Concrete Geodesic Dome

[TaffGoch]

Hi, all,

Has anyone seen any information on this dome?

I found the image on a pinterest website, with no associated description.

It was a Russian webpage, so that's all I can guess at.

-Taff

(aka, TaffGoch, David Price)

[Gravity]

I believe I remember this guy posting about this dome, himself, in this very group.

Adam Winter

The Bootstrap Music Co.

206-430-4425

bootstrapmusic.com

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[Dick Fischbeck]

I think the spherical block man, Peter Roberts in upstate NY, Alfred Station,  works this way.

https://www.masonrymagazine.com/blog/2020/07/21/spherical-block-llc-a-new-way-to-build-with-masonry/

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[valle....@gmail.com]

hi. I'm looking to find more info too. only found the webpage but don't have info only the pic.

here ....    http://ivgazeta.ru/read/14988 

[RC]

I think those are molded aerated concrete. They are lightweight and insulating.  Also, window openings can be cut out using a hand saw.

[Pedro valle]

Hi

following this tutorial [Make Abstract Models with Geometry Nodes - YouTube] (https://www.youtube.com/watch?v=K5eNdCpoRD4&t=346s)
manage to build an ok dome. in Blender. made the dome, then export to OBJ and open in Pepakura and unfold the dome, but I can't make the hexagons small like the pic above, so my molds are maybe doble the size, the dome is 10 mts on diameter. I'm planning to build this dome with aircrate in summer Ill post my progress on June. thanks   to these group for all the help. 

On Tuesday, June 7, 2022 at 3:51:51 PM UTC-7 TaffGoch wrote:

[RC]

Your hexagon sizes are bigger because it is not the same model as the picture. Compare with the marked up picture.

[Ashok Mathur]

From a theory point of view Concrete Geodesic Domes:

Concrete has near-zero tensional strength.

It can not be used to make tensgrity structures.

Here the term concrete is being used as a shorthand for Reinforced Cement and Concrete Structures.

RCC structures have good tensional strength and can be used to make tensegrity structures.

Secondly, tensegrity requires islands of compression elements surrounded by continuous tensional structures.

By casting RCC in hexagonal blocks, tensional elements no longer surround the compressional  elements.

You end up making a compressional dome which stays together by interlocking and inertia.

Regards

Ashok

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[lemondealc]

Is there any cast shape that would have both the islands of compression and the tension? 

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[Pedro valle]

Hi  Robert .. 

I changed the subdivisions from 3 to 4 and  I thing I get   the same dome as the pics. the only thing Blender don't make all the hexagons the same size, they are not symmetrical. is another software like maybe solidworks that will generate a dome with hexagons symmetrical? or all the 3D software make these dome the same way? I'm traying to make only 3 molds 

 

[Pedro valle]

Hi Ashok

thank you for your reply .so building a dome with aircrate is no safe? will be better wood or metal?  appreciate you advise.   

[Ashok Mathur]

Dear Pedro

I am not saying that a aircrete concrete structure is unsafe.

I was merely trying to distinguish between the mathematics 

and physics of a structure.

Deriving coordinates of vertices from a geodesic dome alone 

does not make the subsequent structure geodesic in nature. 

You have used only the mathematics of the geodesic dome 

but cut out the physics by casting the structure in separate 

disjointed blocks.

You can make an aircrete geodesic dome by casting the panels

of aircrete but joining the retaining structures till they form a continuous 

tension ring around the aircrete.

By the way, I spent the third week of October in Istanbul

surrounded by about 100 compressive spherical dome that soar

over 100 feet in the sky. Properly made spherical compressive domes

are also long lasting structures.

Some such domes at Istanbul are about two milineum  

old while others are several hundred centuries old.They transfer their 

weight from the circular shape to a square foundation using a unique in

 between called squinch.

Persian domes transfer their weight to an octagonal base using another in between.

Compressive spherical domes of aircrete are also possible 

but will use a maths different from geodesic.

Regards

Ashok

Sent from my iPhone

On 17-Nov-2022, at 7:53 AM, Pedro valle <escap...@gmail.com> wrote:



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[Pedro valle]

Hi Ashok

                     still working on the design, I'm using Blender 3D   but the shapes are no symmetrical so if I made the molds, I'll need to make a lot because no all the hexagons are the same, I will try different method in blender I just saw another way to make a dome, I would like to travel and see all those places like Istanbul and Rome to see the Pantheon.

last summer June and July, I made a roof with almost 2,000 bricks, using a wood mold to hold the bricks till they dry then I put a wire mesh and then plaster. I made 3  Bóvedas, in Spain I think they call them  Bóvedas  catalanas o Bóveda de cañón, don't know how they call them in English.

and I will have off June and July again. so, I will try to build a dome, with concrete because is way cheaper that metal or wood.    

[Dx G]

Interesting discussion. Please keep in mind that some parts of a dome can actually be in tension, rather compression, depending on the design.  Many of the older domes have a tension band/ring/belt, buttresses or some mass at key points near the base to prevent any bulging.  Often they are not visible.

DxG

[RC]

Pedro,

Near the equator of a spherical dome, the compression forces transition to tensional hoop stresses.  Perhaps you could build a light-weight metal tube geodesic dome as a supporting structure first.  Then, assemble hexagon blocks like the one shown below onto each hub.  These have grooves molded into them that straddle the metal conduit.  The hex blocks could be sized so that their is a small mortar gap between blocks.  The gaps between blocks would take care of tolerance build up errors, and would be filled with mortar.  The concrete blocks will be great for enclosing the structure and handling compression forces.  And, the metal conduit dome frame will resist tensional forces.  (had to repost again because I drew the block incorrectly)

[Pedro valle]

Hi, Robert C. 

that's a great idea, in the pandemic, my employer sends me home for 3 months, so I bought a welder machine and learn the basics of welding no like a pro, but I can build the metal structure. and on top I can put the metal mesh and plaster. my plan is built cylindrical shape 8 feet tall with a door maybe some windows and in top the dome. fells like that way dont waste space and making a door in the dome is kind hard. 

[RC]

Sounds like you made good use of your three months off!  The grooves of the blocks would face inward.  I was imagining a support dome framework made of flattened end tubes bolted together.  The grooves would be fairly deep so that you could mortar over the embedded tubes.

[Pedro valle]

oh. even better because will be faster to build. I saw some videos how to build domes that way. mean just the flattened end tubes technique, no with concrete.  just need to find the design.   thank you. 

[Pedro valle]

found how to make domes with the flattened end tubes 

4V Geodesic Dome Calculator Software in Feet and Inches for Calculating Strut Lengths - Geodesic Dome Plans (ziptiedomes.com)

4V 5/12 Kruschke geodesic dome calculator » Domerama

Make a Geodesic Dome From Steel Tubing : 13 Steps (with Pictures) - Instructables 

I need to find metal pipes, the cheapest, I found was $11    1/2''x10' steel. but the cooper ones are $13  

[Erich Nolan Bertussi]

this was a great thread to read guys.

i wonder in your scenarios are you utilising the single exact same irregular hex and single exact same irregular pent shape for the dome?

I hope to be doing something like this and pre-fabing passivhaus compliant hexes and pents...

so this thread is awesome for me!

To view this discussion on the web visit https://groups.google.com/d/msgid/geodesichelp/5E5A45E1-2CFB-4868-9288-30144C59BAE0%40gmail.com.

[Patrick McDonnell]

Hi,

I am interested in making this style of dome using concrete molds. Obviously trying to keep the number of different molds to a minimum.

I have tried a number of times to model the geometry in CAD from first principals but I usually end up with the outside being ok but the panels not lining up properly on the inside.

I would like to be able to model this in CAD with 'prefect' geometry from first principals, is this possible? I have done extensive googling and searching through this forum and the closest thing I have found is mention of Goldberg division but could not find enough information to make this work.

Any pointers would be much appreciated.

Cheers,

Paddy

[Levente Likhanecz]

Chris Kitrick has this hexagonal dome. Its obviously looks higher frequency. All hexa is flat, and all vertices on sphere

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[Dx G]

Patick,
  You said

"Obviously trying to keep the number of different molds to a minimum."

Bingo.  

An important part of choosing the right design of a dome (or almost anything, for that matter) is keeping the variation in parts and part size to a minimum.  This reduces cost, increases simplicity and makes success more likely.

 Firstly, with respect to concrete block domes, if you look, you will specifically see that this was partly addressed in Robert's patent.

https://patents.google.com/patent/US10487494B1/en

 He discusses using mortar to fill the space between blocks when they are used for a larger or smaller dome, as this changes the dihedral angles. There are other ways to do that, but its a start.

