Class-II, Method-3

[TaffGoch]

I finally reproduced Duncan Stuart's chord factors, given in Domebook2, Hugh Kenner's book, and Joe Clinton's NASA report, for an 8v, Class-II, Method-3 geodesic sphere.

The textual descriptions in the documents are garbage. They don't help much, at all! You can not get the chord factor results, following the descriptions.

The problem: All other method descriptions employ a plane triangle, which is 1/6th of an icosahedron face. Method-3 actually requires that the guidelines be "drawn" on a spherical triangle -- NOT a plane triangle. This is somewhat depicted in an illustration in Kenner's book, but the text doesn't mention the use of a spherical triangle. 

I reproduced the chord factors by using plane rotations & translations, and their lines-of-intersection, producing a perfect match. (The Kenner illustration gave me the clue to use plane rotation/translation intersections.) The method is reminiscent of that employed by Temcor/Richter.

I'll be correcting the 3D Warehouse model, to change the Method-3 spheres, after I've modeled a sphere for each frequency.

-Taff

[TaffGoch]

The 3D Warehouse model has been updated, to introduce the correct Method-3 geodesic spheres:

http://sketchup.google.com/3dwarehouse/details?mid=cadb8096cff1ceccd169a1cbcd45a8f8

The Method-3 subdivision produces a chord-count equal to the frequency. The chord factors now fit the description from Hugh Kenner's book, "Geodesic Math and How to Use It":

"...each side-edge division will propagate itself by zigzagging all the way down the main triangle, and each horizontal division will recur as a set of parallel reflections." (page 67)

The Domebook2 description is wholly inadequate. Credit for the method to Duncan Stuart, but I still can't see how anyone could generate the correct geometry from the Domebook2 description.

________________

Extra tid-bit: If you look at the Synergetics diagram, from the recently-posted PDF file, you'll see that the chord-factors "zigzag" down, just as described above. Coincidence? Perhaps, a clue as to what method was used?

-Taff

[TaffGoch]

Here are the steps I used to construct the class-II, method-3 sphere.

Image 01: Start with a 1/6th symmetry right triangle, from a unit-radius icosahedron primary polyhedron triangle (PPT)

Image 02: Produce the 3 triangle-boundary planes, through the center of the icosahedron

Image 03: Rotate copies of the plane defined by the short right triangle edge, using the long-edge plane as the plane of rotation, around the centerpoint of the icosahedron

Image 04: Produce unit radial lines along the plane intersections (the boundary planes and the rotated planes)

(to be continued)

[TaffGoch]

Image 05: Translate parallel copies of the long-edge boundary plane, to each of radial-line endpoints

Image 06: Produce the first two "internal" chords, from the radial line endpoints. Note that they will be equal length

Image 07: Rotate the two aforementioned chords, using the long-edge boundary plane as the plane of rotation, and the icosahedron centerpoint as the axis

Image 08: Produce the next new pair of equal chords, from the left-most radial-line endpoints on the second & third parallel planes

(continuing...)

[TaffGoch]

Image 09: Rotate the left-most, previously-produced chord, using the same rotation plane/axis as before

Image 10: Produce the next pair of equal chords, from the radial-line endpoints between the third & fourth parallel planes

Image 11: Produce the next chord, from the radial-line endpoints between the fourth plane and the top-most right-triangle apex

Image 12: Produce the remaining chords, connecting the primary boundary radial-line endpoints

(continuing....)

[TaffGoch]

Image 13: The faces produced

Image 14: Delete internal radial lines, as they are no longer needed. Keep the three "corner" radial lines, for use as rotation axes

Image 15: Mirror and rotate the 1/6th construction, to produce the icosahedron-face-boundary shell/cap

Image 16: Produce the "half-panel" triangles, along the edges of the icosahedron-face boundary

(continuing....)

[TaffGoch]

Image 17: The completed 8v, class-II, method-3 geodesic sphere

_________________________

Note that the procedure can be applied to class-I tessellations, as well.

Also note that method-4 employs the same procedure, except that the long-edge plane (of the initial right-triangle) is used for the rotated-plane copies, and the short-edge plane is used to produce the parallel plane copies

-Taff

[TaffGoch]

Addendum: When I refer to the short-edge and long-edge, of the right-triangle, I refer to the right-angle edges, not the hypotenuse.

Sorry, if my failure to make that distinction confused the procedure description.

-Taff

[TaffGoch]

I just tested Method-4, and the segmented boundary-crossing arcs look much smoother than Method-3.

Still only 8 chord-factors for an 8v, class-II, method-4 tessellation.

Color me impressed....

-Taff

[Hector Alfredo Hernández Hdez.]

Whats about chord factor layout?

have you ever use this ways in octahedron?

thanks


PD: I'm surprised by 30 years of geodesic calculations before and after the mac / pc, you are my idol.

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[TaffGoch]

Hector,

I haven't tried an octahedral, method-3, tessellation. I've been too busy with the icosa subdivisions. (I, now, have to add method-4 to my 3D Warehouse model!)

So far, I've left chord factors to the end user, measuring the edge lengths in the SketchUp model. I may, as I did with your subdivision method, eventually produce chord-factor images, based on the method-3 & method-4 models.

So much to do. So little time....

___________________

You are too kind, in your compliment.

I'm just sharing my exploration of currently-documented methodology (however inadequately-documented.) I haven't produced anything that is actually new or revolutionary. I consider myself a fairly-skilled mimic, though, replicating other people's work.

I can only hope that my explorations are beneficial to others. (SketchUp makes a pretty good 3D exploratory tool.)

-Taff

[TaffGoch]

Hector,

Here are diagrams of chord factors, from both my model, and from Hugh Kenner's book, "Geodesic Math and How to Use It"

Note that my diagram shows a triacontahedron face, not an icosahedron. Kenner's illustrations are of the top-half of a triacontahedron face.

-Taff

[TaffGoch]

To preclude confusion (from inadequate description,) I produced an illustration of the construction geometries for Methods 3 and 4

[Hector Alfredo Hernández Hdez.]

what about use brow planes of method 3 and grey plans of method 4?
take your time...

On Thu, Oct 20, 2011 at 9:40 AM, TaffGoch <taff...@gmail.com> wrote:

To preclude confusion (from inadequate description,) I produced an illustration of the construction geometries for Methods 3 and 4

[Gerry in Quebec]

Thanks, Taff,
Much learned from this exercise. I can see how some confusion might
arise from Kenner's descriptions (plural ... p. 67 & Appendix 1, p.
163). With the help of spherical trig, which is the most comfortable
way for me to get at the numbers, the Method 3 exercise now seems
pretty straightforward.
Cheers,
- Gerry

>  Method3Chords.png
> 167KViewDownload
>
>  Kenner_Illustration_10.12.png
> 95KViewDownload
>
>  Kenner_Illustration_10.11.png
> 117KViewDownload

[TaffGoch]

Gerry,

I suspect that you already know that, in the "Data" section of his book, Kenner tabulates results, for Class-II,Method-3, providing...

