By Conway notation, it is acD, a rectified chamfered dodecahedron. The terminology "rectified" is by Norman Johnson, student of Coxeter. Conway uses a for ambo, same as rectified. This application can make it operationally: levskaya.github.io/polyhedronisme/?recipe=A10acD Conway notation is here: en.wikipedia.org/wiki/Conway_polyhedron_notation You can rectified any of the Goldberg polyhedra to make similar appearances. en.wikipedia.org/wiki/Goldberg_polyhedron
Dear Paul, I only having some of the background required to follow your work. With that limitation and from my perspective it looks like a profound discovery. Most of my work has been in the construction of free standing tension structures. My pieces that are somewhat associated with your work and also associated with Buckminsterfuller's and Kenneth Snelson"s are tension structures like theirs that follow the outlines of an icosadodecahedron. If you extend the 30 rod elements of their tension structures and join their ends and you suspend them from each other the result is 6 woven hoop pentagons and a free standing tension structure. You can change the pentagons into star shapes. The last star shape one I tried to make had star points of 18 degrees and valleys of 90 degrees. I don't know where I got off course. That design is hard to assemble. The orientation is easier when the star shape is closer to a pentagon. In the assembly the rule I was using is most obvious when the first 2 stars are joined. That rule is "star points above valleys and at 180 degrees to that valleys below star points". As stars are added that rule is in play for every valley and point that come into proximity. When you'r about to add the last star you can see it's path on the already assembled stars. My assembly also satisfied the rule of completion in that when the last piece is added all the valleys have points above them. There must be another rule of assembly that I don't know for what the result is may interest you. For viewing what polyhedron this extreme shape follows you should be guided by the points of the stars. This completed piece has 2 let's say polar hexagons (not pentagons) and a wide equator of squares and triangles. I haven't counted them yet. Ed Gorczyca gorczycaed@yahoo.com Show less REPLY
If you were to divide the hexagons into six triangles and the the pentagons into five triangles, you would have the topology of a 4 frequency dome, just with flattened hexagonal and pentagonal faces. Leaving these triangles out is pleasing to the eye, but less strong, since the 2 equatorial geodesics that would otherwise run straight through the centre of each hexagon are all absent.
Hello Paul, and thank you so much for the informative videos. A must watch for all dome builders, methinks. Have you considered building a dome using a catenary shape? I was doing some research, and while it adds height and structural integrity, it also seems to make the build more complicated due to the shapes changing towards the top of the dome. What do you think?
Yes it does mean a more complex build, I have a design that could do catenary shape but in a tunnel: ua-cam.com/video/nv8xIO8ulmU/v-deo.html You can build anything on a rectangular footprint with this method but I feel a round dome is pretty difficult. Or you could do a hexagon footprint with this method: ua-cam.com/video/aFfYkWCaCws/v-deo.html I'll try and upload something catenary soon.
At about 5:50 you talk about going to 50% truncation, which then becomes rectification of the polygon, but you say you're not sure why. I suppose you could be referring to two things there - you're unsure of why the actual word "rectify" was chosen to represent that particular function, or you're unsure about the reason the function gets a different name when it reaches that specific 50% mark. Now, I'm fairly certain you know more about the subject of polyhedra than I do, so I wouldn't presume to be the one giving _you_ an answer to anything, really. But if it is the case that what you're unsure about is something like the second option I spoke about, and if it's something you're interested in knowing the answer, I believe I could explain it adequately enough. That just may spare you the burden of looking it up yourself, at least. And I'd be happy to do so - I think I'll wait for you to affirm such a request before I type it all out though. That's the prudent course of action, I think..
It does look very much like it but it's not. The face angles on a rhombic triacontahedron art 63.4° and 116.6° Chamfering, rectification or truncation will never change the 116.6° to 120° which is what you need to make a regular hexagon face. So the difference is that the orange face is a regular hexagon in this polyhedron but the rectified truncated triacontahedron has irregular hexagons. Check this link: en.wikipedia.org/wiki/Chamfered_dodecahedron You can see that the hexagon is not regular.
I'm sorry to spoil your discovery for you, but I have seen this before, and it dates back at least as far as the 70s. It's no surprise you couldn't find it, though, because the people who are interested in geodesic dinners don't seem to be interested in the building techniques that this structure lends itself to. I saw this design in some instructions for how to build a geodesic dome simply by lashing together bamboo poles. This shape is perfectly suited for that purpose because, by adjusting the parameters, you can make the vertices line up into 12 (I think) perfectly circular rings. You can imagine how this would be ideal for a tent made with fibreglass poles, with the lashing points sewn into the sheet that you drape over it. I built a model using coffee stirrers and elastic bands, and it works well. Then I noticed that the "waist" of hexagons has a cross-cross pattern which, with some extra struts, could be turned into a lattice like the wall of a yurt / ger, which suggests very interesting possibilities.