However, there are also some options in design that are very frequently (and sadly) overlooked.  Some have been discussed in this forum under various topics.  I've been looking at these myself for the creased diamond panel domes I've been evaluating.  So let me mention a few here.

1) Pentakis Dodecahedron
This is similar to a 2 frequency Class 2 Method 3 (triacon). The nice part is like the triacon, although all the panels are the same,  in the Pent Dod, all the dihedral angles are the same. There are some limits on dome size, but the edge-up configuration provides a nice truncation.  Gerry posted some really good material on this.

2) Edge up 3f
A dome that has been popular for decades is the 3f ico alternate. However, one approach few people seem to consider is the edge-up configuration, which provides a flat truncation for a hemisphere.  People often bring up the Kruske dome, which has its assets, but low variation in parts is not one of them.

3) Catalan solids
This is an entire class of polyhedrons with this description:

"Catalan solids are a group of thirteen convex polyhedra that are the duals of the Archimedean solids. They are characterized by having faces that are all the same type, constant dihedral angles, and specific symmetry properties, but their vertices are not symmetric."

 Excellent resource.  There are lots of them if you look.
https://dmccooey.com/polyhedra/
https://dmccooey.com/polyhedra/Catalan.html

https://www.qfbox.info/4d/catalan3d

https://mathworld.wolfram.com/CatalanSolid.html

There is more out there, but this is a good start for you and others that understand the value of reducing variation in the structure.

Remember, every success begins with the decision to try.
 John F. Kennedy

Let us know if this helps and what else might.

Dx G

[mcdonn...@gmail.com]

Thanks Levente,

 

I had looked at Chris model a few years ago but I could not figure out how to re-create this from first principals and adapt to different frequency.

 

This definitely would meet my objectives, when panelised the panels are very close to being perfect with only tiny error.

 

Does anybody know how to go about creating a model like this from first principals?

 

 

 

Very minor “step” at internal faces.

 

 

Cheers,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Levente Likhanecz
Sent: Thursday, 1 January 2026 8:39 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

Chris Kitrick has this hexagonal dome. Its obviously looks higher frequency. All hexa is flat, and all vertices on sphere

On Wed, Jun 8, 2022 at 12:51 AM TaffGoch <taff...@gmail.com> wrote:

Hi, all,

 

Has anyone seen any information on this dome?

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvED%3DoUWMQ70a8TEWSExGvHDYN1HapHSSz-d4EYfjD6BHadg%40mail.gmail.com.

[mcdonn...@gmail.com]

Thanks for the response Dx G,

 

I am aware of Roberts work and patent but this is not really what I am after.

 

The other geometry you mention is interesting but I could not find any higher frequency (smallest panel size) divisions of this type of geometry?

 

What Chris Kitrick modelled in sketchup looks perfect but I would like to understand how to replicate this from first principals.

 

Cheers,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Dx G
Sent: Friday, 2 January 2026 1:50 AM
To: Geodesic Help Group <geodes...@googlegroups.com>
Subject: Re: Concrete Geodesic Dome

 

Patick,

Chris Kitrick has this hexagonal dome. Its obviously looks higher frequency. All hexa is flat, and all vertices on sphere

 

On Wed, Jun 8, 2022 at 12:51 AM TaffGoch <taff...@gmail.com> wrote:

Hi, all,

 

Has anyone seen any information on this dome?

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/134974a5-0470-4888-a864-e8d6c8c025f3n%40googlegroups.com.

[Dx G]

Paddy,

 Yes, that is not a trivial point.  For a big dome, a higher frequency is needed to keep the part size down, but its not a great plan if one loses the uniformity of the lower frequency.  I haven't yet come across an easy fix for that, but I'm early in my study of this and will remain hopeful something useful lies in waiting to be discovered...or invented. 

  You might want to look at some of the Class 3 domes as well, although I suspect you already did if you were looking at Goldberg Polyhedra. They are problematic for flat truncations, but do offer some interesting options. I especially like the ones that use a proliferation of equilateral triangles, as that is about as uniform as you can get.

Dx G

[Levente Likhanecz]

Its a 9V, next best i would go with 6V, just guessing

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/000001dc7b98%247717b4c0%2465471e40%24%40gmail.com.

[Levente Likhanecz]

i think so, on higher frequency you can play with the mortar when gluing together the blocks. some quick drying foam, aligning the blocks on template

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/001601dc7b9a%24d9d6b820%248d842860%24%40gmail.com.

[TaffGoch]

Patrick,

It is, apparently, an 8-frequency (version-II) subdivision, 8v(4,4)

The unique "brick" count, if I recall, is 10, with one brick mirrored, so 11 molds. 

This applies to this particular SketchUp model. Other subdivision methods may produce lower count.

Scaling in SketchUp can be tricky. Don't expect the "push/pull" tool to provide much help with producing brick thickness. 

The "scale" tool provides better results, but takes more patience in modeling.

-Taff

(David Price)

[mcdonn...@gmail.com]

Hi Levente,

 

I am not sure I follow your responses, I am trying to understand how to construct this type of geometry from first principals.

 

Looking back through the posts, from the example SketchUp file that you provided it unless I am mistaken looks like Chris provided guidance on the geometry:

Right

Spherical

Triangle           a                      b                      c                       A                      B                      C

----------------------------------------------------------------------------------------------------------

t[00]       3.553711440822  2.578525175352  4.389661472179 54.080004313636 35.999999999999 90.000000000000

t[01]       6.834457847956  2.578525175352  7.302536379205 69.425611380982 20.728385258139 90.000000000000

t[02]       6.084041403983  4.046366665742  7.302536379205 56.494384305383 33.720741822183 90.000000000000

t[03]       6.391704213641  3.539091538968  7.302536379205 61.142976945229 29.054694637655 90.000000000000

t[04]       6.723588499164  3.539091538968  7.594356471029 62.362638749389 27.845319767203 90.000000000000

t[05]       6.514490666570  3.911735747030  7.594356471029 59.145366825843 31.077340116398 90.000000000000

t[06]       6.933642498040  4.046366665742  8.022995142563 59.875534090283 30.369702011965 90.000000000000

t[07]       7.010223195830  3.911735747030  8.022995142563 60.979099083874 29.260595976070 90.000000000000

t[08]       7.037872451809  4.046366665742  8.113092799663 60.248931819435 30.000000000000 90.000000000000

 

Do you know how to calculate this or what the theory is behind this?

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDnBpTyUMprekpwMK0wNR29i8%2BShYV7GD5%2BGpYPPtAW6Jg%40mail.gmail.com.

[mcdonn...@gmail.com]

Thanks Taff,

 

I am trying to create this type of flat panel geometry myself. I would probably want higher frequency than this.

 

I am fine modelling regular domes from first principals but it seems like there must be a special trick required to get geometry similar to the Chris Kitrick example provided by Levente.

 

Cheers,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of TaffGoch
Sent: Friday, 2 January 2026 5:36 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

Patrick,

It is, apparently, an 8-frequency (version-II) subdivision, 8v(4,4)

The unique "brick" count, if I recall, is 10, with one brick mirrored, so 11 molds. 

This applies to this particular SketchUp model. Other subdivision methods may produce lower count.

Scaling in SketchUp can be tricky. Don't expect the "push/pull" tool to provide much help with producing brick thickness. 

The "scale" tool provides better results, but takes more patience in modeling.

 

-Taff

(David Price)

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[Levente Likhanecz]

i don't know. he (Chris) provided some datas on another website. from there i extracted strut lengths and manually rebuilded the 9V hexadome.
it needs only 5 molds, 1 penta and 4 hexagon, those extracted on the left side

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/000001dc7be7%247e4eaf60%247aec0e20%24%40gmail.com.

[Chris Kitrick]

For those interested here are 2 links on the generation of geodesic geometry composed of plane hexagons:

Plane Hexagonal Geodesic Tessellations

Plane Snub Tessellations

Happy New Year,

Chris

[mcdonn...@gmail.com]

Thanks Chris,

 

I did find those references and tried to follow them but alas I could not figure out how following the calculations arrived at the table below.

 

Do you do these calculations by hand or use excel or a computer program to assist.

 

Do you have a worked example that you could share?

 

Your response is much appreciated.

 

Best regards Paddy

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/bf397da0-1e87-4326-8a30-381592e2c6c1n%40googlegroups.com.

[Bryan L]

Hi Paddy, 

the method I used to create a hex-pent from a given geodesic dome was to create a tangent plane (to the sphere origin) at each vertex and intersect the resultant planes. The line of intersection of the planes became the edge of the hex-pent structure.

