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How does Stella4D determine the dual?
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adrian



Joined: 04 Nov 2014
Posts: 5

PostPosted: Sun Aug 23, 2015 3:29 pm    Post subject: Reply with quote

guy wrote:
Going back to the canonical construction of the solid and its dual, we can readily see that your Chestahedron/ellipsoid construction is a simple stretching of the canonical construction and that symmetries relating to the stretching axis are preserved. In particular, the relative volumes and proportions of the polyhedron and its dual are preserved: they remain congruent. That is, the dual of your Chestahedron, obtained by reciprocation in an edge-tangent ellipsoid, is congruent to the Chestahedron - in short, it is self-dual about the ellipsoid.


The canonical triangular antihermaphrodite isn't the only form with a midsphere (consider the form with four equilateral triangles and the dihedral angle of an icosahedron at the base) . If you scale the chesterhedron model to make the edge-tangent ellipsoid into a sphere it won't necessarily transform the chesterhedron into the canonical form. To check the case here, the antihermaphodite has vertices at two heights above the base, and I make it that the ratio of these two heights in the canonical form is different to the chesterhedron, and so I don't believe they can be transformed into each other by scaling.

Also, consider that reciprocating in the ellipsoid will produce perpendicular dual edges at the same tangency points. For the chesterhedron to be self-dual in the ellipsoid, the upper and lower ring of tangency points would need to have the same radius, and this is not the case from the image posted.
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guy



Joined: 11 Feb 2008
Posts: 75
Location: England

PostPosted: Sun Aug 23, 2015 4:27 pm    Post subject: Reply with quote

I see what adrian means, I had forgotten that possibility. Sad
If we take the polyhedron with its mid-ellipsoid and squash it down to make the ellipsoid spherical, the construction will still be projective but will the polyhedron necessarily be canonical?
Given that there are many morphs with edge-tangent spheres and our starting point (equal areas) is essentially arbitrary, the answer is probably not.
The key to the solution is whether or not the mean position of the vertices (their centre of gravity if you like) is also the sphere centre. This is a distinguishing feature of the canonical form.
Another way of saying this is that many morphs have an edge-tangent sphere but only that of the canonical form is concentric. And only a concentric edge-tangent sphere will yield a congruent dual.
We can also observe that the edge tangency points lie in three parallel circles on the ellipsoid. For self-dual (ie canonical) construction, the two outer circles must be the same size. The illustration previously shown suggests that this is not the case.
So it looks as if I gave the wrong answer first time round. Sorry about that.
And it leaves open the question, what is the geometry of the dual in the ellipsoid?

To answer your original question, if you compute the average vertex position and then construct a concentric sphere, this will give the more standard dual of the figure - this is how Stella does it.
Varying the size of the sphere will yield similar duals of different sizes, but I don't know how Stella picks the size.
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Swami Polytope



Joined: 14 Aug 2015
Posts: 20

PostPosted: Sun Aug 23, 2015 5:31 pm    Post subject: Reply with quote

Hi Guy,

Thx for the great replies. I have read all three. I will address them in turn, in time. As for the first one (on duality/projective geom)...

Yes, I have come to understand that "duality" is a many faceted thing;)Smile And, I have been aware that projective geometry is a beast unto itself. None of this is a problem, though. Instead, they are fascinations, and things to be grokked... after much contemplating!
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Swami Polytope



Joined: 14 Aug 2015
Posts: 20

PostPosted: Sun Aug 23, 2015 5:42 pm    Post subject: Reply with quote

FYI, I have started to read your fine paper, "Vertex figures". The intro addresses much of what I have encounter, and intimated. Looking forward to reading in its entirety. I'm sure it will be helpful in my understanding in these matters. Thx.

I, of course, in my research, encountered the vertex figure, and the related Dorman-Luke construction. In fact, I started to implement it in my Virtual Chestahedron, but encountered some practical issues, so opted for a modified attempt. Here are two graphics explaining such... (FYI, they clearly do not result in a proper dual, but are interesting in their own right:)) BTW, that first attempt involved constructing a sphere at the vertex to define equal length intersection points on the edges.





Last edited by Swami Polytope on Sun Aug 23, 2015 7:45 pm; edited 2 times in total
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Swami Polytope



Joined: 14 Aug 2015
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PostPosted: Sun Aug 23, 2015 7:00 pm    Post subject: Reply with quote

Hi Guy,

As to your second, recent reply: "Going back to the canonical construction of the solid and its dual..."

