Re: 7V class-I geodesic made from 6 unique isosceles triangles using 6 different strut lengths.

[Gerry in Quebec]

Hi Robert,

Nice work on the 5v & 7v layouts. These have fewer strut counts than the Mexican method and another method described by Hugh Kenner (author of Geodesic Math and How to Use It). In both those cases the strut count equals the frequency. You've also managed to reduce the number of unique triangular panels & eliminate scalene triangles. I will take a closer look at these. Thanks.

- Gerry in Quebec

On Wednesday, August 12, 2015 at 2:38:20 PM UTC-4, Robert Clark wrote:

I'm trying to find the minimum required number of unique triangles required to create a 5V, 6V, 7V, or 8V geodesic dome. Wondering what solutions others have come up with.  My best efforts have been a 5V created with 4 isosceles triangles. And a 7V with 6 isosceles triangles. Interested in making a greenhouse using Solawrap bubble greenhouse plastic. It's supposed to last up to 25 years, have an R1.7 value, and costs about $1 US dollar per square foot. See Harvest Pathway online.

Also, I created a 12 foot diameter plywood dome based on a 4V class-II geodesic using 3 unique shapes each with bi-lateral symmetry.

[Hector Alfredo Hernández Hdez.]

Are all the vertices are on the sphere?

If not they are, what close are of Ri=1?

Can you share the  coordinates of the vertex?

Thanks

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[Hector Alfredo Hernández Hdez.]

I know how to obtain coordinates of vertices from length of strusts

If you give me a lenght struths with maximun acurate

I will try

(7V of course)

one way is through solve equations system

another is by CAD manipulations

Last one is more easy...

2015-09-29 21:16 GMT-07:00 Robert Clark <clark.rob...@gmail.com>:

On the 7V, all vertices are on the unit sphere, but on the 5V they are just slightly above and below as shown in the attached image. I'll need to look around to find dimensions for the 7V.

I was laid off recently and a lot of my files were still on my company computer, so those are gone. And, consequently, I no longer have access to 3D modeling software (SolidWorks).

- Rob

[Hector Alfredo Hernández Hdez.]

Thank you, 

if you're right, I'll surprise so much.

And I will try to prove mathematically

2015-09-29 23:03 GMT-07:00 Robert Clark <clark.rob...@gmail.com>:

Hector,

Here is the dxf for the 7V sized for a 1 unit sphere.

- Rob 

 

[Hector Alfredo Hernández Hdez.]

I token e this measures of 7v

                                                                               0.14414718
                                                                                      a1
                                                               0.17132600               0.17132600 
                                                                       b1=g2                      b1

                                                   0.17944727      0.17509141     0.17944727
                                                            c1=f2              c2                   c1

                                      0.17944727      0.17944727    0.17944727      0.17944727
                                           d1=f2                    d1                 d1                      d1

                       0.17132600       0.17944727      0.17944727    0.17944727      0.17944727    0.17132600
                               e1=g2           e2=f2                  e2                  e2                      e2                  e1

         0.15274613    0.17944727    0.17944727       0.17944727     0.17944727       0.17944727  0.15274613
              f1                        f2                   f2                       f2                     f2                         f2                f1
 
0.12285083        0.17132600      0.17132600        0.17509141       0.17132600        0.17132600     0.12285083
    g1                           g2                     g2                        g4                    g2                          g2                 g1

And asume thats rigth

6 different lengths

I am ok?

[Hector Alfredo Hernández Hdez.]

Hello Robert
I have good and bad news about your job.

I have not finished calculating the inverse vertices, but is enough to say that


all effectively Ri = 1, but not all
the lengths of struts that should be equal, are equal


For Practical Usesthe differences tare very small,..., almost zero

but different from zero for theoretical purposes ...

add files, which you can verify which indicated ...

(some struts should obtain free but fail....)

In an old work between Gerry and Taff, the reasons for
mexican method  begins to fail for n = 6
It is that the system of nonlinear equations has more variales than degrees of freedom...



However, I recognize the importance of practical solutions, that solve real problems

2015-09-30 10:37 GMT-07:00 Robert Clark <clark.rob...@gmail.com>:

Yes. Those 6 lengths all seem correct.

-Rob

[Hector Alfredo Hernández Hdez.]

Robert thank you very much,

To know that lengths must be equal is enough
to reproduce your results.

I will raise and will solve a set of nonlinear algebraic equations of the form D ^ 2 (i, j) = D ^ 2 (p, q)
where D ^ 2 (i, j) is the length squared between node i and node j, I will use Newton's method.

It might require using least squares to reduce the numerical errors.

See you

2015-10-01 7:02 GMT-07:00 Robert Clark <clark.rob...@gmail.com>:

I checked again and the difference in values that should have been identical were on average .0000003. When doing parametric modeling in solidworks I used constraints extensively. The struts were set to always be equal. And the vertices were constrained to the surface of the unit sphere. Essentially, I set a bunch of rules and constraints and the software would either successfully fit it or fail. I think the pattern logic is still valid, but round off error was introduced during export as a dxf.

[Hector Alfredo Hernández Hdez.]

Hello Robert

Tell me if I'm okay.


I counted 7 different lengths to V7

9 equations (e_i)       one by each D^2(i,j) = D^(p,q)

8 degrees of freedom (red pen) , I am using a lot of simetry (like you)

That means that if you are lucky one or more equations
be redundant.

