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Is there a formula to determine which polyhedra will tessellate in 3D without any spaces?

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without any spaces? – jspecter Oct 14 '11 at 14:10
without voids between the forms. i.e. perfectly tessellate. – Carly Mallon Oct 14 '11 at 14:37

You can find a "survey" of information and references about space filling polyhedra here:

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what is the point of posting links? – abel Feb 6 '15 at 1:47

There were no formulas until I just developed them. they are based on closest packing of unit radius spheres and synergetics by R. Buckminster Fuller.

Fuller discovered that the cubo octahedron aka the vector equilibrium could be created by closest packing unit radius (same size) spheres in accord with the formula 10F to the 2nd power + 2.

This gives the total of the outer layer of closest packed spheres in the VE (vector equilibrium).

For the total of all the layers of closest packed spheres we use the formula I developed.

It is 20F to the 3rd power + 30F to the 2nd power + 10F, and then you divide this total by 6 and add 2F + 1.

F = Frequency and relates to the number of layers in the Omni-directionally expanding agglomeration of closest packed unit radius spheres.

When all of the spheres are replaced by closest packed rhombic dodecahedron we may continue to expand these closest packed Rhombic dodecahedrons ad infinitum. The rohombic dodecahedron is an all-space filling polyhedron.

This formula is in my copyrighted book which is not published yet. It is called "Symmetry, Prime Number Patterns and the Unified Field of Einstein."

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