bastardized 6v icosa layout

[Gerry Toomey]

Paul,

Here's a simple model of your scenario A of the 6v icosa, the one with 6 strut lengths. I couldn't get Excel to find a solution for either of the other two scenarios you mentioned,  B or C.

- Gerry

[Hector Alfredo Hernández Hdez.]

All vertixes are on sphere exacly more than 6 decimal points?

2013/12/14 Gerry Toomey <toomey...@gmail.com>

Paul,

Here's a simple model of your scenario A of the 6v icosa, the one with 6 strut lengths. I couldn't get Excel to find a solution for either of the other two scenarios you mentioned,  B or C.

- Gerry

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[Gerry in Quebec]

Hola Hector:

Yes... Here are the chord factors of Paul Robinson's 6v design, scenario A -- to 8 decimal places, without any variation in the radii of vertices. The colours refer to Paul's initial diagram. Unfortunately I spelled (spelt?) one of the subject words with a "z" (bastardized) instead of with an "s" (bastardised). Thus was born a second thread.

0.16055585	black
0.18270499	pink
0.18813556	gold
0.19912873	green
0.20834789	blue
0.20950044	red

Gerry in the Quebec freezer

[Hector Alfredo Hernández Hdez.]

Great is a good new :)

See you.

2013/12/14 Gerry in Quebec <toomey...@gmail.com>

[Hector Alfredo Hernández Hdez.]

Do you want tahi I make list of all posibilities of scenaries to try have 6

 legth strut of frecuency 6?

2013/12/14 Hector Alfredo Hernández Hdez. <hecto...@gmail.com>

[Hector Alfredo Hernández Hdez.]

Do you want that I make list of all posibilities of scenaries to try have 6

[Gerry Toomey]

Sure, Hector. Here's another one to add to your list. I haven't made a visual model of it, so I don't know yet whether it looks nice.

Chord factors
0.165328	dark pink
0.176450	yellow
0.177937	black
0.199129	green
0.208348	blue
0.209500	red

- Gerry

[Gerry Toomey]

Here's what it looks like....

[Hector Alfredo Hernández Hdez.]

Green numbers on vertix means freedom grade. There are 5 freedom grades.

Pool's desing is perfectely feasible, becase he have 5 equations only.

while "mex" 6 equations.

This is the reason :)

2013/12/15 Gerry Toomey <toomey...@gmail.com>

--

[Hector Alfredo Hernández Hdez.]

I have a mistake ..., pool's desing have 6 equations k=d is other

however strut d is independient to the others... because neibor struts are free...

is so interesting and say that the comparation of Fredom Grade Vs # equations
is not enought to conclude someting...

maybe can be generated other match betwen a,b,d or e....

 

2013/12/15 Hector Alfredo Hernández Hdez. <hecto...@gmail.com>

[Hector Alfredo Hernández Hdez.]

"b" is single struth...

[Paul Robinson]

Very interesting Hector, I had another think about it and I may have found a solution:

Black vertices I think intersect with sphere, not sure about red ones...

I've tried lots of different solutions, but this is the only one that has 5 strut lengths, hope it works.

Paul

On 16 Dec 2013, at 01:54, Hector Alfredo Hernández Hdez. wrote:

"b" is single struth...

2013/12/15 Hector Alfredo Hernández Hdez. <hecto...@gmail.com>

I have a mistake ..., pool's desing have 6 equations k=d is other

however strut d is independient to the others... because neibor struts are free...

is so interesting and say that the comparation of Fredom Grade Vs # equations
is not enought to conclude someting...

maybe can be generated other match betwen a,b,d or e....

 

2013/12/15 Hector Alfredo Hernández Hdez. <hecto...@gmail.com>

Green numbers on vertix means freedom grade. There are 5 freedom grades.

Pool's desing is perfectely feasible, becase he have 5 equations only.

while "mex" 6 equations.

This is the reason :)

2013/12/15 Gerry Toomey <toomey...@gmail.com>

Here's what it looks like....

<6v-icosa-classI-6-strut-lengths.jpg>

[Paul Kranz]

You can tell if the red vertices intersect with the sphere if the face angles of all the vertices add up to 720 degrees. The icosahedron us the parent polyhedron of this six-frequency geodesic. By definition a six frequency icosa 12 points on the icosa vertices, 150 points on the icosa edges (not including the vertices) and 120 points on the icosa faces, for a total of 282 vertices. This means that in order for each vertex to touch the same sphere, the angle of deficiency at each vertex needs to be 720 / 282, or 2,553191 degrees. So, if all of the face angles of the triangles that meet at each vertex is (360 - 2.553191 =) 357.446809 degrees, then all of the vertices are on the same sphere.

Paul sends...

Very high regards,

 

Paul sends...

ReviveTheFlame.com

[Hector Alfredo Hernández Hdez.]

The best way is to check if all vertix have Ri==1
   ....

2013/12/16 Paul Kranz <pa...@revivetheflame.com>

[Gerry in Quebec]

The 6v icosa, class I, has 362 vertices, and the angular deficit varies from one vertex type to another, but the cumulative deficit is 720. For example, for the method 1 layout, the angular deficit at 5-way vertices, i.e., centres of the pentagons, is 1.38 degrees, while the deficit at the 6-way vertices at the centres of the regular hexagons is 2.33 degrees.

- Gerry 

[Gerry Toomey]

The blue dome follows Paul's 5-strut arrangement shown on the left. Radii to the 5 corners of each pentagon are 0.99029 when all other radii are 1 unit in length. Pairs of triangles sharing red edges have valleys between them.... so the dome is not fully convex. Using Excel's Solver function, I found several other combinations of radii and vertex positions that yield the 5-strut layout, but all result in concave sections around the pentagons.

   - Gerry

[Hector Alfredo Hernández Hdez.]

Insted of we can think about it, no fully convex dome can be strong too :)

2013/12/17 Gerry Toomey <toomey...@gmail.com>

[Paul Robinson]

Thanks Guys for all the feedback, I think I'll use a 6 strut layout because some of the panels in the 5 strut would have different dihedral angles for the same size panel. I have more than enough ideas to put something together, I'll post it when I'm done.

Thanks again for all your help.

All the best,

                     Paul

On 17 Dec 2013, at 12:52, Hector Alfredo Hernández Hdez. wrote:

Insted of we can think about it, no fully convex dome can be strong too :)

2013/12/17 Gerry Toomey <toomey...@gmail.com>

The blue dome follows Paul's 5-strut arrangement shown on the left. Radii to the 5 corners of each pentagon are 0.99029 when all other radii are 1 unit in length. Pairs of triangles sharing red edges have valleys between them.... so the dome is not fully convex. Using Excel's Solver function, I found several other combinations of radii and vertex positions that yield the 5-strut layout, but all result in concave sections around the pentagons.
   - Gerry

<5-strut-dome-with-valleys.jpg>