From memory, starting structures with the least inventory (strut lengths) created hex-pents with the most variance in hex types and vice versa - from memory, I might be wrong on that.

The resultant hex-pent vertexes have varying radii.. Also from memory, I think someone created a hex-pent with equivalent radii (possibly not planar hexes). I haven't read Chris's papers yet.

Bryan

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/000c01dc7d0e%2432499710%2496dcc530%24%40gmail.com.

[Levente Likhanecz]

just i pulled a hexapen on 6V icosahedron.

the red-yellow-green-blue all flat (hope so), all vertices pulled to sphere. (with kruschke method)
the blue uni strut, all same
in skp everybody can check / reproduce.

steps to build:
1. the blue is the original icosahedral projection.

2. the blue i rotate 72 degrees around sphere center-icosa triangle tip axle.
3. a connect the 4 vertices. to get the green plane plane

4. the green / magenta plane(= icosa triangle side), intersection, with sphere radius -> intersect the +2 vertices of the gree hexagon:

5. once i have the green hexa, then i have the yellow plane.
i extend the yellow plane, to have intersection line with magenta plane.
do the "kruschke" vertice projection to have the remaining vertices of yellow:

6. with that i have the pentagram vertices as well.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di6b97ae-cMGQJ%3D%3D1A%3DYcxSSeOyAm95C3Cm1cuX%2BWe0BBg%40mail.gmail.com.

[Levente Likhanecz]

[mcdonn...@gmail.com]

Thanks Levente!

 

I really appreciate the time you took to do this, I followed your simple instructions and was able to replicate this method with success.

 

Do you think this method would work on a 9V? I will try later when I get time.

 

 

Thanks again.

 

Paddy

 

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Levente Likhanecz
Sent: Monday, 5 January 2026 6:10 AM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

just i pulled a hexapen on 6V icosahedron.

the red-yellow-green-blue all flat (hope so), all vertices pulled to sphere. (with kruschke method)
the blue uni strut, all same
in skp everybody can check / reproduce.

steps to build:
1. the blue is the original icosahedral projection.

2. the blue i rotate 72 degrees around sphere center-icosa triangle tip axle.
3. a connect the 4 vertices. to get the green plane plane

4. the green / magenta plane(= icosa triangle side), intersection, with sphere radius -> intersect the +2 vertices of the gree hexagon:

5. once i have the green hexa, then i have the yellow plane.
i extend the yellow plane, to have intersection line with magenta plane.
do the "kruschke" vertice projection to have the remaining vertices of yellow:

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDn7_pX45OO3VKVhyVbK3-5nQH6Cedwzj98oQtyy4Ysvvg%40mail.gmail.com.

[mcdonn...@gmail.com]

Hi Bryan,

 

I had tried this method in the past and it did not get the level of standardisation that I was after.

 

Chris’ method seems like the ultimate solution but alas with the information available figuring out the maths is beyond my abilities.

 

What Leventhe came up with for 6V looks very promising if the methods can be applied to any frequency.

 

Cheers,

 

Paddy

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di6b97ae-cMGQJ%3D%3D1A%3DYcxSSeOyAm95C3Cm1cuX%2BWe0BBg%40mail.gmail.com.

[Bryan L]

Paddy,

I realise I was off the mark with that method - something I used a long time ago before trying to keep the vertices at unity.

I have read Chris's paper and generally understand what he has described. I think from what he said a {3,5+}3,3 would have no degrees of freedom and I interpret that to mean it should be able to be created geometrically.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/002b01dc7df3%244c6a6bc0%24e53f4340%24%40gmail.com.

[Levente Likhanecz]

you have freedom with anything inside the edges of the icosahedral triangle. the trouble comes at the connecting border, what is entering on the left must leave on the right.xxs

inside the triangle we can adjust any vertices for hexagon flatness by drawing a "projection helper plane" like the magenta on the side, just it has to be great circle plane (must touch the center of the sphere). may be this way we can draw a 9V geometrically, if i take into consideration what Bryan said about 3,3 

one more thought, let say you have a non flat hexa, you can check which 3 vertices form a plane where the remaining "out of plane" vertices the least out of plane, that way the shifted vertices should stay closer to their original, less distorsion.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/002b01dc7df3%244c6a6bc0%24e53f4340%24%40gmail.com.

[Bryan L]

There are many solutions but if we have a goal of minimising hex types (and therefore moulds for the OP), following Chris's method is better than basing a 2,2 off an icosa centre hexagon. If you look at Chris's Figure 4 diagram, the hexagons that lie on the short b leg of the Schwartz triangle are equivalent (defined by the 30 / 60 degree right triangles labelled 1). That will give just two different hexagons for a 2,2, and 3 for a 3,3. Currently we have 3 with the 2,2 and probably there will be 6 with the 3,3. I am just trying to model a 2,2 using Chris's method and if time permits will do a 3.3.

Here are the wiki entries for each

https://en.wikipedia.org/wiki/List_of_geodesic_polyhedra_and_Goldberg_polyhedra#/media/File:Conway_polyhedron_dkt5daD.png

https://en.wikipedia.org/wiki/List_of_geodesic_polyhedra_and_Goldberg_polyhedra#/media/File:Conway_polyhedron_dkdktI.png

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDm5p5px5vX84cEkQ5rmjT9D7WKjWdZcpJuRG7aTdk0Tpg%40mail.gmail.com.

[Levente Likhanecz]

unfortunately math and me like dumb and dumber

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di52D41pg%3DkZWUo-F7EyYt22Dy1fs%3DG0ciJ2Vr8G4tD8Sg%40mail.gmail.com.

[Levente Likhanecz]

i succeded to geometrically design a 9V hexa dome, all flat all vertices unit radius
following the previous method, just i had to select planes to intersect little different.

1. phase the blue is the standard projection. then next, the green hex i had to adjust the 2 blue circled vertices to lay on the red handdrawn trapezoid flat (intersect with icosa trianle side flat).

2. once i had the green (and blue) i rotated 72 degrees to have a neighboring icosa triangle (on the right side of the pic.)
from there the red triangle across the edge of 2 icosa triangle. on the (red triangle --- icosa side) intersect i circle out the 4th vertice of the red plane.

then the 2 red plane i intersect, from their intersection line i draw a new plane from sphere center, on that plane a compass out the sphere radius, to intersect the radius arc with the 2 redplanes intersection line.
this point will bi the "5." vertice

the last vertice i find from the point 5. i mirror to get the orange hexagon (because it is already flat)
the bluecircled red triangle plane i extend (rotate a copy) to have intersecting line with the icosa side plane.
on the side plane compass out the sphere radius intersect with the intersection line, and have the last vertice which scribe out the pentagon later.

like this i have not 5, but 6 uniq (5+1), but possible to reconstruct purely geometric

On Mon, Jan 5, 2026 at 1:15 PM Bryan L <bhla...@gmail.com> wrote:

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di52D41pg%3DkZWUo-F7EyYt22Dy1fs%3DG0ciJ2Vr8G4tD8Sg%40mail.gmail.com.

[Levente Likhanecz]

[mcdonn...@gmail.com]

Hi Levente,

 

Than you again for spending the time to do this and provide the great explanation of your method. This really helped me get past by mental block.

 

This has finally clicked for me and I am comfortable modelling these. I ended up developing my own method which is slightly different than yours.

 

 

 

Cheers,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Levente Likhanecz
Sent: Monday, 5 January 2026 10:56 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

i succeded to geometrically design a 9V hexa dome, all flat all vertices unit radius

following the previous method, just i had to select planes to intersect little different.

1. phase the blue is the standard projection. then next, the green hex i had to adjust the 2 blue circled vertices to lay on the red handdrawn trapezoid flat (intersect with icosa trianle side flat).

2. once i had the green (and blue) i rotated 72 degrees to have a neighboring icosa triangle (on the right side of the pic.)
from there the red triangle across the edge of 2 icosa triangle. on the (red triangle --- icosa side) intersect i circle out the 4th vertice of the red plane.

 

then the 2 red plane i intersect, from their intersection line i draw a new plane from sphere center, on that plane a compass out the sphere radius, to intersect the radius arc with the 2 redplanes intersection line.
this point will bi the "5." vertice

 

the last vertice i find from the point 5. i mirror to get the orange hexagon (because it is already flat)
the bluecircled red triangle plane i extend (rotate a copy) to have intersecting line with the icosa side plane.
on the side plane compass out the sphere radius intersect with the intersection line, and have the last vertice which scribe out the pentagon later.

 

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDmkXRsiyMNq3dwXbY-j9CM8aDu3Gvp%3Dor5cJBt8DETh-w%40mail.gmail.com.

