## Johnsonian Polytopes

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

### Re: Johnsonian Polytopes

Alright. So here are the renders for the square biantiprismic ring:

This is a projection centered on one of its two square antiprisms. As you can tell, the cells aren't very easy to see, so to help you out here's the other square antiprism highlighted:

And here's the cubical cell, the 3rd member of the ring:

Keiji asked for a cube-first projection, so here it is:

As you can see, the projection envelope actually isn't a cube; some of the tetrahedral cells the square pyramids protrude outside (the square antiprisms are technically hidden from view here, but I turned off visibility clipping so that the structure of the whole thing is clearer). For your benefit, the next image highlights one of the square antiprisms:

And no polychoron is complete without coordinates, so here they are:

Cube:
(±1, ±1, ±1, 0)
Square:
(±sqrt(2), 0, 0, H)
(0, ±sqrt(2), 0, H)
where H = sqrt(2*sqrt(2) - 1).

Verifying that the edge lengths are all correct is left as an exercise for the reader. ;-)
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### Re: Johnsonian Polytopes

As far as I see, only 14 Johnsonian polyhedra may form pyramids and bipyramids:
- 5 regular polyhedra;
- 4- and 5-antiprisms;
- 3- and 5-prisms (and 5-prism is very close to failure: it has radius of outscribing sphere about 0.9867 of edge length)
- 4- and 5- pyramids;
- 3 diminishing icosahedra.
There is a very small chance that among other bodies there will be some the happen to be inscribed in the sphere. But I think that there is no such things.
Mrrl
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### Re: Johnsonian Polytopes

Mrrl wrote:[...]
(and 5-prism is very close to failure: it has radius of outscribing sphere about 0.9867 of edge length) [...]

But actually, that's a good thing, because it means that the pentagonal prism pyramid will be very shallow, and a large number of duoprisms containing pentagonal prisms will have "stellated" (or rather "pyramidized") forms.
quickfur
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### Re: Johnsonian Polytopes

Are you saying that wintersolstice's list included some polyhedra that it shouldn't have? For example, the snub disphenoid (for reference the list is on CRF polychora discovery project)

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### Re: Johnsonian Polytopes

Keiji wrote:Are you saying that wintersolstice's list included some polyhedra that it shouldn't have? For example, the snub disphenoid (for reference the list is on CRF polychora discovery project)

Yes. As I see, it has different angles between faces that means that it is not inscribed in the shpere. But if you form Johnsonian pyramid of the body (keeping it in 3D without distortions), then distances from the apex to vertices of the body are equal => distances from the projection of apex on the base to vetices are equal => body is insribed in the sphere. Contradiction
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### Re: Johnsonian Polytopes

Oh, I see - if you can't inscribe a polyhedron in a sphere, then it means you can't find a point in 3D space which is equidistant from all the original points, so it follows that you can't find a point in 4D space which is either, because moving the point perpendicular to the original 3D space multiplies all the lengths by the same amount. Therefore, you can't make a pyramid with equal edge lengths. Is this all correct?

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### Re: Johnsonian Polytopes

For the icosidodecahedron and the pentagonal orthobirotunda (J34), I think those should be able to be inscribed in a sphere. In these cases, I take it they can not form CRF pyramids because the distance from each point to the center (of the sphere) is already longer than the edge length?

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### Re: Johnsonian Polytopes

quickfur wrote:
Mrrl wrote:[...]
(and 5-prism is very close to failure: it has radius of outscribing sphere about 0.9867 of edge length) [...]

But actually, that's a good thing, because it means that the pentagonal prism pyramid will be very shallow, and a large number of duoprisms containing pentagonal prisms will have "stellated" (or rather "pyramidized") forms.

