Clinton's Equal Central Angle Conjecture

[TaffGoch]

Joseph D. Clinton's paper has been mentioned in a few discussions, in this group and in other geodesic forums. I've always intended to produce 3D digital models of the icosahedral examples from the paper, but never seemed to get around to it ('til now.)

I've posted a SketchUp model that includes the paper's 9 icosahedral spheres:

http://sketchup.google.com/3dwarehouse/details?mid=8d844ae2415824c0c901553684b98881

It seems that some of the angle data, from the paper, are off (as mentioned, previously, by Gerry & Adrian.) My models, therefore, are a bit "off" in a couple of the spheres, but I hope to resolve those discrepancies at a future date (again, when I can get around to it.)

You can find the PDF-format version of Clinton's paper at Dick Fischbeck's "Randome" site, and at Domerama. I was surprised that it's not posted in more places, online, so I've also attached it below.

[Gerry Toomey]

Here are central angles for 3 of the icosa spheres.

- Gerry

[TaffGoch]

Thanks, Gerry,

Are these the values mentioned before (in other discussion)? I was
planning to search through old discussions, to glean values from your
and Adrian's posts. This may save me the trouble. I do recall the
I{2,2} being off by a 1/100th-degree, and you commenting that it may
have been a "fat finger" typo.

Have you (or are you planning) "Solver-ed" values for any of the other
icosa systems from the paper?

I'm revisiting the model, now, to try-out your Excel values....

[Hector Alfredo Hernández Hdez.]

Hi all,

What keeps our relationship work "MX-method"?

See you

2013/6/2 TaffGoch <taff...@gmail.com>

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[TaffGoch]

I've incorporated Gerry's three revised values into the 3D model, and recalculated the two simplest icosa systems, to higher precision.

So far, 5 of the 9 icosa spheres employ "confirmed" central angle values.

-Taff

[Gerry in Quebec]

Taff,
For icosa sphere I{3,3}, Solver gives a central angle of 7.719953
degrees, very close to Clinton's value.

The three angles I posted earlier were from my own Excel files; I
didn't systematically search the earlier discussion threads.

I'll take a look at the remaining three polyhedra and my old files.
- Gerry

>  ClintonEqualEdge; Table 1(rev).png
> 59KViewDownload

[Paul Kranz]

Gerry:

 

Does this mean that these do not follow the phi rule of the icosahedron (i.e. the centers of four planar pentagons form the golden rectangle)?

 

Paul sends...

[Gerry in Quebec]

Hi Paul,
The spherical pentagon faces of these Goldberg-Clinton polyhedra,
unlike most of their spherical hexagon faces, can be flattened. And
the centres of those flat pentagons will all be equidistant from the
centre of the polyhedron. So, the icosa symmetry remains intact and if
you connect up an appropriate "quartet" of pent centres, you should
end up with icosa golden rectangle.

- Gerry

[TaffGoch]

Paul,

The icosahedron vertices are "fixed," serving as centerpoints for the pentagons, which remain regular. The golden ratio geometry holds true 

[TaffGoch]

Thanks, Gerry,

Modeling, successfully, in SketchUp, I can confirm that 6 of the 9 icosa examples are verified.

[Paul Kranz]

Gerry and Taff:

 

If the ployhedra all form a golden rectangle, than how can their central angles be different? Doesn't the central angle of the icosa have to be 63.4349488° because of the rectangle?

 

Paul sends...

 

 

 

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Very high regards,

 

Paul sends...

ReviveTheFlame.com

[TaffGoch]

Paul,

The golden-rectangle references apply to the icosahedron ONLY.

The central-angle references, of the conjecture, and of the tables posted by Gerry & me, refer to the arcs on the surface of the spherical polyhedra, and the angle formed by the lines drawn from the arc endpoints to the center of the sphere. Yes, they are, indeed, central angles, but they aren't the icosahedron central angle.

[TaffGoch]

Gerry,

For the I{2,1} central angle, my modeling produces 15.168713°

Have you "Solvered" the {2,1} icosa, yet, for comparison?

