On the 30-Rhombus Symmetric Solid Object

by Frank C. Fung ( 1st published in February, 2013. )

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Section VI - Structural Dimensioning of the Icosahedron

Summary for the Section :

• In this Section , we shall go thru the Structural Dimensioning process for the Icosahedron .

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Bringing-in the Icosahedron :

• Let us first bring-in this Icosaheron for our use in this Section :

  

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Bisecting the Icosahedron :

• Let recall that :
  • there are always 12 Nodal-Points and 30 Edges associated with the Icosahedron ,

    and

  • the straight-line passing thru the mid-points of a pair of opposite Edges is always an [ Axis of Bi-Symmetry ] .

• Let us now select a [ pair of Opposite Edges ] ,
  • as marked-off in [ white-color ] in the 3 diagrams below ;

  

  and :

  • pass a straight-line thru the mid-points of the said [ pair of opposite edges ] :
    • to act as an [ Axis of Bi-Symmetry ] ,
      • as per the [ red-color ] straight-line as shown in the 3 diagrams below .

  

• Let us now split the the Icosahedron in half , via :
  • a [ bisecting plane ] perpendicular to the [ Axis of Bi-Symmetry ] passing thru Centroid of the Icoshedron ;
    • as shown below .

  

  We then bring-in this set of 3 diagrams below ,

  • with the diagram in-the-middle showing the Cross-Section of the Icosahedron on the [ Bisecting Plane ] .

  

  Let us now take a closer look at the diagram on-the-right above , and we see here that :

  • the 4 triangles [ Triangle A-K-J ] / [ Triangle E-J-K ] / [ Triangle D-M-L ] / [ Triangle B-L-M ]
    • are split-in-half during the Bisection Process .

  As such ,

  • the point [ Point F ] is the mid-point of [ Line Segment J-K ] , and

  • the point [ Point C ] is the mid-point of [ Line Segment L-M ] .

• Let us now denote the [ length-of-Side ] for the 20 identical Equilateral Triangles of the Icosahedron
  • as the distance [ S ] ;
    • same notation as in the last Section .

  Then , refering to the diagram on-the-right above , we can now establish the following :

  • Based on the equilateral triangle [ Triangle A-K-J ] :
    • the length of [ Line-Segment A-F ] is [ 0.866025 * S ] ;

  • Based on the equilateral triangle [ Triangle E-J-K ] :
    • the length of [ Line-Segment E-F ] is [ 0.866025 * S ] ;

  • Based on the equilateral triangle [ Triangle D-M-L ] :
    • the length of [ Line-Segment D-C ] is [ 0.866025 * S ] ;

  • Based on the equilateral triangle [ Triangle B-L-M ] :
    • the length of [ Line-Segment B-C ] is [ 0.866025 * S ] ;

  As a result , we now have this set of 3 diagrams below ,

  • with the diagram in-the-middle showing the lengths for the 6 line-segments in the [ Cross-Section Planar View ] .

  

• Let us now also identify , for our calculation puposes below :
  • the point [ Point G ] as the mid-point of [ Line-Segment A-B ] ,

    and

  • the point [ Point H ] as the mid-point of [ Line-Segment E-D ] ;
    • as shown in the diagram on-the-left below .

  

  Let us now also create 2 new line-segments :

  • [ Line-Segment G-H ] and [ Line-Segment F-C ] respectively ,
    • as marked-off in [ red-color ] in the diagram in-the-middle above .

  And we take note here that :

  • [ Line-Segment G-H ] is a line joining the mid-points of a pair of [ Opposite Edges ] ,
    • namely [ Edge A-B ] and [ Edge E-D ] respective ;

  • [ Line-Segment F-C ] is also a line joining the mid-points of a pair of [ Opposite Edges ] ,
    • namely [ Edge J-K ] and [ Edge M-L ] respective .

  And based on this information , we can now come to the conclusion that :

  • First :
    • [ Line-Segment G-H ] and [ Line-Segment C-F ] are of-equal-length ,

      i.e. :

      • 

        arising directly from the symmetric nature of the Icosahedron .

      And that-is-to-say :

      • the [ mid-point to mid-point distance ] for any pair of [ Oppiste Edges ] is invariant-and-constant throughout the Icosahedron ,
        • regardless of which particlular pair of [ Opposite Edges ] is selected .

  • Second :
    • [ Line-Segment G-H ] and [ Line-Segment C-F ] do intersect one-another at-right-angles and at their mid-points .

      This is because :

      • [ Line-Segment G-H ] is an [ Axis of Bi-Symmetry ] and :
        • when the Icosahedron is rotated thru an angle of 180 degrees using [ Line-Segment G-H ] as the axis-of-rotation ,
          • [ Point F ] and [ Point C ] are a pair of corresponding points exchanging positions .

      • [ Line-Segment F-C ] is also an [ Axis of Bi-Symmetry ] and :
        • when the Icosahedron is rotated thru an angle of 180 degrees using [ Line-Segment F-C ] as the axis-of-rotation ,
          • [ Point G ] and [ Point H ] are a pair of corresponding points exchanging positions .