You can erase one equation and obtain same length,...
(I dont think so)

Experience shows that, although the system of nonlinear equations has no solution, BUT there is
a very good approximation sufficient for practical purposes, the same happened in the Mexican method for n> = 6

check draw..

Anyway, you know, I love practical solution too.

[sergioco...@gmail.com]

Robert and others:  I really apreciate this job, I think will have a lot of value in practical uses. Im working with panelized geos and can tell about doing hundreds of different sized panels.

Im giving a try on the V4 with 3 different panels (was published in this group), already did a sketchup design and thinking seriously to make a real model.

Im kind of good with sketchup (no equations just design). Wathever I can help...

blessings, Sergio

El jueves, 1 de octubre de 2015, 15:18:46 (UTC-3), Robert Clark escribió:

When I came up with my pattern scheme, I started with the following image. I put as many equilateral triangles onto the surface of the unit sphere as I could. I think there is only one solution for the length of the sides on these equilateral triangles. I found a curious coincidence that the magenta lines were the same length. I then added the blue lines creating isosceles triangles.

If I had access to solidworks I would try to recreate this pattern. Perhaps one of the members can also give it a try for fun.

- Rob

[Hector Alfredo Hernández Hdez.]

Thanks Bob,

looks like no equal to DXF

See you.

2015-10-09 6:30 GMT-07:00 Robert Clark <clark.rob...@gmail.com>:

Hector,

Very good news. I was able to recreate my 7V geodesic model and have verified that all 6 unique triangular faces are exactly isosceles and all vertices lie exactly on the surface of a 1 unit radius sphere. I can't explain mathematically why it works out, but just that it does. I've included two images showing the results and the lengths of the 6 different struts. Almost 1/3 of the faces are equilateral triangles and the rest are isosceles. A dome using just 4 foot long struts would create a dome 44.5 feet in diameter.

- Rob Clark

[Hector Alfredo Hernández Hdez.]

Hi Rob.

Good news

I have rebuilt your DXF as promised.

I am sending files
struts that have a small circle have slight discrepancies
the measure should struth 0.1712999 (red)
0.17129487 measures

the strut should measure 0.17942002 (gray)
0.17929083 measures

Discrepancies very small and negligible for practical purposes.

I can assure you that it is impossible to cut back these small discrepancies are due to the problem (modeling) rather than numerical errors


By the way in the Mexican method was only one discrepancy 0.000000000000001

(by numerical errors)

:)

[Hector Alfredo Hernández Hdez.]

Excuse Robert !

I have very very good news !!

  I found an error in the reconstruction
All you indicated you're absolutely right.

(It was fixed)

the smallest discrepancy found is
of 0.000000001.
I am pleasantly surprised


This research is cause for me and my students.


Thank you so much for sharing with us

(Gerry take note :) )

[Hector Alfredo Hernández Hdez.]

Yes I am teacher.
And doctoral student too.

Thanks

El oct 9, 2015 9:09 p.m., "Robert Clark" <clark.rob...@gmail.com> escribió:

Glad to share! Are you a teacher?

- Rob

[Gerry in Quebec]

Rob & Hector,
Based on Rob's arrangement of the 6 chord lengths for the 7v icosa, I ran the numbers through an Excel Solver scenario, similar to the ones used for various Mexican Method layouts. I'm happy to confirm what you both already know -- all Rob's chord factors are good and all radii are equal.

 

Nice geodesic layout, Rob.
- Gerry in Québec

[Hector Alfredo Hernández Hdez.]

I thanks Gerry.

Now I am thinking if there are or not others layout with same number of different length... topic for bachilier tesis...  

See you

El oct 1, 2015 10:59 a.m., "Robert Clark" <clark.rob...@gmail.com> escribió:

Here is a colored pattern that shows which struts are equal by legnth. I hope that helps. I'm not that good with math equations just modeling.

- Rob 

[Hector Alfredo Hernández Hdez.]

I want to share a very good Idea with you.

It is a exaustive plan to try to found absolutelly

all layouts with a certain number of different length struth

for frecuency=5 icosahedron based dome.

After of this, make a automatic evaluations  about each solution.

Please open the atachment files, to follow my explanation.

The png files

 cointaining a minimun set of struths to obtain a complete set by simetry.

there are 9 different struths (maximum)

if i want 5 differents leght strut, 4 must be equals to others.

I make a program to generate a complete set of posibilities

(.dat file)

next i ll explain the codification

these are the firts 5 rows


   1  1 1 1 4 1 1 7 8 9  5
   2  1 1 1 4 1 4 7 8 9  5
   3  1 1 1 4 1 6 1 8 9  5
   4  1 1 1 4 1 6 4 8 9  5
   5  1 1 1 4 1 6 6 8 9  5

for example

row 4 says:

4  1 1 1 4 1 6 4 8 9 

    1 2 3 4 5 6 7 8 9
4  1 1 1 4 1 6 4 8 9 

this means that

S2=S1

S3=S1

S5=S1

S7=S4

Another

row 1226
           1 2 3 4 5 6 7 8 9

 1226  1 2 1 4 2 6 7 7 2  5

S3=S1

S5=S2

S8=S7

S9=S2

The main idea is write automatically a mathematical model

(No lineal equations) and try to solve.

(Newton's method is a good way)

make a report of solutions and quasi-solutions

wth vertix coordinates (rectangular)

and others measures

Obviously all need be programated, maybe I wiill use a old codes...

but I dont know when :(

What dou you think about it?