Ah you said "perfectly circular rings" I am aware of this design, it has great circles producing pentagons hexagons and triangles frames just as mine but there is a difference. Mine has regular hexagons. Polyhedrons are mathematically constructed solids so even though the shape you refer to probably looks almost identical it is in fact very different mathematically.
Paul Robinson I see. To me, that's a distinction without a difference, but of course that depends on your mathematical framework. If you need the hexagons to be regular, I believe I've seen that on Wikipedia where it's listed as the ambo derivative of the chamfered dodecahedron. I haven't studied Conway polyhedrons in any depth, so I could be wrong about that. Either way, I've never seen it used in a geodesic dome.
I don't think it is possible to tell from just looking at an image if the hexagon is regular. Here's the link to the wiki page: en.wikipedia.org/wiki/Chamfered_dodecahedron I am aware of the rectified truncated triacontahedron, which again looks the same but this has hexagons with equal edge length but slightly compressed between two vertices. Here's another example, the hexagons are obviously not regular but it looks like the triangles are all the same: levskaya.github.io/polyhedronisme/?recipe=A10acD Oh I also noticed that the hexagons are not planar. From a geodesic perspective if the hexagon is regular then you can triangulate with 6 identical triangles, if it is not regular/planar you need at least 3 different triangles. I did the video to demonstrate a possible new polyhedron and to show how I made it. Many of the polyhedra on wikipedia don't show face angles, chord factors etc so it's impossible to compare. I think we can agree that there are hundreds of possible solutions to produce a polyhedron with same look but I believe there is only one solution that has regular hexagons.
Paul Robinson I mostly agree with you, but I believe the requirement that the hexagons be regular does not fully constrain the chord factors. You need one more constraint, which in your case is the implicit constraint that the vertices intersect with a sphere.
How many points does it have in total? This could be a Carbon Molecule. Ever heard about Nano-tubes? How about Carbon C-76 looking exactly like DNA because it comes Left and Right Handed? It reminds me of a Carbon C-60 Molecule.
Yes but if you make a triangulated 4v from this you only have 4 different frames (3 Isosceles one equilateral) other 4v domes have 6 different frames and some are scalene triangles.
Hi Paul. I'm really enjoying your channel so far, and just subscribed. On the topic of new dome designs, I came across this one the other day. m.ua-cam.com/video/js3Bp_3glvY/v-deo.html It was particularly interesting to me for the prospective ease of manufacturing. He states that every part is cut to the same length, and at the same angles (which means a single jig can be used, and would make both the parts fabrication and their assembly a nearly foolproof endeavor) In the comments of the linked video he gives some of the angles and dimensions by which each of the parts should be cut, but it's pretty obscure and incomplete... I love how precise you are, and your in depth explanation of the design details, including why things are the way they are. If this design interests you as well, I would love for you to plug the figures into one of your programs and clear up the obscurities not revealed in that guy's video.
Math and geometry is forever .Thank you so much
Very cool!
Well done!
Reminds me of the "bent pyramid" notion of the 5, 6 relationship.
By Conway notation, it is acD, a rectified chamfered dodecahedron. The terminology "rectified" is by Norman Johnson, student of Coxeter. Conway uses a for ambo, same as rectified. This application can make it operationally:
levskaya.github.io/polyhedronisme/?recipe=A10acD
Conway notation is here:
en.wikipedia.org/wiki/Conway_polyhedron_notation
You can rectified any of the Goldberg polyhedra to make similar appearances.
en.wikipedia.org/wiki/Goldberg_polyhedron
Yep, rectified chamfered dodecahedron confirmed
Congratulations! I'll just call it a Paulrobinsonyhedron.
It's a rectified chamfered dodecahedron. I think I'll try to make one with Polydron.
Paulyhedron?
@@MelindaGreen yes
@@lyrimetacurl0polydron
I did same thing with Rhombic Dodecahedron, don't think it has been done before?