[Gerry in Quebec]

Hi Taff,
For the I{2,1}, I get a central angle of 15.305866º using Solver. More
to come later on the remaining 2 spheres.
- Gerry

>  ClintonEqualEdge; Table 1 (Rev).png
> 61KViewDownload

[Paul Kranz]

Taff:

 

Wonderful, thanks!

 

Paul sends...

[Gerry in Quebec]

Taff,
I tweaked the Solver scenario for the I{2,1} central angle. The new
value is 15.297679º.
- Gerry

[Gerry Toomey]

Here's a jpg of one of Taff's virtual models of sphere I{2,0}, planar version, accompanied by a photo of a physical model, 3.5 ft in diameter, made of plastic pipe. An interesting trait of this Goldberg-Clinton structure is that, with just one chord length, it can have two distinct shapes: a wire-frame sphere with all vertices equidistant from the centre and with one central angle for all chords, but non-planar hexes; or a polyhedron with two distinct radial lengths and two distinct central angles, with all faces planar.
- Gerry

[Paul From Geo-Dome]

Would this be more commonly know as a truncated rhombic triacontahedron?

On 4 Jun 2013, at 15:07, Gerry Toomey <toomey...@gmail.com> wrote:

Here's a jpg of one of Taff's virtual models of sphere I{2,0}, planar version, accompanied by a photo of a physical model, 3.5 ft in diameter, made of plastic pipe. An interesting trait of this Goldberg-Clinton structure is that, with just one chord length, it can have two distinct shapes: a wire-frame sphere with all vertices equidistant from the centre and with one central angle for all chords, but non-planar hexes; or a polyhedron with two distinct radial lengths and two distinct central angles, with all faces planar.
- Gerry

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<I{2,0}-Clinton-Goldberg-models-virtual+physical.jpg>

[Gerry in Quebec]

Yes.
http://en.wikipedia.org/wiki/Truncated_rhombic_triacontahedron
- Gerry

On Jun 4, 2:23 pm, Paul From Geo-Dome <ad...@geo-dome.co.uk> wrote:
> Would this be more commonly know as a truncated rhombic triacontahedron?
>

> On 4 Jun 2013, at 15:07, Gerry Toomey <toomey.ge...@gmail.com> wrote:
>
>
>
>
>
>
>
> > Here's a jpg of one of Taff's virtual models of sphere I{2,0}, planar version, accompanied by a photo of a physical model, 3.5 ft in diameter, made of plastic pipe. An interesting trait of this Goldberg-Clinton structure is that, with just one chord length, it can have two distinct shapes: a wire-frame sphere with all vertices equidistant from the centre and with one central angle for all chords, but non-planar hexes; or a polyhedron with two distinct radial lengths and two distinct central angles, with all faces planar.
> > - Gerry
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[TaffGoch]

Gerry,

I asked about the I{2,1} system, because it confounded me, for awhile. I could get many values to serve, which worked perfectly.

I realized that my symmetry criteria were not strict enough. The other system examples have mirror symmetry "built-in," so it is easy to overlook. With the {2,1} example, the proper rotational symmetry must be maintained, which I was not doing.

The attached image shows:

1) The "target" red box, where the arc endpoints must eventually coincide,

2) The icosa-face-center axis of rotation for 1st arc,

3) The icosa-edge-midpoint axis of rotation for 2nd arc,

4) The icosa "peak" vertical axis of rotation for the pentagon AND 3rd arc.

I was rotating the 3rd arc, exclusive of the pentagon. My mistake. I should have been rotating the pentagon & 3rd arc together. Note that the surface angles that the 3rd arc makes with the pentagon arcs must be congruent. This will produce the shortest arc length possible, and produces only ONE correct solution, rather than multiple "workable" solutions.

__________________________

I am assuming that you are, mathematically, approaching the solution, using the same rotational concept as I, when I initially attacked the problem -- i.e,; not rotating the pentagon and "protruding" arc, together. Am I correct?

__________________________

(Complicating the exercise, is the fact that changing the "test" arc length also changes the size of the pentagon, forcing me to adjust the position of the "attached" endpoint of the 3rd arc, before rotating the pentagon & 3rd arc "set.")