...spherical coordinates, θ and φ.

-Taff

[TaffGoch]

Hector,

Your suggestion will produce a valid subdivision, but there are 17 chord factors, for an 8v frequency.

(I don't know if you realize that your "Mexican" method can be applied to a triacontahedron half-face, producing 8 chord factors.)

-Taff

[Gerry in Quebec]

Yes, I've looked at Kenner's coordinates and chord factor tables for
class II, method 3 "icosa" (or howevever you wish to call it). I
thought I might do one exercise in Excel, 16v, to demonstrate the
math. The only problem is that the spherical trig is expressed in
Excel notation, which means it's not very transparent for other people
if I post it. Same problem with the 6v Temcor calculations I did that
Dondalah was asking about. It's a matter of converting "Excelese" to
standard trig notation (as in Kenner's book).... A bit like converting
Quebecois jouale to European French!

Tedious exercise. Probably a waste of time considering Kenner gives
all the necessary data for frequencies 4, 6, 8, 12 and 16.
- Gerry

[Hector Alfredo Hernández Hdez.]

Thanks Taff.

I would to make draws like yours.
But I dont know how to start. I know some toos of Autocad.
will be easy to you say to me what is the beginig?

MAYBE ISNOT A GOOD IDEA, only that 8 chord factor  will in a more undertabla layout, let me reproduces last ones ..

See you.

[Gerry in Quebec]

Good question, Hector. What's the best starting point? I'm in a
similar situation!
- Gerry

On Oct 20, 6:24 pm, Hector Alfredo Hernández Hdez.

<hectorh...@gmail.com> wrote:
> Thanks Taff.
>
> I would to make draws like yours.
> But I dont know how to start. I know some toos of Autocad.
> will be easy to you say to me what is the beginig?
>

> On Thu, Oct 20, 2011 at 2:35 PM, TaffGoch <taffg...@gmail.com> wrote:
> > Hector,
>
> > Your suggestion will produce a valid subdivision, but there are 17 chord
> > factors, for an 8v frequency.
>
> > (I don't know if you realize that your "Mexican" method can be applied to a
> > triacontahedron half-face, producing 8 chord factors.)
>

> *MAYBE ISNOT A GOOD IDEA, only that 8 chord factor  will in a more

> undertabla layout, let me reproduces last ones ..
>

> See you.*

>
>
>
>
>
> > -Taff
>
> > --
> > You received this message because you are subscribed to the "Geodesic Help"
> > Google Group
> > --
> > To unsubscribe from this group, send email to
> > GeodesicHelp...@googlegroups.com
> > --
> > To post to this group, send email to geodes...@googlegroups.com
> > --

> > For more options, visithttp://groups.google.com/group/geodesichelp?hl=en- Hide quoted text -
>
> - Show quoted text -

[Hector Alfredo Hernández Hdez.]

I am undertood...
this is the motive to use a C Language programming..., but can be a headhache too . . . :)
A good program should work fine for "any" n value, however we need to be a good programmer.

last one is the second step.

The first are CAD exploring and Calculation in spreadshet for low "n".

We are in RIGHT way.

Please accept my apologies for not being able to apply more time to this work.
well a little bit, sometimes.

[TaffGoch]

Learning Google SketchUp became a little easier for my when I realized that "stick-building" was not the way to go.

What worked better, for me, was to produce planes, whether square or circular, for subsequent rotations and intersections.

Some stick-building can be used, but I do so only to produce planar faces, and to produce lines-of-intersections between planes. I stick-build vertical and horizontal lines, which are easy, because SketchUp will snap to x,y,z axes. I can then rotate these lines, around any plane I desire, but you have to produce the plane-of-rotation beforehand.

YouTube has a lot of video tutorials, from beginner-to-expert. That's a really good starting resource. Google produced a lot of the tutorials, and have their own YouTube channel:

http://www.youtube.com/user/SketchUpVideo#g/u

Individuals have also produced a lot of YouTube videos, which you can find by using YouTube search.

There is a SketchUp Help Group, which has compiled a list of learning, video & tip resources:

https://sites.google.com/site/sketchupsage/resources

(In fact, I started providing geodesic help at the SketchUp Help Group, before initializing the Geodesic Help Group.)

-Taff

[TaffGoch]

By the way, I start all my geodesic tessellations by building on an icosahedron "template"

I've read a lot of SketchUp Help Group posts, regarding how hard it is to build an accurate icosahedron. I respond to all of them, by pointing to a model that I used to build a buckyball. I first constructed an absolutely-precise icosahedron, by using the golden section "trick" that is documented in many places. Here's that buckyball model:

http://sketchup.google.com/3dwarehouse/details?mid=3af9af9384d2097858f8f3aa29de1c06

Note that Google's own icosahedron model is not precise. It's close, to a few decimals, but the edges are all NOT equal. Don't use it! Since I needed the 6-decimal precision (SketchUp's maximum,) I chose to correctly build my own.

____________________

Attached is my starting icosahedron "template" model. Note that the icosahedron "group" is locked, since I use it to "draw" new lines on it, rather than modifying it. Whenever you don't need the template, or it "gets in the way," you can turn off it's visibility by unchecking it's layer, in the Layer dialog box (which I keep displayed all the time.)

Note that SketchUp layers are NOT like Autocad layers. SketchUp layers control visibility only. They do not separate entities. Instead, use groups & components to separate entities.

Components differ from groups, in that copies of components are all "linked." Change the internal entities of one component, and all the component copies will share the change. This is not true of groups. Personally, I use components almost exclusively. One of component's other advantages is that you can change the orientation of the component's internal axes. Groups don't provide this feature. Components have the advantage, in my opinion, so why use groups?

You will find, for example, that the icosahedron model is a group, composed of components. Each internal component is an icosahedron face. If you build your models this way, you can create a tessellation of a face component, and all the others will automatically mimic it.

That's enough, for a start. By the way, there is an official online SketchUp Users Guide that describes all the tools and entity descriptions:

http://sketchup.google.com/support/bin/answer.py?answer=116174

-Taff

[Dondalah Proust]

Taff, your exercise makes Method III look like a cinch.  Kenner's Illustration 10.11 makes it plain.

Can you make a drawing like 10.11 for Method IV?  I can't quite translate Illustration 10.11 into a Method IV graphic.

Although Illustration 10.11 is for Class II Method III, I don't see why it couldn't apply equally well to Class I, provided that the frequency be even.  Do you agree?