  • Third :
    • We can now state that :
      • [ Line-Segment A-B ] intersects [ Line-Segment G-H ] at right-angles ,

        and

      • [ Line-Segment E-D ] also interesects [ Line-Segment G-H ] at right-angles .

      And this is because :

      • [ Line-Segment G-H ] is an [ Axis of Bi-Symmetry ] and :
        • when the Icosahedron is rotated thru an angle of 180 degrees using [ Line-Segment G-H ] as the axis-of-rotation ,
          • [ Point A ] and [ Point B ] are a pair of corresponding points exchanging positions ;

            and

          • [ Point E ] and [ Point D ] are also a pair of corresponding points exchanging positions .

• Let us now summarise our findings thus far with this set of 3 diagrams below :

  

  And with this information in-hand ,

  • we are now ready to do the calculations for the [ distance from Centroid to Nodal-Point ] for the Icosahedron , next .

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Calculating the distance from Centroid to Nodal-Point :

• Let us now bring-in this set of 3 diagrams below and :
  • identify [ angle G-O-B ] as the unknown angle [ Angle PHI ] that we would like to solve for ,
    • as per the diagram in-the-midle below .

  

  And for this diagram in-the-middle , let us just state that :

  • 

    arising from :

    • 

      with [ H-sub-Y ] and [ H-sub-Z ] being the half-distances of [ L-sub-Y ] and [ L-sub Z ] respectively .

• Let us now drop a perpendicular line from [ Point B ] to the [ Line-Segment F-C ] ,
  • as per the diagram on-the-right above ;

  so that :

  • the said Line intersects [ Line-Segment F-C ] at the point [ Point P ] as shown .

• Let us now set up the unknown variable [ u ] , being the length of [ Line-Segmemt C-P ] .

  We then have immediately this relation :

  • 

  And since we have already established immediately above :

  • 

    it then follows that :

    • 

• Let us now write-out this Pythagorus Equation for the right-angle-triagle [ Triangle B-P-C ] :
  • 

    where :

    • 

  And we recall , at this point , that :

  • 

  Substituting this and the value for [ H-sub-Z ] into the Pythagorus Equation above then yield us this :

  • 

  Consequently :

  • 

• And we take note here that :
  • the above equation is a Quadratic Equation based on [ S ] .

  The roots to the equation here are then :

  • 

    yielding :

  • 

    or :

  • 

  For our evaluation purposes here ,

  • we would be interested only in the positive root of [ u ] .

  Consequently , we take :

  • 

    or :

  • 

  And on numerical evaluation, we have :

  • 

• Let us now do the calculations for the [ tangent of PHI ] , and we have :
  • 

    based on the diagram in-the-middle above .

  Since we have already estabished above that :

  • 

  we can now write :

  • 

    or :

  • 

  And on numerical evaluation :

  • 

  Consequently , the value of [ PHI ] is :

  • 

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Constructing our Standard Model for the Icosahedron :

• Let us now construct our Standard Model for the Icosahedron based on this criterion :
  • 

    as established above .

  

  And the { X-Y-Z Co-ordinates } for the 20 Nodal-Points in our Standard Model for the Dodecahedron

  • are then as follows :

    Nodal Point I.D.		Co-Ordinates				Distance from Centroid to Nodal-Point		Comments
    		X	Y	Z
    ********	*	***********	***********	***********	*	***********	*	***************
    A	*	+1.000000	+0.618034	0.000000	*	1.175571	*	forms Edge A-B
    B	*	+1.000000	-0.618034	0.000000	*	1.175571	*
    C	*	-1.000000	-0.618034	0.000000	*	1.175571	*	forms Edge C-D
    D	*	-1.000000	+0.618034	0.000000	*	1.175571	*
    E	*	0.000000	+1.000000	+0.618034	*	1.175571	*	forms Edge E-F
    F	*	0.000000	+1.000000	-0.618034	*	1.175571	*
    G	*	0.000000	-1.000000	-0.618034	*	1.175571	*	forms Edge G-H
    H	*	0.000000	-1.000000	+0.618034	*	1.175571	*
    J	*	+0.618034	0.000000	+1.000000	*	1.175571	*	forms Edge J-K
    K	*	-0.618034	0.000000	+1.000000	*	1.175571	*
    L	*	-0.618034	0.000000	-1.000000	*	1.175571		forms Edge L-M
    M	*	+0.618034	0.000000	-1.000000	*	1.175571	*

• We then have this set of 3 diagrams below ,
  • showing the positions of the 12 Nodal-Points , i.e. :
    • Nodal-Points [ A ] thru [ M ] , forming 6 edges .

  

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go to the next section : Section VII - A Quick Review of the Rhombus

go to the last section : Section V - A Quick Review of the Equilateral Triangle

return to the The 30-Rhombus Symmetric Solid Object HomePage

Original dated 2013-2-04