[Gerry in Quebec]

Hi Hector,

Given the huge numbers of possible strut-length combinations to consider, this would be a nice contribution to geodesic design. Would you do it for other frequencies too? This would reduce time wasted by people trying to come up with more practical layouts based on trial-&-error or on semi-systematic methods.

 

I understand each combination of nine struts must have separate lengths for nos. 1 & 4 because this is required by the symmetry of the underlying polyhedron (e.g., icosahedron, octahedron, tetrahedron). But many strut combinations that respect the 1 & 4 restriction seem to be absent from your data file. For example (illustration attached):

 1 2 3 4 5 5 5 5 5

 

Does your data file include restrictions other than the interdependence of strut lengths 1 & 4?

- Gerry

[Hector Alfredo Hernández Hdez.]

S1 = S4 only and only if frequency = 1.

Surely there are other restrictions
hard to characterize but I have only included the S1 diff S4

you  combination is in the list

 5901 1 2 3 4 5 5 5 5 5      5
(I just using  SEARCH IN THE FILE :) )

You can try any.

i have a mistake in draw, 4 is not always in mobile
The main idea is to generate all possible designs to be evaluated and each person to choose the one.

Obviusly we are seaching a desings with less different lenghts too  :)

and we will to considerate another frecuencies, since 3 to

9 maybe.


Depending on the results, this could lead us to write a book, as you can see my English is very bad. (Saint google traslator helps me a lot)

--

[Hector Alfredo Hernández Hdez.]

S1 = S4 if and only if frequency = 1.

Sorry.

[Hector Alfredo Hernández Hdez.]

by the way, for other polyhedrons (tetra and octahedron)

we need to change the original vertixes of main polyhedron only.

[Gerry in Quebec]

Yes, I see now that combination 1 2 3 4 5 5 5 5 5  is indeed in the list. When I searched for it earlier I couldn't find it. Maybe when I was manipulating your data file, I accidentally cut off the bottom section of entries. So, that combination, the last line, couldn't be found. Sorry about that!

 

One other useful restriction that may be easy to code for is that the 12 struts forming a hexagonal set of 6 triangles must not all be the same length. If they were all equal in length, all 6 triangles making up the hex shape would lie in the same plane, which is not permitted in a geodesic dome..

- Gerry

[Hector Alfredo Hernández Hdez.]

I understand what the hexagonal array, do not worry, probably the vast majority of the chances will be discarded for other reasons.

Each case Takes 1, 2 or 3 seconds to be cheked by the program.

Right now the most important thing is that  the list of possibilities is very long, to try to make one by one.

Thanks for the comments, we need to maintain a critical eye.

See you.

[Hector Alfredo Hernández Hdez.]

Let some radios not be equal to 1 (exactly), practical solution is very good, I like.

For larger frequencies may occupy hours or days of computing, while we can wait for,  it will be fine.

I am also planning to automatically (or almost) generate
the DWG file of the solutions or cuasi-solutions.

I am so happy. A long time ago I wanted to do the program that generated the list of possibilities, without repeating any missing elements or.

See you.

2015-11-02 10:59 GMT-07:00 Robert Clark <clark.rob...@gmail.com>:

Hector & Gerry,

That is a good logical way to solve it. When I modeled my 5V in SolidWorks I used almost the same layout method.  But, to find solutions, I had to use trial and error to find combinations that worked. Sometimes, I found a solution that was very very close. So, I allowed some vertices to float above or below the surface of unit sphere. I did this for my 4 strut solution to  5V class I geodesic.

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[Hector Alfredo Hernández Hdez.]

Where is Taff?

Join us !

[Gerry in Quebec]

Here are four layouts of the 5v icosa, class I, each with five strut lengths and equal radii. I've fit them into Hector's numbering scheme. Maybe later I'll post strut maps and chord factors for each. Note: none of these sits flat at the 6/15, 7/15, 8/15 or 9/15 truncations.

 

Does anyone know of other 5v layouts with only 5 chord factors?

 

- Gerry in Québec

On Monday, November 2, 2015 at 12:59:25 PM UTC-5, Robert Clark wrote:

Hector & Gerry,

That is a good logical way to solve it. When I modeled my 5V in SolidWorks I used almost the same layout method.  But, to find solutions, I had to use trial and error to find combinations that worked. Sometimes, I found a solution that was very very close. So, I allowed some vertices to float above or below the surface of unit sphere. I did this for my 4 strut solution to  5V class I geodesic.

On Mon, Nov 2, 2015 at 11:27 AM, Gerry in Quebec <toomey...@gmail.com> wrote:

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[Hector Alfredo Hernández Hdez.]

Good Job, Gerry, what possibilities are you discarted?

many are discarted quicly :(

[Hector Alfredo Hernández Hdez.]

   65  1 1 1 4 5 6 1 1 9  5

neither sit flat I thinkk so, big windowd in the middle :)

[Hector Alfredo Hernández Hdez.]

by the way dome's Kenner is so easy to get by manipulations CAD :)

I was done until frecuency 10.

[Hector Alfredo Hernández Hdez.]

65 desing, sit flat?

maybe 7/15 ...

[Gerry in Quebec]

Hi Hector,

I haven't discarded anything yet from your data file ... just compiling what we already know from various sources. For a complete 5v icosa, class I inventory (for, say, 4 struts and 5 struts), it would be good to include the number of unique panels, as this is an important aspect of efficiency of construction. Among these 4-strut and 5-strut layouts, I don't think we will find any that sit flat at any useful truncation. But I may be wrong. I think they would all make efficient spherical cages for fish farming  :-)

 

I've attached some strut maps and an updated file for the 5 layouts.