Dear Paul,
I only having some of the background required to follow your work. With that limitation and from my perspective it looks like a profound discovery. Most of my work has been in the construction of free standing tension structures. My pieces that are somewhat associated with your work and also associated with Buckminsterfuller's and Kenneth Snelson"s are tension structures like theirs that follow the outlines of an icosadodecahedron. If you extend the 30 rod elements of their tension structures and join their ends and you suspend them from each other the result is 6 woven hoop pentagons and a free standing tension structure. You can change the pentagons into star shapes. The last star shape one I tried to make had star points of 18 degrees and valleys of 90 degrees. I don't know where I got off course. That design is hard to assemble. The orientation is easier when the star shape is closer to a pentagon. In the assembly the rule I was using is most obvious when the first 2 stars are joined. That rule is "star points above valleys and at 180 degrees to that valleys below star points". As stars are added that rule is in play for every valley and point that come into proximity. When you'r about to add the last star you can see it's path on the already assembled stars. My assembly also satisfied the rule of completion in that when the last piece is added all the valleys have points above them. There must be another rule of assembly that I don't know for what the result is may interest you. For viewing what polyhedron this extreme shape follows you should be guided by the points of the stars. This completed piece has 2 let's say polar hexagons (not pentagons) and a wide equator of squares and triangles. I haven't counted them yet.
Ed Gorczyca
gorczycaed@yahoo.com
Show less
REPLY
Nicely done!
can you cut that to give a flat base for a dome? might like to try buiding one
It is similar to Electra model designed by David Mitchell. There are different possibilities of adding hexagonal faces as described by Tom Hall.
Ok I thought this was a new Johnson solid and surprised. Still, this was very cool.
If you were to divide the hexagons into six triangles and the the pentagons into five triangles, you would have the topology of a 4 frequency dome, just with flattened hexagonal and pentagonal faces. Leaving these triangles out is pleasing to the eye, but less strong, since the 2 equatorial geodesics that would otherwise run straight through the centre of each hexagon are all absent.
DEGREES • FACES • EDGES • VERTICES
Triangle:
* Degrees: 180°
* Each interior angle: 60°
* Faces: 1 (triangle)
* Edges: 3
* Vertices: 3
Square:
* Degrees: 360°
* Each interior angle: 90°
* Faces: 1 (square)
* Edges: 4
* Vertices: 4
Pentagon:
* Degrees: 540°
* Each interior angle: 108°
* Faces: 1 (pentagon)
* Edges: 5
* Vertices: 5
Hexagon:
* Degrees: 720
* Each interior angle: 120°
* Faces: 1 (hexagon)
* Edges: 6
* Vertices: 6
Tetrahedron:
* Degrees: 720
* Each interior angle: 60°
* Faces: 4 (equilateral triangles)
* Edges: 6
* Vertices: 4
Heptagon:
* Degrees: 900
* Each interior angle: 128.57°
* Faces: 1 (heptagon)
* Edges: 7
* Vertices: 7
Octagon:
* Degrees: 1080
* Each interior angle: 135°
* Faces: 1 (octagon)
* Edges: 8
* Vertices: 8
Nonagon :
* Degrees: 1260
* Each interior angle: 140°
* Faces: 1 (Nonagon)
* Edges: 9
* Vertices: 9
Decagon:
* Degrees: 1440°
* Each interior angle: 144°
* Faces: 1 (Decagon)
* Edges: 10
* Vertices: 10
Pentagonal Pyramid
* Degrees: 1440
* Each interior angle: 108°
* Faces: 6 (5 triangles, 1 pentagon)
* Edges: 10
* Vertices: 6
Octahedron:
* Degrees: 1440
* Faces: 8 (equilateral triangles)
* Edges: 12
* Vertices: 6
Stellated octahedron:
* Degrees: 1440
* Faces: 8 (equilateral triangles)
* Edges: 12
* Vertices: 6
Pentagonal Bipyramid
* degrees: 1800
* Faces: 10 (10 triangles)
* Edges: 15
* Vertices: 7
Hexahedron (Cube):
* Degrees: 2160
* Faces: 6 (squares)
* Edges: 12
* Vertices: 8
Triaugmented Triangular Prism:
* Degrees: 2520
* Faces: 10 (6 triangles, 4 squares)
* Edges: 20
