Quadrature Imprinting

With all the new discoveries from the Source field, to Space, to NMR, to Spinning nodes,  it is time to revisit the Joe Cell, and see what may be possible using 4 nodes on each tube.

Cell Photo

It would seem that tubes 2 and 3 are closely aligned for a self sustaining field on quadrature configuration, having approximately [Sqrt 2] diameter relationship.
One touches inside the square and one touches on the outside tips of the square.

Measurements in mm

Tube         Diameter
1 Inner        22.58       24.45
2                 48.44       50.95
3                 72.55       76.34
4                 98.53       101.94
5 Outer                       129.20

Calculations in mm

Calculations Diagram

Tube 3 Square Calculation

Segment length = Diameter * sin (1/2 angle)
SL = D * sin(45)

Diameter Range Tube 3
ID         72.550 mm
Median  74.445 mm
OD        76.340 mm

SL  =   51.30 mm
SL  =   52.64 mm 
SL  =   53.98 mm

If we imprint something in this range on tube 3, will the 4 nodes be stable?
Will the 4 nodes also come up on tube 2 between the ones on tube 3 showing the sqrt of 2 SSF anchored into the cell.

Diameter Range Tube 2
ID           48.440  mm
Medium   49.695 mm
OD          50.950 mm

SL  =  34.25 mm
SL  =  35.14 mm
SL  =  36.03 mm

36.03 * sqrt 2 = 50.95 mm  So technically the form will not quite reach between the two tubes.
This may be why JC work is not easy or simple to bring up a cell in SSF modes that will sustain strong voltages.

We will likely have better results using a 2x square side length with a 4x increase in frequency, but the sqrt (2) fractals may not directly hit any of the tubes...

Tubes 3 and 4 look more promising.

The Electric Self Sustaining Form

Can we now imprint the spinning nodes ring that will be supported by the stationary nodes?
Will it create a voltage differential between tubes 2 and 3?

Spinning node ring

Find the diameter of the spinning node ring shown in red.
Area = median square side ^2

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