On Symmetry Through-put linking 3-4-5-6-7 Dimension's ,
Octo-Symmetry for the { 4-Dimension Quasi-Octahedron } ,
and 2 Identical { 8-D Structures } oriented Symmetrically

by Frank C. Fung ( 1st published in July, 2017. )

Top Of Page

Section XVI - The Octo-Symmetric Nature of the { 4-D Quasi-Octahedron }

Summary for the Section :

• In this Section , we shall :
  • construct the 96-triangles { 4-Dimension Quasi-Octahedron } ,

    and

    point out the very special Octo-Symmetric nature of the { 4-Dimension Quasi-Octahedron } .

go to Top Of Page

Recalling the 24 Squares of the { 4-D Quasi-Cube } :

• Let us recall that :
  • we have identified the 24 Squares for the { 4-Dimension Quasi-Cube } in Section IX ,
    • and the co-ordinates of the Center-Points of 24 Squares are :

    Square I.D.	* * *	Position Vector for Center-Point of the Square					* * *	Square I.D.	* * *	Position Vector for Center-Point of the Square
    	*	Vector I.D.	Co-ordinates				*		*	Vector I.D.	Co-ordinates
    *******************	*	***********	***	***	***	***	*	*******************	*	***********	***	***	***	***
    Square A1-B1-A3-B3		Vector D1	+1	+1	0	0	*	Square A2-B2-A4-B4		Vector D2	-1	-1	0	0
    Square A5-B5-A7-B8		Vector D3	+1	-1	0	0	*	Square A6-B6-A8-B7		Vector D4	-1	+1	0	0
    Square A5-B2-A8-B3		Vector D5	0	0	-1	+1	*	Square A6-B1-A7-B4		Vector D6	0	0	+1	-1
    Square A1-B5-A4-B7		Vector D7	0	0	+1	+1	*	Square A2-B6-A3-B8		Vector D8	0	0	-1	-1
    *******************	*	***********	***	***	***	***	*	*******************	*	***********	***	***	***	***
    Square A1-B3-A5-B5		Vector E1	+1	0	0	+1	*	Square A2-B4-A6-B6		Vector E2	-1	0	0	-1
    Square A3-B1-A7-B8		Vector E3	+1	0	0	-1	*	Square A4-B2-A8-B7		Vector E4	-1	0	0	+1
    Square A3-B3-A8-B6		Vector E5	0	+1	-1	0	*	Square A4-B4-A7-B5		Vector E6	0	-1	+1	0
    Square A1-B1-A6-B7		Vector E7	0	+1	+1	0	*	Square A2-B2-A5-B8		Vector E8	0	-1	-1	0
    *******************	*	***********	***	***	***	***	*	*******************	*	***********	***	***	***	***
    Square A3-B3-A5-B8		Vector F1	+1	0	-1	0	*	Square A4-B4-A6-B7		Vector F2	-1	0	+1	0
    Square A1-B1-A7-B5		Vector F3	+1	0	+1	0	*	Square A2-B2-A8-B6		Vector F4	-1	0	-1	0
    Square A1-B3-A8-B7		Vector F5	0	+1	0	+1	*	Square A2-B4-A7-B8		Vector F6	0	-1	0	-1
    Square A3-B1-A6-B6		Vector F7	0	+1	0	-1	*	Square A4-B2-A5-B5		Vector F8	0	-1	0	+1

• It is our contention here that :
  • the 24 nodal-points associated with the 24 Position Vectors in the above table are , in fact :
    • the 24 nodal-points for the 96-triangles { 4-Dimension Octahedron } ,
      • with the centroid of each of the 96 triangles being equi-distant from the Point-Of-Origin .

go to Top Of Page

Using { Set A / B / C } instead of { Set D / E / F } for our construction :

• Let us recall that 24 Position Vectors above belongs to :
  • { Set D } , { Set E } and { Set F } ;
    • and the length of each of the 24 Position Vectors above is the { Square-root of [ 2 ] } .

