How to make the dual of the 2v icosahedron more regular?

[FLWQ]

Hi,

I don't know, but I think I used the simplest method to make a dual of
the 2v icosahedron (I created planes at the vertexes of the
icosahedron, that end when they intersect with the planes originating
from the other vertexes).

That results in it's dual (2v dodecahedron?).

When I let the icosahedron have a radius of 1, the sides of the
(regular) pentagon in the dodecahedron are 0.413. And the sides of the
(irregular) hexagon are 0.413, 0.334, 0.334, 0.413, 0.334, 0.334.

The pentagon has an area of 0.027 Sqm.
The hexagon has an area of 0.031 Sqm.

To me the hexagons look a bit too irregular. Is there a way to make
the sides more the same length?

[TaffGoch]

Alfred,

 

I played with this idea, myself, a while back. I got pretty close, but didn't complete my objective.

 

I suspect that this can be further refined, until all edges are equal, but the hexagons won't have 120-degree angles between edges. Perhaps my reasoning is off, but I'm pretty sure equal edges are possible.

 

The attached model has everything in separate layers, so you can turn them on/off, to examine the different structures.

 

Tonight, I'll give it some more thought, and see what I can come up with, as far as equal edges go.

 

-Taff

[TaffGoch]

Alfred,

 

You should note that the hexagon & pentagon face planes are not at equal distance from the newest polyhedron centerpoint (far-right polyhedron.)

 

This question is related to the Clinton Equal-edge Conjecture, and I'm pretty sure it's been "solved" already, for this 2v-dual configuration.

 

-Taff

[TaffGoch]

Alfred,

 

It was nagging at me -- that I'd pursued this to a satisfactory conclusion, already. I dug through "my stack of stuff," and found the model I'd experimented with.

 

Attached is a model with equal edges, and the hexagons look pretty "regular." (The hexagon face angles are about 4-degrees off, from 120-degrees.)

 

-Taff

[FLWQ]

Hi Taff,

Thanks :)

The differences I could discover in your (far-right polyhedron) are:

--The angles of the hexagon where the hexagon doesn't touch the
pentagon are 116. Where they do touch the pentagon they are 121.
--Also the distance between the pentagon and the centerpoint is 1.
Between the hexagon and the centerpoint it is 0.98.

I guess all you've done is moved the hexagons closer to the
centerpoint by 0.02? Probably not, because I don't understand how
that would result in those 116 degree angles.

I'll look into the Clinton Equal-edge Conjecture.

(It's more important to me that the result be regular and symmetrical
than that the edges are the same length. So I need to understand how
the different figures were created.)

[FLWQ]

Hi Taff,

I wrote the reply before this one before I saw your latest reply.

Now I'm off to bed :)

[TaffGoch]

 

Well, if you try to use regular pentagons AND hexagons, you wind up with something like the attached image.

 

You, ultimately, have to compromise, somewhere.

 

-Taff

[FLWQ]

I've tried understanding http://www.freewebtown.com/randome/ClintonEqualEdge.pdf
but that is over my head.

I've also found this http://hexdome.com/essays/isometric_struts/index.php
which lists the advantages and disadvantages of this method.

I've also looked at your the "Clinton conjecture" in your first file
(2v dual) again. In that model the vertexes are all a distance of 1
from the centerpoint, the lines are all also equally long. The only
thing I can discover is that, when you delete the lines to the center
of a hexagon and try to make a new face, you can't.

You can however draw two lines through the hexagon that cross in the
middle and end at the cornerpoints of the two adjacent pentagons,
which creates two planes. Then you can draw a line connecting the
points that don't have lines starting from them yet, which creates two
additional planes. At the center of the hexagon the distance of the
first two planes to the second two planes is 0.0044.

That seems to me a much smaller distortion than the other solutions.
Am I missing something?

[FLWQ]

How did you make your last example? I think I like it, and also might
like to use it in higher frequency structures ...

[TaffGoch]

On Dec 4, 5:30 am, Alfred van Dijk wrote:
>
> I've also looked at your the "Clinton conjecture" in your first file
> (2v dual) again. In that model the vertexes are all a distance of 1

> from the centerpoint, the lines are all also equally long....
>
> ... creates two planes. Then you can draw a line connecting the

> points that don't have lines starting from them yet, which creates two
> additional planes. At the center of the hexagon the distance of the
> first two planes to the second two planes is 0.0044.
>
> That seems to me a much smaller distortion than the other solutions.
> Am I missing something?

________________________

Alfred,

While the Clinton-conjecture spheres do have equal struts, and equal
central-radii, the "faces" are not planar, and are composed of six
triangles (peaked in the middle.) In the model, the Clinton-conjecture
hexagonal faces are representations of "dished" faces, to depict a
spherical surface/face (like a wok.)

You discovered that the hexagonal face "planes" are pretty close to
flat. This is what the Hexdome essay was referring to, when it
mentions non-flat faces. They're probably "flat enough" for plywood
sheathing, but not glass (as mentioned in the essay.)

If you modify vertices, to make the hexagonal faces flat, you have to
change the strut lengths or the central radius, producing the results
depicted in my previously-posted images/models. Like I said -- you
have to compromise, somewhere.

No, you're not missing anything -- just realizing the irregularities
of hexagonal geodesic domes.

-Taff

[TaffGoch]

On Sat, Dec 4, 2010 at 5:52 AM, Alfred van Dijk wrote:
>
> How did you make your last example?

__________________

 

Alfred,

 

Here are the steps I used, starting with a precise icosahedron. I hope that I've included enough steps for comprehension.

 

-Taff

[FLWQ]

Hi Taff,

Thank you!

I can't get over how much patience you have making these models :)

[TaffGoch]

Alfred,

 

It doesn't strike me as requiring patience. SketchUp provides a precise 3D "canvas" or "chalkboard" for the geometry concepts & hypotheses I see with my "mind's eye." SketchUp is a great 3D spacial-perception and visualization tool!

 

By the way, the final polyhedron is a special (equal-edges) version of a truncated triacontahedron. I've posted a modified model to the 3D Warehouse, which now includes an additional hidden layer, depicting the relation to the rhombic triacontahedron:

 

http://sketchup.google.com/3dwarehouse/details?mid=3ded5b0c2983fe05313cf1c2c0ee38cb

 

-Taff