# THE GRID

## THE VOLUME UNIT

The range defining the omni-symmetrical distortion of a volume unit extends from its form as a sphere to that of a tetrahedron. Its least energetic form (in terms of surface area) is the sphere, while the tetra-form marks the volume unit’s greatest surface potential through omni-symmetrical distortion alone.

For modeling sake, imagine the spherical form of the volume unit as being comprised of an immense number of radians extending outward from the center and impinging uniformly upon the underside of its surface. As the volume unit becomes more energetic by focusing its radians into fewer and fewer points, surface area correspondingly increases. Finally, at the opposite end 0f the range, the volume unit assumes definitive characters: an icosahedron (12 points); cube (8 points); octahedron (6 points); and ultimately the tetrahedron (4 points).

By omni-symmetrical distortion alone, the 1.0 volume-unit’s surface area increases from its 4.83597… surface units as a sphere to its 7.205621… surface units in the form of a tetrahedron.

7.205621…/ 4.83597… = 1.490007…/ 1.000…

It can be extrapolated from the equation above, that if the surface of the original sphere be transformed into a square equal in area to the sphere’s area, and then subdivided into “100th parts”, then this maximum increase in surface area as a tetrahedron can likewise be represented by a square comprised of 49 such “100th parts”. The surface area representation “grids” that follow illustrate the surface of the spherical form of the volume unit (fig. 1) and then as the surface of the tetrahedronal form (figure 2). The surface of the original spherical form of the volume unit has been rendered into the area of the square depicted above. Since the edge of the square is subdivided into 10/10, its area divides into 100/100. Thus each of the small squares represents a hundredth part of the original spherical surface (or, 0.0483597…).

The grid below represents the surface of the tetrahedronal form. It is a combination of two squares: The 100/100 from its form as a sphere plus the 49/100 gained by distorting to the tetrahedron. Thus far the range of distortion and its affect on surface area has been easily modeled by the previous grids. But when the sub-units for the surface of the tetra-form are arranged differently a new accounting comes into focus. This is the grid illustrated (below) in figure 3. It too measures 149/100 and accurately models the surface of the tetrahedron resulting from the maximum distortion of the spherical volume unit. But unlike the grid comprised of the two squares, this grid shouts out with at least two very distinct features. First, it is nearly a complete arrangement; and second, it lacks just one subunit of account from being complete.

In fact, the grid in figure 3, on closer inspection, is implying a structuring plateau which is even far simpler than first suspected. For if the missing 100th part was filled in then the transformation of surface from sphere to tetrahedron would be modeled by the three rectangles shown in figure 4. In figure 4, two of the three rectangles is the equivalent of the sphere’s 100/100 surface and in this case is more precisely expressed as 2/2 (two “halves” rather than one hundred “hundredths”). The three rectangles together represent the ideal surface quantity toward which the transformation through distortion of volume strives to attain (as the tetrahedron’s surface) yet falls short of the implied goal by one exquisitely scaled sub-unit (again, 1/100 of the volume 1.0 sphere’s surface). This implied ideal transformation is most clearly described by:

2/2 _______ 3/2

and can be read as “two halves transforming into three halves”. The original sphere’s surface area of 2/2 nearly increases to 3/2 except for it lacking one small increment of surface. And most important, this missing increment is obviously scaled to the natural sub-divisions inherent to the mathematics of this specific geometry.

Note: had the sphere’s surface increased from “one-hundred hundredths” to say for example 1.491637… as the tetra-form’s surface instead of its actual 1.490007… there would be no need for the previous grids, nor would they be possible given that hypothetical quantity. It is because of the three zeros after the 49 that we are able to regard the quantity as a discrete unit. It is comprised of 49 “one-hundredths” whether the measured quantity be required to the hundredths, thousandths, ten-thousandths, or hundred-thousandths decimal place.

## THE IDEAL PATTERNS

It is obvious that 1.49 is implying in rounder tolerances 1.5 which written another way is

3/2

So in this sense we can imagine the sphere’s initial surface area trying to go from 2/2 to an ideal 3/2 surface unit by focusing its volume into the form of a tetrahedron. That’s what the previous grids have shown us. If it “could” actually do so, it would reach important definite geometric structural plateaus which can be easily seen in the following models.

