (top and bottom along the axis with hemi-tetragonal symmetry) with the surrounding triangles. Such sharp angles means that it's likely that augmentation with a triangular prism pyramid will still remain convex.

On a related note, I've been thinking recently about many of our previous failed attempts to generalize the construction of unusual Johnson solids to 4D. An idea that occurred to me is, instead of looking at actual Johnson solids for inspiration, we should look to near-miss Johnsons instead, or other partly-Johnson partly-non-CRF polyhedra. The reason is that the non-CRF faces in the 3D case in some cases can be substituted by a 4D CRF when generalizing to 4D (e.g., the square ursahedron is non-CRF in 3D because of the √2 edge, but the octahedral ursachoron is CRF because the CRF square pyramid plays the role of the triangle with √2 edge in 4D). Also, near-miss Johnsons can often be interpreted as projections from 4D, where the non-unit edges can be considered as hints of curvature into 4D, e.g. D4.8.x. We may have better luck finding CRFs this way than trying to generalize directly from 3D, which in many cases doesn't work because the 3D construction depends on a configuration that can only work in 3D.

The augmented snub disphenoid prism seems like it might be possible, due to the sharp angles between the top and bottom pairs of triangles (top and bottom along the axis with hemi-tetragonal symmetry) with the surrounding triangles. Such sharp angles means that it's likely that augmentation with a triangular prism pyramid will still remain convex.

On a related note, I've been thinking recently about many of our previous failed attempts to generalize the construction of unusual Johnson solids to 4D. An idea that occurred to me is, instead of looking at actual Johnson solids for inspiration, we should look to near-miss Johnsons instead, or other partly-Johnson partly-non-CRF polyhedra. The reason is that the non-CRF faces in the 3D case in some cases can be substituted by a 4D CRF when generalizing to 4D (e.g., the square ursahedron is non-CRF in 3D because of the √2 edge, but the octahedral ursachoron is CRF because the CRF square pyramid plays the role of the triangle with √2 edge in 4D). Also, near-miss Johnsons can often be interpreted as projections from 4D, where the non-unit edges can be considered as hints of curvature into 4D, e.g. D4.8.x. We may have better luck finding CRFs this way than trying to generalize directly from 3D, which in many cases doesn't work because the 3D construction depends on a configuration that can only work in 3D.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

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.

Thanx for recommendation.

In fact, it was written as to be an easy to be grasped intro into 4D "visualization".

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

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