Here's a new Jupyter Notebook undertaking to compare the 4D IVM tetravolume measuring system vs-a-vis the 3D XYZ cubic volume system we all learn in school.

By 4D I don't mean "four mutual perpendiculars" but four rays from the origin through the corners of the topologically minimum container, the most primitive "cage" enclosing volume, a "simplex" of Euclidean dimension: Dim=3.

Think of the 4D as a "branding mark" signalling a different approach to 3rd powering. The Notebook expresses this difference in a way similar to the way Bucky does in

Imagine two sticks of varying length with a common origin, at any angle to start, and imagine an operation that connected the two tips with a 3rd segment, thereby marking off the area within.

This is "closing the lid" in the case of a triangle (Dim=2). However, leaving the angle unspecified fails to nail down the numeric results of the computation. We need to decide on a fixed angle to always get the same result.

Fixing the included angle defines a specific triangle (SAS).

In XYZ multiplication, that included angle is 90 degrees, upon which, after we "close the lid", we then multiply by 2, placing two of the resulting triangles on opposing sides of a shared hypotenuse, thereby defining a rectangle.

That's our standard algorithm for area. |A||B| is a rectangle, a square if |A| = |B|. We all know this cold by elementary school.

In the IVM (different scaffolding), the included angle is 60 degrees and upon closing the lid, we do not proceed further to double the resulting area.

That internal original area, of the triangle, is simply defined to be |A| x |B|, where |A| and |B| are the stick lengths. |A||B| is an equilateral triangle if |A| = |B|.

Moving on to volume (Dim=3)...

In XYZ we start with 3 sticks from a common origin (0,0,0) fanning out in a mutually orthogonal arrangement. Closing the lid begets a right tetrahedron that is only 1/6th of the total volume we come up with, by forming a rectilinear parallelepiped from these sticks.

|A||B||C| is a brick, maybe a cube.

In the IVM, three sticks from a common origin fan out along the imaginary edges of a regular tetrahedron, which establishes relative direction.

The magnitudes are variable, so the resulting tetrahedron may not be regular. The angles at this corner remain fixed, as they do in the XYZ case.

We simply "close the lid" and call that the volume. The numbers stay the same. 2 x 4 x 5 = 40, same as before, it's just that 40 is tetrahedron-shaped, and obtained from adding three sticks to the initial three, thereby forming a six-edged, four faceted shape, not a hexahedron as before.[1]

How might we bridge these two operations?

Taking our cue from Bucky, we construct a tetrahedron from four unit-radius balls, closest-packed, edges 1/2 D where D = ball diameter. Saying the edges are 1 and the volume is 1, makes the cube in which said tetrahedron inscribes = 3 tetravolumes.

That's a fixed and known relationship: the tetrahedron inscribed in a cube as face diagonals has 1/3rd the volume (this generalizes to any parallelepided).

However in XYZ this same cube of edges √2 (in R units) and face diagonals 2, will have a volume of 2.828427... or √2 √2 √2 i.e. "√2 cubed".

So there's our conversion constant: 3/2.828.. or √(9/8).

Another way to think about it: the XYZ scaffolding or matrix consists of unit cubes of edges R. It wasn't originally developed with sphere packing in mind (unlike the IVM, which is the scaffolding associated with the FCC or CCP).

The unit R (2R = D) gives us R * R * R = 1 or R-cubed in XYZ (cube shaped), whereas D * D * D givers us 1 in the IVM (tetrahedron shaped).

Using our conversion constant is like converting between currencies or energy values (joules, calories) where the R-edged XYZ cube is about 6% bigger than the D-edged IVM tetrahedron. The two units of volume are somewhat close (like the Canadian and American dollar) but not the same.

If you know the volume in XYZ cubic units, multiply by √(9/8) to get the same volume in IVM tetrahedron units, or use √(8/9) to go the other way.

Why bother? What's the pay-off?

The tetrahedron of edges D divides thrice into a cube of face diagonals D (as we've seen) four times into an octahedron of edges D (its space-filling complement in the IVM), six times into a rhombic dodecahedron (space-filling CCP ball encasement), twenty times into a cuboctahedron (from 12-balls-around-1 and connecting corners).