## The 2-Dimensional Model

Using the basic “unit” of the previous grid patterns, i.e. a 1/100 part of the 1.0 volume-unit’s surface, we can illustrate the difference between the actual results of the transformation as well as its implied ideal to which it approaches. These are depicted below at left. The arrangement on the right in figure 5 is constructed from the square grid made from the sphere’s surface (above the circle’s diameter); and below it are two squares, each made from the 49 sub-units created in the transformation (this relates to data derived from previous figures 2 and 3). Notice that the small squares come very close but do not touch the surrounding circle. By contrast, the arrangement on the left depicts squares below made from 50 sub-units, the implied ideal (and relates to data derived from previous figure 4). This arrangement is “complete” in that the smaller squares do share points on the surrounding circle. It is also illustrating a 1 dividing into 2 transformation since the area of the two smaller squares equals the area of the larger square and the circle itself is divided in two by its diameter.

## The 3-Dimensional Model

Another model measuring this transformation of surface quanta can be constructed in the following way. The spherical volume transforming into the tetrahedronal form increased its surface area in the same measure as it would had the sphere’s volume increased from 1.0 unit to 1.818783… This is the volume of a sphere with the same surface area as the 1.0 volume unit tetrahedron.

When this change in surface quantity is equated to a “puffing-up” of original volume into this 1.818783… spherical volume, the same ideal ratio appears

3/2

in this model, but just out of reach. Once again, as we will see, the ideal is a 1/100 volume part greater than what the tetrahedron’s surface area contains as a hypothetical spherical volume quantity.

The model showing these relationships is constructed in the following manner (refer to Figure 6, below). The 1.0 volume-unit sphere is placed within the confines of the smallest vector equilibrium (cube-octahedron) such that the sphere’s surface is tangent to the vector equilibrium’s eight triangular face planes. This spherical volume is contained volume, contained within the confines of the vector-equilibrium. Analogously speaking, it represents the volume unit in its lowest energetic state. Here it should be noted that the vector equilibrium models the minimum nucleated system (with respect to closest-packed uni-radiused spheres) while the tetrahedron models the minimum system. Now if “released” volume can be modeled by the sphere growing larger and larger until it is finally free of the vector equilibrium altogether, then the point at which it “escapes” is modeled by the smallest sphere in which the vector-equilibrium itself is perfectly embraced. This ideal sphere’s volume is:

[(3/2)^3]^1/2

and can otherwise be expressed as:

1.8371173…

In Figure 6, the Volume 1.0 sphere and the Volume 1.8371173… sphere are placed respectively “in” and “around” the vector- equilibrium. The Volume 1.818783… sphere, if represented on the scale of this illustration, would be undistinguishable from the Volume 1.8371173… sphere since their radii are very nearly the same: 0.757… vs. 0.759… This shows us just how close the tetrahedron’s surface really does come to this “ideal escape surface quantity”.

The relationship between the sphere having the tetrahedron’s surface area and the ideal “escape surface” sphere just outside the vector equilibrium is best revealed by comparing their volumes:

1.818783…/1.837117…   =   0.9900…/1.000

The simple equation above shows that the volume of the ideal sphere is 1/100 volume unit larger than the volume of the sphere with its surface equal to the surface of the 1.0 volume unit tetrahedron.

It is obvious from the previous equation that a “grid” can also be used to illustrate these volumes exactly like the previous grid was used to model the surface transformations.

In Figure 7 below, the cube on the right is the 1.837117… volume quantity. Its’ spherical form has been transformed by distortion into a perfect cube. Furthermore, its’ edge-length has been subdivided into 10 sub units (like the edge of the square made previously from the surface area of our original 1.0 volume unit sphere). This subdivides the cube’s volume into 1000/1000. Now by comparison, the 1.818783… volume equals 990/1000 of the ideal’s 1000/1000. This is modeled by the cube on the left shown clearly missing an exquisite 1/100 part of the ideal’s volume. In this sense it is incomplete, and is analogous to the “incompleteness” seen previously in the surface area grid of the tetra-form of the 1.0 volume unit (refer to figure 3). Once again, it is essential for the reader to understand that none of these models would be possible were it not for the two zeros following the 0.99 quantity (and is analogous to the three zeros following the 1.49 surface transformation). The zeros are what make it 99/100 of the ideal’s volume, whether tolerances are required to the hundredths, thousandths, or ten-thousandths number place the quantity remains the same. The remainder after the second zero is so small as to be inconsequential to the scale of these models.

# Modeling the Sphere and Cylinder

A sphere bores a cylindrical path through space-time. Since the sphere is always in motion through time, with respect to this consideration, its domain must be a cylinder. Specifically, at any given point in time this cylindrical domain is modeled on “The Cylinder of Maximum Volume”. Its circular top and bottom surface planes align perpendicularly to the sphere’s direction in time.