* Vertices: 14
Octadecagon (18-sided polygon):
* Degrees: 2880
* Faces: 1 (octadecagon)
* Edges: 18
* Vertices: 18
Icosagon (20-sided polygon):
* Degrees: 3240
* Faces: 1 (icosagon)
* Edges: 20
* Vertices: 20
Truncated Tetrahedron
* Degrees: 3600
* Faces: 8 (4 triangles, 4 hexagons)
* Edges: 18
* Vertices: 12
Icosahedron:
* Degrees: 3600
* Faces: 20 (equilateral triangles)
* Edges: 30
* Vertices: 12
Cuboctahedron or VECTOR EQUILIBRIUM
* Degrees: 3600
* Faces: 14 (8 triangles, 6 squares)
* Edges: 24
* Vertices: 12
3,960 DEGREES
88 x 45 = 3,960
44 x 90 = 3,960
22 x 180 = 3,960
11 x 360 = 3,960
Rhombic Dodecahedron
* Degrees: 4,320
* Faces: 12 (all rhombuses)
* Edges: 24
* Vertices: 14
* Duel is Cuboctahedron or vector equilibrium
Tetrakis Hexahedron:
* Degrees: 4320
* Faces: 24 (isosceles triangles)
* Edges: 36
* Vertices: 14
Icosikaioctagon (28-sided polygon):
* Degrees: 4680
* Faces: 1 (icosikaioctagon)
* Edges: 28
* Vertices: 28
Triacontagon:
* Degrees: 5040
* Each interior angle: 168°
* Faces: 1 (30-sided polygon)
* Edges: 30
* Vertices: 30
Octagonal Prism:
* Degrees: 5040°
* Each interior angle: 135°
* Faces: 10 (2 octagons, 8 rectangles)
* Edges: 24
* Vertices: 16
5400 DEGREES
5,760 degrees
6,120 degrees
Dodecahedron:
* Degrees: 6480
* Faces: 12 (pentagons)
* Edges: 30
* Vertices: 20
7560 DEGREES
6840 DEGREES
7,200 DEGREES
7560 DEGREES
Rhombicuboctahedron:
* Degrees: 7920
* Faces: 26 (8 triangles, 18 squares)
* Edges: 48
* Vertices: 24
Snub Cube:
* Degrees: 7920
* Faces: 38 (6 squares, 32 triangles)
* Edges: 60
* Vertices: 24
Trakis Icosahedron:
* Degrees: 7920
* Faces: 32 (20 triangles, 12 kites)
* Edges: 90
* Vertices: 60
8,280 DEGREES
8640 DEGREES
9000 DEGREES
9,360 degrees
9,720 degrees
Icosidodecahedron:
* Degrees: 10,080°
* Faces: 30 (12 pentagons, 20 triangles)
* Edges: 60
* Vertices: 30
Truncated Icosidodecahedron:
* Degrees: 10,080°
* Faces: 32 (12 regular pentagons, 20 regular triangles)
* Edges: 60
* Vertices: 30
? 10,440 degrees
Rhombic Triacontahedron:
* Degrees: 10,800
* Faces: 30 (rhombuses)
* Edges: 60
* Vertices: 32
11160 DEGREES
11,520 DEGREES
11,880 DEGREES
12,240 DEGREES
12,600 DEGREES
12960 DEGREES
END OF 360 DEGREE POLAR GRID
Small Triambic Icosahedron:
* Degrees: 14400
* Faces: 20 (intersecting non-regular hexagons)
* Edges: 60
* Vertices: 32
Small Ditrigonal Icosidodecahedron:
* Degrees: 16,560
* Faces: 50 (12 pentagons, 20 triangles, 18 squares)
* Edges: 120
* Vertices: 60
Truncated Cuboctahedron or Great Rhombicuboctahedron
* Degrees: 16,560
* Faces: 26 (12 Squares, 8 Hexagons, 6 Octagons)
* Edges: 72
* Vertices: 48
Small Rhombicosidodecahedron
* Degrees: 20,880
* Faces: 62 (20 triangles, 30 squares, 12 pentagons)
* Edges: 120
* Vertices: 60
Rhombicosidodecahedron
* Degrees: 20,880
* Faces: 62 (30 squares, 20 triangles, 12 pentagons)
* Edges: 120
* Vertices: 60
Truncated Icosahedron:
* Degrees: 20,880
* Faces: 32 (12 pentagons, 20 hexagons)
* Edges: 90
* Vertices: 60
Disdyakis Triacontahedron:
* Degrees: 21600
* Faces: 120 (scalene triangles)
* Edges: 180
* Vertices: 62
Deltoidal Hexecontahedron
* Degrees: 21,600
* Faces: 60 (kites)
* Edges: 120
* Vertices: 62
Ditrigonal Dodecadodecahedron:
* Degrees: 24480
* Faces: 52 (12 pentagons, 20 hexagons, 20 triangles)
* Edges: 150
* Vertices: 60
Great Rhombicosidodecahedron
* Degrees: 31,680
* Faces: 62 (12 pentagons, 20 hexagons, 30 squares)
* Edges: 120
* Vertices: 60
Small Rhombihexacontahedron:
* Degrees: 31,680
* Faces: 60 (12 pentagons, 30 squares, 20 hexagons)
* Edges: 120
* Vertices: 60
Pentagonal Hexecontahedron:
* Degrees: 32,400
* Faces: 60 (pentagons)
* Edges: 120
* Vertices: 62
That's a near-miss johnson solid. I call it "The Super-Football"
Hello Paul, and thank you so much for the informative videos. A must watch for all dome builders, methinks. Have you considered building a dome using a catenary shape? I was doing some research, and while it adds height and structural integrity, it also seems to make the build more complicated due to the shapes changing towards the top of the dome. What do you think?