Gerry, if you send me the Excel language for the geometry, I can help you translate that into standard math notation.  Then we can share it with the group as standard math.  I like the approach, and I think you do too, of  creating the vertices in double precision floating point with 18 digit accuracy, outputting the vertices in OFF format, and using Adrian's off_to_3ds program to translate the output into a Sketchup model.  This way, we get beyond the 6 decimal accuracy limitation, and we make sure the radius is one in the model.

Hector, give us time, and we'll have you fluent in C.  This would be a good project to start on, because you already understand the subject.

As all of you may know, the co-founder of the C language left this Earth just a couple of days ago, Dennis Ritchie.  He'll be missed, along with Steve Jobs.  The Kernighan and Ritchie book, called The C Language, was and is a classic.

Dondalah

---

From: TaffGoch <taff...@gmail.com>
To: geodes...@googlegroups.com
Sent: Wednesday, October 19, 2011 9:33 PM
Subject: Re: Class-II, Method-3

[Adrian Rossiter]

Hi Dondalah and All

On Thu, 20 Oct 2011, Dondalah Proust wrote:
> do too, ofￜ creating the vertices in double precision floating point

> with 18 digit accuracy, outputting the vertices in OFF format, and using
> Adrian's off_to_3ds program to translate the output into a Sketchup

The Antiprism programs also accept coordinate input, and if
the input doesn't look like OFF it will be processed as if
it was a text coordinate file.

The processing is fairly relaxed. Any line containing three
numbers separated by commas and/or whitespace is taken as a
set of coordinates; any other line is ignored. CSV is fine,
but the numbers should not be enclosed in quotes. If this is
unavoidable the CSV file can be processed using a search and
replace technique to convert quotes into spaces. (A filter
like the sed program is good, but loading the file into a
text processor and doing it by hand works too).

The conv_hull program can recreate the faces of many geodesic
models given only the vertices (higher frequency tetrahedral
ones will have incorrect faces near the 3-way vertices). If
the base model is a tiling unit the convex hull will add
extra faces to close the inner side of the unit, but these
can be removed with off_util '-D f,', or within Sketchup.

As an example, I have included some coordinates below. Save
the whole text of the email as msg.txt and run it through

conv_hull msg.txt | off_to_3ds > cube.3ds

The result should be a 3ds format cube (with triangulated faces).

,,,1 1 1
1, 1, -1
1,-1,1
1 -1 -1
jklavhjg
-1 1 1
-1 1 -1
-1 -1 1
-1 -1 -1

Adrian.

P.S. I released a new Antiprism snapshot last night

http://tech.groups.yahoo.com/group/antiprism/message/6167
--
Adrian Rossiter
adr...@antiprism.com
http://antiprism.com/adrian

[Dondalah Proust]

Gerry,

Here's one for you.  The picture Icosa_Face.png shows the face of an icosahedron.  Tau is the mid-point at theta 58 deg.  According to Kenner's Illustration 10.11, tau is divided into 8 layers.  In the illustration, you only see the top 4 layers.  If you divide tau by 8, you should get the value of "a" in the picture, shouldn't you?  That gives the value of "a" to be 7.285 degrees in Illustration 10.11, or 0.127 in radians.

Now if you take sin(36) * 2 you should get Kenner's value for "b" in Illustration 10.12 (or Taff's green chord), if you multiply it by the hypotenuse.  To get the hypotenuse, you take cos(36) and divide it into the length of "a" in Illustration 10.11.  This "a" in radians is 0.127.  Cos(36) is 0.809.  Therefore 0.127 divided by 0.809 is 0.157, which should be the hypotenuse.  But Kenner gives the chord for the hypotenuse (the red chord in Taff's illustration) to be 0.170.

Spherical trig is not my forte, therefore I can't get the calculation for the red chord and the green chord correct.  Can you or anyone help?

I added a second picture, Top_Triangle.png, to represent the top triangle.  I'm trying to get the phi-theta values for point B, and the length of AB and BC, given that AC should be 0.127 in radians.  Point A should be phi 0 theta 0.  Point C should be phi 36 deg and theta 7.285 deg.  And the angle BAC is 36 degrees.

We know that the horizontal sides of the triangle follow the great circle.  Therefore we know that the theta value for point B is going to be greater than the theta value for point C.  Therefore it is false to assume that theta for B is equal to theta for C.

Please help, thanks,

Dondalah

---

From: Dondalah Proust <donda...@yahoo.com>
To: "geodes...@googlegroups.com" <geodes...@googlegroups.com>
Sent: Friday, October 21, 2011 12:29 AM
Subject: Re: Class-II, Method-3

[Dondalah Proust]

Hi Adrian,

I have used cartesian coordinates with antiview to illustrate isometric diagrams with great success.  It's a terrific program.

Kenner does his calculations in polar coordinates, so I have followed his example and done all my calculations in polar coordinates, too.  I think that Gerry in Quebec does his calculations in polar coordinates as well.

As a result of using polar coordinates, I have always converted polar to cartesian in order to use antiview.  And it works like a charm.

Just as you posted this, I posted a problem with method 3 calculations.  So you can see that I'm not looking for a black box solution, but am trying to get an understanding of the math behind Kenner's method 3.  As you can see, I struggle with the math side of the application.  Any insight you have on the math behind his method 3 will be most welcome.

Cheers,

Dondalah

---

From: Adrian Rossiter <adr...@antiprism.com>
To: "geodes...@googlegroups.com" <geodes...@googlegroups.com>
Sent: Friday, October 21, 2011 4:30 AM
Subject: Re: Class-II, Method-3

Hi Dondalah and All

On Thu, 20 Oct 2011, Dondalah Proust wrote:

> do too, of  creating the vertices in double precision floating point with 18 digit accuracy, outputting the vertices in OFF format, and using Adrian's off_to_3ds program to translate the output into a Sketchup

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[Dondalah Proust]

Okay, I think I see the fallacy in my calculations.  I forgot that we were in Class 2.  In Class 1, my figures for tau should be correct, but that doesn't pertain to Class 2.  It's late, I'm tired, so I'm going to have to go back to the drawing board and re-run the numbers for Class 2.

Thanks to all for putting up with this trial and error,  more this evening,

Dondalah

---

From: Dondalah Proust <donda...@yahoo.com>
To: "geodes...@googlegroups.com" <geodes...@googlegroups.com>

Sent: Friday, October 21, 2011 5:00 AM
Subject: Re: Class-II, Method-3

[Gerry in Quebec]

Once I finish the class II, method 3 modeling in Excel using spherical
trig, I'll clean up the presentation and get back to you.
- Gerry

[Hector Alfredo Hernández Hdez.]

I want to see the maths, maybe you can scan hand papers...

for a particular frecuency I can try using Autocad, this give us more decimal than Scekchpad...

[Dondalah Proust]

Hector and Gerry,

I'm attaching a spreadsheet, and PDF, of some calculations that don't quite come out right.  Can you guys read .ODS spreadsheets?  If not, can you read the .XLS spreadsheet okay?  I did the spreadsheet in Libre Office, which saves in .ODS format.