This cylinder (see Figure 8) snuggly holding the sphere within is

3/2

times greater than both the volume and surface area of the enclosed sphere. In fact, the sphere’s surface area is equal to the cylinder’s side area. This relationship has been well known for many centuries and is the geometrical basis for the popular Mercator map projection. But what is probably not so well known (if at all) is how the relationship between the sphere and cylinder is modeled by the same previous grids. For example, regardless of “size”, if the surface of the sphere within the cylinder is modeled by the 100/100 square grid of Figure 1, then the surface of the cylinder is 150/100 and can be equally represented by the three complete rectangles comprising Figure 4. If the volume of the sphere within is scaled to the 1.0 volume unit, then the surface of the cylinder reveals itself to be the same as the surface of the 1.837117… volume sphere . . . i.e. that ideal “escape surface” quantity. And when the surface of the cylinder is scaled to equal that crucial 1/100 part from the volume 1.0 sphere’s surface, the volume of the sphere within the cylinder is

1/1837117…*

##### * and is recognized as the same proportioning describing the mass-ratio between the electron and average nucleon.

Again, using the previous grids of Figures 3 and 4 as surfaces of “cylinders of maximum volume” and the volumes of the spheres captured within are respectively 0.9900… and 1.000… If we read these quantities as “990 thousandths” and “1000 thousandths” then we can again use the cubes of Figure 7 as comparative models of these spherical volumes. The cube volume with the missing edge row of sub-units is the spherical volume within the cylinder made from the tetra-forms surface. The whole cube to the right is the spherical volume within the cylinder made from the ideal “escape surface quantity”, which is the 1.837117… spherical volume’s surface.

Emerging from the previous data is an unmistakable geometric system of discrete quantities of both surface and volume. Central to this system are the two spherical volumes of 0.9900… and 1.000. These volumes represent the “actual” verses the “ideal” toward which the omni-symmetrical distortion of the spherical volume unit strives to attain.

These volume quantities differ by 1/100 of the “ideal’s” volume. The surfaces of the cylinders holding each as a spherical volume within likewise differ by 1/100 of the 1.0 volume unit’s spherical surface (which we know that as a cylinder’s surface will hold a sphere with volume 1/1837.117…). And when we compare these spherical surfaces (of the 1.000… and 0.9900… volume units) as shown below

4.835975…   –   4.803747…  =  0.0322280…

we arrive at a surface difference (0.o32228) which if modeled as a sphere encompasses a volume of

1/1838.1259…*

##### * and mirrors the mass relationship between the electron and the neutron.

So far the geometric patterning (resulting from the range of omni-symmetrical distortion of the volume unit) has exposed a spherical volume on the scale of an electron’s mass** to be associated with bridging the difference between the actual and the ideal geometric patterns. The spherical 1/1838… volume’s surface is the difference between the spherical v0lumes’ 1.000 and 0.9900…

surfaces (which volumes model the proportioning between the actual and the ideal). The surface area of the 1.837117… volume sphere is the “escape surface” quantity that distortion alone slightly fails to achieve due to a lack of 1/100 surface unit ; and a spherical volume of 1/1837.117…is held within the cylinder made from the same crucial 1/100 part of the surface of the 1.0 volume unit sphere.

# Modeling Hydrogen

There are other ways of modeling this original spherical volume transforming and somehow achieving that special “escape” size. These also show a volume particle on the scale of an electron making that crucial difference, allowing the “actual” to become the “ideal”. One of these modeling methods bears a striking resemblance to the hydrogen atom, the simplest and most abundant element in the universe.

Imagine the original sphere representing 1.0 volume unit ejecting a portion of its volume. If it could throw off just the right amount, the now two separate spherical volume units could focus their volumes into the tetrahedronal form thereby achieving “escape surface” status. Remember, the single sphere is the most efficient* geometric form for packaging volume. So now there are twospheres which together equal the original sphere’s volume but with a greater initial *i.e., requiring the least surface area surface area. When focused into the tetrahedronal form, their surfaces together will just equal the “escape surface” quantity.

To accomplish this 1/1668.514… of the original sphere’s volume is ejected as another spherical volume. The original sphere’s surface area is diminished slightly (from 4.835975… to 4.834043…), but the second sphere brings surface area with it as well (0.034376700…). The two spheres together now total 4.86842012… surface units. When each of these spheres transform their volumes into tetrahedrons their combined surfaces total 7.25396379… and as the surface of a single sphere would encompass a volume of 1.837117… achieving “escape surface” and the status of volume released.

