The polygons pictured both have the same perimeter. If you spin them, the outer diameter you now see will appear to be different on each.

The outer corners of the polygon will generate the points of vibration, thus diameter is with reference to these points. The fractal segment length is the distance between two adjacent nodes. If the tube is cut to a multiple, it's entire surface will now vibrate up in sections of [squares] or a grid of vibrational equidistant nodes equaling the fractal segment length.

Diameter is related in an irrational way to resonant segment length on flat polygons other then the Hex form, so cannot link energy directly from it as a harmonic in these two forms. The ratio of segment to diameter is not a whole number ratio but an infinite series of digits. What this means is that on a resonating tube, diameter will not be resonant at an 8 or 4 node resonance at the center point between them.

The limiting function as we alter segment counts on the polygon is the circle having the same perimeter. This is the base diameter for perimeter coupling formula.

Within the distance or thickness of the polygons points to the diameter having the same perimeter will lie all the higher frequency octaves possible for the form to couple to using this method.

There will be a harmonic relationship between diameter and segment vibration fractal. This unique form is found in the vectorequilibium at the core. Buckminster Fullers Most important geometrical discovery.

Thus placing a HEX form at the central core of a Joe Cell will set up an inwards vibration to the very center that is harmonically coupled. Likewire setting up a square or a triangle using the vector equlibrium distances rather then segment length resonance is a different technique.

With octagon resonance the frequency will be about 2x of the square. If placed on the same tube however, on the same diameter circle, the frequency will not be exactly 2x. This is because as you increase the segment count on a fixed diameter circle, you increase the perimeter distance. The frequency will drift lower, because the segments on the octagon will no longer be exactly half the squares. The two forms will loose cross coupling of energy on the same tube at any one fixed diameter.

The solution to "frequency drift" as one adds more segments and nodes, is to reduce the diameter of the tube by some small "correction factor." Then make the tube thicker to encompass both nodal rings, and then maybe also good extra running inwards to the circle limit to gain higher frequencies also and the center vector equlibrium.

The goal is to make all the sides, or the perimeter be exactly the same total length, and at this point light taking either path will circle it in the same exact time interval. The two systems will couple and exchange a resonant energy pulse. Energy will be able to move either inwards or outwards, and from high frequency to low, or low frequency to high. If we can establish simultaneous propagation of longitudinal pulses around both perimeters, we will have a coupled system.

This will generate a coherent field across frequency, and is very probably the model of a Light Node. It speaks to tube thickness, and a down shift of the energy to the tubes cut length.

We now use our tube cutting formula only this time solve for Diameter to see what happens as we force the segments perimeters to stay equal on the polygons.

We will assign the square a side length of 2" and the octagon 1" to keep both perimeters equal. Frequency will now be harmonically linked between two octaves.

Diameter = Fractal Length / (sine (1/2 the angle))

Diameter (octagon) = 1" / (sine (360/8 / 2) = 2.6131"

Diameter (square) = 2" / (sine (360/4 / 2) 2.82842"

These two forms will have a harmonic frequency link, the diameter will drop on the higher frequency.

Diameter of larger one to smaller one = 1.0823922

Now as the fractal Length approaches zero, at infinity segment count, the form becomes a circle with the same perimeter as all the polygons.

Perimeter of the (circle) = 1 * 8 = 8"

Diameter of the circle = 8 / pi = 2.54647"

If we build a tube with ID [inside diameter] of 2.54647" and OD [outside diameter] of 2.82842"

We can now cut a length that will support fully nearly an infinite number of fractal octaves off the base frequency of the cut.

It will support resonance from a square to infinity segments.

They should all be happy to set on the tube simultaneously, and exchange energy.

Any polygon with an octave link to the square will now cross couple energy into the system.

On the great pyramid, the torsional bubble that forms rings from the center vertical axis, will couple energy to its spherical field from the base using perimeter entrainment.

The tip of the pyramid will receive a layered ring of the spheres field, vibrating with the same perimeter as the base but at a much higher frequency. It will form 1/2 a sphere above ground, best guess. If one now entered the vibrational layer they could cut a length that would light up with vibration based on a higher fractal.

Within this bubble is another fractal in operation one from the vector equlibrium that will focus all to a centeral point in space.

A boundary layer that intersects the apex. A strong node will appear at that point.

With the Giza pyramid there will be two of these, because we have cross coupled two diameters using perimeter entrainment.

Diagram shows a Hex or base vector equlibrium at the center of a system on a tube cut to a fractal length of its radius.

The perimeter of the squares formed on the surface grid of the tube, apear to be generating strong nodes in space around the tube at their new radius.

This is the radius of a circle having the same perimeter as the new nodes on its surface grid.

This appears to be the strongest method of coupling between tubes but comes out a bit less then a standard Joe Cell.

Also the main nodes of energy are located lower in space then the end of the inner tube.

If we design the next tube to wrap these points with 12 nodes, that all do this same thing towards the center, having the correct diameter, the two systems will mesh nodes on at least 6 points.

A complex computation, however this will lay all the stress of the system to one center, and also be able to down shift all the higher frequencies from the larger tubes down towards the Hex form at the center.

There is also a node that appears directly at the end length between the nodes as well but is a little higher, and secondary coupling point for a 2D system of linking the energy. Shown in the diagram on the left. The triangles are now equlateral and flat to the tubes ends.

The information we will need to mesh these is the distance the new node will stand off the tubes surface.

On the next tube we will need to know the inner distance it will stand for a 12 node ring. For a perfect match we then add these together to discover the next tubes diameter.

The next tube will be dropped down by the correct distance for nodes to mesh perfectly. There will now be a full set of nodes existing in the gap between the two tubes and energy will be able to shift frequency moving through them. It is the downshift of frequency that results in energy gain.

Diagram shows a method of linking energy inwards to the center point of the system at its lowest frequency, the Hex pattern.

2nd tube is a 12 node ring set up to have 6 nodes of equlateral triangles hitting the inner tubes 6 outer extended nodes in the same plane.

We will see a 2 dimensional compression forming on all the rings of the inside tube. Energy moves through the triangles at near the same frequency. No pyramids are used.

If one has a model of the vector equlibrium and sites down it so as to see two squares align along a tube, they may notice that now 4 external node points emerge, as well as a central point of compression at the center of the tube. It is formed by three hex forms wraping it at various angles.

Also if we run the tube through the triangles then we get an external hex pattern emerging outside the tube.

This demonstrates there are ways to work the vector equlibrium and come up with various tube designs using 3D coupling, fully convergent on a center point in space.

Tube would have to be cut to contain the vector equlibrium pattern on the surface and see if it will manifest.

The fractal segment length will be the diameter of a tetrahedron. This should force the square to go onto the ring of the tube.

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