[mcdonn...@gmail.com]

Thank Bryan,

 

I am very interested to see what you come up with using Chris’ method. I would really like to understand it but I read and re-read his papers today and to me (without background in spherical maths) there is just not enough information there for me to figure it out.

 

Thanks for the wiki links.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di52D41pg%3DkZWUo-F7EyYt22Dy1fs%3DG0ciJ2Vr8G4tD8Sg%40mail.gmail.com.

[Levente Likhanecz]

I found geometrically doable 9V 5 mold solution. 

the trick is i dont use the standard central hexagon. instead i split the space for 2 similar (but smaller).

on the icosa triangle i mark the weight point (from tip to opposit side middle point)
from icosa triangle's centerpoint i split the arc from sphere center 3 part (i draw a 3 segment arc between icosa center and bottom line)
the top segment (pie) lower edge intersect with the icosa triangle face, there will be the modified central hexagon bottom side.
this bottom side line rotate around 60-60-60...
will cut out the red colored new central hexagon.
now i can project out the red hexagon to the sphere radius, getting the blue hexas.
this modified hexa i can put all four pieces.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/002f01dc7eee%24ebe01640%24c3a042c0%24%40gmail.com.

[mcdonn...@gmail.com]

Thanks yet again Levente,

 

I can follow the method to get the blue hexagons but it looks like the other hexagons are also not standard projections, how did you derive the geometry for these?

 

Cheers,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Levente Likhanecz
Sent: Wednesday, 7 January 2026 1:12 AM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

I found geometrically doable 9V 5 mold solution. 

the trick is i dont use the standard central hexagon. instead i split the space for 2 similar (but smaller).

 

on the icosa triangle i mark the weight point (from tip to opposit side middle point)
from icosa triangle's centerpoint i split the arc from sphere center 3 part (i draw a 3 segment arc between icosa center and bottom line)
the top segment (pie) lower edge intersect with the icosa triangle face, there will be the modified central hexagon bottom side.
this bottom side line rotate around 60-60-60...
will cut out the red colored new central hexagon.
now i can project out the red hexagon to the sphere radius, getting the blue hexas.
this modified hexa i can put all four pieces.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDkxMO%3Dd029jwcF2scXhWiD3hquMrEYHnXVovGeWGfzYZw%40mail.gmail.com.

[Bryan L]

I have just completed a GP2,2 using Chris's method

I was able to work it out using only plane trig - if a little convoluted...

The solution involves finding two different relationships for the unknown radius that is the midpoint of the centre hex edge adjacent to the icosa edge (labelled r3 in the diagram). One in terms of a and the other in terms of b. I substituted the a, b terms to be in terms of c, equated the two relationships and solved for c.

This relied on the fact that the angle between face centre radius and a is 90 degrees as is the angle between the mid edge radius and b. It is also necessary to find the angle subtended at the origin between the face centre and the icosa edge centre (calculated as 2 * 10.452578726 degrees - I assume the designers of SketchUp only worked in Euclidean space and not Spherical space as the Protractor tool only has 3 decimals precision - nowhere near enough for these sort of operations).

r3 = sqrt(1 - b^2)

r3 = a / sin(10.452578726)

then find r2, the face centre radius which = sqrt(1 - c^2)

Because the centre hex and the one straddling the icosa edge are equivalent, the radius on the icosa mid edge is the same as r2.

I had to find another angle and edge length where the 2nd hex intersects with the pentagon.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/002f01dc7eee%24ebe01640%24c3a042c0%24%40gmail.com.

[mcdonn...@gmail.com]

Hi Bryan,

 

I can follow your calcs but when I put the geometry in my CAD program it will not allow R3 to be entered as calculated.

 

A0	10.4525787239447
c	0.208347891104592
a	0.180434566521490
b	0.104173945552296
r2	0.978054781836002
r3	0.994559092798446
r3	0.994559092798446
r3	0.994559092798446

 

I am sure I have done something wrong but it has me stumped.

 

 

Cheers,

 

Paddy

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di4%2BLevoi9soj2QNeBen82EyrAQOLykxLUWfM4a_gjWs2Q%40mail.gmail.com.

[Bryan L]

Hi Paddy,

Your numbers look correct.

I didn't specifically enter in R3. I drew r2 and then created a plane tangent to r2. Then I drew 6 spokes at length c on the plane. Then drew the hex edges and deleted the spokes. Then r3 should just be there as the midpoint on the one edge. Maybe the tolerance is tight on your CAD program? But you don't really need it. Draw another r2 on the icosa edge midpoint and then you can create the plane of the half hex along there and create the two other edges.

The workflow in SketchUp is completely different to other CAD programs as you can draw so much from inference is SU - not so easy in the other CADs. What are you using?

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/003101dc7f8c%241c3e0d90%2454ba28b0%24%40gmail.com.

[mcdonn...@gmail.com]

Hi Bryan,

 

Thanks, I think I over complicated it trying to draw it all in 3D sketch.

 

I am using a work computer with SolidWorks. The problem with SolidWorks is that it is always trying to be “helpful” adding constraints but often, and as it turned out in this case puts in constraints that you do not want and over constrains things. Sometimes it is more work trying to find the hidden constraint that breaks things.

 

Followed you approach and geometry is fine and all dimensions check out with your calcs.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di6dxyN0itPfJqmBaKc0YLMTVmcG1j%3DnJrZVF5-RDmaB%2BQ%40mail.gmail.com.

[Tensegrity Wiki]

https://community.wolfram.com/groups/-/m/t/3599683

At: The Wolfram Demonstration Project Stewart Dickson (2022), "Non-Spherical Geodesic Structures"
https://demonstrations.wolfram.com/NonSphericalGeodesicStructures/
In a 1991 Graphics Gallery of the Mathematica Journal,
S. Dickson, Graphics Gallery: "Many-Handled Surfaces," The Mathematica Journal, 1(4), 1991 pp. 51–58.
we demonstrated a system for building "Many-Handled Surfaces" modeled after chemical molecular bonding geometry extending techniques developed by Richard Buckminster Fuller. The Wolfram Demonstration is an interactive version which assembles structures of triangulated surface patches along backbones of tetrahedral or octahedral lattice topologies.

The construction method is modular such that the construction components can be "thickened" and composed for 3D printing. Stewart Dickson (2011), "Thickening a Polygon Mesh for Rapid Prototyping (3D Printing)" Wolfram Demonstrations Project. https://demonstrations.wolfram.com/ThickeningAPolygonMeshForRapidPrototyping3DPrinting/

I think that this naturally draws one to imagine constructing these objects at architectural scale.

- TensegrityWiki
Help create the Encyclopedia of Tensegrity
Visit http://tensegritywiki.com

[Levente Likhanecz]

ahhh. once i have the 4 blue i rotate a copy (of the four) 72 degrees (to have the neighboring icosahedral trinagle).
form there same procedure, like the previous version.

like tihs, just now you have the blues instead of the greens.
(between the greens the red planes... and so on, exactly as before)

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/000d01dc7f70%247600a250%246201e6f0%24%40gmail.com.

[mcdonn...@gmail.com]

Got it, thanks again Levente

 

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Levente Likhanecz
Sent: Wednesday, 7 January 2026 8:04 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

ahhh. once i have the 4 blue i rotate a copy (of the four) 72 degrees (to have the neighboring icosahedral trinagle).

form there same procedure, like the previous version.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvED%3DVa_6M_Y2W0Xv-yCoPja3Vrrdy1p89NY3jstFyk%2B9e3w%40mail.gmail.com.

[Bryan L]

Goldberg {5+,3}3,3

The starting row is more or less the same except instead of dividing the central angle between face centre and icosa mid edge by 2, this time it's by 3 - because there are two hexagons the same instead of 1 1/2.

Result 4 unique hexagons

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/001301dc7fdb%243d73d900%24b85b8b00%24%40gmail.com.

[Levente Likhanecz]

i can see the pattern, here is, how i draw geometrically the next piece:　

looks like the division of the arc from icosa triangle center point to side is making the equal hexagon belt:　

where the bottom edge of top equal segment marks (light orange-orange) projected back on the icosa triangle face.
draw a paralel with icosa triangle bottom line through this intersection on the icosa triangle face.　

rotate 6x 60 degrees, then ther you will have the central hexagon. 

then i project the hexagon vertices to the sphere, result is the blue

after i populate the blues:

then i would continue with the intersection of the red planes:　

i located the vertices for the green hexa, then i continue with the yellow plane, on the magenta icosa side plane:

the yello by measure i see some symmetry let see if it works:

then the yellow mirrored:　

i open another mail, because have some size limit i dont wanna exceed

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di4MNTzACrt47cy1u8SZAHN-EorZAyXcU_6%3DAE%3DeA2pKOw%40mail.gmail.com.