Yes it does mean a more complex build, I have a design that could do catenary shape but in a tunnel: ua-cam.com/video/nv8xIO8ulmU/v-deo.html You can build anything on a rectangular footprint with this method but I feel a round dome is pretty difficult. Or you could do a hexagon footprint with this method: ua-cam.com/video/aFfYkWCaCws/v-deo.html I'll try and upload something catenary soon.
At about 5:50 you talk about going to 50% truncation, which then becomes rectification of the polygon, but you say you're not sure why. I suppose you could be referring to two things there - you're unsure of why the actual word "rectify" was chosen to represent that particular function, or you're unsure about the reason the function gets a different name when it reaches that specific 50% mark.
Now, I'm fairly certain you know more about the subject of polyhedra than I do, so I wouldn't presume to be the one giving _you_ an answer to anything, really. But if it is the case that what you're unsure about is something like the second option I spoke about, and if it's something you're interested in knowing the answer, I believe I could explain it adequately enough. That just may spare you the burden of looking it up yourself, at least. And I'd be happy to do so - I think I'll wait for you to affirm such a request before I type it all out though. That's the prudent course of action, I think..
Looks like Fuller's Bamboo Dome in Dome Book 2...but don't know if it is exactly the same...
combination 1
4 triangles: 4 x 180 = 720
3 Squares: 3 x 360 = 1080
4 pentagons: 4 x 540 = 2160
2 hexagons: 2 x 720 = 1440
Total = 5400
Combination 2
6 triangles: 6 x 180 = 1080
2 Squares: 2 x 360 = 720
4 pentagons: 4 x 540 = 2160
2 hexagons: 2 x 720 = 1440
Total = 5400
Combination 3
12 triangles: 12 x 180 = 2160
2 Squares: 2 x 360 = 720
2 pentagons: 2 x 540 = 1080
hexagons: 2 x 720 = 1440
Total = 5400
Combination 4
10 triangles: 10 x 180 = 1800
3 Squares: 3 x 360 = 1080
2 pentagons: 2 x 540 = 1080
2 hexagons: 2 x 720 = 1440
Total = 5400
Combination 5
8 triangles: 8 x 180 = 1440
4 Squares: 4 x 360 = 1440
4 pentagons: 4 x 540 = 2160
2 hexagons: 2 x 720 = 1440
Total = 5400
Combination 6
6 Triangles: 6 x 180 = 1080
5 Squares: 5 x 360 = 1800
2 pentagons: 2 x 540 = 1080
2 Hexagons: 2 x 720 = 1440
Total = 5400
Combination 7
4 Triangles: 4 x 180 = 720
6 Squares: 6 x 360 = 2160
2 pentagons: 2 x 540 = 1080
2 Hexagons: 2 x 720 = 1440
Total = 5400
Combination 8
8 triangles = 1440
2 squares = 720
2 pentagons = 1080
3 hexagons = 2160
Total = 5400
thats a rectification of the order-5 truncated rhombic triacontahedron (or chamfered dodecahedron)
It does look very much like it but it's not. The face angles on a rhombic triacontahedron art 63.4° and 116.6° Chamfering, rectification or truncation will never change the 116.6° to 120° which is what you need to make a regular hexagon face. So the difference is that the orange face is a regular hexagon in this polyhedron but the rectified truncated triacontahedron has irregular hexagons. Check this link: en.wikipedia.org/wiki/Chamfered_dodecahedron
You can see that the hexagon is not regular.