Kenner doesn't explain this in Illustration 10.11, but his "a" value is in radians.  This makes sense, because his triangle is curved, and he's looking down on the center line.  As the result of using spherical trig for the value of "a", you have to translate to cartesian to get the chord lengths.

At the top, the spreadsheet gives the theta value for the bottom end point of each "a" segment.  The center line of Kenner's illustration is phi 36 degrees.

The golden ratio is calculated as (sqrt(5) + 1) / 2

This is the same as cos(36) * 2

Tau is the arctan of the golden ratio.

The center of the side pentagon is 180 degrees minus (tau*2).  This is 63.434949 degrees.

d2r is conversion factor going from degrees to radians

r2d is conversion factor going from radians to degrees

Once you have calculated 63.434949 degrees as the bottom of the center line, you can divide 63 degrees by 8 to give you the length of Kenner's value for "a".

The top triangle is an isosceles triangle with the length of "a" as the center line.  One half of the bottom line is one half of chord(b), Taff's green chord.  All we know at this point is the arc length of "a" and the top angle of 72 degrees divided by two.  From that we have to calculate Taff's red and green chords.

Chord(a) is Taff's red chord 0.170287

Arc(a) is the spherical distance of the arc from the top of the triangle to the left corner of the first triangle, where chord(a) intersects with chord(b), Taff's green chord.

x1,y1,z1 are the cartesian coordinates for the top of the triangle.

x2,y2,z2 are the cartesian coordinates for the intersection between chord(a) and chord(b)

From these two points, we get the distance of chord(a), the square root of x^2 + y^2 + z^2.

Once I get the length of arc(a), I can calculate the length of arc(b) using two times sin(36) times the length of arc(a).  I then have to calculate the green chord, chord(b), using the distance formula based on the arc.

After all this was done, the numbers are close, but still not correct.

Taff's chord(a) is 0.170287, my calculation comes out as 0.170855.

Taff's chord(b) is 0.199457, my calculation comes out as 0.200759.

Can you guys see the hole in my calculations?

Thanks,

Dondalah

---

From: Hector Alfredo Hernández Hdez. <hecto...@gmail.com>
To: geodes...@googlegroups.com
Sent: Friday, October 21, 2011 9:18 AM
Subject: Re: Class-II, Method-3

[Gerry in Quebec]

Hi Dondalah,
Yes, exactly right. The class II parent triangle is quite a bit
different from the class I parent triangle. For the class II, 16v, the
arc value of what Kenner calls the vertical median is 31.71747441
degrees. That's "[arctan 2] / 2". So you divide that by 8 (namely the
class II frequency divided by 2), to get the arc length of the top "a"
arc in Kenner's Diagram 10.11. That works out to 3.96468430 degrees.

Spherical trig is a bit sticky at times. But stick with it!

I myself have run into a snag in getting my numbers for the 16v class
II method 3 to agree with Kenner's.
I'll keep you and Hector posted.

- Gerry

On Oct 21, 6:27 am, Dondalah Proust <dondalah...@yahoo.com> wrote:
> Okay, I think I see the fallacy in my calculations.  I forgot that we were in Class 2.  In Class 1, my figures for tau should be correct, but that doesn't pertain to Class 2.  It's late, I'm tired, so I'm going to have to go back to the drawing board and re-run the numbers for Class 2.
>
> Thanks to all for putting up with this trial and error,  more this evening,
>
> Dondalah
>
> ________________________________

> From: Dondalah Proust <dondalah...@yahoo.com>

> To: "geodes...@googlegroups.com" <geodes...@googlegroups.com>
> Sent: Friday, October 21, 2011 5:00 AM
> Subject: Re: Class-II, Method-3
>
> Gerry,
>
> Here's one for you.  The picture Icosa_Face.png shows the face of an icosahedron.  Tau is the mid-point at theta 58 deg.  According to Kenner's Illustration 10.11, tau is divided into 8 layers.  In the illustration, you only see the top 4 layers.  If you divide tau by 8, you should get the value of "a" in the picture, shouldn't you?  That gives the value of "a" to be 7.285 degrees in Illustration 10.11, or 0.127 in radians.
>
> Now if you take sin(36) * 2 you should get Kenner's value for "b" in Illustration 10.12 (or Taff's green chord), if you multiply it by the hypotenuse.  To get the hypotenuse, you take cos(36) and divide it into the length of "a" in Illustration 10.11.  This "a" in radians is 0.127.  Cos(36) is 0.809.  Therefore 0.127 divided by
>  0.809 is 0.157, which should be the hypotenuse.  But Kenner gives the chord for the hypotenuse (the red chord in Taff's illustration) to be 0.170.
>
> Spherical trig is not my forte, therefore I can't get the calculation for the red chord and the green chord correct.  Can you or anyone help?
>
> I added a second picture, Top_Triangle.png, to represent the top triangle.  I'm trying to get the phi-theta values for point B, and the length of AB and BC, given that AC should be 0.127 in radians.  Point A should be phi 0 theta 0.  Point C should be phi 36 deg and theta 7.285 deg.  And the angle BAC is 36 degrees.
>
> We know that the horizontal sides of the triangle follow the great circle.  Therefore we know that the theta value for point B is going to be greater than the theta value for point C.  Therefore it is false to assume that theta for B is equal to theta for C.
>
> Please help, thanks,
>
> Dondalah
>
> ________________________________

> From: Dondalah Proust <dondalah...@yahoo.com>

> To: "geodes...@googlegroups.com" <geodes...@googlegroups.com>
> Sent: Friday, October 21, 2011 12:29 AM
> Subject: Re: Class-II, Method-3
>
> Taff, your exercise makes Method III look like a cinch.  Kenner's Illustration 10.11 makes it plain.
>
> Can you make a drawing like 10.11 for Method IV?  I can't quite translate Illustration 10.11 into a Method IV graphic.
>
> Although Illustration 10.11 is for Class II Method III, I don't see why it couldn't apply equally well to Class I, provided that the frequency be even.  Do you agree?
>
> Gerry, if you send me the Excel language for the geometry, I can help you translate that into standard math notation.  Then we can share it with the group as standard math.  I like the approach, and I think you do too, of  creating the vertices in double precision floating point with 18 digit accuracy, outputting the vertices in OFF format, and using Adrian's off_to_3ds program to translate the output into a Sketchup model.  This way, we get beyond the 6 decimal accuracy limitation, and we make sure the radius is one in the model.
>
> Hector, give us time, and we'll have you fluent in C.  This would be a good project to start on, because you already understand the subject.
>
> As all of you may know, the co-founder of the C language left this Earth just a couple of days ago, Dennis Ritchie.  He'll be missed, along with Steve Jobs.  The Kernighan and Ritchie book, called The C Language, was and is a classic.
>
> Dondalah
>
> ________________________________

> From: TaffGoch <taffg...@gmail.com>

> To: geodes...@googlegroups.com
> Sent: Wednesday, October 19, 2011 9:33 PM
> Subject: Re: Class-II, Method-3
>
> Hector,
>
> Here are diagrams of chord factors, from both my model, and from Hugh Kenner's book, "Geodesic Math and How to Use It"
>
> Note that my diagram shows a triacontahedron face, not an icosahedron. Kenner's illustrations are of the top-half of a triacontahedron face.
>
> -Taff
> --
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>
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[Hector Alfredo Hernández Hdez.]