-Taff

[Gerry in Quebec]

Taff,
I had the same difficulty with the rotational aspects, particularly
the relationship between pentagon size and the skew angle between
pentagons. Rather than try to calculate the relationship, I decided to
simply fill in the known spherical coordinates -- for the corners and
centres of the spherical icosa triangle -- and then guesstimate the
coordinates of the remaining vertices in Joe Clinton's symmetry
diagram. In that diagram there are three pentagons and three hexagons.
I then used Solver to find a solution (i.e., adjust the estimated
coordinates) whereby all central angles would be the same. I also
included some other constraints based on the internal symmetry of
Clinton's diagram. But I got more than one "solution" which I found
confusing.

Just now I added a "sledgehammer" constraint to the Solver
scenario.... I calculated the area of one spherical pentagon and one
spherical hexagon. Multiplying the pent area by 12 and the hex area by
60 gives the area of the spherical polygon. The constraint was that
the area of the 72 polygons be exactly 4 pi, namely the area of a
sphere of unit radius. Solver found a new solution: central angle =
15.169069º. I think this gets around the rotational problem. And it's
significantly closer to the value you posted.

Did you find a new value too?
- Gerry
P.S. I can see why class III topology (is that the right word?)
doesn't get as much attention as classes I & II. It's really hard on
the old noodle.

>  Equal_Central_Angle_{2,1}.png
> 173KViewDownload

[Hector Alfredo Hernández Hdez.]

Ya me está dando mucha curiosidad el problema que mencionan.

quisiera entenderle, me hace falta tiempo para hacer

matematica interesante.....

Saludos

2013/6/4 Gerry in Quebec <toomey...@gmail.com>

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[Adrian Rossiter]

Hi Taff and Gerry

On Tue, 4 Jun 2013, TaffGoch wrote:
> I realized that my symmetry criteria were not strict enough. The other
> system examples have mirror symmetry "built-in," so it is easy to overlook.
> With the {2,1} example, the proper rotational symmetry must be maintained,
> which I was not doing.

...

> I was rotating the 3rd arc, exclusive of the pentagon. My mistake. I should
> have been rotating the pentagon & 3rd arc together.

It isn't necessary for preserving the rotational symmetry. If
you fix all the the vertices not on the pentagons and rotate
the pentagons consistently then the model will still have
icosahedral rotational symmetry.

> Note that the surface
> angles that the 3rd arc makes with the pentagon arcs must be congruent.
> This will produce the shortest arc length possible, and produces only ONE
> correct solution, rather than multiple "workable" solutions.

This is a separate point, and it seems reasonable, as the equalising
of angles occurs as a feature of minimising "perimeter", but it would
be nice to see something more formal.

A couple of examples are the Fermat point on the plane or sphere

http://www.jstor.org/discover/10.2307/3618527?uid=3737952&uid=2129&uid=2&uid=70&uid=4&sid=21102074905983

Or Fagnano's problem in the plane

http://www.math.uoc.gr/~pamfilos/eGallery/problems/Fagnano.html

Which also appears to hold on the sphere

http://books.google.es/books?id=fhoAAAAAMAAJ&pg=PA76&lpg=PA76&dq=inscribed+spherical+triangle+of+minimum+perimeter&source=bl&ots=iugdhefjtE&sig=bElvAqaGQeRJjIX-ZxMaor0i07Y&hl=en&sa=X&ei=QO2uUfmUAbGg7Aa5vYGIBA&ved=0CC8Q6AEwAA#v=onepage&q=inscribed%20spherical%20triangle%20of%20minimum%20perimeter&f=false

Adrian.
--
Adrian Rossiter
adr...@antiprism.com
http://antiprism.com/adrian

[Gerry in Quebec]

I simplified the Solver routine, but retained the "area of sphere"
constraint. I also increased the accuracy of the known icosa
coordinates. The central angle I now get is 15.168726º, very close to
15.168713, the value you first posted, Taff.
- Gerry

[TaffGoch]

Gerry,

Looks good enough (to me, anyway) to "confirm" the I{2,1} sphere. So, that's "7-of-9" icosa values verified, thus far.