  For our construction of the 96-triangles { 4-Dimension Quasi-Octahedron } to follow :

  • it would be more convenient for us to use :
    • { Set A } , { Set B } and { Set C } instead ;
      • with the length of each of the 24 Position Vectors here being equal to [ 2 ] ;

    noting here of course that :

    • the relative positioning of { Set D } / { Set E } / { Set F } within the group

      are essentially the same as

    • the relative positioning of { Set A } / { Set B } / { Set C } within the group
      • as previously discussed in Section XII , Section XIII and Section XIV .

go to Top Of Page

Constructing the 96-triangles { 4-Dimension Quasi-Octahedron } :

• Let us now bring-in :
  • the 24 Position Vectors for { Set A / B / C } .

    Line I.D.	* * *	Position Vector for 1st end-point of the Line					* * *	Position Vector for 2nd end-point of the Line
    	*	Vector I.D.	Co-ordinates				*	Vector I.D.	Co-ordinates
    *********************	*	***********	***	***	***	***	*	***********	***	***	***	***
    Cross-Diagonal Line 1	*	Vector A1	+1	+1	+1	+1	*	Vector A2	-1	-1	-1	-1
    Cross-Diagonal Line 2	*	Vector A3	+1	+1	-1	-1	*	Vector A4	-1	-1	+1	+1
    Cross-Diagonal Line 3	*	Vector A5	+1	-1	-1	+1	*	Vector A6	-1	+1	+1	-1
    Cross-Diagonal Line 4	*	Vector A7	+1	-1	+1	-1	*	Vector A8	-1	+1	-1	+1
    *********************	*	***********	***	***	***	***	*	***********	***	***	***	***
    Cross-Diagonal Line 5	*	Vector B1	+1	+1	+1	-1	*	Vector B2	-1	-1	-1	+1
    Cross-Diagonal Line 6	*	Vector B3	+1	+1	-1	+1	*	Vector B4	-1	-1	+1	-1
    Cross-Diagonal Line 7	*	Vector B5	+1	-1	+1	+1	*	Vector B6	-1	+1	-1	-1
    Cross-Diagonal Line 8	*	Vector B7	-1	+1	+1	+1	*	Vector B8	+1	-1	-1	-1
    *********************	*	***********	***	***	***	***	*	***********	***	***	***	***
    Axial-Direction Line 1	*	Vector C1	+2	0	0	0	*	Vector C2	-2	0	0	0
    Axial-Direction Line 2	*	Vector C3	0	+2	0	0	*	Vector C4	0	-2	0	0
    Axial-Direction Line 3	*	Vector C5	0	0	+2	0	*	Vector C6	0	0	-2	0
    Axial-Direction Line 4	*	Vector C7	0	0	0	+2	*	Vector C8	0	0	0	-2

• For the construction of the 96-triangles { 4-Dimension Quasi-Octahedron } ,
  • let us now select 8 { Anchor Nodal-Points } as the core base-points for our construction ,

    namely :

    • { Point C1 } , { Point C2 } being the 2 end-points of { Axial-Direction Line 1 } ,
    • { Point C3 } , { Point C4 } being the 2 end-points of { Axial-Direction Line 2 } ,
    • { Point C5 } , { Point C6 } being the 2 end-points of { Axial-Direction Line 3 } , and
    • { Point C7 } , { Point C8 } being the 2 end-points of { Axial-Direction Line 4 } .

  We shall now attempt to identify a pair of nodal-points :

  • with one nodal-point from { Set A } and the other nodal-point from { Set B } ;

    so that :

    • the 2 Position Vectors for the 2 said Nodal-Points do intersect one-another at the angle of [ 60 degrees ] ;

      and

    • each of the 2 said Position Vectors above will also intersect the Position Vector for the { Anchor Nodal-Point in question }
      • at the angle of [ 60 degrees ] ;

    noting here that :

    • whenever 3 position-vectors intersect one-another at the angle of [ 60 degrees ] ,
      • the end-points of the 3 said position-vectors do form an equilateral triangle .

  And the results are then as shown in this table below :

  • with the successful pairs of Nodal-Points being marked-off with a capital [ X ] .

