On the 30-Rhombus Symmetric Solid Object

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

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Section IV - Structural Dimensioning of the Dodecahedron

Summary for the Section :

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

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Bringing-in the Dodecahedron :

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

  

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Setting up the Cube within the Dodecahedron :

• Let recall that there are exactly 20 Nodal-Points associated with the Dodecahedron and :
  • we shall now select a specific set of eight (8) Nodal-Points , so that :
    • the said 8 Nodal-Points do form a cube .

  

• Let us quickly state why the said formation , as shown above , is necessarily a Cube .
  • First :
    • For the 12 edge-lines of the said Cube :
      • each edge-line is in fact a cross-diagonal of one of the 12 Pentagons , and :
        • the 12 pentagons here are identical .

      Therefore , the 12 edge-lines here are necessarily of equal-length .

  • Second :
    • Each Straight-Line passing thru an opposite pair of Nodal-Points is always an [ Axis of Tri-Symmetry ] .

  

  As such :

  • Each of the three (3) Pentagons sharing the same Nodal-Point are always ranked on par with respect to one-another ;
    • as per the diagram on-the-left below.

  

  Consequently ,

  • each set of 3 corresponding Cross-Diagonals with one Cross-Diagonal residing in each of the 3 Pentagons ,
    • must necessarily be also ranked-on-par with respect to one-another ;
      • as per the diagram in-the-middle and on-the-right above .

  As a result :

  • The [ intersecting angles ] for the 3 corresponding Cross-Diagonals at the said Nodal-Point , when taken in pairs ,
    • are always equal .
  • Third :
    • Since the Tri-Symmetric Effect depicted immediately above is applicable in replica to all 20 Nodal-Points ,
      • and in particular , our [ 8 Nodal-Points in question ] ;

      we can now safely conclude that :

      • all [ intersecting angles ] for corresponding Cross-Diagonals in pairs at the 8 Nodal-Points are indeed the same .

• And based on the above , we can now conclude that :
  • our [ 8 Nodal-Points Structural Frame ] in-question here is indeed a Cube .

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Calculating the distance from Centroid to Nodal-Point :

• Now that we have established the [ Cube ] within the Dodecahedron ,
  • we can now use :
    • the distance from the [ Centroid of the Cube ] to any [ one of its 8 Nodal-Points ] ,

    to estimate :

    • the distance from the [ Centroid of the Dodecahedron ] to any [ one of its 20 Nodal-Points ] ;

      which in effect are the one-and-the-same distance .

  

• Let us now denote the distance from the [ Centroid ] to a [ Nodal-Point ] as the distance [ rho ] ,
  • and we have , based on the configuration of the [ cube ] :
    • 

  Since we have already established that :

  • each side of the [ Cube ] is a Cross-Diagonal of identical Pentagons ;

  we can now write :

  • 

    where :

    • 

• Let us recall that we have previously established this relation for the Pentagon in the last Section :
  • 

    where :

    • [ R ] is the distance from the [ Centroid of the Pentagon ] to any [ one of its 5 Nodal-Points ] .

  Thus , we can now express [ rho ] in terms of [ R ] via this equation below :

  • 

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Dimensioning the Dodecaherdon :

• Let us now bring-in this set of 3 diagrams below , and identify :
  • the nodal-points [ Point A ] and [ Point B ] as 2 nodal-points forming an edge ,
    • as per the diagram on-the-left below ;

    and

  • the point [ Point O ] as the [ Centroid of the Dodecahedron ] ,
    • as per the diagram in-the-middle below .

  

  The diagram on-the-right above then gives the planar view of isosceles triangle [ Triangle A-O-B ] ;

  • and we shall now mark-off the point [ Point M ] as the mid-point of [ Line-Segment A-B ] , as shown .

• We shall now attempt to solve for the value of [ Angle PSI ] ,
  • being the angle [ Angle A-O-M ] as shown above .

  Let us now denote the distance of [ Line-Segment O-M ] as the unknown distance [ L-sub-U ] ,

  • i.e. :
    • 

  Then , on applying the Pythagorus Theorem to the right-angle-triangle [ Triange A-M-O ] ,

  • we can now write :
    • 

      yielding :

    • 

• Let us bring-in , at this point , this relation from the last Section on the Pentagon ,
  • i.e. :
    • 

      where :

      • 

      • [ R ] is the distance from the [ Centroid of the Pentagon ] to any [ one of its 5 Nodal-Points ] .

  Substituting this and our expression for [ rho ] above into the expression for [ L-sub-U ] immediately above ,

  • we can now write :
    • 

      yielding :

    • 

• Let us now bring-in this expression for the [ tangent of PSI ] ,
  • based on the Planar Diagram on-the-right above :
    • add AM over OM .

  Substituting therein the expressions we have already established for [ L-sub-S ] and [ L-sub-U ] above ,

  • we then have :
    • 

      yielding :

    • 

• Let us now bring-in this expression for the [ Sine of 36 degrees ] :
  • 

  Substituting this into the expression immediately preceeding for the [ tangent of PSI ]

  • then yield us this :
    • 

      yielding :

    • 

  And finally this expression for the [ Tangent of PSI ] :

  • 

• And upon full numerical evaluation of the above , we have :
  • 

  Consequently :

  • 

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Constructing our Standard Model for the Dodecahedron :

• Let us now construct our Standard Model for the Dodecahedron 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	0.000000	0.381966	*	1.070466	*	forms Edge A-B
    B	*	+1.000000	0.000000	-0.381966	*	1.070466	*
    C	*	-1.000000	0.000000	-0.381966	*	1.070466	*	forms Edge C-D
    D	*	-1.000000	0.000000	+0.381966	*	1.070466	*
    E	*	+0.381966	+1.000000	0.000000	*	1.070466	*	forms Edge E-F
    F	*	-0.381966	+1.000000	0.000000	*	1.070466	*
    G	*	-0.381966	-1.000000	0.000000	*	1.070466	*	forms Edge G-H
    H	*	+0.381966	-1.000000	0.000000	*	1.070466	*
    J	*	0.000000	+0.381966	+1.000000	*	1.070466	*	forms Edge J-K
    K	*	0.000000	-0.381966	+1.000000	*	1.070466	*
    L	*	0.000000	-0.381966	-1.000000	*	1.070466		forms Edge L-M
    M	*	0.000000	+0.381966	-1.000000	*	1.070466	*
    N1	*	+0.618034	+0.618034	+0.618034	*	1.070466	*	8 Nodal-Points forming a Cube
    N2	*	+0.618034	+0.618034	-0.618034	*	1.070466	*
    N3	*	+0.618034	-0.618034	-0.618034	*	1.070466	*
    N4	*	+0.618034	-0.618034	+0.618034	*	1.070466	*
    N5	*	-0.618034	+0.618034	-0.618034	*	1.070466	*
    N6	*	-0.618034	+0.618034	+0.618034	*	1.070466	*
    N7	*	-0.618034	-0.618034	+0.618034	*	1.070466	*
    N8	*	-0.618034	-0.618034	-0.618034	*	1.070466	*

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

  

• We also have this set of 3 diagrams below ,
  • showing the positions of the remaining 8 Nodal-Points , i.e. :
    • Nodal Points [ N1 ] thru [ N8 ] , forming a Cube .

  

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

go to the last section : Section III - A Quick Review of the Pentagon

return to the 30-Rhombus Symmetric Solid Object HomePage

Original dated 2013-2-04