I coudnt hope.

thanks to TAFF

(my bad english make to me work a lot of)

about you question Taff...,my experience says that the minimun different struts for "mexican" method will be 9 struts for 8 frecuency (equalizing by "levels")  I wannt try... (but exist the possibility that no intentional two strust have the same legth, but thiss will be a casuallity :( )


(please convert to SP format, to share)

2011/10/21 Hector Alfredo Hernández Hdez. <hecto...@gmail.com>

[TaffGoch]

Hector,

I couldn't find a uniform solution, either.

I did produce 8 chord factors, but with one vertex having a central-radial of 1.000057 (red-green crossing vertex.)

-Taff

[TaffGoch]

Complete class-II, 8v, Mex sphere:

[Hector Alfredo Hernández Hdez.]

Looks like so BEATIFULL.

see that ranges of factors chords is more small that others. that means that this polyhedro is more close to be regular than others and more strong and easy to ensamble (almost with out plans).

You are writting history . . .

thanks a lot

--

[TaffGoch]

Hector,

Similar to your AutoCad image, I can squeeze-out 9 decimals from SketchUp, by scaling-up the model by 1,000. Subsequently, all I have to do is take measurements, and then move the decimal point to the left, 3 places.

Confirms your data results, for class-II, 8v, method-3....

-Taff

[TaffGoch]

Hector,

I had two digits transposed. The red-green crossing vertex has a central radius of 1.000075...

On Fri, Oct 21, 2011 at 5:53 PM, TaffGoch wrote:

...with one vertex having a central-radial of 1.000057 (red-green crossing vertex.)

[Hector Alfredo Hernández Hdez.]

the mine all radios are ONE.

[Hector Alfredo Hernández Hdez.]

WHAT DO YOU THINK ABOUT THIS RARE MONSTER?

[Gerry in Quebec]

Hector & Taff,
Very cool variation on the original Mexican theme. For a 6v class II
Mexican, are all the radii equal?
- Gerry

>  Mexican_8v_class-II chords.png
> 60KViewDownload

[Dick Fischbeck]

like

2011/10/21 Hector Alfredo Hernández Hdez. <hecto...@gmail.com>

Looks like so BEATIFULL.

[Adrian Rossiter]

Hi Dondalah

On Fri, 21 Oct 2011, Dondalah Proust wrote:
> Just as you posted this, I posted a problem with method 3 calculations.ￜ

> So you can see that I'm not looking for a black box solution, but am

> trying to get an understanding of the math behind Kenner's method 3.ￜ As
> you can see, I struggle with the math side of the application.ￜ Any

> insight you have on the math behind his method 3 will be most welcome.

I haven't read Kenner's book, but I see from your followup message
that you have found the problem.

Adrian.

[Hector Alfredo Hernández Hdez.]

YES :)

[Gerry Toomey]

Finally.... here's my Excel rendition of the 16 frequency, class II triacon geodesic sphere, method 3. It lays out the calculation of the spherical & cartesian coordinates and chord factors (16) using spherical trigonometry. I've translated only some of the Excel trig equations into more or less regular math notation.

 

The full geodesic sphere has 3840 triangles. I have provided data for a single repeating cluster of 64 triangles.

 

Dondalah, if you can turn the Cartesian coordinates into an OFF file for use in Adrian's Antiprism program or maybe SketchUp, that would be great.

 

The calculations are based on just four numbers: the frequency (16) and the three internal spherical angles of the class II triacon parent triangle (aka PPT), namely 72, 60 and 60 degrees. All coordinates and chord factors agree with Kenner's numbers in Geodesic Math.

 

Cheers,

- Gerry in Quebec

 

On Oct 21, 5:12 pm, Gerry in Quebec <toomey.ge...@gmail.com> wrote:
> Hi Dondalah,
.......

[Hector Alfredo Hernández Hdez.]

UFF it is a LOT of WORK. I did had problems with my wife, I hope no be in similar circustances.

 

By the way, I have C codes for Icodome Class I and for Octadome any frecuency.

 

And including the possibilty of homogenice the border of original polyhedro.

 

Is easy to traslate to Excel , I think so.

 

I did wrote because are the two first steps to make C code for "mexican" method.

I guest that Gerry wrote this too.. (Creo adivinar que Gerry ya escribio esos calculos en Excel)

 

See you.

--

[Dondalah Proust]

Hi Gerry,

You do tremendous work.  That is one fine spreadsheet that you made.  It will take some time to digest it all.

The antiview picture doesn't quite look what you want, but that's why antiview is such a useful tool.

Gerry_16v_Method_3.png is the antiview picture.

I copied your vertices at line 330 into a new spreadsheet with 15 decimal precision.

Then I converted the phi-theta vertices to radians.

After that I converted the phi-theta vertices in radians to x,y,z format.

Then I added the 64 triangles underneath the x,y,z vertices.

Line 2 is #vertices, #triangles, #edges (0)

Line 1 is always "OFF"

We're almost there.  Now that you see how I created the OFF file, you can merge my spreadsheet to the end of yours and test it, after you make a few corrections.

Dondalah

---

From: Gerry Toomey <toomey...@gmail.com>
To: geodes...@googlegroups.com
Sent: Saturday, October 22, 2011 5:24 PM
Subject: Class-II, Method-3

[Dondalah Proust]

Gerry,

My mistake.  Now it looks like it should.

You do good work.

Dondalah

---

From: Dondalah Proust <donda...@yahoo.com>
To: "geodes...@googlegroups.com" <geodes...@googlegroups.com>
Sent: Saturday, October 22, 2011 9:48 PM
Subject: Re: Class-II, Method-3

[Gerry in Quebec]

Hi Dondalah,
Thanks for the .off file and Antiview image of the 16v. I've
incorporated both into the Excel file of trig calculations. I'll send
the new version in a separate post.

Two questions about .off files...

First, is there an easy way to convert the tabular information in
Excel to the correct format with spaces in between the numbers? I did
it a while back for the Temcor layout but it required a lot of
fiddling to convert the numbers to a a simple text format with the
proper spacing. In effect, I had to make extensive use the search-&-
replace function. (Given the number of dumb typing errors I make, this
might be better referred to as the search-&-destroy function.)