[Dx G]

Greetings all,

  It has been interesting following the  discussion regarding hexagons.  I don't want to change the subject, but did have a question. It seems like a key point is to minimize the number of "molds".  So I'm wondering if you have already looked at, and perhaps rejected, the Triacon, Class 2 Method 3.  For quite some time, its been a popular choice for some rather large domes, partly because the variation in part dimensions does not rise as fast as it does for the Class 1 Alternate as frequency is increased.

  One interesting spin on the Triacon is the use of creased diamond panels, although one can use diamond panels (a sort of narrow parallelogram) without the crease, and just make them solid.  For example one layout only has two diamond panels

A) Long center crease chord factor .546, all 4 sides are chord factor .336

B) Center .616, two sides are .363, and the other two sides are .336

If the hexagons you are working with have much lower chord factors, I can see an advantage if you are working with very large domes and want to minimize the part size.  However, if the chord factors above are not prohibitive, it seems to me you would only need two "molds" for the entire dome, and you will notice the preponderance of repeat sizes in some.

If this route would clearly be unsatisfactory, I'd be interested in what those faults are.  That would help better focus the radar on this.

Dx G

[Levente Likhanecz]

all the hexas selected and rotated over to have the connecting icosa triangle.
and continue with the next pairs of red planes, with the magenta side plane:

made that 2 new hexas, the orange and light magenta.
comes the last hexa to draw the gray plane to intersect with icosa side plane:

then the final hexa/penta. unfortunately the hexa next to the penta a little streched:

[mcdonn...@gmail.com]

Thanks Levente,

 

It looks like you may have solved this, great work!

 

 

Cheers,

 

Patrick

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Levente Likhanecz
Sent: Thursday, 8 January 2026 6:49 AM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

all the hexas selected and rotated over to have the connecting icosa triangle.

and continue with the next pairs of red planes, with the magenta side plane:

 

made that 2 new hexas, the orange and light magenta.
comes the last hexa to draw the gray plane to intersect with icosa side plane:

then the final hexa/penta. unfortunately the hexa next to the penta a little streched:

On Wed, Jan 7, 2026 at 9:10 PM Levente Likhanecz <likh...@gmail.com> wrote:

i can see the pattern, here is, how i draw geometrically the next piece:　

looks like the division of the arc from icosa triangle center point to side is making the equal hexagon belt:　

where the bottom edge of top equal segment marks (light orange-orange) projected back on the icosa triangle face.
draw a paralel with icosa triangle bottom line through this intersection on the icosa triangle face.　

rotate 6x 60 degrees, then ther you will have the central hexagon. 

 

then i project the hexagon vertices to the sphere, result is the blue

 

after i populate the blues:

then i would continue with the intersection of the red planes:　

 

i located the vertices for the green hexa, then i continue with the yellow plane, on the magenta icosa side plane:

 

the yello by measure i see some symmetry let see if it works:

 

then the yellow mirrored:　

i open another mail, because have some size limit i dont wanna exceed

On Wed, Jan 7, 2026 at 4:02 PM Bryan L <bhla...@gmail.com> wrote:

Goldberg {5+,3}3,3

 

The starting row is more or less the same except instead of dividing the central angle between face centre and icosa mid edge by 2, this time it's by 3 - because there are two hexagons the same instead of 1 1/2.

 

Result 4 unique hexagons

 

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvED%3Dob%2BSDpCwC57g4MiE0x6KORTQMnSr7yngfYD6so7CZRA%40mail.gmail.com.

[mcdonn...@gmail.com]

I can confirm that the Levente method works for 15V as well.

 

I ended up with some interesting shapes but that is almost certainly due to operator error.

 

 

Regards,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Levente Likhanecz
Sent: Thursday, 8 January 2026 6:49 AM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

all the hexas selected and rotated over to have the connecting icosa triangle.

and continue with the next pairs of red planes, with the magenta side plane:

 

made that 2 new hexas, the orange and light magenta.
comes the last hexa to draw the gray plane to intersect with icosa side plane:

then the final hexa/penta. unfortunately the hexa next to the penta a little streched:

On Wed, Jan 7, 2026 at 9:10 PM Levente Likhanecz <likh...@gmail.com> wrote:

i can see the pattern, here is, how i draw geometrically the next piece:　

looks like the division of the arc from icosa triangle center point to side is making the equal hexagon belt:　

where the bottom edge of top equal segment marks (light orange-orange) projected back on the icosa triangle face.
draw a paralel with icosa triangle bottom line through this intersection on the icosa triangle face.　

rotate 6x 60 degrees, then ther you will have the central hexagon. 

 

then i project the hexagon vertices to the sphere, result is the blue

 

after i populate the blues:

then i would continue with the intersection of the red planes:　

 

i located the vertices for the green hexa, then i continue with the yellow plane, on the magenta icosa side plane:

 

the yello by measure i see some symmetry let see if it works:

 

then the yellow mirrored:　

i open another mail, because have some size limit i dont wanna exceed

On Wed, Jan 7, 2026 at 4:02 PM Bryan L <bhla...@gmail.com> wrote:

Goldberg {5+,3}3,3

 

The starting row is more or less the same except instead of dividing the central angle between face centre and icosa mid edge by 2, this time it's by 3 - because there are two hexagons the same instead of 1 1/2.

 

Result 4 unique hexagons

 

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvED%3Dob%2BSDpCwC57g4MiE0x6KORTQMnSr7yngfYD6so7CZRA%40mail.gmail.com.

[mcdonn...@gmail.com]

Hi Bryan,

 

I can follow the calcs (same as GP2,2  but /3) and get as far as the centre hex, could share how you did the rest?

 

Cheers,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Bryan L
Sent: Thursday, 8 January 2026 1:02 AM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

Goldberg {5+,3}3,3

 

The starting row is more or less the same except instead of dividing the central angle between face centre and icosa mid edge by 2, this time it's by 3 - because there are two hexagons the same instead of 1 1/2.

 

Result 4 unique hexagons

 

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di4MNTzACrt47cy1u8SZAHN-EorZAyXcU_6%3DAE%3DeA2pKOw%40mail.gmail.com.

[Bryan L]

Hi Lev,

how did you determine the last vertex of the yellow hex? The one that joins the orange and light magenta hexes.

From Chris's paper, that is the first degree of freedom. 

I am up to that point and about to do some guesswork...

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvED%3Dob%2BSDpCwC57g4MiE0x6KORTQMnSr7yngfYD6so7CZRA%40mail.gmail.com.

[mcdonn...@gmail.com]

Dx G,

 

One of the potential issues I see with triangles\diamonds vs hex is the robustness of the ‘corners’ and also possibly complete filling of the molds.

 

I am not familiar with the Triacon Class 2 Method 3 however so thanks for the suggestion and I will have a look at this in more detail.

 

Thanks,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Dx G
Sent: Thursday, 8 January 2026 6:45 AM
To: Geodesic Help Group <geodes...@googlegroups.com>
Subject: Re: Concrete Geodesic Dome

 

Greetings all,

  It has been interesting following the  discussion regarding hexagons.  I don't want to change the subject, but did have a question. It seems like a key point is to minimize the number of "molds".  So I'm wondering if you have already looked at, and perhaps rejected, the Triacon, Class 2 Method 3.  For quite some time, its been a popular choice for some rather large domes, partly because the variation in part dimensions does not rise as fast as it does for the Class 1 Alternate as frequency is increased.

  One interesting spin on the Triacon is the use of creased diamond panels, although one can use diamond panels (a sort of narrow parallelogram) without the crease, and just make them solid.  For example one layout only has two diamond panels

 

A) Long center crease chord factor .546, all 4 sides are chord factor .336

B) Center .616, two sides are .363, and the other two sides are .336

 

If the hexagons you are working with have much lower chord factors, I can see an advantage if you are working with very large domes and want to minimize the part size.  However, if the chord factors above are not prohibitive, it seems to me you would only need two "molds" for the entire dome, and you will notice the preponderance of repeat sizes in some.

 

If this route would clearly be unsatisfactory, I'd be interested in what those faults are.  That would help better focus the radar on this.

 

Dx G

 

 

 

 

 

 

 

 

On Wednesday, January 7, 2026 at 2:10:33 PM UTC-6 Levente Likhanecz wrote:

i can see the pattern, here is, how i draw geometrically the next piece:　

looks like the division of the arc from icosa triangle center point to side is making the equal hexagon belt:　

where the bottom edge of top equal segment marks (light orange-orange) projected back on the icosa triangle face.
draw a paralel with icosa triangle bottom line through this intersection on the icosa triangle face.　

rotate 6x 60 degrees, then ther you will have the central hexagon. 