The five-fold symmetric Icosahedron, Pentagonal Dodecahedron, and Rhombic Triacontahedron (RT) all fit in here as well, incommensurably in most cases.

The Jitterbug Transformation is used to connect the Cuboctahedron of edges D to a corresponding icosahedron of edges D.

Said Icosahedron of volume ~18.51, combined with its dual, define a rhombic triacontahedron that, if shrunk down by by 1/Phi (linearly) gives the RT of 120 E-mods (60 left, 60 right-handed).

The RT sharing vertexes with the RD of volume 6 has volume 7.5 exactly (IVM tetravolumes) and shrunk by 2/3 gives the RT of volume 5 exactly, and the 120 T-mods, same shape as the E-mods but a tad smaller, by about.001%.

What David Koski does is phi-scale the E-mods to express volumes as a linear combination of mods of different size (same shape).

Quoting from Koski's paper:

Quoting again from his paper [2]:

That's about all the mathematics one needs to know, to understand about tetravolumes. It's not that hard.

We also get to think again about foundational matters, such as what basic assumptions we might vary to produce interesting new flavors of mathematics.

[1] Just for fun, lets compute the lengths of the "lid" edges and feed these numbers to our Pythonic tetrahedron volume computer, same one as in the Jupyter Notebook. Given edges a=2, b=4, c=5 I get d=3.46410, e=4.5825, f=8943540673 for a volume of...

[2] http://4dsolutions.net/synergetica/RevisitingS&Emodules.pdf

By 4D I don't mean "four mutual perpendiculars" but four rays from the origin through the corners of the topologically minimum container, the most primitive "cage" enclosing volume, a "simplex" of Euclidean dimension: Dim=3.

Think of the 4D as a "branding mark" signalling a different approach to 3rd powering. The Notebook expresses this difference in a way similar to the way Bucky does in

*Synergetics*982.44-47.Imagine two sticks of varying length with a common origin, at any angle to start, and imagine an operation that connected the two tips with a 3rd segment, thereby marking off the area within.

This is "closing the lid" in the case of a triangle (Dim=2). However, leaving the angle unspecified fails to nail down the numeric results of the computation. We need to decide on a fixed angle to always get the same result.

Fixing the included angle defines a specific triangle (SAS).

In XYZ multiplication, that included angle is 90 degrees, upon which, after we "close the lid", we then multiply by 2, placing two of the resulting triangles on opposing sides of a shared hypotenuse, thereby defining a rectangle.

That's our standard algorithm for area. |A||B| is a rectangle, a square if |A| = |B|. We all know this cold by elementary school.

In the IVM (different scaffolding), the included angle is 60 degrees and upon closing the lid, we do not proceed further to double the resulting area.

That internal original area, of the triangle, is simply defined to be |A| x |B|, where |A| and |B| are the stick lengths. |A||B| is an equilateral triangle if |A| = |B|.

Moving on to volume (Dim=3)...

In XYZ we start with 3 sticks from a common origin (0,0,0) fanning out in a mutually orthogonal arrangement. Closing the lid begets a right tetrahedron that is only 1/6th of the total volume we come up with, by forming a rectilinear parallelepiped from these sticks.

|A||B||C| is a brick, maybe a cube.

In the IVM, three sticks from a common origin fan out along the imaginary edges of a regular tetrahedron, which establishes relative direction.

The magnitudes are variable, so the resulting tetrahedron may not be regular. The angles at this corner remain fixed, as they do in the XYZ case.

We simply "close the lid" and call that the volume. The numbers stay the same. 2 x 4 x 5 = 40, same as before, it's just that 40 is tetrahedron-shaped, and obtained from adding three sticks to the initial three, thereby forming a six-edged, four faceted shape, not a hexahedron as before.[1]

How might we bridge these two operations?

Taking our cue from Bucky, we construct a tetrahedron from four unit-radius balls, closest-packed, edges 1/2 D where D = ball diameter. Saying the edges are 1 and the volume is 1, makes the cube in which said tetrahedron inscribes = 3 tetravolumes.

That's a fixed and known relationship: the tetrahedron inscribed in a cube as face diagonals has 1/3rd the volume (this generalizes to any parallelepided).