  Anchor Nodal-Point in Question		* * *	32 Pairs of Nodal-Points with the Position Vectors for each pair of Nodal-Points intersecting one-another at the angle of [ 60 degrees ]																																			* * *
  Nodal- Point I.D.	Position Vector for Nodal-Point		A 1 * B 7	A 2 * B 8	A 3 * B 6	A 4 * B 5	A 5 * B 2	A 6 * B 1	A 7 * B 4	A 8 * B 3	* * *	A 1 * B 5	A 2 * B 6	A 3 * B 8	A 4 * B 7	A 5 * B 3	A 6 * B 4	A 7 * B 1	A 8 * B 2	* * *	A 1 * B 3	A 2 * B 4	A 3 * B 1	A 4 * B 2	A 5 * B 5	A 6 * B 6	A 7 * B 8	A 8 * B 7	* * *	A 1 * B 1	A 2 * B 2	A 3 * B 3	A 4 * B 4	A 5 * B 8	A 6 * B 7	A 7 * B 5	A 8 * B 6
  ********	**********	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*
  Point C1	Vector C1	*									*	X		X		X		X		*	X		X		X		X		*	X		X		X		X		*
  Point C2	Vector C2	*									*		X		X		X		X	*		X		X		X		X	*		X		X		X		X	*
  ********	**********	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*
  Point C3	Vector C3	*	X		X			X		X	*									*	X		X			X		X	*	X		X			X		X	*
  Point C4	Vector C4	*		X		X	X		X		*									*		X		X	X		X		*		X		X	X		X		*
  ********	**********	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*
  Point C5	Vector C5	*	X			X		X	X		*	X			X		X	X		*									*	X			X		X	X		*
  Point C6	Vector C6	*		X	X		X			X	*		X	X		X			X	*									*		X	X		X			X	*
  ********	**********	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*	**	**	**	**	**	**	**	**	*
  Point C7	Vector C7	*	X			X	X			X	*	X			X	X			X	*	X			X	X			X	*									*
  Point C8	Vector C8	*		X	X			X	X		*		X	X			X	X		*		X	X			X	X		*									*

  And we do observe here that :

  • For each of the 8 { Anchor Nodal-Points } ,
    • we have now identified 12 pairs of points
      • with each pair hosting one point from { Set A } and one point from { Set B } ,

      to form 12 equilateral triangles associated with the said { Anchor Nodal-Point in-question }

      • with :
        • the said { Anchor Nodal-Point in-question } serving as the 3rd-and-final nodal-point for each of the 12 triangles .

  Thus , we now have 96 distinct equilateral triangles ,

  • i.e. : [ 96 ] = 8 x 12 .

    because none of the 96 triangles are duplicated .

• And for the sake of clarity and precision ,
  • let us now list-out the 96 triangles of the { 4-Dimension Quasi-Octahedron } .

  The 96 Triangles are then :

  • For the { Anchor Nodal-Point C1 } , the 12 triangles are :
    • { C1-A1-B5 } , { C1-A3-B8 } , { C1-A5-B3 } , { C1-A7-B1 } ,
      { C1-A1-B3 } , { C1-A3-B1 } , { C1-A5-B5 } , { C1-A7-B8 } ,
      { C1-A1-B1 } , { C1-A3-B3 } , { C1-A5-B8 } , { C1-A7-B5 } .

  • For the { Anchor Nodal-Point C2 } , the 12 triangles are :
    • { C2-A2-B6 } , { C2-A4-B7 } , { C2-A6-B4 } , { C2-A8-B2 } ,
      { C2-A2-B4 } , { C2-A4-B2 } , { C2-A6-B6 } , { C2-A8-B7 } ,
      { C2-A2-B2 } , { C2-A4-B4 } , { C2-A6-B7 } , { C2-A8-B6 } .