[TaffGoch]

Hector,

It is, indeed, an interesting math exercise.

It is particularly curious that a child could produce any of these, physically, using equal-length pipe-cleaners.  Something that initially seems so simple can be so mathematically-complex (or, at least, quite confusing.)

-Taff

____________________________

[TaffGoch]

Thanks, Adrian,

I'm definitely going to dig into the Fermat-point details -- looks promising.

Thanks, particularly, for the link to "The Mathematical Repository" ebook. I keep collecting math links from Google Books, and find the old, historic math books to be founts of information. Some of the grade-school math books are stunningly advanced. It puts our contemporary high-school curriculum to shame.

-Taff

[Hector Alfredo Hernández Hdez.]

Thanks Taff,

There are many issues that seem easy and are hard, I like the that seem difficult and are easy :)

The pictures help a lot, but the equations are inevitable.


I want to learn to solve systems of equations as does Gerry, as there are many interesting problems.

See you soon

2013/6/5 TaffGoch <taff...@gmail.com>

--

[Gerry in Quebec]

Taff et al,
For the Goldberg-Clinton I{5,0} sphere, the "Solvered" value I get for
the central angle is 8.023548º -- very similar to Joe Clinton's
number: 8.0237.

Solver was much more cooperative on this one than on the class III
sphere.
- Gerry in soggy Quebec

>  ClintonEqualCentralAngle; Table 1.png
> 66KViewDownload

[TaffGoch]

Table filling in, nicely.

Class-III subdivision models okay, so long as I use method-1. Any other method, though, is, truly, "uncooperative."

Since method-1 produces the most-divergent chord lengths, the Class-III subdivision I don't generally construct by modeling. I use calculations for those, employing a "charge-repulsion" method (Thomson problem):

http://thomson.phy.syr.edu/thomsonapplet.php

...selecting "ico{m,n}" and "edge springs"

It didn't produce equal central angles for the I{2,1}, though, so I did model that particular tessellation. I wouldn't want to have to model any higher-frequency Class-III....

-Taff

[Gerry in Quebec]

Taff,
For the I{4,0} sphere, the central angle via Solver is 10.029939º --
again, pretty close to Clinton's value (10.0277).
- Gerry

>  ClintonEqualCentralAngle; Table 1 (Rev).png
> 76KViewDownload

[Gerry in Quebec]

Taff,
Would a set of Cartesian coordinates help you with the modeling work
for any of these spheres?
In the case of the class III sphere (I{2,1}, I have the coordinates
related to the Solver exercise, but I don't know whether these would
be enough to create a SketchUp model. I haven't tried to produce any
visuals of these spheres using OFF files in Antiview. If I did, the
class I structures would be the easiest to model.

- Gerry

On Jun 6, 3:57 pm, TaffGoch <taffg...@gmail.com> wrote:

[TaffGoch]

Gerry,

Thanks. I have been able to model easily, using only rotations, so I don't really need the cartesian coordinates. The only reason to use them would be to 3D model them to double-check the "fit" of the central-angle arcs on the sphere. (Hmmm, perhaps that would be a good idea, anyway.)

I fine-tuned my I{4,0} sphere model, and got precision down to 10.030282°. While I do know that value is a "little" too big, I don't know how "little." Sending the cartesian coordinates would permit a double-check on your value, as well as mine, so go ahead and attach a file containing the coordinates.

-Taff

[TaffGoch]

The precision of my modeling of the I{4,0} central angle is now 10.030225°....

[Gerry Toomey]

Hi Taff,

Here's an Excel file with the XYZ coordinates and a vertex diagram of I{4,0}.

- Gerry

[TaffGoch]

Gerry,

Confirmed your values, by virtue of 3D modeling of the cartesian coodinates:

10.029939848° central angle

0.1750555 arc length (unit sphere radius)

[TaffGoch]

I think the icosa system model is finished...

http://sketchup.google.com/3dwarehouse/details?mid=8d844ae2415824c0c901553684b98881

...unless you spot differences from your values (and, I think there are potentially, a couple, at least.)