It should be noted that in this transformation the amount of newly created surface area is equal to the surface area of the newly ejected volume particle minus the surface loss of the original sphere which (it is supposed) exited as part of the new volume particle’s surface. This newly created surface area amounts to 0.032(44425…), and is the surface area of an electron-size sphere since its volume equals

1/1819.782…

This image of a sphere ejecting a volume “particle” in this size range with respect to itself mirrors the process known as beta decay when a neutron ejects an electron and becomes a proton/electron system known as hydrogen.

Another less dynamic way of arriving at this same amount of surface area needing to be created is by modeling the “escape surface” quantity as a tetrahedron’s surface. Its volume is 1.010080… compared to the original 1.000… unit. When it relaxes back into the form of a sphere its volume stays the same, but its surface area becomes 4.86842011… which is again 0.032(44425…) more than the 4.835975… surface of the original volume 1.0 sphere.

Now if we model this newly created surface area as being in the form of a sphere with accompanying volume, and then focus this spherical volume into a tetrahedron, its surface grows to 0.0483420…, which quantity we recognize as a 1/100th part of the original sphere’s surface after it ejected the volume particle. Again that critical one-(h)underth part.

So for the tetrahedron to generate enough surface area for its 1.0 volume unit to actually attain “escape surface” status it has to create additional volume amounting to

0.010080…

which just happens to be a 1/100th part of the same numbers patterning the atomic weight of hydrogen (i.e., 1.0080).

The previous modeling revealed that 1.010080… is a special volume unit directly related to achieving “escape surface” status. And it too is easily represented by “the grid”. Its volume of 1010/1000 is the cube of Figure 9; and, the 151 sub-unit grid to its right is its associated cylindrical surface domain. Whether these models of the workings of geometry have any relation in actuality as to the “why” of the hydrogen atom’s atomic weight would be mere speculation. However, The Geometry of Form identifies this 0.010080… volume quantity as special, and in a way model of some idealized yet missing “potential”. This we glean from the transformational models thus far presented. But there is another model showing this volume quantity being related by scale to a kind of actualized potential resulting from “spinning” geometry’s fundamental units of surface and volume.

# The Energy of Spin

The equilateral triangle is the form of geometry’s minimum surface unit*. This triangle of surface area can be spun into a circle of surface area. By assigning 1.0 unit to the area of the entire circle, the area of the triangle is calculated to be 0.4134966… This means that by imparting the triangle with the motion of “spin” an additional amount of surface area is created or released from potential. We arrive at this amount by subtracting the triangle’s area from the circle’s area:

1.000 0.4134966… = 0.58650(3328…)

##### Adding a fourth point in a tetrahedronal closest packing delineates geometry’s minimum volume.

Likewise, when the form of geometry’s minimum volume unit (the tetrahedron) is imparted with spin, this same portion again defines the amount of volume “created” by ” by spinning the tetrahedron into a cone. This 0.58650… newly created quantity of surface area or volume is simply a manifestation of the circular motion imparted to the geometric forms.

If we model the surface area component as being the surface of a sphere, its radius is 0.2160(38080…). And the “cubic root”, or edge length of a cube with a volume of 0.010080(26…) is also 0.2160(18355…) showing that both geometric forms are scaled to a common unit. This relationship is graphically depicted in Figure 10. It shows the spherical 0.58650… surface unit nestled within a cubical system comprised of eight sub-cubes each with a 0.010080… volume. Now as a volume quantity, 0.58650… also models as a cube. Its edge-length is 0.837060…. and is again directly related to geometry modeling the energy of spin. In fact, this 0.837060… unit of length is the geometric constant used to calculate the “spin quantum” for any regular tetrahedron.

The existence of a geometric “spin quantum” can be illustrated in the following manner. If a regular tetrahedron is “spun” into a cone there is (as noted above) an increase in volume. This newly created volume is also modeled as a spinning cone with a tetrahedron within (that is actually doing the spinning). The edge-length of this tetrahedron will always be 0.837060… times the edge-length of the original spun tetrahedron. This is its “spin quantum”. Every tetrahedron of any size has its own specific “spin quantum”. But their edges will always relate as 1 : 0.8370… And if we read the decimal point as “and” we’ll arrive at

### 1.837

which looks conspicuously familiar to the “escape surface quantum”.

# The Cleavage Tetrahedron

Another way of modeling the 1.0 volume unit achieving “escape surface” status is by cleaving, or truncating its tetraform. If it is cleaved at just the right place, the two newly exposed triangular planes together will just equal the missing escape surface quantum of 0.0483420… This results in one very large but slightly truncated tetraform and one comparatively small regular tetrahedron. In The Geometry of Form this smaller tetrahedron is the “cleavage tetrahedron”.