 

- Gerry

[Hector Alfredo Hernández Hdez.]

5271 makes Gerry :)

[Hector Alfredo Hernández Hdez.]

5271 done by Gerry

[Hector Alfredo Hernández Hdez.]

You are Right Robert Number of differents panel is good idea.

There are another measures to take...

[Hector Alfredo Hernández Hdez.]

choises discarted


   1  1 1 1 4 1 1 7 8 9  5   imposible a los extremos de s5 y s6 les falta longitud para llegar a la diagonal
   2  1 1 1 4 1 4 7 8 9  5 pasa lo mismo no llega bien a la diagonal

[Hector Alfredo Hernández Hdez.]

Remember: we are looking for solutions

and cuasi-solutions too :)

[Gerry in Quebec]

Okay, we're making good progress. Seven analyzed & only 5894 to go  ;-)

[Gerry in Quebec]

Hi Hector,

Your 5v icosa "layout 3" in my list (row 65 in your 5-strut data file) doesn't sit flat at any truncation between the upper and lower pentagons. Antiprism/Antiview image attached.

- Gerry

[Hector Alfredo Hernández Hdez.]

Robert, your job with ico-f5 with 4

different leght struths  like to me so much.

Hardly you will find another best :) ...
(the present seaching is to be secure about it),

the main problem is that doesnt sit flat.

but......

you can try with octahedron like original polyhedron insted of icosahedro.

Half dome will have 100 triangles, good number (half dome icosahedron based f5 have 125 aprox, doesnt exit 1/2 dome really ).

Gerry was done another version (like yours), it like to me too.

would you like to try?

[Hector Alfredo Hernández Hdez.]

Yes I am undertood very well.

With mathematical model working, we will save a lot time :)

2015-11-05 10:33 GMT-07:00 Robert Clark <clark.rob...@gmail.com>:

Hector, I have not tried octahedron. I have only done icosahedron. Maybe, when I have a little time, I will explore the octahedron.

I wish the 5V with 4 struts had a flat base. It looks very close. There is always compromise. If the base is not flat, then you just have to put a little more work into building a stem wall that will fit. This picture shows that. 

I like the research you have done on finding solutions.  Good job.  Let me know how I can help more.

- Robert

[Hector Alfredo Hernández Hdez.]

I'm working on the program to implement the mathematical model that calculates all possible designs.

I send you a taste of the automatic generation of the DXF file

frequency 100.

Have you ever seen before? :)

[Gerry in Quebec]

Hector,

Is your idea to calculate all the possible combinations of struts for a given frequency (e.g., 9 possible strut lengths for 5v) AND produce a computer model of any desired combination of those struts that works (e.g., a 5v with only 5 struts)?

 

Here is another example of a 5v icosa geodesic sphere with 5 strut lengths, with all radii equal in length. It's row 3 in your 5v data file. This brings to 6 the number of 5-strut layouts I've seen.

 

Hasta luego.

- Gerry in Québec

[Hector Alfredo Hernández Hdez.]

Hi Gerry.

I had not seen the case 3, nor tried.

You notice that the fraction 6/15 is close sit flat ?,


by the way, how long did you find it and plot it ?.

My guess is that we will have many solutions :)

For 5v I would really get all the solutions for 8,7,6,5 0r 4 struts.

9 struts there are an infinite number of solutions, even considering all dihedral symmetry

At the time consider that not all radios are 1. I can also generate all combinations (some free Rii), especially for quasi-solutions. And cheking.


The plan is :
 for other frequencies do something similar.

[Hector Alfredo Hernández Hdez.]

6/15 & 7/15 too

sorry.

[Hector Alfredo Hernández Hdez.]

SO sorry again

is 7/15 & 8/15

close to sit flat.

[Gerry in Quebec]

Hi Hector,

The particular spreadsheet I'm using does not rotate the vertices to determine whether a dome might sit flat at useful truncations. As a general rule, a flat truncation requires a large number of chord factors (8 or 9?). Also, I have not included the radii as variables in the Excel Solver scenario, though this is pretty easy to do, in order to find useful quasi-solutions. (We did this, of course, for the various Mexican layouts.)

 

It took me about 25 minutes to revise the spreadsheet and make images of this latest geosphere using Antiprism. See the attached image -- row 4 in your 5v data file. I think you're right: there are probably lots of 5-strut solutions for the 5v icosa (and octa).

- Gerry

P.S. I tried to semi-automate the Solver function, but I don't think it saves any time compared with manually inputting a different set of constraints/target for each scenario in your data file.

[Gerry in Quebec]

Hi Hector, Rob & others,

Here are two more 5v icosa geospheres with only 5 chord factors. These are the combinations given in rows 5 & 7 of Hector's 5-strut data file. No. 7 shows a big variation in triangle size.

 

This brings to 9 the number of 5-strut layouts I've seen for the 5v icosa. I may take a quick look at some of the 4-strut combinations Hector listed in the second data file he posted. Maybe something useful will pop up.

 

In my earlier post on 5v icosa strut combinations, I thought 8 or 9 different lengths might be needed to ensure a flat base at some useful cut-off point. Looking back through my files, I see that the flat-base, 5v layout used for a 42-foot, 7/15 greenhouse in northwestern USA has only 6 chord factors and all triangles are either isosceles or equilateral (no scalenes).

[Hector Alfredo Hernández Hdez.]