@@Geo-Dome ohhhhh ok
I'm sorry to spoil your discovery for you, but I have seen this before, and it dates back at least as far as the 70s. It's no surprise you couldn't find it, though, because the people who are interested in geodesic dinners don't seem to be interested in the building techniques that this structure lends itself to.
I saw this design in some instructions for how to build a geodesic dome simply by lashing together bamboo poles. This shape is perfectly suited for that purpose because, by adjusting the parameters, you can make the vertices line up into 12 (I think) perfectly circular rings. You can imagine how this would be ideal for a tent made with fibreglass poles, with the lashing points sewn into the sheet that you drape over it.
I built a model using coffee stirrers and elastic bands, and it works well. Then I noticed that the "waist" of hexagons has a cross-cross pattern which, with some extra struts, could be turned into a lattice like the wall of a yurt / ger, which suggests very interesting possibilities.
Ah you said "perfectly circular rings" I am aware of this design, it has great circles producing pentagons hexagons and triangles frames just as mine but there is a difference. Mine has regular hexagons. Polyhedrons are mathematically constructed solids so even though the shape you refer to probably looks almost identical it is in fact very different mathematically.
Paul Robinson I see. To me, that's a distinction without a difference, but of course that depends on your mathematical framework.
If you need the hexagons to be regular, I believe I've seen that on Wikipedia where it's listed as the ambo derivative of the chamfered dodecahedron. I haven't studied Conway polyhedrons in any depth, so I could be wrong about that. Either way, I've never seen it used in a geodesic dome.
I don't think it is possible to tell from just looking at an image if the hexagon is regular.
Here's the link to the wiki page: en.wikipedia.org/wiki/Chamfered_dodecahedron
I am aware of the rectified truncated triacontahedron, which again looks the same but this has hexagons with equal edge length but slightly compressed between two vertices.
Here's another example, the hexagons are obviously not regular but it looks like the triangles are all the same: levskaya.github.io/polyhedronisme/?recipe=A10acD
Oh I also noticed that the hexagons are not planar.
From a geodesic perspective if the hexagon is regular then you can triangulate with 6 identical triangles, if it is not regular/planar you need at least 3 different triangles.
I did the video to demonstrate a possible new polyhedron and to show how I made it. Many of the polyhedra on wikipedia don't show face angles, chord factors etc so it's impossible to compare. I think we can agree that there are hundreds of possible solutions to produce a polyhedron with same look but I believe there is only one solution that has regular hexagons.
Paul Robinson I mostly agree with you, but I believe the requirement that the hexagons be regular does not fully constrain the chord factors. You need one more constraint, which in your case is the implicit constraint that the vertices intersect with a sphere.
Peter McArthur .
For what it's worth, the first shape is what you get when you combine the soccer ball with it's dual.
How many points does it have in total? This could be a Carbon Molecule. Ever heard about Nano-tubes? How about Carbon C-76 looking exactly like DNA because it comes Left and Right Handed? It reminds me of a Carbon C-60 Molecule.
Can you try Hexakis Octahedron
l beleave l saw this in the whole world catalog some place?
It is nothing but truncated icosahedrons if you join centers of triangles.
didn't he cover that in the video?
Try triacontahedron, you cut the edge you got 12 pentagon, you cut the edge again you got the polyhedron you wish
It is much like a 4v geodesic full dome.
Yes but if you make a triangulated 4v from this you only have 4 different frames (3 Isosceles one equilateral) other 4v domes have 6 different frames and some are scalene triangles.
That's what i was wondering, why geodome makers complicate things with so many strut sizes.
I made such mini geodome from carton.. cant attach d pic though
its a truncated truncated icosohedron
No that's got decagons. This looks more like a chamfered dodecahedron with triangles in between (rectified?)
I will call it the hecatoicosduoherdron
Hi Paul. I'm really enjoying your channel so far, and just subscribed. On the topic of new dome designs, I came across this one the other day.
m.ua-cam.com/video/js3Bp_3glvY/v-deo.html
It was particularly interesting to me for the prospective ease of manufacturing. He states that every part is cut to the same length, and at the same angles (which means a single jig can be used, and would make both the parts fabrication and their assembly a nearly foolproof endeavor) In the comments of the linked video he gives some of the angles and dimensions by which each of the parts should be cut, but it's pretty obscure and incomplete... I love how precise you are, and your in depth explanation of the design details, including why things are the way they are. If this design interests you as well, I would love for you to plug the figures into one of your programs and clear up the obscurities not revealed in that guy's video.
Maybe it's a octadecahedron