The size of this special tetrahedron is determined in the following manner. The surface of the sphere modeling “volume released” is the known ideal 7.253963788… surface quantity. But the maximum surface possible for the 1.0 volume unit is its tetraform’s surface of 7.205621731. The difference between these two surfaces is 0.0483420…, the missing “escape surface” quantum.

To expose this amount of new surface area by truncating one edge of the tetrahedron means that the two newly exposed triangular planes together must total this difference. Therefore each of the two newly exposed triangular planes measure (0.0483420…/2) in area. Four of these triangles form the 0.096684114… surface of the cleavage-tet. The 0.0015542665 volume of this special tetrahedron when converted into the form of a sphere has a (0.032444254… × 2) surface area which is easily recognized as being twice the amount of newly created surface area when the original 1.0 volume unit “ejected” the spherical volume particle.

Another view of this special tetrahedron arises by modeling its system of four vertices as the centers of four spherical domains. Since this tetrahedron’s edges measure 0.236263… then half this length is the radius of each sphere. The volume of these spheres (0.00690…) show them to be within a 99.7% correspondence to the volume (0.00688530…) of “half-transit-tetrahedrons”, or geometry’s form which accounts for surface and volume differences when dividing the spherical 1.o surface unit.

# The Electron Tetrahedron

The surface area of the “electron tetrahedron” is half the surface area of the “cleavage tetrahedron”. Both of these tetrahedrons arise as solutions allowing the 1.0 volume unit to achieve “escape surface” status.

In The Geometry of Form the “electron tetrahedron” references a narrow range of electron-size volumes which have arisen as solutions in geometric transformations. This range begins with the 1/1819.782… volume of the newly created surface area (on the ejected volume particle). It includes the 1/1832.82… volume of the 1.0 lineal unit in the form of a tetrahedron’s edges; and the 1/1837.11…volume which as a sphere occupies a space-time domain equivalent to the missing one-(h)underth part (refer to page 11); and 125 of these units as a single cube has a 1.0 unit surface area. It is also the actual ratio between the mass of the average nucleon and electron mass. Capping off this range of electron-size volumes is the 1/1838.1259… quantity resulting from the surface difference between spheres volume 1.0 and volume 0.9900…

All of these electron-size volumes are the same volume to a 99% degree of accuracy. If we modeled any one of these volumes with a basketball size sphere, the others proportionally scaled would be indistinguishable from one another, even on this exaggerated scale.

As tetrahedronal volumes, they all very closely model “The Fundamental Unit of Length” in the form of a tetrahedron’s edges. The largest measures 1.00(238…) and the smallest measures 0.999(037…). Of course the 1/1832.82… volume, as a tetrahedron, perfectly models the 1.0 lineal unit since it is the exact sum of its edges.

Even more remarkable is that this range of electron-size quantities of surface and volume emerging from “The Grid” and its associated geometric transformations is also itself able to be modeled by this very same grid. For the two volume quantities anchoring each end of this range compare in the same way as a volume 1.0 and volume 0.9900…:

(1/1838.12…) / (1/1819.78…) = 0.9900… / 1.000

And once again, the two cubes in Figure 7 model this range of electron-size volumes which are built-in to the very fabric of geometry.

# The 49 One-(h)Underth Units

On the backdrop of what has transpired thus far, one shouldn’t be surprised at discovering that further modeling of the 49 one-hundredth parts (of the 1.0 volume unit’s surface area) reveals data consistent with previous results.

For example, when this special surface area is defined as the surface of a “cylinder of maximum volume”, which has been shown to be the form of a sphere’s four dimensional space-time domain, then the volume of the corresponding sphere within is 0.186705552… Now if this sphere’s volume is equally rewritten as (343 X .0005443…) it’s modeling capability will start to show through. For 343 can be a cubical system comprised of 343 sub-cubical-units (seven sub-cubes on an edge). And the volume of each of these sub-units is

1/1837.117…

or again the familiar mass ratio between electron and average nucleon. Proof that this is one of geometry’s fundamental volumetric building blocks is that 1.0 surface unit in the form of a cube has a volume equal to exactly 125 of these 1/1837.117… units.

This volume (.186705552…) of the sphere within the cylinder can also be written as (8 X .023338…). This .023338… volume quantity is more revealing written as

(1/1835.969…)^1/2

showing itself to be scaled to a measure which is the same size ratio as the masses of an electron relative to a proton. And if this 0.023338… volume is packaged in the form of a cube, the surface of this cube is

0.490007…

or that amount of surface which the volume unit is capable of creating through omni-symmetrical distortion alone.