First, by simple inspection, I did not imagine the Chestahedron to be self-dual if reciprocated in the ellipsoid.

You wrote:
Quote:
"If you scale the chesterhedron model to make the edge-tangent ellipsoid into a sphere it won't necessarily transform the chesterhedron into the canonical form."

- I will have to check...

You wrote:
Quote:
"I make it that the ratio of these two heights in the canonical form is different to the chesterhedron"

- You are correct.

You wrote:
Quote:
"consider that reciprocating in the ellipsoid will produce perpendicular dual edges at the same tangency points."

- Please, clarify. Thx.

Your wrote:
Quote:
"For the chesterhedron to be self-dual in the ellipsoid, the upper and lower ring of tangency points would need to have the same radius, and this is not the case from the image posted."

- Correct, they do not have the same radius.
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Swami Polytope



Joined: 14 Aug 2015
Posts: 20

PostPosted: Sun Aug 23, 2015 7:19 pm    Post subject: Reply with quote

Hi Guy,

As to your third, recent reply: "I see what adrian means,..."

The centroid, for triangle side length of 1, is ~0.419 (according to Mathematica), but ~0.544 (according to Antiprism). Not sure what I've done differently, or am missing. Anyhow, neither of these is the midsphere center, or midellipsoid center. In fact, these centers "appear" to be the same. (See image below). I can't say for certain, because both have been determined experimentally/interactively. Stella4D does not export in a format from which the midsphere can be gotten. Antiprism neither. So, I "adjusted" things, until it looked right - in the case of the midsphere, and the midellipsoid. I believe them to be "close", but definitely not exact.


(The small sphere is the midsphere center; the cube is the midellipsoid center.)

You wrote:
Quote:
"We can also observe that the edge tangency points lie in three parallel circles on the ellipsoid. For self-dual (ie canonical) construction, the two outer circles must be the same size. The illustration previously shown suggests that this is not the case."

- You are correct, those circles are not the same size.[/img]
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guy



Joined: 11 Feb 2008
Posts: 75
Location: England

PostPosted: Sun Aug 23, 2015 7:28 pm    Post subject: Reply with quote

It was adrian who wrote:
Quote:
consider that reciprocating in the ellipsoid will produce perpendicular dual edges at the same tangency points

I do not think this is true here.
It is generally true that dual edges will be at right angles for reciprocation about any sphere: the symmetry of the sphere forces a symmetry of the edge-crossing angles, i.e. right angles. For example this condition is met for any canonical dual pair.
But for the ellipsoid we have projectively stretched everything and messed up most of the angles. Metric relationships seldom survive more than the most basic projective transformations (such as translations).
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Swami Polytope



Joined: 14 Aug 2015
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PostPosted: Sun Aug 23, 2015 7:47 pm    Post subject: Reply with quote

I see. (And thx for the correction on the misattributed quotation.)

So, the new edge does not have to be perpendicular to the original, and, in fact, is only so for certain specific midsurfaces and symmetrical cases. The dual is a tricky little fellow!Smile
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Swami Polytope



Joined: 14 Aug 2015
Posts: 20

PostPosted: Sun Aug 23, 2015 8:52 pm    Post subject: Reply with quote

P.S. I have acquired the book, 'Dual Models' (MWenninger), and lectures, 'Lectures on Polytopes' (GunterMZiegler). That should keep me busy for a while! Very Happy ...
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Swami Polytope



Joined: 14 Aug 2015
Posts: 20

PostPosted: Sun Aug 23, 2015 8:53 pm    Post subject: Reply with quote

P.P.S. Finished reading your paper, "Vertex figures" (for the first time, at least). Very interesting. And, quite well written - in line with your ability of expression demonstrated in this thread. Glad to know you:)
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adrian



Joined: 04 Nov 2014
Posts: 5

PostPosted: Sun Aug 23, 2015 9:34 pm    Post subject: Reply with quote

Swami Polytope wrote:

The centroid, for triangle side length of 1, is ~0.419 (according to Mathematica), but ~0.544 (according to Antiprism). Not sure what I've done differently, or am missing.

