ETPV

Memo to Faculty

School of Tomorrow

Kirby Urner
4 min readNov 23, 2024

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Say I have two lattices:

1. all XYZ cubes packed together, ye olde XYZ coordinate system… Cartesian, Fermatian…

2. octet truss, Bell kite: all tets and octs (twice as many tets, octs 4x tet volume) w/ vertices in CCP conformation.

There’s no inherent size relationship between them; I might have itty bitty XYZ cubes next to an octet truss of relatively gargantuan proportions.

However, as a designer-inventor, as a Bucky type, I’m free to establish a size relationship should I wish to.

Thought process: why not have the relationship be such that the XYZ cube and IVM tetrahedron are fairly close in size? What would be a natural way to do this?

Notice: if we define an IVM ball to have radius R and diameter D, then our 4-ball IVM tetrahedron has edges D (= 2R).

Put a D-edged tetrahedron next to an R-edged cube. Which is bigger, volume-wise? Answer: the cube, but only by about 6%.

The Synergetics Constant S3 is:

(XYZ cube of edges R)/(IVM tetrahedron of edges D) = ~1.06066

Either the R-cube or D-tet might play the role of unit volume. In XYZ, R-cube is the unit. In the IVM, D-tet is the unit.

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