-Taff

[Blair Wolfram]

If the central angles are all equal, couldn't the paper just as well be called, "Equal Edge Conjecture" without changing any context?

Blair

--

[TaffGoch]

Yep. In fact, I changed my filenames from "EqualEdgeConjecture..." to "EqualCentralAngleConjecture..." to match Joe Clinton's article title, to both correspond and avoid questions from novices.

-Taff

[Gerry in Quebec]

Taff,
Thanks for posting the SketchUp model of the Clinton Conjecture
examples. I see a couple of very minor discrepancies. For I{5.0}, the
tape measure gives two chord lengths depending which edge you measure:
0.139917 and 0.139926. For I{2,1}, there are also two chord lengths:
0.263971 and 0.263972. The values I get with Solver are, respectively,
0.139923 (central angle = 15.168726º) and 0.263972 (central angle =
8.023548º). Of course, for building physical models, the SU values are
fine.
- Gerry

On Jun 6, 9:47 pm, TaffGoch <taffg...@gmail.com> wrote:
> I think the icosa system model is finished...
>

> http://sketchup.google.com/3dwarehouse/details?mid=8d844ae2415824c0c9...

>
> ...unless you spot differences from your values (and, I think there are
> potentially, a couple, at least.)
>
> -Taff
>

>  ClintonEqualCentralAngle; Table 1 (Rev).png
> 78KViewDownload

[Dick Fischbeck]

Is anyone trying to figure out a proof? That would be big.

[Gerry Toomey]

 

 

Here are a couple of errata in Joe Clinton's "Conjecture" paper that I think are worth pointing out. Images of pages 1 & 4 are attached with corrections.

 

Coincidentally, in Geodesic Math and How to Use It (page 39), Hugh Kenner makes the same error that Clinton makes on page 1, regarding octahedral polyhedra.... Understandable, I suppose, because the number 8 is so strongly associated with octahedral symmetry.

 

- Gerry

[Gerry in Quebec]

Gee, Dick. I thought YOU were working on it :-)

From page 8 of Clinton's Conjecture paper:

"Since first proposing the conjecture stated herein in a paper to the
5th International Conference
on Space Structures, held at the University of Surrey, Guildford, UK
on 19-21 August 2002, it
has come to my attention that Dick Fischbeck Ref 16 has been working
on similar problem.
He randomly arranges cones of consistent diameter on the surface of a
sphere based on the
spherical excess angle of a polyhedron. He has also proposed a
variation where the central
angles of the cones may consistently generate random tetrahedral
pyramids that will tessellate
the sphere. Perhaps a careful study of his RanDomes will provide a
solution to the Goldberg
polyhedra equal central angle problem."

- Gerry

On Jun 7, 11:12 am, Dick Fischbeck <dick.fischb...@gmail.com> wrote:
> Is anyone trying to figure out a proof? That would be big.
>

[TaffGoch]

I revised the model, to double-check and confirm and incorporate your values.

The table, included in the model, is depicted below:

[TaffGoch]

As suggested by Dick Fischbeck's comment:

[Paul Kranz]

Are the platonic solids the only polyhedra that represent a network verticies that all equidistant from each other?

 

Paul sends...

 

On Sat, Jun 8, 2013 at 4:18 PM, TaffGoch <taff...@gmail.com> wrote:

As suggested by Dick Fischbeck's comment:

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Very high regards,

 

Paul sends...

ReviveTheFlame.com

[TaffGoch]

No. The Platonic solids are not only equal-edged, but equal-angled. Each face is a regular polygon.

If you remove the equal-angle criterion (which we see in these subject examples,) then seemingly-unlimited possibilities present. In fact, that's part of Clinton's "conjecture" -- is there an limit?

[TaffGoch]

Missing from the paper's listing: Icosa{3,1}

This is not THE solution; merely, ONE possible solution.

Edge central angle: 11.131850°

Edge linear length: 0.193983

Edge arc length: 0.194288

[Gerry Toomey]

Taff,
For comparison, here are my I{3,1} results from Solver:
Edge central angle: 11.135312º 
Edge linear length: 0.194042 
Edge arc length: 0.194348
XYZ coordinates attached.