 

then i project the hexagon vertices to the sphere, result is the blue

 

after i populate the blues:

then i would continue with the intersection of the red planes:　

 

i located the vertices for the green hexa, then i continue with the yellow plane, on the magenta icosa side plane:

 

the yello by measure i see some symmetry let see if it works:

 

then the yellow mirrored:　

i open another mail, because have some size limit i dont wanna exceed

 

On Wed, Jan 7, 2026 at 4:02 PM Bryan L <bhla...@gmail.com> wrote:

Goldberg {5+,3}3,3

 

The starting row is more or less the same except instead of dividing the central angle between face centre and icosa mid edge by 2, this time it's by 3 - because there are two hexagons the same instead of 1 1/2.

 

Result 4 unique hexagons

 

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/a065f9a5-2a99-44b6-9de6-d6ec1cdd5efdn%40googlegroups.com.

[Bryan L]

Hi Paddy,

I am not quite sure where you are stuck.

In the first row, from the face centre across to the icosa edge, all the hexes are the same. So after I create the first in the centre, I just rotate a copy about the hex mid edge and sphere origin. And if there is a half one on the edge, rotate one there as well.

After that, just as you must be doing following Lev, it is creating new hex planes from any two adjacent ones and progressing from there.

Is that what you needed?

On Thu, 8 Jan 2026 at 16:36, <mcdonn...@gmail.com> wrote:

Hi Bryan,

 

I can follow the calcs (same as GP2,2  but /3) and get as far as the centre hex, could share how you did the rest?

 

Cheers,

 

Paddy

 

[mcdonn...@gmail.com]

Hi Bryan,

 

Ok, no worries, this is what I have been doing. I thought you might have been calculating more.

 

I looks like you still have to deal with DOF by making assumptions. Does Chris’ method come up with a number of discrete solutions for these by calculation rather than educated guess?

 

For Lev’s 12V I think setting the two edges on the yellow hex below to be equal resolved for me. That was my interpretation of Lev’s symmetry comment.

 

For the 15V I must have made an assumption that resulted in the irregular shapes but they are still planar and everything works out ok.

 

 

Cheers,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Bryan L
Sent: Thursday, 8 January 2026 6:26 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

Hi Paddy,

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[Bryan L]

Hi Paddy,

Not operator error. Any hex that is completely contained within the Schartz triangle has freedom in the upper triangles. Their dimensions will influence the model from that point.

It is interesting that in your 4,4 model the hex adjacent to the pent seems quite regular compared to the one in Lev's model that is a bit strecthed. That tells me you did something a little different.

See section 5.1 in Chris's paper especially the last part and figure 6

[Levente Likhanecz]

at 12V i'v seen that the stone joining the pentagon forced out of proportions, would be interesting, how Bryan's math-way workout comes to the 15V.

but my estimation, that without iterative optimalization the blue belt forcing linearly the geometry.

what is another diff, that Chris paper describes class 3 and we went by class 1.xxs 

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/003901dc805f%2489960130%249cc20390%24%40gmail.com.

[mcdonn...@gmail.com]

Hi Bryan,

 

OK now re-reading after doing it is starting to make more sense.

 

After my initial battle with the CAD software and it’s automatic constraints I decided to embrace this and drive the geometry purely from the constraints (but turning off automatic and just adding them all myself) until there are minimum DOF left. So I probably have not been following Lev’s instructions exactly if what I try works.

 

Cheers,

 

Patrick

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Bryan L
Sent: Thursday, 8 January 2026 6:41 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

Hi Paddy,

 

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[Levente Likhanecz]

can we compare dimensions of the stone next to 12V penta?

it is 10000 radius sphere, so just shift the decimal point.

(sketchup has a 6 decimal digit limit, so i use to go higher radius to see more digit)

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/001601dc8042%241437f280%243ca7d780%24%40gmail.com.

[Bryan L]

On Thu, 8 Jan 2026 at 19:37, <mcdonn...@gmail.com> wrote:

Hi Bryan,

 

Ok, no worries, this is what I have been doing. I thought you might have been calculating more.

After the first row, I am calculating the length of the hex edges along the icosa edge to get a precise value. In the plane of the Icosa edge and sphere origin, I work out the angle between the last vertex and the line of the new hex plane and then calculate the exact edge length. I am not sure what Lev is doing there because that edge may differ from the two known below it - see figure 4 in Chris's paper.

 

I looks like you still have to deal with DOF by making assumptions. Does Chris’ method come up with a number of discrete solutions for these by calculation rather than educated guess?

No he doesn't come up with any discreet solutions although he does mention educated guess which I assume comes from practice.

Remember that in the 2,2 and 3,3 geometry dictated everything. 4,4 is the first with a DOF and 5,5 has 2 DOF - labelled in figure 4. 

 

For Lev’s 12V I think setting the two edges on the yellow hex below to be equal resolved for me. That was my interpretation of Lev’s symmetry comment.

Yes I guessed that was what he meant.

 

 For the 15V I must have made an assumption that resulted in the irregular shapes but they are still planar and everything works out ok.

Yes it is a solution but playing with the dimensions of those upper triangles may hopefully lead to insights into what leads to fairly regular hexes.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/006601dc8079%24f93a54e0%24ebaefea0%24%40gmail.com.

[mcdonn...@gmail.com]

This is what I have @ 10000R:

 

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Levente Likhanecz
Sent: Thursday, 8 January 2026 7:03 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

can we compare dimensions of the stone next to 12V penta?

it is 10000 radius sphere, so just shift the decimal point.

(sketchup has a 6 decimal digit limit, so i use to go higher radius to see more digit)

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDnJ83x5QiAovc7x_G4sFz13Qq6Rrn1pO5%3DR8%2BF2myBiQA%40mail.gmail.com.

[mcdonn...@gmail.com]

I had another go at this and paid a bit more attention to any constraints I enforced.

 

There are 2 manual dimensions entered that control the geometry. I could probably put an equation in to optimise for a desired result.

 

 

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Bryan L
Sent: Thursday, 8 January 2026 6:41 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

Hi Paddy,

 

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[Levente Likhanecz]

may be the CAD adjusted something for you, or just some place selected another arbitrary symmetry for planes.

meanwhile i made a 15V version. upto some point the geometry has forced intersections.

her till the the black encircled vertice of the light magenta hex everything is linear, no choice.

from there i selected to create the 6th vertice by the blue hexes red marked plane.
symmetry is a bit off but the final ston will be less distorted.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/009a01dc8084%24e90c5450%24bb24fcf0%24%40gmail.com.

[mcdonn...@gmail.com]

I got it dialled in fairly well, it is quite sensitive to very small changes so required a bit of iterative tweaking.

 

I chose to force a couple of dimensions as it gives a lot of control.

 

 

 

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Levente Likhanecz
Sent: Thursday, 8 January 2026 9:39 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

may be the CAD adjusted something for you, or just some place selected another arbitrary symmetry for planes.

meanwhile i made a 15V version. upto some point the geometry has forced intersections.

her till the the black encircled vertice of the light magenta hex everything is linear, no choice.

from there i selected to create the 6th vertice by the blue hexes red marked plane.
symmetry is a bit off but the final ston will be less distorted.

 

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvED%3D%2BhaPdTSOJy1EUDSVAAbieUmr_xmYpyXFU2bT8R7qF5w%40mail.gmail.com.

[Levente Likhanecz]

in sketchup (2017) i still have it mostly manual. intersecting is assisted

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/00c901dc8095%2456d9d0b0%24048d7210%24%40gmail.com.

[mcdonn...@gmail.com]

In most modern parametric software in including some of the free or low cost options such as Onshape or Fusion360, the sketch tool is quite a powerful solver. It took your guidance to get me there but once it clicked for me I realised you can build the geometry as desired with all manual constraints and then deal with the DOF when you can’t find of any more true constraints.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDkNwEnCp%2BoMt5ZOSqpeyj-C%3DQ1_5WGNAd3dh-og6_JXYg%40mail.gmail.com.

[Levente Likhanecz]

i have some new thought about unassisted hand drawing. the first vertices where we have freedom to move/place a vertice (by educated guess), we can use "circumscribed circle" tool.

this i have in sketchup as well.
because the flat cut out a lesser circle from the sphere, and all vertices of the hexas is on this circumscribed circle.　

soooo, we can shift vertices around its circumference.

we cannot move "locked by geometry" vertices, e.g. on icosa triangle sideplane it is locked.　

this is the tool

the tool calculates  the center of the circumscribed circle, then we can use it to rotate away any vertice.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/00de01dc8098%24165cb130%2443161390%24%40gmail.com.