Second, once I have a simple .txt file with the OFF numbers neatly
arranged, how do I "convert" it to a .off file, i.e., so that the new
file extension is .off instead of .txt?

I'm keen to learn Antiview and SketchUp for design work.... However,
there's quite a bit of "computer knowledge" inertia to overcome in my
case (read computer ignorance).

[Adrian Rossiter]

Hi Gerry

On Sun, 23 Oct 2011, Gerry in Quebec wrote:
> Two questions about .off files...
>
> First, is there an easy way to convert the tabular information in
> Excel to the correct format with spaces in between the numbers? I did

I was able to copy and paste The OFF table in your spreadsheet
into a text document. (The process worked fine and the face count
was correct but there were redundant faces after the 64th.)

> Second, once I have a simple .txt file with the OFF numbers neatly
> arranged, how do I "convert" it to a .off file, i.e., so that the new
> file extension is .off instead of .txt?

It is already an OFF file. You can give it an .off extension
by renaming it.

Antiprism can repeat the tile to make a sphere in the following
way. The 5-fold vertex of the tile is (0,0,1) and needs to be
aligned with an Antiprism icosahedral 5-fold axis e.g. (0,1,phi).
The 3-fold vertex of the tile is (1,0,0) and needs to be aligned
with an appropriate Antiprism icosahedral 3-fold axis e.g. (1,1,1).

The command below looks complex but just says align the vectors
(0,0,1) and (1,0,0) with (0,1,phi) and (1,1,1), in that order,
and then repeat to have icosahedral symetry. I pasted your OFF
data into tri.off and the sphere is written to sph.off

off_trans -R 0,0,1,1,0,0,0,1,1.61803398874989484,1,1,1 tri.off | poly_kscope -s I > sph.off

I have attached a raytraced image of the result.

The duplicate vertices along the tile edges can be merged with
off_util and the -M and -l options.

[Gerry in Quebec]

Thanks Adrian. Once I learn the Antiprism software I hope to be able
to import data, manipulate them and produce nice clear images like
you've done with this 16v sphere (and of course dynamic 3D graphics).
In the meantime, could you convert another .off file for me into an
antiprism png image... and let me know whether the formatting is
acceptable. I'll post separately for that, from gmail. It will be
a .off file and an png to illustrate what I'm up to.
- Gerry

>  sph.png
> 420KViewDownload

[Gerry Toomey]

Hi Adrian,

The two attachments are a .off file and a png image of the dome layout I have in mind. Five of the 40-triangle clusters will produce a dome at the 7/12ths truncation, i.e., 7 rows of triangles.... one more row than a hemisphere.

 

The more conventional 4v icosa class I, method 1 dome doesn't sit flat at either the 5/12ths or 7/12ths truncation (only the equator), has a lot of scalene triangles, and has 6 unique chord lengths.

 

The Fuller-Kruschke version is an improvement in that it sits flat at the 5/12ths and 7/12ths cutoff lines, as well as the equator. (Fuller and/or colleagues came up with that design, but Dave Kruschke, who worked for Temcor for a year, was the first person to publish the info as far as I know.) But that version also has 6 chord factors and lots of scalene triangles.

 

Hector Hernandez's 4v design, which we've been calling the Mexican Method (MM), doesn't sit flat at any of the 3 useful truncations and has lots of scalene triangles. But it is a very nicely proportioned dome and has only 4 chord factors.

 

The dome design in the attached diagram has also has only 4 chord factors. But it also has only 4 triangle types and no scalene triangles. While I don't expect it will be as handsome as the Mexican version, I sure would like to see what it actually looks like. This is one layout in a series I did.... Taff modeled an earlier one of these last year in SketchUp, "solution 3c".

Cheers,

- Gerry

 

 

Adrian's earlier message today, Sunday, 23 Oct 2011:

[Dondalah Proust]

Hi Gerry,

Once again, you do excellent work.  You may have caught this already, but here are two changes you can make to the spreadsheet.

Delete rows 438 to the end.  These are redundant lines.  Up to row 437, you have 64 triangles.

Normal convention is not to have a blank row between the vertices and the triangles.  So, for cosmetic reasons, you could eliminate row 373.

I haven't run Adrian's script, yet, because I have to upgrade to his latest release of antiprism.  I'm getting errors, when I run his script.

I'll let you know when I install his latest release.

Good job, Gerry,

Dondalah

---

From: Gerry Toomey <toomey...@gmail.com>
To: geodes...@googlegroups.com

Sent: Sunday, October 23, 2011 7:17 AM
Subject: Class-II, Method-3

Here's a new version of the Excel file showing trig calculations for the 16v, class II triacon geodesic sphere, method 3.

 

This update includes two items provided by Dondalah: all the data in .off format, plus an Antiview screen shot of the 64 triangles. Sixty of these triangle clusters form a full geodesic sphere.

 

Sorry for the large file size.

[Adrian Rossiter]

Hi Gerry

On Sun, 23 Oct 2011, Gerry Toomey wrote:
> The dome design in the attached diagram has also has only 4 chord factors.
> But it also has only 4 triangle types and no scalene triangles. While I
> don't expect it will be as handsome as the Mexican version, I sure would
> like to see what it actually looks like. This is one layout in a series I
> did.... Taff modeled an earlier one of these last year in SketchUp,
> "solution 3c".

The OFF file was pretty much fine, but the vertex count should have
been 30, and one of the faces was, I think, 10 15 16 but should
have been 11 15 16.

For reference here is the command that produces the POV-Ray
file. It is long, but most of the command is just setting
display options

poly_kscope -s C5 sol3d.off | off2pov -R -75,0,0 -e 0.01 -F 0.6,0.7,1,0.5 -E grey50 -v 0.02 -B white -D 2.3 -C0,-0.2,0 > sol3d.pov

I ran the result through POV-Ray like this

povray +a +H600 +W800 sol3d.pov

The raytraced image is attached.

Adrian.
--
Adrian Rossiter
adr...@antiprism.com

http://antiprism.com/adrian

[Gerry in Quebec]

Hi Adrian,
Thanks, Hawkeye, for spotting those two glitches in the OFF file. All
corrected now. I'm starting to get the hang of it. The 29 vs 30 error
was me forgetting to change the vertex count because I normally label
from 1 forward, not 0 forward. The other error was a typo during
transposition of the triangle vertices.

Also, thanks for posting the raytraced image of the dome. It's really
great for me to see what, up till now, has been just an idea in my
head, about how to build a more economical dome (more with less) --
using some of the the conventional subdivision methods merely as
starting points, not as faits-accomplis.
- Gerry

On Oct 23, 3:47 pm, Adrian Rossiter <adr...@antiprism.com> wrote:
> Hi Gerry

....