I like a 7 number, because, we can have a beatifull window in the middle triangle.

All
1 1 1 s4 s5 s6 s7 s8 s9 desings

 like so much, because we can combine with other of the same family :)

[Bryan]

I painstakingly modeled Robert's V5 in sketchup.

Not sure I am game to tackle his V7.

But, noting his 6 chords in the V7, did anyone ever try to see if something similar could be achieved in higher frequencies?

[Bryan]

A point to note, I scaled up to 100 for his 8 decimals but then had to go one higher to split some differences and get them to round accordingly...

The 0.24940854 struts were actually 0.249408535, 249408536, 0.249408537 and 0.249408541

[Hector Alfredo Hernández Hdez.]

If you let leave to vertices a little of sphere you can have a nice measures :)

El ago 10, 2016 4:27 AM, "Bryan" <bhla...@gmail.com> escribió:

A point to note, I scaled up to 100 for his 8 decimals but then had to go one higher to split some differences and get them to round accordingly...

The 0.24940854 struts were actually 0.249408535, 249408536, 0.249408537 and 0.249408541

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[Hector Alfredo Hernández Hdez.]

Also frecuencies of the form f=6k can have interesting results, by simetry.....

[Bryan]

Here is the V7

[Chris Kitrick]

If your symmetrical constraints are relaxed you can create some alternative grids that are very efficient with respect to component count (edges, faces). Below is a layout of a V5 ( {3+,5}5,0 ) configuration where you maximize the number of equilateral triangles. Typically you can achieve 50% of the dome to having the same panel. Of course there are some additional complexities introduced: the overall grid is chiral. All vertices lie on the sphere.

Layout options for equilateral triangles (across two icosahedral faces):

The result for the middle layout option is attached. 

On Wednesday, August 10, 2016 at 11:08:02 PM UTC-7, Bryan wrote:

Here is the V7

[Hector Alfredo Hernández Hdez.]

wow, too interesting job

[Bryan L]

I found this V5 while playing with the Temcor design applied to a sphere (I will start another thread when I can collate / present everything).

It doesn't sit flat. 3 green struts do form a great circle but two pinks are off by about 7mm / 1m R

[Bryan L]

In my original post I right but wrong. The great circle is not a truncation plane, which needs to be a lesser circle...

Anyway, here is a V5 truncatable at 2/5, 3/5 with 7 struts.

1 2 3 3 4 5 6 7 6

[Bryan L]

Thanks Robert,

I will be posting others in my last thread

[Bryan L]

The dat file enrtry is 1 2 3 4 2 5 3 3 3.

The pic is wrong. I was not using Hector's numbering scheme...

[Eric Marceau]

If I understand correctly the goal, which is to identify the dimensions for struts such that they are all of the same length, then I would like to share my insights into what I believe is relevant to the problem at hand.

Referencing the book "Polyhedra and Geodesic Structures" by Vincent J. Matsko, Revised 2011, there is some guiding information there.

Assuming that the icosahedron is the starting point, and that we use "great circles" to establish intersections on spherical cords, the angle at spherical centre between 2 vertexes on the circumscribed sphere is shown to be   acos ( 1 / SQRT(5) ), which is 63.434948823... degrees (page 47, table 3.1).

Given that

- each of the triangular edge projections onto the sphere is by means of a great circle, and

- each subdivision for tesselation is obtained by intersections with rotations of each of the three planes thru the central axis/point of the sphere,

- spherical symmetry dictating that each triangle must be identical,

if the we wish values for a Class I 7v design,

then the angle of sweep for each of the triangular planes is   ( 63.434948823 degrees ) / 7 =  9.062135546 degrees.

That in turn dictates that each of the straight struts forming the edges of triangles must all be of the same chord length, namely

S = 2 * {SPHERICAL RADIUS} * sin ( 9.062135546 / 2)

notably with NO variation, regardless of the strut placement in the grid.

Given that this was derived from geometry and intersections, I believe this method of calculation is universally applicable to all sizes/frequencies of domes with Class I tesselation, overcoming the limitations of some approaches dictating their applicability to only even-numbered frequencies.

I hope that I have not erred in my reasoning and would appreciate feedback.

________________________________________________________________

B.Eng. McGill '80, retired

On Saturday, 18 January 2020 at 08:20:41 UTC-5 clark.rob...@gmail.com wrote:

Bryan,

The 123345555 solution is very elegant - all isosceles triangles!  very nice.

Robert

[Eric Marceau]

To reduce my earlier posting to a single line,

for a dome based on Icosahedron projection onto a sphere,

where all intersections and interpolations are based on rotations of planes for great circles on their axis line thru the sphere's circle,

the resulting Class I dome of frequency N should use a single, ideal strut length calculated as follows:

S = 2 * {SPHERICAL RADIUS} * sin [ acos ( 1 / SQRT(5) ) / ( N * 2) ]

The tabulated results are as follows (couldn't attach my original spreadsheet):

Again, I am looking for feedback to confirm that I have not erred in my original logic in arriving at that formula.

Naturally, if the dome is to be truncated to form a planar base for bolting onto a flat foundation or wall, the bottom struts would need to be adjusted to force their anchoring on the foundation's plane.

Thank you in advance for your feedback.

________________________________________________________________

B.Eng. McGill '80, retired

[Ashok Mathur]

Dear McGill,

I admire the original work you have done but believe it is mid-directed.

You have gone on an unproven myth in geodesic world that the strength of a geodesic dome comes from the fact that it’s triangles are close to equilateral ones.