Augmented duoprisms? Yes, you can do it with duoprisms 5,N for N<20 by setting pyramids on every subset of 5-prism cells, and for N<39 - for subsets that have no adjacent elements. It gives finite family of many millions bodies (about hundred millions, I guess)
Last edited by Mrrl on Tue Nov 22, 2011 9:41 pm, edited 1 time in total.
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### Re: Johnsonian Polytopes

Keiji wrote:For the icosidodecahedron and the pentagonal orthobirotunda (J34), I think those should be able to be inscribed in a sphere. In these cases, I take it they can not form CRF pyramids because the distance from each point to the center (of the sphere) is already longer than the edge length?

Correct. If you have at least one face or a planar regular contour of edges with 6 or more sides (like in triangular cupola or cuboctahedron), you can't build pyramid because of large radius. Actually, the only body that required manual check of radius was 5-prism: existance of the pyramid for icosahedron is proven by 600-cell.
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### Re: Johnsonian Polytopes

Mrrl wrote:[...]
What if you take two n,3 skew duoprisms and glue the together by the prismatic cells? Is there a chance to get convex body? Also we may try to glue them by antiprismatic cells (and get two new bodies).

OK, so i've been doing a little thinking along this line of gluing our bipolyhedral rings to make new CRFs.

I think it's possible with the triangular biantiprismic ring; I'll calculate coordinates and try it later.

For the square biantiprismatic ring, I think gluing by the cubical facets won't give you a convex shape -- even in cube-centered projection you can see the square pyramids protrude outwards, which means the dihedral angle is > 90°, so the result of gluing will be non-convex. However, I think it might be possible to glue two of them by their antiprismic cells; if i'm not mistaken, it should still be convex even though the angle between some of the cells will be very shallow. Unfortunately this is the harder case to investigate. For gluing by the cube face we can just mirror the coordinates, but gluing by the antiprisms is harder to calculate.

EDIT: Actually, I just checked: attaching two copies of square biantiprismic rings by their square antiprism cells is also non-convex. The tetrahedral cells protrude from the square antiprismic envelope of the parallel projection centered on the square antiprism, which means their dihedral angle > 90°. And actually, in general, I think the shape of an n-gonal antiprism will always cause some of the dihedral angles to be > 90°, so the only convex possibilities are the biantiprismic ring family itself. (Not 100% sure on this one, but I suspect this is true. )
Last edited by quickfur on Tue Nov 22, 2011 10:03 pm, edited 1 time in total.
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### Re: Johnsonian Polytopes

Mrrl wrote:[...] Augmented duoprisms? Yes, you can do it with duoprisms 5,N for N<20 by setting pyramids on every subset of 5-prism cells, and for N<39 - for subsets that have no adjacent elements. It gives finite family of many millions bodies (about hundred millions, I guess)

Yes! Augmented duoprisms, that's what they are. But are you sure about the millions of members? Don't forget the rotational symmetry of the duoprisms, so not all of the combinations are distinct.
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### Re: Johnsonian Polytopes

I remember that. There is about 1 million of possible noncongruent sets for N=38 (I'll check it later), and we have geometric progression for less N, so we get about 4 millions in total. Slightly less than hundred millions
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### Re: Johnsonian Polytopes

I've just rewritten much of the project page and the summary on the Polytope page in light of the new information about pyramidal forms.

If there are really millions of possible augmented duoprisms, could we count them separately from the others (like how the infinite families are kept separate), so that it doesn't dwarf the more interesting shapes?

Keiji

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### Re: Johnsonian Polytopes

Keiji wrote:[...] If there are really millions of possible augmented duoprisms, could we count them separately from the others (like how the infinite families are kept separate), so that it doesn't dwarf the more interesting shapes?

Well, obviously we're not going to be writing a big long list of all the CRF's now that it has millions of members. We should still include the augmented duorisms in the final count of CRFs that don't belong to infinite families, though.
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### Re: Johnsonian Polytopes

Yes, I know that I was just saying we should focus articles more on the smaller families, and we should have a list of those families' members.

