# 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.

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

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?

Find the diameter of the spinning node ring shown in red.

Area = median square side ^2