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|Triangles=|Squares=|Pentagons=|Rhombic=|Rods=60+90|Spheres=32|Author=[[User:Amafirlian|Amafirlian]] 00:58, 9 August 2007 (UTC)}} |
|Triangles=|Squares=|Pentagons=|Rhombic=|Rods=60+90|Spheres=32|Author=[[User:Amafirlian|Amafirlian]] 00:58, 9 August 2007 (UTC)}} |
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− | This support structure is amazingly close. It consists of [[Five Intersecting Tetrahedra|five intersecting tetrahedra]], connected to the spheres in the rhombic triacontahedron with valency 3 |
+ | This support structure is amazingly close. It consists of [[Five Intersecting Tetrahedra|five intersecting tetrahedra]], connected to the spheres in the rhombic triacontahedron with valency 3 (there are twenty of them). The length of the tetrahedral edges is 3 rods. |
== Building Instructions == |
== Building Instructions == |
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# Build the lower halve of the rhombic triacontahedron. |
# Build the lower halve of the rhombic triacontahedron. |
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# Start putting support rods in (3 connected rods). |
# Start putting support rods in (3 connected rods). |
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− | #* Basically it suffices to connect one side to a sphere of valency 3, and see to which sphere of valency 3 the other side leads you. |
+ | #* Basically it suffices to connect one side to a sphere of valency 3, and see to which sphere of valency 3 the other side leads you. |
+ | #* Count edges: support rods always span four edges of the triacontahedron. |
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#* Also keep a look out for the five intersecting tetrahedra that develop. |
#* Also keep a look out for the five intersecting tetrahedra that develop. |
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# Complete the upper halve of the rhombic triacontahedron. |
# Complete the upper halve of the rhombic triacontahedron. |
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# Put in the remaining support rods. |
# Put in the remaining support rods. |
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− | * |
+ | * As an unexpected bonus, this construction method also provides the easiest way of building the [[Five Intersecting Tetrahedra|five intersecting tetrahedra]]! |
{{clr}} |
{{clr}} |
Revision as of 11:51, 9 August 2007
Rhombic Triacontahedron Near Miss | |
' | |
Type | Near Miss |
Rods | 60+90 × |
Spheres | 32 × |
Author | Amafirlian 00:58, 9 August 2007 (UTC) |
This support structure is amazingly close. It consists of five intersecting tetrahedra, connected to the spheres in the rhombic triacontahedron with valency 3 (there are twenty of them). The length of the tetrahedral edges is 3 rods.
Building Instructions
- Build the lower halve of the rhombic triacontahedron.
- Start putting support rods in (3 connected rods).
- Basically it suffices to connect one side to a sphere of valency 3, and see to which sphere of valency 3 the other side leads you.
- Count edges: support rods always span four edges of the triacontahedron.
- Also keep a look out for the five intersecting tetrahedra that develop.
- Complete the upper halve of the rhombic triacontahedron.
- Put in the remaining support rods.
- As an unexpected bonus, this construction method also provides the easiest way of building the five intersecting tetrahedra!
Different Views
Calculations
This model is indeed a very near miss.
The support struts may be seen as "step three diagonals" of the dodecahedron formed by the points of the intersecting tetrahedra.
This strut is itself the diagonal of a square whose sides are formed by the inscribed pentagram on the dodecahedral faces, which have length the golden ratio, phi (=1.618approx) [this is a "root-two-phi bar" see ???]
So, the support strut is sqrt(2) times phi times the length of the dodecahedral edge.
The dodecahedral edge is the short diagonal of the golden rhombus of the rhombic triacontahedron: this length is 4/sqrt(10+2*sqrt(5)) (1.051 approx).
It's the same as the inverse of the circumradius of the icosahedron, see Mathworld
Multiplying all these together gives us the geometric ratio between the rhombic triacontahedron edge length and the support strut.
- Geometric ratio = 2.406
- Actual geometric distance = 93.45mm
- Length of strut = 3 bars + 0.61mm
We know that once the error gets down to around half a millimetre, the structure can usually absorb it.
A very small smile is permitted.
Related Links
Voted: --Karl Horton 09:10, 9 August 2007 (UTC)