Keiji

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### Re: Johnsonian Polytopes

quickfur wrote:EDIT: Actually, I just checked: attaching two copies of square biantiprismic rings by their square antiprism cells is also non-convex. The tetrahedral cells protrude from the square antiprismic envelope of the parallel projection centered on the square antiprism, which means their dihedral angle > 90°. And actually, in general, I think the shape of an n-gonal antiprism will always cause some of the dihedral angles to be > 90°, so the only convex possibilities are the biantiprismic ring family itself. (Not 100% sure on this one, but I suspect this is true. )

Have you checked both ways of connecting? What you said forbids sequence "prism,antiprism,antiprism,prism" is the ring, but what about "prism,antiprism,prism,antiprism"? There you combine different kinds af dihedral angles, so their sum still may be < 180°even when one of them > 90°.
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### Re: Johnsonian Polytopes

Doesn't "prism, antiprism, prism, antiprism" just yield the prism of an antiprism of an n-gon anyway? Or are you guys talking about something else entirely?

Keiji

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### Re: Johnsonian Polytopes

Keiji wrote:Yes, I know that I was just saying we should focus articles more on the smaller families, and we should have a list of those families' members.

You may even drop away all bodies that can be splitted into two smaller Johnsonian bodies... What we need is to mark all ridges that have angle between cells less than 90°, and later check all combinations of combining bodies by cells of equal form. Remember that you have to check augmentations of all known bodies, starting with prisms of Johnsonain polyhedra...
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### Re: Johnsonian Polytopes

Keiji wrote:Doesn't "prism, antiprism, prism, antiprism" just yield the prism of an antiprism of an n-gon anyway? Or are you guys talking about something else entirely?

Not exactly. Intuitively, antiprismatic prism has square section somewhere (right angles between prism and antiprism). We say about conbinations of two truangular rings, so we'll get some irregular tetragon in section...
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### Re: Johnsonian Polytopes

Mrrl wrote:
quickfur wrote:EDIT: Actually, I just checked: attaching two copies of square biantiprismic rings by their square antiprism cells is also non-convex. The tetrahedral cells protrude from the square antiprismic envelope of the parallel projection centered on the square antiprism, which means their dihedral angle > 90°. And actually, in general, I think the shape of an n-gonal antiprism will always cause some of the dihedral angles to be > 90°, so the only convex possibilities are the biantiprismic ring family itself. (Not 100% sure on this one, but I suspect this is true. )

Have you checked both ways of connecting? What you said forbids sequence "prism,antiprism,antiprism,prism" is the ring, but what about "prism,antiprism,prism,antiprism"? There you combine different kinds af dihedral angles, so their sum still may be < 180°even when one of them > 90°.

What I meant by "gluing" is not attaching cells to each other via their 2D ridges, but to glue the 4D shapes together like fitting two Lego blocks together. The mating cells must be the same shape, otherwise the result is obviously nonconvex. After the gluing the mated cells are internal to the resulting polytope and no longer show up on the surface. A simple 3D example is gluing two square pyramids to make an octahedron: after gluing, the square face is no longer on the surface.
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### Re: Johnsonian Polytopes

What about gluing together some of the bicupolic rings?

For example, glue the triangular orthobicupolic ring to another copy of itself by one of the triangular cupolic cells.

Keiji

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### Re: Johnsonian Polytopes

Keiji wrote:What about gluing together some of the bicupolic rings?

For example, glue the triangular orthobicupolic ring to another copy of itself by one of the triangular cupolic cells.

That's certainly something worth trying. From what I can tell, the result, if it's CRF, would have a 4-membered ring: cupola-cupola-prism-prism. Hmm, how come we haven't considered that possibility before?
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### Re: Johnsonian Polytopes

quickfur wrote:What I meant by "gluing" is not attaching cells to each other via their 2D ridges, but to glue the 4D shapes together like fitting two Lego blocks together. The mating cells must be the same shape, otherwise the result is obviously nonconvex. After the gluing the mated cells are internal to the resulting polytope and no longer show up on the surface. A simple 3D example is gluing two square pyramids to make an octahedron: after gluing, the square face is no longer on the surface.