  • For the { Anchor Nodal-Point C3 } , the 12 triangles are :
    • { C3-A1-B7 } , { C3-A3-B6 } , { C3-A6-B1 } , { C3-A8-B3 } ,
      { C3-A1-B3 } , { C3-A3-B1 } , { C3-A6-B6 } , { C3-A8-B7 } ,
      { C3-A1-B1 } , { C3-A3-B3 } , { C3-A6-B7 } , { C3-A8-B6 } .

  • For the { Anchor Nodal-Point C4 } , the 12 triangles are :
    • { C4-A2-B8 } , { C4-A4-B5 } , { C4-A5-B2 } , { C2-A7-B4 } ,
      { C4-A2-B4 } , { C4-A4-B2 } , { C4-A5-B5 } , { C2-A7-B8 } ,
      { C4-A2-B2 } , { C4-A4-B4 } , { C4-A5-B8 } , { C2-A7-B5 } .

  • For the { Anchor Nodal-Point C5 } , the 12 triangles are :
    • { C5-A1-B7 } , { C5-A4-B5 } , { C5-A6-B1 } , { C5-A7-B4 } ,
      { C5-A1-B4 } , { C5-A4-B1 } , { C5-A6-B5 } , { C5-A7-B7 } ,
      { C5-A1-B1 } , { C5-A4-B4 } , { C5-A6-B7 } , { C5-A7-B5 } .

  • For the { Anchor Nodal-Point C6 } , the 12 triangles are :
    • { C6-A2-B8 } , { C6-A3-B6 } , { C6-A5-B2 } , { C6-A8-B3 } ,
      { C6-A2-B3 } , { C6-A3-B2 } , { C6-A5-B6 } , { C6-A8-B8 } ,
      { C6-A2-B2 } , { C6-A3-B3 } , { C6-A5-B8 } , { C6-A8-B6 } .

  • For the { Anchor Nodal-Point C7 } , the 12 triangles are :
    • { C7-A1-B7 } , { C7-A4-B5 } , { C7-A5-B2 } , { C7-A8-B3 } ,
      { C7-A1-B5 } , { C7-A4-B7 } , { C7-A5-B3 } , { C7-A8-B2 } ,
      { C7-A1-B3 } , { C7-A4-B2 } , { C7-A5-B5 } , { C7-A8-B7 } .

  • For the { Anchor Nodal-Point C8 } , the 12 triangles are :
    • { C8-A2-B8 } , { C8-A3-B6 } , { C8-A6-B1 } , { C8-A7-B4 } ,
      { C8-A2-B6 } , { C8-A3-B8 } , { C8-A6-B4 } , { C8-A7-B1 } ,
      { C8-A2-B4 } , { C8-A3-B1 } , { C8-A6-B6 } , { C8-A7-B8 } .

go to Top Of Page

The Octo-Symmetric Nature of the { 4-D Quasi-Octahedron } :

• It is our contention here that :
  • the structural shape for each { 12-Triangles Structure associated with an [ Anchor Nodal-Point ] } is the same ,

    and therefore :

    • the 96-triangles { 4-Dimension Quasi-Octahedron } is deemed to be Octo-Symmetric in nature .

  But more confirmation on this matter later in the next Section ,

  • when we do the structural analysis on the { 4-Dimension Quasi-Octahedron } .

go to Top Of Page

Some Features of the { 4-D Quasi-Octahedron } :

• The { 4-Dimension Quasi-Octahedron } therefore has :
  • 24 nodal-points ,

  • 96 sides or edges comprising of :
    • 32 lines arising from pairing { Set A / Set B nodal-points } ,
    • 32 lines arising from pairing { Set C / Set A nodal-points } , and
    • 32 lines arising from pairing { Set C / Set B nodal-points } ;

    and

  • 96 triangles .

go to Top Of Page

What's Next :

• Let us do the Structural Analysis on the 96-triangles { 4-Dimension Quasi-Octahedron } , next .

go to Top Of Page

go to the next Section : Section XVI - Structural Analysis for the { 4-D Quasi-Octahedron }

go to the last Section : Section XV - A Special Issue on Swapping { 4 Orthogonal-Lines Structures }

return to the Symmetry Through-put HomePage

Original dated 2017-7-23