But reality is that the type of triangles do not matter at all as the strength comes from the fact that  a geodesic dome is a tensgrity where compression and tension are balanced. If it is tensgrity, the sides and included angles do not matter at all.

I request you to give some thought to this and simultaneously find proof that near equal sided triangles are more stronger than other variants in a tensgrity.

As almost all dome failures occur at hubs and not at struts at all. We should look at creating better hubs rather than making better struts.

Regards

Ashok

Again I salute for the effort made.

Sent from my iPhone

On 12-Mar-2022, at 2:25 AM, Eric Marceau <eajma...@gmail.com> wrote:



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[Eric Marceau]

My submission was in direct response to the stated goal presented by Robert Clarke in the first posting to this discussion, in his own words:

"I'm trying to find the minimum required number of unique triangles required to create a 5V, 6V, 7V, or 8V geodesic dome."

Regardless of any other perspectives regarding preferences or various other optimizations, including tensgrity as you have indicated, his goal was the minimum number of unique triangles.

The solution I proposed offers a single, uniform triangle shape for all triangular faces.

Is that not the optimized minimum which Robert requested?

For my own sake, I had planned to pursue further study, as a thought exercise, to offer the strut length deviations/adjusted to fit a specified "foundation" plane of intersection, optimized for close match on tesselation of uniform triangles of the "ideal" sphere already identified.  Not sure when I will be able to complete that, but I will post my findings for this further study when complete it.

P.S.  My name is not "McGill".

[Ashok Mathur]

Dear Eric

Sorry for the goof up in your name.

Ashok 

Sent from my iPhone

On 12-Mar-2022, at 10:50 AM, Eric Marceau <eajma...@gmail.com> wrote:



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[Ashok Mathur]

Dear Eric

Yes you have found a good solution to the challenge posed.

Regards

Ashok

Sent from my iPhone

On 12-Mar-2022, at 10:57 AM, Ashok Mathur <ashokch...@gmail.com> wrote:

Dear Eric

[Adrian Rossiter]

Hi Eric

On Fri, 11 Mar 2022, Eric Marceau wrote:
> My submission was in direct response to the *stated goal presented by
> Robert Clarke* in the first posting to this discussion, *in his own words*:
>
> "I'm trying to find the *minimum required number of unique triangles*

> required to create a 5V, 6V, 7V, or 8V geodesic dome."
>
> Regardless of any other perspectives regarding preferences or various other
> optimizations, including tensgrity as you have indicated, his goal was the
> minimum number of unique triangles.
>
> The solution I proposed offers a single, uniform triangle shape for all
> triangular faces.

I'm not really following. Perhaps you could sketch a 2-frequency model
and mark the strut lengths on it so that all the triangles are the same
shape and the vertices lie on a sphere. Or are you proposing something
else?

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

[Gerry in Quebec]

Hi Eric,
Your formula for calculating those "ideal strut lengths" is correct. However, those equal lengths apply only to the subdivided edges of the overall icosahedron. The edges inside each of the 20 subdivided faces of a geodesic sphere based on the icosahedron will have lengths that differ from that of the subdivided edges (though there are exceptions).

Consider a 4-frequency icosahedral geodesic sphere. It has 320 faces, 162 vertices and 480 edges. Your ideal length of 0.275904 is correct for 120 of those edges. (Of course there are other subdivision methods in which those edges are not all equal.)

Let's say you were dividing up one of the 20 equilateral faces of an icosahedron into 16 smaller triangles (4-frequency). If you do NOT project the resulting vertices onto the surface of the imaginary sphere surrounding the icosahedron, then all the edge lengths will be 0.262866, a little shorter than chord length 0.275904. But when you project ALL the vertices to the surface of the sphere, you enter the realm of spherical geometry which differs substantially from planar Euclidian geometry. In spherical geometry/trigonometry, for example, the interior angles of a spherical triangle don't add up to 180 degrees.... the sum is always greater than that.

In the attached diagram by Taff Goch, to which I've added a few items, triangle 1-2-3 can be considered a spherical triangle on the surface of a sphere whose radius = 1. We know from your formula that chord 1-2 has a length of 0.275904. So what is the length of chord 2-3? We know that arc 1-11 is acos [ 1 / sqrt (5) ], or 63.43495 degrees. And arc 1-2 is one-quarter of arc length 1-11, namely 15.85874 degrees. We also know that spherical angle 2-1-3 (not the planar angle) is 72 degrees, namely 360 degrees / 5, because an icosahedron has 5 equilateral triangles converging at each vertex.

Arc 2-3 =  2 x arcsin [sin arc (1-2) x sin spherical angle (2-1-3) / 2) ]
Arc 2-3 = 18.48601 degrees
Chord length 2-3 = 2 x sin arc (2-3) / 2 = 0.321244

I hope this helps.

Cheers,
- Gerry in Québec

[Eric Marceau]

I apologize for the fact that I do not have a CAD system that allows me to create a model of what I am trying to explain.  (I am ramping up on FreeCAD, but it does not seem to want to work with my approach to things, specifically obtaining curves of intersection between planes and spheres.)

Attached is a sketch, as requested, to clarify what I am talking about.

Adrian asked me to show that on a 2v Icosahedron, but the problem does not reveal itself at that level.  You must have frequency 3v+ to see the impact of what I am trying to point out.