Yes, of course, you take antiprismatic cells of two bodies. But these cells have no-equivalent bases: base A of the cell each body is a ridge between two antiprisms and base B is a ridge between prism and antiprism. When you glue two bodies together, you can glue base A1 to A2 and B1 to B2 (and get "prism,antiprism,antiprism,prism" ring), or glue A1 to B2 and B1 to A2 - and you get "prism,antiprism,prism,antiprism". First one is not convex - you proved that sum of angles at base A1+A2 is more than 180°. But what about the second - where we should consider angles at A1+B2?
Last edited by Mrrl on Wed Nov 23, 2011 12:07 am, edited 2 times in total.
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### Re: Johnsonian Polytopes

Oh I see what you mean.

Hmm... I'm not sure, but it seems to be still impossible to get a convex shape. The problem is that the cells surrounding the antiprisms have shadow that fall outside of the base (if you consider the antiprisms lying on a horizontal hyperplane: the shadows fall outside the base). As far as I can tell, the corresponding shadows from the two copies of the polychoron will always overlap each other in the area that falls outside the base, so no matter which way you connect them, the result is non-convex.
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### Re: Johnsonian Polytopes

If I'm not mistaken, angle between antiprism and tetrahedron is acos((sqrt(2)-2)/2/sqrt(sqrt(8)))=100.03°, and angle between antiprism and 4-pyramid is acos((H-sqrt(2)+1/sqrt(3*sqrt(8)))=71.21°. If it's correct, than angles around this antiprism after combining of bodies are 171.24°<180° - that means that the ring is convex! It has 2 cubes, 2 4-antiprisms, 8 tetrahedra and 8 4-pyramids
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### Re: Johnsonian Polytopes

quickfur wrote:Oh I see what you mean.

Hmm... I'm not sure, but it seems to be still impossible to get a convex shape. The problem is that the cells surrounding the antiprisms have shadow that fall outside of the base (if you consider the antiprisms lying on a horizontal hyperplane: the shadows fall outside the base). As far as I can tell, the corresponding shadows from the two copies of the polychoron will always overlap each other in the area that falls outside the base, so no matter which way you connect them, the result is non-convex.

There are two kind of cells around the antiprims: shadows of tetrahedra fall outside, but shadows of 4-pyramids fall inside. And question is about the relations between dihedral angles.
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### Re: Johnsonian Polytopes

Hmm you're right, I checked the projections again, the pyramids don't overhang over the base on one square face of the antiprism; only the tetrahedra cast a shadow on the side of the other square face.

So if we fit the polytopes together by having the tetrahedral shadows fall on opposite sides of the joined antiprism base, then the result should be uniform... hmmmm... how to calculate the coordinates for this though?
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### Re: Johnsonian Polytopes

quickfur wrote:So if we fit the polytopes together by having the tetrahedral shadows fall on opposite sides of the joined antiprism base, then the result should be uniform... hmmmm... how to calculate the coordinates for this though?

Easy:

Cube:
(±1, ±1, ±1, 0)
First square:
(±sqrt(2), 0, 0, H)
(0, ±sqrt(2), 0, H)
Second square:
(±sqrt(2), 0, 2, H)
(0, ±sqrt(2), 2, H)
where H = sqrt(2*sqrt(2) - 1).
Second square is the top of the cube standing on the first square

And I hope that it is not uniform - otherwise we get 4-antiprismatic prism that is not interesting.
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### Re: Johnsonian Polytopes

Wait, are you talking about joining the rings via their cubical cells?
I was talking about joining them via their antiprism cells.

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### Re: Johnsonian Polytopes

Unfortunately, it is uniform.

You can see this is just the prism of the 4-antiprism.
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