- Gerry

 

>  Icosa {3,1}.png
> 127KViewDownload

[TaffGoch]

I'm sure the Solver results are preferred. I shouldn't have posted 6-decimal precision, because I had only worked precision down to about 3 points. (I know better than to do that. Shame on me.)

Have you adjusted (your head) to the class-III solutions? No more noodle harshness?

________________________

I looked for the Solver add-in for my Excel 2007, and found plenty of references of how to add it, but no luck. Turns out, I have "Excel 2007 Professional Plus," which is distributed through educational systems, and doesn't come with add-ins, except for the one primary statistical library add-in, "analys32.xll" 

I even updated Excel 2007, through the Microsoft update download, to Service-Pack-3, and that didn't do the trick. (Apparently, it checks to see if you have the long-fancy-named "crippled" version, and won't include add-ins.)

I haven't tried to find a copy of the "solver.xlam" or "solver.xll" file, to see if I can download/acquire just it, and put in the right directory for loading into Excel. That may be my next pursuit, even if it is only to see if the procedure will work.

-Taff

[TaffGoch]

Correction:
"...one more fixed rotation axis, that being the centerpoint of edge 8/9 (as well as 5/6 and 20/21."

Should read:
"...one more fixed rotation axis, that being the centerpoint of edge 8/9 (as well as 5/11 and 20/21.)"

[TaffGoch]

Interesting comparison. I'm glad I didn't strive for higher precision.

For those following along, note that the solutions involve (in fact, require) rotations around certain fixed points. Using the Toomey image notations, vertex 13 is fixed, as is the central axis of the pentagon, vertex 1. The Excel iterative solution employs these as essential criteria.

For my modeling, using "manual" manipulation/iteration, I employed one more fixed rotation axis, that being the centerpoint of edge 8/9 (as well as 5/6 and 20/21.

I also rotated the pentagon AND it's 5 "protruding" edges as a whole, which produces congruent surface angles for the protruding arcs. Gerry did not require that limitation, and the protruding arcs are free to rotate around vertices 2 & 3. The resulting difference can be seen in the side-by-side image. (I am surprised that it made such little difference in our central-angle results.)

Note that I can not model with the pentagon protruding arcs being free. That introduces too much complexity, and can only be "solved" using computer successive mathematical iteration (as far as I can tell.)

-Taff

[TaffGoch]

Gerry,

This may be of interest to you:

[Paul Kranz]

This first time I saw this pattern was in the dimple pattern of a golf ball. But, I can't remember which one.

 

Paul sends...

 

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[Paul Kranz]

Taff:

 

If you were to place 20 points equidistant from each other on the surface of sphere, you would end up with the verticies of an icosa. Am I correct that the highest number of points that you can place equidistant from each other on a sphere is 20?

 

Paul sends...

On Sun, Jun 9, 2013 at 5:55 PM, TaffGoch <taff...@gmail.com> wrote:

No. The Platonic solids are not only equal-edged, but equal-angled. Each face is a regular polygon.

If you remove the equal-angle criterion (which we see in these subject examples,) then seemingly-unlimited possibilities present. In fact, that's part of Clinton's "conjecture" -- is there an limit?

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[Adrian Rossiter]

Hi Taff

On Wed, 5 Jun 2013, TaffGoch wrote:
> I'm definitely going to dig into the Fermat-point details -- looks
> promising.

The Fermat point is interesting with regard to polyhedra as it
is used in the construction of the uniform snub polyhedra from
their Wythoff symbols (e.g. when a snub dodecahdron is constructed
from the symbol '|5 3 2').

Adrian.
--
Adrian Rossiter
adr...@antiprism.com
http://antiprism.com/adrian

[Adrian Rossiter]

Hi Paul

On Tue, 11 Jun 2013, Paul Kranz wrote:
> If you were to place 20 points equidistant from each other on the
> surface of sphere, you would end up with the verticies of an icosa. Am I
> correct that the highest number of points that you can place equidistant
> from each other on a sphere is 20?