> The OFF file was pretty much fine, but the vertex count should have
> been 30, and one of the faces was, I think, 10 15 16 but should
> have been 11 15 16.
>
> For reference here is the command that produces the POV-Ray
> file. It is long, but most of the command is just setting
> display options
>
>     poly_kscope -s C5 sol3d.off | off2pov -R -75,0,0 -e 0.01 -F 0.6,0.7,1,0.5 -E grey50 -v 0.02 -B white -D 2.3 -C0,-0.2,0 > sol3d.pov
>
> I ran the result through POV-Ray like this
>
>     povray +a +H600 +W800 sol3d.pov
>
> The raytraced image is attached.
>
> Adrian.
> --
> Adrian Rossiter
> adr...@antiprism.comhttp://antiprism.com/adrian
>

>  sol3d.png
> 377KViewDownload

[Gerry in Quebec]

Hi Dondalah,
Thanks for spotting those two boo-boos. I seem to be making errors in
pairs today! (Two errors in the 4v off file as well that Adrian
caught). I corrected the redundant numbers which Adrian mentioned as
well (I must have hit the paste button twice) and eliminated the blank
line between vertex coordinates and triangle vertices. I'm getting
there!

Cheers,
- Gerry in blue-skyed Quebec

On Oct 23, 3:47 pm, Dondalah Proust <dondalah...@yahoo.com> wrote:
> Hi Gerry,
>
> Once again, you do excellent work.  You may have caught this already, but here are two changes you can make to the spreadsheet.
>
> Delete rows 438 to the end.  These are redundant lines.  Up to row 437, you have 64 triangles.
>
> Normal convention is not to have a blank row between the vertices and the triangles.  So, for cosmetic reasons, you could eliminate row 373.
>
> I haven't run Adrian's script, yet, because I have to upgrade to his latest release of antiprism.  I'm getting errors, when I run his script.
>
> I'll let you know when I install his latest release.
>
> Good job, Gerry,
>
> Dondalah
>
> ________________________________

> From: Gerry Toomey <toomey.ge...@gmail.com>

> To: geodes...@googlegroups.com
> Sent: Sunday, October 23, 2011 7:17 AM
> Subject: Class-II, Method-3
>
> Here's a new version of the Excel file showing trig calculations for the 16v, class II triacon geodesic sphere, method 3.
>  
> This update includes two items provided by Dondalah: all the data in .off format, plus an Antiview screen shot of the 64 triangles. Sixty of these triangle clusters form a full geodesic sphere.
>  
> Sorry for the large file size.
> - Gerry
> --
> You received this message because you are subscribed to the "Geodesic Help" Google Group
> --
> To unsubscribe from this group, send email to GeodesicHelp...@googlegroups.com
> --
> To post to this group, send email to geodes...@googlegroups.com
> --
> For more options, visithttp://groups.google.com/group/geodesichelp?hl=en
>

>  Gerry_16v_Method_3_Redundant.png
> 199KViewDownload
>
>  Gerry_16v_Method_3_Cosmetic.png
> 269KViewDownload

[Dondalah Proust]

Gerry,

Looks good.

Thanks, Adrian.

Dondalah

---

From: Gerry in Quebec <toomey...@gmail.com>
To: Geodesic Help Group <geodes...@googlegroups.com>
Sent: Sunday, October 23, 2011 4:17 PM
Subject: Re: Class-II, Method-3

> To unsubscribe from this group, send email to GeodesicHelp+unsub...@googlegroups.com

> --
> To post to this group, send email to geodes...@googlegroups.com
> --
> For more options, visithttp://groups.google.com/group/geodesichelp?hl=en
>
>  Gerry_16v_Method_3_Redundant.png
> 199KViewDownload
>
>  Gerry_16v_Method_3_Cosmetic.png
> 269KViewDownload

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You received this message because you are subscribed to the "Geodesic Help" Google Group
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[TaffGoch]

Method 4 employs an equal-angular radial subdivision of the HYPOTENUSE, not the short side of the right-triangle.

Depiction of my retraction (of my prior post):

[Dondalah Proust]

Hi Gerry,

An idea came to me during sleep about Class 2 Method 4.  What if you rotate the sphere so that the hexagon is at the top?  There would be 3 pentagons around the side, so that angle A would be 120 degrees.  Angles B and C would be 36 degrees.  Does this match Taff's description of Method 4?  It seems like most of your math would remain the same if you treat it this way.  What do you think?  You'd end up with twice as many triangles on the "face" this way.  It's really two faces in one.

Dondalah

---

From: TaffGoch <taff...@gmail.com>
To: geodes...@googlegroups.com
Sent: Monday, October 24, 2011 9:23 PM
Subject: Re: Class-II, Method-3

Method 4 employs an equal-angular radial subdivision of the HYPOTENUSE, not the short side of the right-triangle.

Depiction of my retraction (of my prior post):

[TaffGoch]

Dondalah,

Rotating the PPT (principle polyhedron triangle) to face upwards, centered on the z-axis, should make calculations much easier (although, I don't think the computer "cares" a whole lot!)

It would certainly make the coding logic easier to follow, for someone other than the author.

_______________

By the way, your image is not Method-4 -- It is Temcor's method. (That's my fault. Sorry for my previous mis-identification.)

-Taff

[Paul Kranz]

Taff:

Is there someway to tell which method is which by just looking at them? They look the same to me.

Paul sends...

--

[TaffGoch]

Paul,

Most are VERY similar. You have to compare the chord factors. It's not hard to visually differentiate between Method-1 and Method-2, for example, but other methods often exhibit visually-indistinguishable differences.

So, why so many methods?

Some are more uniform, regarding divergence in strut lengths.

Some are more uniform, regarding area of panels.

Some appear more "smooth," in that arcs, crossing PPT boundaries, don't "dog-leg" as much.

Some have greater economy of struts (fewer chord factors.)

Some have greater economy of panels (manufacture of a small set of panels, "cloned" across the PPT.)

______________

Wouldn't it be great, if all could be combined into one method? That's the trick, and that's why the originators developed many methods, looking for the "ideal" subdivision. One method, having ALL of the best attributes, doesn't exist (or hasn't been found.)

You have to pick the attributes you want, then select the method that best provides for that objective.

-Taff

[Paul Kranz]

Taff:

Thank you for your thorough reply!

For the lower frequencies, though, aren't they uniform? I mean how many different places can you put the vertexes for a 2V icosa on a sphere?

Paul sends... 

-Taff

--

[TaffGoch]

Your observation is correct...

...which explains one of the models I have posted at the 3D Warehouse:

[TaffGoch]

I completed a Method-4 subdivision, and am somewhat disappointed. The only self-similar chord-factors are those along the hypotenuse, so all chord-factor "savings" (of the Methods 3 & Temcor) are lost. There are not 8 chord-factors for an 8v -- there are 13.