[Levente Likhanecz]

technically this circuscribed/lesser circles method can replace the Kruschke method

[Dx G]

Levente, 

  That would be of some interest.  One of the features of the Kruscke dome is its flat truncation.  Would this alternative approach you refer to provide that, and reduce the variation in component size we see for the Kruscke dome?

Dx G

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[Levente Likhanecz]

i refer in my method as Kruschke method, how he pull a vertice to a plane. where i have to intersecting plane. then i set a great circle plane on this intersection line of the 2 intersecting plane, then with a compass i draw a circle on this plane on a given radius.

something like this. just this part of the K method.

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[Levente Likhanecz]

okay, it is very similar to the K method, just i work with a single hexa as a lesser circle, not a belt on the full sphere

[Bryan L]

That is a good observation Lev. It led me to discover an error in an earlier 4,4 model I made (I had a vertex <> 1.0)

The only issue I have is the way SU draws arcs / circles with a user defined number of segments. You can't then reliably intersect with those segments and actually be on the circle unless the segment end happens to magically be the point you need - and with trig / circles / spheres that is highly unlikely.

To summarize, we can't change the radius of the hex - that is defined by the intersection of the two preceding hexes. But we can alter the face angles at the centre (the B angle and therefore b edge) of unknown pair(s) of the little right triangles.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDkq2QE%3D%2BbZ__SLQ3NJUv8Qf0%3D-F8mhuQwfStK%3DTU5Ob5Q%40mail.gmail.com.

[Bryan L]

Paddy, you asked about calculations I used. I have a more clear picture in my mind now

After the initial calculation of the radii and centre hex radius, there will be times when we have an intersection of two planes, and therefore a defined edge (line) - anchored at a vertex - but we don't know how long it needs to be.

I use plane trig to find it. I measure the radius from the end of the temporary edge and then we have 3 sides (edge, measured r and r of vertex (= 1)) and can use the Cosine rule to find the angle between the edge and the anchored vertex (in the plane defined by the edge and the origin). Once I have that angle I can calculate the edge length with simple right angle trig,

This or a similar method will be used to find the edge length of hex edges either on the icosa edge or inside the Schwartz triangle (subsequent to creation of a floating vertex for example).

A geometric method I just thought of (saves the calculations) is to find the radius tangent to the edge (defining a 90 angle). The edge length required will be two times the distance between the anchor vertex and the tangent point. SU can find the tangent by inference and I assume other sketching type CAD could as well.

HTH,

Bryan

[mcdonn...@gmail.com]

Hi Bryan,

 

Could I trouble you for a sketch to go with this description please?  I can’t quite follow it.

 

Regards,

 

Paddy

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Bryan L
Sent: Friday, 9 January 2026 12:11 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

Paddy, you asked about calculations I used. I have a more clear picture in my mind now

 

After the initial calculation of the radii and centre hex radius, there will be times when we have an intersection of two planes, and therefore a defined edge (line) - anchored at a vertex - but we don't know how long it needs to be.

 

I use plane trig to find it. I measure the radius from the end of the temporary edge and then we have 3 sides (edge, measured r and r of vertex (= 1)) and can use the Cosine rule to find the angle between the edge and the anchored vertex (in the plane defined by the edge and the origin). Once I have that angle I can calculate the edge length with simple right angle trig,

 

This or a similar method will be used to find the edge length of hex edges either on the icosa edge or inside the Schwartz triangle (subsequent to creation of a floating vertex for example).

A geometric method I just thought of (saves the calculations) is to find the radius tangent to the edge (defining a 90 angle). The edge length required will be two times the distance between the anchor vertex and the tangent point. SU can find the tangent by inference and I assume other sketching type CAD could as well.

 

HTH,

 

Bryan

On Fri, 9 Jan 2026 at 12:21, Bryan L <bhla...@gmail.com> wrote:

That is a good observation Lev. It led me to discover an error in an earlier 4,4 model I made (I had a vertex <> 1.0)

 

The only issue I have is the way SU draws arcs / circles with a user defined number of segments. You can't then reliably intersect with those segments and actually be on the circle unless the segment end happens to magically be the point you need - and with trig / circles / spheres that is highly unlikely.

 

To summarize, we can't change the radius of the hex - that is defined by the intersection of the two preceding hexes. But we can alter the face angles at the centre (the B angle and therefore b edge) of unknown pair(s) of the little right triangles.

 

On Fri, 9 Jan 2026 at 05:18, Levente Likhanecz <likh...@gmail.com> wrote:

i have some new thought about unassisted hand drawing. the first vertices where we have freedom to move/place a vertice (by educated guess), we can use "circumscribed circle" tool.

this i have in sketchup as well.
because the flat cut out a lesser circle from the sphere, and all vertices of the hexas is on this circumscribed circle.　

soooo, we can shift vertices around its circumference.

we cannot move "locked by geometry" vertices, e.g. on icosa triangle sideplane it is locked.　

this is the tool

the tool calculates  the center of the circumscribed circle, then we can use it to rotate away any vertice.

 

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di6wBHqKoi5k3DExtmdJeANCwp-dS4jpAZtuUgNgNQwyng%40mail.gmail.com.

[Levente Likhanecz]

Hi Bryan,　

this "circle intersect" plugin very-very good tool. it can intersect 2 arcs on !same plane, or an arc+a line. need some settling of the usage, because the arc cannot be part of a group.　

the other thing with the segmented arc (circumscribed), that it has a center. you can rotate on the plane, around its center. or draw a full circle around this circumscribed center (mark with a line end the center then  discard the circumscribed arc), for quick check once you have the circle 1 select it, and i add 1 - to 24, in the segments property, to have 124 segments.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di7iusWdm-7YSW%3DA-gT_OLe%3D_4JgtjGXhcBzv-fJ4ENA7g%40mail.gmail.com.

[Levente Likhanecz]

with the arc tool (same as many other tools in SU) you can hoover on a plane, press down cursor to lock it onto that plane. then the edge of that plane and the SU cursor change to pink.
with the arc tool then you click down the center, then select the beginning point, there before you click and lock the beginning  of the arc you can even type on the keyboard the requested radius+enter. (like you can project/strech a straight line by keyboard for a given length).
the arc tool will automatically display with what passing near, with what you will have the intersect. if i remember, it even can intersect passing through a plane.

On Fri, Jan 9, 2026 at 2:22 AM Bryan L <bhla...@gmail.com> wrote:

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di7iusWdm-7YSW%3DA-gT_OLe%3D_4JgtjGXhcBzv-fJ4ENA7g%40mail.gmail.com.

[Bryan L]

On Fri, 9 Jan 2026 at 20:12, <mcdonn...@gmail.com> wrote:

Hi Bryan,

 

Could I trouble you for a sketch to go with this description please?  I can’t quite follow it.

 

Regards,

 

Paddy

I was dreading this as I'm not the best at these sort of images. But you must have been doing something similar...

Lets say you have the first construction done.

Then you can establish the plane of the next hexagon.

We need the edge that follows the Icosa edge so we can extend the hex plane and intersect it with the plane of the Icosa edge,

Leaving us the correct line but how to determine its length?

Extending the edge to the origin gives a triangle where we have 3 known sides 

We can solve for the angle A and then find x.

Or, constructing a tangent to line b gives us x / 2

Having found the length of that edge, we can move onto the intersection with the adjacent hexagon and follow exactly the same procedure.

If you were doing something differently I am all ears..

[mcdonn...@gmail.com]

Thanks Bryan,

 

I get those points with hex co-planar and on the edges those planes also perpendicular to icons planes as Lev did, the first DOF that I get is this point:

 

 

I can’t follow the construction of your grey planes?

 

How would you calculate or determine this point from similar constraints?

 

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Bryan L
Sent: Friday, 9 January 2026 8:45 PM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

On Fri, 9 Jan 2026 at 20:12, <mcdonn...@gmail.com> wrote:

Hi Bryan,

 

Could I trouble you for a sketch to go with this description please?  I can’t quite follow it.

 

Regards,

 

Paddy

 

I was dreading this as I'm not the best at these sort of images. But you must have been doing something similar...

 

Lets say you have the first construction done.

 

 

Then you can establish the plane of the next hexagon.

 

We need the edge that follows the Icosa edge so we can extend the hex plane and intersect it with the plane of the Icosa edge,

 

Leaving us the correct line but how to determine its length?

 

Extending the edge to the origin gives a triangle where we have 3 known sides 

 

We can solve for the angle A and then find x.