The 20 points from your example [dodecahedron] aren't all
equidistant from each other. Only certain pairs of points
have a common separation (the edges have the minimum
seperation). However, if prisms and antiprisms are
acceptable then the number of points is unlimited.

[TaffGoch]

Adrian,

I read that, too, BUT...

I haven't been able to hit the snub vertex with a Fermat point. I've tried great circles, spherical triangles, plane triangles, etc. and several are REAL close. They never hit, though, right on the money. 

I have to wonder whether this statement is a "loose" or "approximates" type of definition (which is no definition, at all, since it doesn't DEFINE the point.)

[Adrian Rossiter]

Hi Taff

The vertex isn't the Fermat point (although there is PDF that came
first in my search saying saying it is!)

You can find the vertex by reflecting the Fermat point in the sides
of the Schwarz triangle, to give a new triangle, and the vertex is
the circumcentre of this triangle.

There are problems if the Schwarz triangle has an angle greater
than pi/3, as the Fermat point is no longer equiangular. But it is
really an equianguler point that is of interest rather than a strict
Fermat point, and this still exists and can be used (there are
generally 8 such points if you can join the point to the triangle
vertices with arcs longer than pi).

There is a set of models in Antiprism that are calculated from
the Wythoff symbol, and use the method with the Fermat point.
Unfortunately I don't have an expression for the equiangular
point (it looks possible to derive one, but very long-winded),
and so I find it iteratively.

Here are a couple of examples, of the snub dodecahedron and the
snub tetrahedron (image attached).

antiview wythoff_:5_3_2
antiview wythoff_:3_3_2

[TaffGoch]

Adrian,

It was that first-search-result PDF (of a slide presentation) that has confounded me, because it is SO off-the-mark. The slide in question referred to the snub-dodecahedron & the Fermat point, so that is the case with which I've been experimenting.

I'll try the reflection technique, for triangle circumcenter production. Thanks for the lead. (I have not found this described anywhere!)

-Taff

[Adrian Rossiter]

Hi Taff

On Tue, 11 Jun 2013, TaffGoch wrote:

> I'll try the reflection technique, for triangle circumcenter production.
> Thanks for the lead. (I have not found this described anywhere!)

It is mentioned in Uniform Solution for Uniform Polyhedra by
Zvi HarοΏ½El

http://www.math.technion.ac.il/S/rl/docs/uniform.pdf

Section 4.4 talks about the snubs. The vertex construction
is given in parenthesis!

[Gerry in Quebec]

Taff,
Thanks for the comments and comparison diagram for Icosa {3,1}.
Yesterday I noticed that spherical angle 1-2-6 in my spreadsheet &
numbered symmetry diagram was not exactly equal to angle 1-3-4. They
should be equal, though, to maintain symmetry around the main axis
passing through points 0 and 1. So I added that equivalence constraint
to the Solver scenario. (The counterpart angles at vertices 24 & 25
are automatically adjusted to match.) The resulting central angle of
edges shrinks a bit, from 11.135312º to 11.133932º, a little closer to
your value 11.131850º.

So, I've thrown everything but the kitchen sink into the list of
constraints, and I'm sure there's some redundancy. I also used the
cumulative spherical area constraint, as I did with the other class
III sphere you modeled, I{2,1}. That constraint requires the area of 5
spherical pent triangles plus the area of the 6 spherical hexagons (3
of each type) to equal the area of the spherical icosa parent triangle
(1-24-25), namel 4pi / 20, or pi / 5.

- Gerry

On Jun 10, 7:44 pm, TaffGoch <taffg...@gmail.com> wrote:
> Gerry,
>
> This may be of interest to you:
>

>  Icosa{3,1} compare.png
> 90KViewDownload

[TaffGoch]

Thanks, Adrian,

(The Har'El paper could, seriously, use some diagrams!)

[TaffGoch]

Gerry wrote:
"The resulting central angle of edges shrinks a bit, from 11.135312º
to 11.133932º, a little closer to your value 11.131850º."

___________________

Wait... So, my result remains shorter? That sounds, to me, like a
closer approximation. (Right?)

Isn't that (shouldn't that be) one of the criteria? Relying on just
mental reckoning, I would think so. We all know, however, that such
"logical" conclusions can be wrong. That's why we formulate and define
hypotheses, and then proceed to prove or dis-prove.