I see no advantage to the Method-4 subdivision (which may explain why it is only parenthetically-mentioned in Domebook 2.)

Superiorly, both methods, 3 & Temcor, produce 8 chord-factors, for an 8v subdivision.

I've highlighted the attached image, to depict which plane is radially-subdivided (by equal central-radius angles) in each method.

-Taff

[Gerry in Quebec]

Hi Taff,
Except for the colour schemes, methods 3 and 4 look identical to me.
- Gerry

>  Methods_3,4,Temcor.png
> 124KViewDownload

[Hector Alfredo Hernández Hdez.]

It easy Gerry

Methodo 3    equal angles in horizontal cicle

Methodo 4   equal angles in "hypotenusa circle"


M Temcor  equal agles in vertical circle.   :)

[TaffGoch]

Gerry, Hector has it described properly

Subdividing the hypotenuse angle, by 4 equal divisions, does not produce the same results as subdividing the long right-angle side, by 4 equal divisions.

The multiple rotated & translated planes do, indeed look virtually the same, as the technique of rotations/translations is identical. The difference is in the DEGREE of the rotations and translations.

-Taff

[Gerry in Quebec]

Okay, thanks, Hector and Taff. I understand what you're saying, but
for some reason, I thought the subdivision lines on the sphercial
triangle edge should be perpendicular to the hypoteneuse, which I
guess would create a real mess. I guess it's time to buckle down and
learn SketchUp and Antiprism -- to establish a better visual
reference.... I feel like a colour-blind person being taken on a tour
of the Louvre!
- Gerry

[TaffGoch]

Yeah, I tried the radial-fan, perpendicular to the hypotenuse, and it was, indeed, a mess.

[Gerry in Quebec]

Oh, and thanks for setting a good example with your spelling.....
hypotenuse not hypoteneuse!

[TaffGoch]

Hypotenuse/perpendicular-planes attempt depicted....

[TaffGoch]

I got the impression that "hypoteneuse" was an archaic British spelling.

-Taff :)

[Dondalah Proust]

We're getting bounce backs from the group's email account.  Message included below.

----- Forwarded Message -----
From: Dondalah Proust <donda...@yahoo.com>
To: "geodes...@googlegroups.com" <geodes...@googlegroups.com>
Sent: Sunday, October 30, 2011 2:11 PM
Subject: Re: Class-II, Method-3

Hi Gerry,

Attached is a C program that produces class 2 method 3 icosahedrons from 4v to 16v.

The executable is c2m3icosa.exe.

A batch script for testing this program with antiview is called shwsph.bat.  For example,

shwsph 12

will produce a 12v sphere in antiview.

Cheers,

Dondalah

---

From: TaffGoch <taff...@gmail.com>
To: geodes...@googlegroups.com

Sent: Tuesday, October 25, 2011 6:33 PM
Subject: Re: Class-II, Method-3

I got the impression that "hypoteneuse" was an archaic British spelling.

-Taff :)

[Dondalah Proust]

Hi Gerry,

Google hasn't let me attach source code, or binary code, so I'll try attaching a Base64 version of the executable c2m3icosa.exe.

The parameter is -f frequency, where frequency is 4, 6, 8, ..., 16.

For example "c2m3icosa -f 12" generates a 12v PPT.

Base64 conversion programs are readily available on the Web.  If anyone converts it back to binary, it can be renamed as c2m3icosa.exe.

I'll look into finding Dropbox space for the program.

Cheers,

Dondalah

---

From: TaffGoch <taff...@gmail.com>
To: geodes...@googlegroups.com
Sent: Tuesday, October 25, 2011 6:33 PM
Subject: Re: Class-II, Method-3

I got the impression that "hypoteneuse" was an archaic British spelling.

-Taff :)

[Gerry Toomey]

About a week ago I posted an Excel spreadsheet (filename: 16v-icosa-classII-method-3-revise.xls). It calculates the frequency 16 coordinates step by step, based on spherical trigonometry. Dondalah noticed some errors in my textual description and suggested some formatting changes. The updated file is attached. The calculated spherical and cartesian coordinates and chord factors are identical to those in the earlier file.

I’ve deleted the message and file I posted Oct. 23. Text of original message:

“Here's a new version of the Excel file showing trig calculations for the

16v, class II triacon geodesic sphere, method 3.

This update includes two items provided by Dondalah: all the data in .off
format, plus an Antiview screen shot of the 64 triangles. Sixty of these

triangle clusters form a full geodesic sphere.”

- Gerry

[Gerry in Quebec]

Thanks, Dondalah.
I'm still having trouble understanding how Antiview/Antiprism work, so
it may be a while before I'm ready to test your c2m3icosa.exe program
with it. But I think it will be cool to be able to generate images of
various frequencies of the class II, method 3 icosa, based on the
trig for the 16v.
Cheers,
- Gerry in Quebec (so far spared from the big snow storms)

On Oct 31, 4:26 am, Dondalah Proust <dondalah...@yahoo.com> wrote:
> Hi Gerry,
>
> Google hasn't let me attach source code, or binary code, so I'll try attaching a Base64 version of the executable c2m3icosa.exe.
>
> The parameter is -f frequency, where frequency is 4, 6, 8, ..., 16.
>
> For example "c2m3icosa -f 12" generates a 12v PPT.
>
> Base64 conversion programs are readily available on the Web.  If anyone converts it back to binary, it can be renamed as c2m3icosa.exe.
>
> I'll look into finding Dropbox space for the program.
>
> Cheers,
>
> Dondalah
>
> ________________________________

> From: TaffGoch <taffg...@gmail.com>

> To: geodes...@googlegroups.com
> Sent: Tuesday, October 25, 2011 6:33 PM
> Subject: Re: Class-II, Method-3
>
> I got the impression that "hypoteneuse" was an archaic British spelling.
>
> -Taff :)
>
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>  c2m3icosa-exe.b64
> 70KViewDownload

[Dondalah Proust]

Hi Gerry,

This is the procedure that Adrian gave us.  The example is for a 12v sphere.

c2m3icosa  -f  12   >tri.off
off_trans -R 0,0,1,1,0,0,0,1,1.61803398874989484,1,1,1 tri.off | poly_kscope -s I > sph.off
antiview sph.off -e 0.005

Cheers,

Dondalah

---

From: Gerry in Quebec <toomey...@gmail.com>
To: Geodesic Help Group <geodes...@googlegroups.com>

Sent: Monday, October 31, 2011 4:01 PM
Subject: Re: Class-II, Method-3

> To unsubscribe from this group, send email to GeodesicHelp+unsub...@googlegroups.com

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> To post to this group, send email to geodes...@googlegroups.com
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>
>  c2m3icosa-exe.b64
> 70KViewDownload

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