As referenced in my earlier post,  the angle  ⱷ  shown in this sketch is the angle between vectors from spherical centre and each of the 2 vertexes on the circumscribed sphere at A and E, namely acos ( 1 / SQRT(5) ), which is 63.434948823... degrees.  These are NOT angles in the plane of the pentagon at the base of the Icosahedron's top 5 triangular faces.

Equal division of the Icosahedron edge at AE does not give the same results as the equal division of the arc of the Great Circle that passes thru AE.

That is why I believe that relying on a solution driven by the Icosahedron Planar Faces or their Edges prevents the emergence of the elegant solution that is driven by symmetry at the spherical surface.

I hope that clarifies what I am trying to convey.

On Saturday, 12 March 2022 at 11:29:47 UTC-5 clark.rob...@gmail.com wrote:

French architect Alain Lobel created what are called Lobel meshes.  They are space enclosing forms constructed from all equilateral triangles.  TaffGoch has some of these modeled.  They are a very interesting structure with surprising rigidity.

-Robert

[Eric Marceau]

To try to expand on what I already provided, I will used a modified version of the image provided by Gerry in his recent post.

Referencing the below attached mark-up, a plane (PLANE 1) for a Great Circle passes thru A, B and the centre of the sphere.

Another plane (PLANE 2) passes thru B, G, and the centre of the sphere.

If we rotate Plane 1 on the axis passing thru spherical centre by the angle ⱷ , we would have a new plane (PLANE 3) positioned roughly near the location of C .

Similarly, if we rotated Plane 2 on the axis passing thru spherical centre by an angle of   3ⱷ , that plane (PLANE 4) will also be positioned roughly near the location of C .

Now here is my insight:   Plane 3 and Plane 4 will intersect  "precisely"  on the same radial vector line, thru the spherical centre, that is shared with the intersection of each of those 2 planes (separately) with the plane defining the Great Circle passing thru A and G !  

The resultant new spherical arc, on Plane 4, passing thru C, intersects with the spherical arc of A to B at exactly 1/4 the arc distance from A to B. 

That being the case, the triangle formed with vertices A, C and the third point (sorry, call that Q), will be an isoceles triangle.  More importantly, that triangle will be identical for all other triangles derived from the planar rotations and intersections at the intervals of  ⱷ rotations on their respective axes thru the spherical centre.

I hope that clarifies the mechanics of how I arrive at the results.

I just don't have the tools to build the model myself to demonstrate this.

On Saturday, 12 March 2022 at 22:30:31 UTC-5 clark.rob...@gmail.com wrote:

It reminds me of the distortion you get on a Dymaxion map vs the true layout on a globe.  Also, just a fun note, my habit when creating the angle 63.4349 is I use arctan(2).  Just easier for me to remember.

[Gerry in Quebec]

Hi Eric,

I originally thought your conjecture about intersecting planes was that an icosahedral sphere  could be tessellated  with one unique triangle type, an EQUILATERAL triangle. The calculation of chord length 2-3 in the 4-frequency diagram was intended to show the presence of isosceles triangles, even when all 12 edges forming the perimeter of the icosahedral face are equal in length. Now I see that your conjecture is that a single ISOSCELES triangle could be used for the tessellation. No math, only a visual inspection of the symmetry of the subdivided icosa face, is needed to show that the conjecture is incorrect because, to preserve symmetry, the central triangle of the subdivided parent triangle must be equilateral.

- Gerry

[Ashok Mathur]

Dear Gerry

For the moment suppose that the conjecture is that one equilateral triangle and one isosceles triangle will properly tasselet  the sphere.

Would that conjecture hold?

Lobel frames are shown as being held together by magnetic ball bearings that freely rotate. To me it implies that the best hub is one that creates no barriers to self adjustment. You need not calculate and drill anything in the hub. The forces of tension and compression will ensure a stable structure.

The image is from geomag wiki.

Regards

Ashok

Ashok

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[Gerry in Quebec]

Hi Ashok,

As you know, two-frequency icosahedral and octahedral spheres/domes have one type of equilateral triangle and one type of isosceles triangle. I don't think that simple combination is found at higher frequencies.

The Lobel structures and flexible node designed by the late Monsieur Lobel are interesting because of the single equilateral triangle used. But of course they have multiple "radii", if I can use that term. The architectural applications would be limited in countries like Canada and parts of the USA which get a lot of snow.  This is because the valleys between triangular panels (concavity) can act as traps, accumulating heavy snow and ice loads.

Gerry in Québec 

(where we got another dump of snow yesterday)

[Adrian Rossiter]

On Sun, 13 Mar 2022, Ashok Mathur wrote:
> Lobel frames are shown as being held together by magnetic ball bearings

Just to mention, as there is talk of Lobel frames, that these can
be made with Antiprism with a command like

poly_form -s 25 -n 100000 -y geo_4_5 | antiview

Check the precision of the result

poly_form -s 25 -n 100000 -y geo_4_5 | off_report -C E

[edge_lengths_cnts]
1.0000000000001812 = 1830 (range +/- 4.3998138465894954e-13)

[Adrian Rossiter]

Hi Eric

On Sat, 12 Mar 2022, Eric Marceau wrote:
> *Attached is a sketch*, as requested, to clarify what I am talking about.

>
> Adrian asked me to show that on a 2v Icosahedron, but the problem does not
> reveal itself at that level. You must have frequency 3v+ to see the impact
> of what I am trying to point out.

Equal-arc vs equal-edge-segment-projected does not give different
results with the 2-frequency model, but if you mark some edge
lengths for your solution model then it will make it clear how many
triangle shapes are involved.