However in XYZ this same cube of edges √2 (in R units) and face diagonals 2, will have a volume of 2.828427... or √2 √2 √2 i.e. "√2 cubed".

So there's our conversion constant: 3/2.828.. or √(9/8).

Another way to think about it: the XYZ scaffolding or matrix consists of unit cubes of edges R. It wasn't originally developed with sphere packing in mind (unlike the IVM, which is the scaffolding associated with the FCC or CCP).

The unit R (2R = D) gives us R * R * R = 1 or R-cubed in XYZ (cube shaped), whereas D * D * D givers us 1 in the IVM (tetrahedron shaped).

Using our conversion constant is like converting between currencies or energy values (joules, calories) where the R-edged XYZ cube is about 6% bigger than the D-edged IVM tetrahedron. The two units of volume are somewhat close (like the Canadian and American dollar) but not the same.

If you know the volume in XYZ cubic units, multiply by √(9/8) to get the same volume in IVM tetrahedron units, or use √(8/9) to go the other way.

Why bother? What's the pay-off?

The tetrahedron of edges D divides thrice into a cube of face diagonals D (as we've seen) four times into an octahedron of edges D (its space-filling complement in the IVM), six times into a rhombic dodecahedron (space-filling CCP ball encasement), twenty times into a cuboctahedron (from 12-balls-around-1 and connecting corners).

Volumes Table: Tetrahedron 1 Cube 3 Octahedron 4 Rhombic Dodecahedron 6 Icosahedron 18.51... Cuboctahedron 20 2-Frequency Cube 24

The five-fold symmetric Icosahedron, Pentagonal Dodecahedron, and Rhombic Triacontahedron (RT) all fit in here as well, incommensurably in most cases.

The Jitterbug Transformation is used to connect the Cuboctahedron of edges D to a corresponding icosahedron of edges D.

Said Icosahedron of volume ~18.51, combined with its dual, define a rhombic triacontahedron that, if shrunk down by by 1/Phi (linearly) gives the RT of 120 E-mods (60 left, 60 right-handed).

The RT sharing vertexes with the RD of volume 6 has volume 7.5 exactly (IVM tetravolumes) and shrunk by 2/3 gives the RT of volume 5 exactly, and the 120 T-mods, same shape as the E-mods but a tad smaller, by about.001%.

What David Koski does is phi-scale the E-mods to express volumes as a linear combination of mods of different size (same shape).

Quoting from Koski's paper:

E module denotations: e6 = ((√2)/8)ø ̄⁹ = .002325 e3 = ((√2)/8)ø ̄⁶ = .009851 E = ((√2)/8)ø ̄³ = .041731 E3 = ((√2)/8)ø⁰ = .176766 E6 = ((√2)/8)ø³ = .748838

Quoting again from his paper [2]:

A rhombic triacontahedron with a radius of ø¹, is dubbed the Super RT. The long diagonal of the rhombic face = 2, which is R.B.Fuller’s edge for the tetrahedron, octahedron, cuboctahedron or VE, and the resultant icosahedron from the Jitterbug transformation.

The volume of the Super RT is 15√2 or 21.213203 = 120E3 = 480E + 120e3 [tetravolumes].

The icosahedron with an edge = 2, inscribes within the Super RT. It has a volume of 5(√2)ø² = 18.52295. It has an exact E module volume of 100E3 + 20E = 420E + 100e3. [tetravolumes]

That's about all the mathematics one needs to know, to understand about tetravolumes. It's not that hard.

We also get to think again about foundational matters, such as what basic assumptions we might vary to produce interesting new flavors of mathematics.

[1] Just for fun, lets compute the lengths of the "lid" edges and feed these numbers to our Pythonic tetrahedron volume computer, same one as in the Jupyter Notebook. Given edges a=2, b=4, c=5 I get d=3.46410, e=4.5825, f=8943540673 for a volume of...

a = 2 b = 4 c = 5 d = 3.4641016151377544 e = 4.58257569495584 f = 4.358898943540673 tetra = Tetrahedron(a,b,c,d,e,f) print("IVM volume of tetra:", tetra.ivm_volume())

IVM volume of tetra: 39.99999999999998 (check)

[2] http://4dsolutions.net/synergetica/RevisitingS&Emodules.pdf