(I'll have to re-read Clinton's paper, to see if "shortest length" or
"narrowest angle" is mentioned.)

-Taff

[Gerry in Quebec]

Taff,
I redid the Icosa {3,1} spreadsheet, this time calculating all the
symmetry relationships for the class III geometry using spherical
trig. So, this time Solver was used only to adjust the positions of
the four key vertices required to equalize the central angles.

In tinkering with the size of the pentagons (theta coordinates of the
vertices on the pentagon perimeters), I ended up with 4 feasible
Solver solutions for the central angle, all close in value:
11.133932º
11.139365º
11.143012º
11.145712º

I couldn't get Solver to reproduce your value of 11.131850º, which is
smaller than the ones Solver found.

It's a bit surprising, to me at least, that there are multiple valid
solutions. And there are probably other sets of coordinates that also
yield equal central angles.
- Gerry
P.S. If it's of interest to you or others, I will post the four sets
of cartesian coordinates.

[TaffGoch]

Gerry,

I revised my model and enhanced the results, improving the precision to 6 decimals, for the chord factor:

Edge central angle: 11.130123°

Edge arc length: 0.194257 (geodesic factor)

Edge linear length: 0.193952 (chord factor)

I'm not sure how the linear precision affects the angle; whether it also improves to 6 decimals.

_______________________

I can only achieve one solution, and am wondering whether the criteria I employ makes the difference. I'm still keeping the arcs that "protrude" from the pentagon at congruent angles, with respect to the pentagon. (This keeps the protruding arcs "pointing" directly at the center of the pentagon.) Perhaps, such a criterion would limit your results to one solution, if you can devise a way to incorporate into the spreadsheet.

Is there a way that you can use the above values as a starting layout for your Excel iterations, to see how they proceed from there?

-Taff

[Gerry Toomey]

Taff,

Success.... I think.

 

As you suggested, I altered the spreadsheet to include a new condition: the arcs protruding from the pentagon corners MUST point directly to the centres of the pentagons. I did this by making the phi angle (longitude) of vertex 6 equal to the phi angle of vertex 2 (see the attached vertex map). Symmetry calculations ensure all other protruding arcs in the diagram automatically point to their respective pentagon centres as well. I also used your latest central angle, 11.130123º, as a starting value in the Solver scenario.

 

This time Solver found a solution very close to your central angle: 11.130155º. Variation among the central angle values begins at the 8th decimal place: 11.130155259 versus 11.130155260.

So it looks like Goldberg spherical polyhedron Icosa {3,1} has multiple configurations that respect the equal-central-angle requirement of the Clinton Conjecture, but probably only one solution if the more restrictive symmetry constraint is imposed.

- Gerry

 

 

 

 

 

 

>  Icosa {3,1}.png
> 118KViewDownload

[TaffGoch]

I reckon that all class-III tessellations will present multiple solutions, since "twist" is introduced in the exercise.

Since class-I and class-II have no twist, the constraints are more rigid, readily producing one iterative solution.

I don't know how higher class-III frequencies would work-out, but I'd be willing to bet that some apparently-arbitrary criteria are required. (My "pentagon protruding arc" criterion was logical reasoning, and I have no proof that this produces the shortest arc. It merely seemed reasonable, at the time.)

As before, I suspect that your results are the more-precise solution, since my modeling is based on "manual" manipulation of arc/edge rotations. Iterative, yes, but tedious & time-consuming. I always yield to computer precision & speed.

________________________

I would be interested in visually comparing our results. I have been surprised at how much the surface angles change, with only a minor change in the central-angle parameter. Have you produced a cartesian-coordinate table or "csv" file?

-Taff

[Gerry Toomey]

Here's a csv file of the cartesian coordinates of vertices 0-29, following the vertex map of Icosa {3,1}, posted earlier.

Cheers,

- Gerry in Quebec

[TaffGoch]

Spot on -- No discernable visual difference, at 4-to-5 decimal precision....

[TaffGoch]

Should you want the SketchUp 3D model....