 

Or, constructing a tangent to line b gives us x / 2

 

Having found the length of that edge, we can move onto the intersection with the adjacent hexagon and follow exactly the same procedure.

 

If you were doing something differently I am all ears..

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[Bryan L]

The grey planes in my images are the initial centre hex and first row for a 4,4 (from V12) with the other instances already copy rotated around.

Referring to Chris's paper (it should not be discounted - it is the bible for these Class II Goldberg Polyhedra). "Any hexagon that is wholly within the Schwartz triangle will introduce a degree of freedom" - or words to that effect. Two hexagons within the Schwartz - 2 degrees of freedom etc.

With the single hexagon within the Schwartz triangle in the 4,4, initially I set the two unknown triangles with the same face angle (we know the radius is fixed as per Lev's observation - and is probably mentioned in the paper but I might have missed it...). I have since created one with 59 / 60 degrees face angles to see the result and may do a few more to experiment and find out what appears optimal on that last hexagon next to the pent.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/002201dc8158%2479b06220%246d112660%24%40gmail.com.

[mcdonn...@gmail.com]

Ok, thanks Bryan, it looks like we are doing similar thing just with slightly different methods.

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di4rsat-Ez%2BYVFoM3x0L0g74QS9-JbxMgN5YUacyL6ty1w%40mail.gmail.com.

[Bryan L]

Thanks Lev,

I think I understand what you mean. Is it that we can start an arc from the point we want to intersect and then the geometry is correct?

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDmCt05z4wvj2khojwoJkbUMdwH0XmxBTMRUZuzEAESOwQ%40mail.gmail.com.

[Levente Likhanecz]

this is how "circle intersect" works.　

and how the arc tool.
notice how the blue frame color changes to pink when i press down arrow, it is locking on the plane.　

then in the arc tool how the arc i extend, then suddenly reach the point of intersect -> pops up the "from point" and the center of the arc also changes to show, that from that point

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CABWq%3Di4_fYdnuF3yRz1UBNn4dyPzZipEr%3D9EbQZURtEteX_Xdw%40mail.gmail.com.

[Ashok Mathur]

The interactions between Levente, Bryan , and McDonnell has really been fascinating as the interactions have inspired them to further heights.

Can we have a short pause during which each person makes a short write up on his present methods and software and its effectiveness and limitation.

In addition each person could tabulate and give the minimum numbers of hexagons that are being used and give all other relavent parameters . For instance these could be given 6v, 9v, and 15 v domes.

Perhaps DxG could rewrite this request better.

Regards

Ashok

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAOkvEDnBvsUf%3D%3D3J44X%2B9GL9Z6Yw3Jy015kxWBHE_57inuS7Cg%40mail.gmail.com.

[mcdonn...@gmail.com]

Hi Ashok,

 

Forgive me all if this response is not correct but the way I see it:

 

The initial example that Lev provided was based on a calculated table provided by Chris, which to me indicates that there is a mathematical method to solve this, which includes an optimisation phase to deal with the DOF.

 

What Lev has been able to do utilising his awesome SU skills is come up with a method to ‘mostly’ solve this with geometry.

 

Bryan has contributed with very helpful calculations to add another starting method in addition to Levs but based on maths.

 

All methods at this stage ultimately then seem to end with some form of educated guess to resolve the remaining DOF.

 

I think rather than trying to document the different methods of doing this with educated guess which Chris says potentially “leads to unwanted geometric distortion” it would be great if somebody could figure out how to follow Chris’ method as documented, including doing the “optimization” phase so no matter who does this we end up at the same result, e.g. something like the table that Chris provided:

 

Right

Spherical

Triangle           a                      b                      c                       A                      B                      C

----------------------------------------------------------------------------------------------------------

t[00]       3.553711440822  2.578525175352  4.389661472179 54.080004313636 35.999999999999 90.000000000000

t[01]       6.834457847956  2.578525175352  7.302536379205 69.425611380982 20.728385258139 90.000000000000

t[02]       6.084041403983  4.046366665742  7.302536379205 56.494384305383 33.720741822183 90.000000000000

t[03]       6.391704213641  3.539091538968  7.302536379205 61.142976945229 29.054694637655 90.000000000000

t[04]       6.723588499164  3.539091538968  7.594356471029 62.362638749389 27.845319767203 90.000000000000

t[05]       6.514490666570  3.911735747030  7.594356471029 59.145366825843 31.077340116398 90.000000000000

t[06]       6.933642498040  4.046366665742  8.022995142563 59.875534090283 30.369702011965 90.000000000000

t[07]       7.010223195830  3.911735747030  8.022995142563 60.979099083874 29.260595976070 90.000000000000

t[08]       7.037872451809  4.046366665742  8.113092799663 60.248931819435 30.000000000000 90.000000000000

I have tried but I am clearly not smart enough to figure this out based on the papers. I know there are much smarter people than me in this group 😉

I really appreciate Lev and Bryan’s input as at least we all know how to do this with geometry, even if we could all end up with slightly different results but I think it is crazy trying to push this further when there is obviously a documented mathematical solution.

Cheers,

Paddy 

 

From: geodes...@googlegroups.com <geodes...@googlegroups.com> On Behalf Of Ashok Mathur
Sent: Saturday, 10 January 2026 1:02 AM
To: geodes...@googlegroups.com
Subject: Re: Concrete Geodesic Dome

 

The interactions between Levente, Bryan , and McDonnell has really been fascinating as the interactions have inspired them to further heights.

this is the tool

the tool calculates  the center of the circumscribed circle, then we can use it to rotate away any vertice.

 

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/CAD-a3Ogff9OA7vaHh41nt%3D%2BnftsZw9-2x_XisbOzzQrzBZ0uKA%40mail.gmail.com.

[Dx G]

I do appreciate you guys carrying out your discussions such that we can eavesdrop and follow along. A process like this will sometimes lead to valuable new developments in the design process.

I am, however, a bit fixated on the goal.  For example, I'm curious to know what size, weight, or other metrics we should apply to a definition of success.  For a given dome diameter, is there a limit to how big, heavy, etc. we would like a single block to be?  So for the structures you are investigating, do you have any targets in mind for any of those details?  As much as I think the table Ashok proposes would be useful, given what we currently know (and don't) those metrics might be a good starting place.

Part of the reason I ask, is that I think it would be interesting and potentially useful to run those metrics using the creased diamond panels I brought up earlier.  Although hexagons clearly have several advantages over other shapes, as they say, the smallest good deed is better than the grandest good intention.  I'm not saying you should stop looking, so please continue if you are so inclined.  It's just that the Triacons have been used for large domes for decades, with an excellent track record. I'm not saying they cannot be surpassed, and would be anxious to have superior options, but I would like to see how they compare.  If I knew what targets you were trying to hit, I could estimate what that would look like if we used a Triacon with diamond shaped blocks, especially when you consider that a large dome can be made entirely from just two block sizes.  Seems to me that would make it a contender.  If it won't compete, it would be useful to know what constraints needs to be relieved to advance the effort.

Any comments welcome.

Dx G 

[Bryan L]

Paddy,

I have no reference to that table you posted. Do you have a link?

To view this discussion visit https://groups.google.com/d/msgid/geodesichelp/001b01dc81d9%2438619d80%24a924d880%24%40gmail.com.

[Adrian Rossiter]

Hi Paddy and All

On Sat, 10 Jan 2026, mcdonn...@gmail.com wrote:
> All methods at this stage ultimately then seem to end with some form of
> educated guess to resolve the remaining DOF.
>
> I think rather than trying to document the different methods of doing
> this with educated guess which Chris says potentially “leads to unwanted
> geometric distortion” it would be great if somebody could figure out how
> to follow Chris’ method as documented, including doing the
> “optimization” phase so no matter who does this we end up at the same
> result, e.g. something like the table that Chris provided:

I don't know what Chris's method is, but if the aim is to create a model
with its vertices on a sphere and planar faces, to a given precision,
then you could consider using Antiprism to repeatedly project the model
onto a sphere and then planarize the faces. E.g.

off_util -S geo_3_3_d | poly_form -a p | off_util -S | poly_form -a p |
off_util -S | poly_form -a p | off_util -S | poly_form -a p | off_util -S
| poly_form -a p | off_util -S | poly_form -a p | off_util -S | poly_form
-a p | off_util -S | poly_form -a p | off_util -S | poly_form -a p |
off_util -S | poly_form -a p | off_report -S FD

The final precision is

Planarity
maximum_nonplanarity = 2.4733898877894927e-13

Vertex distance from centre
vert_min = 0.99991922871714811 (516)
vert_max = 1.0001628849585873 (163)

I have attached an image of the model

If you repeat the commands in a script the you can probably get very high
precision.

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