The bottom three vertices are at height 0.0, the second three are at height 0.86..., the centroid will be higher than 0.43... (which is half of that height).

Swami Polytope wrote:

Stella4D does not export in a format from which the midsphere can be gotten. Antiprism neither.


When pol_recip finds a reciprocation sphere iteratively, which is how it finds a midsphere, it will dump the final values to the screen. In the case of the chesterhedron the sphere centre it converges on is reported as having "centre=(1.3139702708748769e-16 -4.2281061283777361e-20 0.4310035395432531)"

Swami Polytope wrote:

You wrote:
Quote:
"consider that reciprocating in the ellipsoid will produce perpendicular dual edges at the same tangency points."

- Please, clarify. Thx.


Sorry, I wasn't clear on this. I was just considering the top and bottom rings of tangency points. These reciprocate into perpendicular lines, which makes it easy to see that the dual is not a chesterhedron.
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Swami Polytope



Joined: 14 Aug 2015
Posts: 20

PostPosted: Sun Aug 23, 2015 9:45 pm    Post subject: Reply with quote

On a coincidental side note, stumbled onto this while perusing the math/art gallery from this year's recent Bridges conference. (Perhaps you attended?)...



""Hyperbolic Catacombs," by Roice Nelson (Austin, TX) and Henry Segerman (Oklahoma State University, Stillwater)
Digital Print, 2014
This picture visualizes the regular, self-dual {3,7,3} honeycomb in the upper half space model of hyperbolic 3-space. The cells are {3,7} tilings and the vertex figure is a {7,3} tiling. The cells have infinite volume: the vertices are "ultra-ideal", living beyond the boundary of hyperbolic space. The intersection of each cell with the boundary is an infinite collection of heptagons, together with a disk. The white ceiling and each red "creature" are isometric cells; for all other cells we only show the intersection with the boundary of hyperbolic space, on the floor of the catacombs. Every disk on the floor containing a {7,3} tiling is associated with an ultra-ideal vertex of the honeycom"

(Bold emphasis mine.)
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adrian



Joined: 04 Nov 2014
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PostPosted: Mon Aug 24, 2015 7:22 am    Post subject: Reply with quote

Swami Polytope wrote:

The centroid, for triangle side length of 1, is ~0.419 (according to Mathematica), but ~0.544 (according to Antiprism). Not sure what I've done differently,


I see the difference now, ~0.419 is the centroid of volume and ~0.544 is the centroid of the vertices.

It is not too difficult to calculate the centroid of volume of a right regular antihermaphrodite from the two vertex heights. You can see that it is a pyramid with the base vertices truncated symmetrically to the base edge midpoints. In the case of the triangular antihermaphrodite the formula is

V_cent_ht = (a^2 - (3/4)*b^2) / (4*(a - (3/4)*b))

Where a is the apex height, and b is the other vertex height.

For your chesterhedron model a=1.25640778312683, b=.862948889958022, giving V_cent_ht=.41860490459199 .

[for the n-gonal antihermaphrodite, it looks like the formula for the centroid height should replace the (3/4) with (sin(pi/n))^2]
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guy



Joined: 11 Feb 2008
Posts: 75
Location: England

PostPosted: Mon Aug 24, 2015 9:11 am    Post subject: Reply with quote

On the hyperbolic honeycomb shown:

We can tell from its Schläfli symbol {3, 7, 3} that it is self-dual because the symbol is symmetrical.

I love the way it brings to life what one might call hyperbolic perspective. Traditionally a hyperbolic plane is represented as a disc, with objects of the same (hyperbolic) size getting smaller and smaller towards the edge. This rather mimics the way that equal-sized objects an ordinary plane, say on a chessboard, look smaller in the distance due to ordinary perspective. We can think of the disc boundary as being "further away" but it is an effort of will. By aligning the hyperbolic vault with a perspective view in this 3D image, the eye is fooled into seeing the hyperbolic squashing as ordinary perspective and automatically compensates for it.
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Swami Polytope



Joined: 14 Aug 2015
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PostPosted: Mon Aug 24, 2015 1:02 pm    Post subject: Reply with quote

Thx Adrian, & Guy. Interesting replies, both:) Such a fascinating topic. Just read up on abstract polytopes. FYI, duals seem so much easier in that context. It's when you try to realize a special case in a classical sense/space that things get complicated.
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