[Bryan L]

Hello Eric

On Wednesday, 9 March 2022 at 9:46:21 am UTC+11 eajma...@gmail.com wrote:

Assuming that the icosahedron is the starting point, and that we use "great circles" to establish intersections on spherical cords, the angle at spherical centre between 2 vertexes on the circumscribed sphere is shown to be   acos ( 1 / SQRT(5) ), which is 63.434948823... degrees (page 47, table 3.1).

then the angle of sweep for each of the triangular planes is   ( 63.434948823 degrees ) / 7 =  9.062135546 degrees.

That in turn dictates that each of the straight struts forming the edges of triangles must all be of the same chord length, namely

S = 2 * {SPHERICAL RADIUS} * sin ( 9.062135546 / 2)

 I think I understand where you are coming from.

What you mention relates to the division of an original icosa chord. Equal angle division instead of chord length division. But doesn't apply to other chords / planes between new vertices. The method has been shown before and I think has been described as Class I method II in Dome Book 2 and method IV by Joseph Clinton. There are pros and cons with that method (reduced strut count but loss of natural equator in even freq domes being some). However the division of new planes doesn't result in similar chord lengths and any group of 3 planes or new vertices don't quite align, requiring an average to get a new point. The most economical method with regard to strut count seen so far is the Mexican, giving N struts, with an insignificant loss of equal radii for N > 6.

I have tried FreeCad and it is hard to learn. The free version of Sketchup (called Make) still takes a bit to learn but the user interface is more intuitive.

Bryan

[Paul Kranz]

Ashok: If you go with the theory that a triangle is a locked hinge, as long as the corners are being held together, the triangle will naturally lock-up.

I tested this idea using hinges and found that free-floating, self-adjusting connectors make just as rigid a structure whether the hinges are applied to connect the struts or the triangular panels themselves.

The trick was to use loose-pin hinges so the structure doesn't start folding-up until you are ready for it to.

Very high regards,

Paul sends...

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[Ashok Mathur]

Dear Paul,

I agree almost fully with you.

The exception is that  what 

you say is true for all triangles not just equilateral triangles.

Regards

Ashok

Sent from my iPhone

On 14-Mar-2022, at 7:17 PM, Paul Kranz <pa...@revivetheflame.com> wrote:



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[Bryan L]

Thanks Robert,

I will have a look one day.

Bryan

On Monday, 14 March 2022 at 2:49:12 am UTC+11 clark.rob...@gmail.com wrote:

Bryan,

I use SolidWorks student edition (I get it free as a veteran) for my geometry modeling.  Another very similar software that is reportedly easier to learn is Fusion 360.  If you are a hobbyist wanting to use it for non commercial applications, you can download Fusion 360 for free.

-Robert

[Walter Venable]

Greetings, all!

This is my first time posting to one of these groups, sorry if I get it mixed up the first time. I had tried one other time to join and got an error, it seems to be working this time.

I am a fellow geodesic dome enthusiast, and I saw this discussion about the subject of minimizing the number of strut types needed for a given frequency of dome. And also the related question of which dome strut patterns have solutions and which do not. It is very interesting!

I was impressed by the elegance of Hector's “Mexican method” and also by Robert Clark who had some even lower-count solutions to some domes.

I have some more thoughts on this which I will post soon, but wanted to suggest one thing which was a different way to label the struts in the "minimal triangle representation of a dome's pattern. I have attached a diagram to show this method, it starts with the 2f dome and always has the strut next to the vertex be "1" and the first cross meter is always "2". These are the struts commonly called "a" and "b" in dome books. Then with the 3f dome you leave struts 1 and 2 labelled the same but add two more, and so on for the higher frequencies. The advantage I think this might have is in allowing better comparison of patterns between different frequencies, for the purpose of generalizing discoveries like Robert's of an interesting pattern.

Do you all think this is a useful change? I'm sorry to suggest this at such a late date, Hector, as you have already been using the other system quite a while. I only recently found this group!

Best to ll of you, and I will write more soon.

-- Walt Venable

You can see some of my dome projects from the 90's at http://www.alt-eng.com/DomePage/DomeIndex.html

On Wednesday, August 10, 2016 at 10:06:36 AM UTC-4 Hector Alfredo Hernández Hdez. wrote:

If you let leave to vertices a little of sphere you can have a nice measures :)

El ago 10, 2016 4:27 AM, "Bryan" <bhla...@gmail.com> escribió:

A point to note, I scaled up to 100 for his 8 decimals but then had to go one higher to split some differences and get them to round accordingly...

The 0.24940854 struts were actually 0.249408535, 249408536, 0.249408537 and 0.249408541

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[Dx G]

Pertaining to the discussion about minimizing strut lengths and triangle variations...has anyone looked at class 3 domes?  They are kind of a disaster for flat truncations, but do have some other virtues. They are reputed to have the highest structural strength of all domes, and several have a minimum of strut lengths and triangle types.

DxG

[Dick Fischbeck]

In the long run, chord factors are irrelevant. 

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[Dick Fischbeck]

That is, if we stick with angles.

[Dx G]

Dick,

  Fair enough. Would you say more?  So why are they irrelevant?  And what *is* relevant?  Angles?  How/why...?

Side note, if you are building domes or somehow involved in making the pieces, having more struts and triangles the same is a huge plus. If we want the world to see more domes in use, we have to do everything we can to make adoption as easy as possible, IMHO...

thx DxG

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