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XYLENE POWER LTD.

SPHEROMAK SHAPE PARAMETER

By Charles Rhodes, P.Eng., Ph.D.

INTRODUCTION:
The spheromak shape parameter:
So^2 = (Rs / Rc)
where:
Rs = outside radius
and
Rc = inside radius
is one of the most important spheromak characterization parameters. Experimental data shows that for plasma spheromaks:
So^2 ~ 4.2

This web page identifies possible values of So^2 for atomic particle spheromaks.

Define:
Nr = Np / Nt
and
R = [((So^2 - 1) Kc) / (So^2 + 1)]

Recall that the spheromak boundary condition gives:
Nr^2 + R^2 = (2 / Pi)^2 Hence
0 < R < (2 / Pi)
and
0 < Nr < (2 / Pi)

The total static field energy contained in an electromagnetic spheromak can be expressed as:
Ett = [(Muo C^2 Qs^2) / 32 A^2] [1 / Ro] [4 So (So^2 - So + 1) / (So^2 + 1)^2]
where:
[1 / Ro]
= [(Pi Fh / C So)[Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2 Kc^2]^0.5

Hence:
Ett = [(Muo C^2 Qs^2) / 32 A^2] [4 So (So^2 - So + 1) / (So^2 + 1)^2]
[(Pi Fh / C So)[Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2 Kc^2]^0.5
 
= [(Muo C Qs^2 Pi Fh) / 8 A^2] [(So^2 - So + 1) / (So^2 + 1)^2]
[Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2]^0.5
 
= [(Muo C Qs^2 Pi Fh) / 8 A^2] [(So^2 - So + 1) / (So^2 + 1)^2]
[Np^2 + [Nt^2 (So^2 - 1)^2 / (So^2 + 1)^2]^0.5 (So^2 + 1)
 

Then:
Ett = [(Muo C Qs^2 Pi Fh) / 8 A^2] [(So^2 - So + 1) / (So^2 + 1)]
[Np^2 + Nt^2 R^2]^0.5

The spheromak static field energy is given by:

Ett= [(Muo C Qs^2 Pi Fh) / 8 A^2] [(So^2 - So + 1) / (So^2 + 1)]
Nt [Nr^2 + R^2]^0.5
 
= [(Muo C Qs^2 Pi Fh Nt) / 8 A^2] [(So^2 - So + 1) / (So^2 + 1)][Nr^2 + R^2]^0.5

However, the web page titled ELECTROMAGNETIC SPHEROMAK shows that the spheromak boundary condition is:
[Nr^2 + R^2] = (2 / Pi)^2

Hence:
Ett = [(Muo C Qs^2 Pi Fh Nt) / 8 A^2] [(So^2 - So + 1) / (So^2 + 1)][(2 / Pi)]

The equation for Ett simplifies to:
Ett = [(Muo C Qs^2 Pi Fh Nt) / 8 A^2] [(So^2 - 1) Kc / (So^2 + 1)][2 / Pi]
{(So^2 - So + 1) / [(So^2 - 1) Kc]}
 
= [(Muo C Qs^2 Fh Nt) / 8 A^2] [R][2 / Pi]{(So^2 - So + 1) / [(So^2 - 1) Kc]}
 
= [(Muo C Qs^2 Fh Nt) R / 4 A^2 Pi]{(So^2 - So + 1) / [(So^2 - 1) Kc]}
 

Make substitution:
Muo C Qs^2 = 2 h Alpha
to get:
Ett = [(2 h Alpha Fh Nt) R / 4 A^2 Pi]{(So^2 - So + 1) / [(So^2 - 1) Kc]}
 

or since:
Ett = h Fh
(1 / Alpha) = [ Nt / 2] [R / A^2 Pi]{(So^2 - So + 1) / [(So^2 - 1) Kc]}

Note that Nt must be a positive integer.

In order for the Planck Constant h and Fine Structure Constant Alpha to be really constant the product:
Nt [(So^2 - So + 1) / (So^2 + 1)]
must be constant at the spheromak operating point.
 

POTENTIALLY STABLE STATE OF EQUAL MAGNETIC VECTORS:
When the magnetic vectors are equal:
Np^2 = Nt^2 R^2 or
Nr = Np / Nt
= (So^2 - 1) Kc / (So^2 + 1)
= R

Recall that the spheromak boundary condition gives:
Nr^2 + R^2 = (2 / Pi)^2

In this state:
Nr = R

The boundary condition gives:
Nr^2 + R^2 = (2 / Pi)^2 or
2 R^2 = (2 / Pi)^2 or
R = 2^0.5 / Pi,BR. = 0.4501581586

R = (So^2 - 1) Kc / (So^2 + 1)
or
R (So^2 + 1) = (So^2 - 1) Kc
or
So^2 (Kc - R) = (Kc + R)
or
So^2 = (Kc + R) / (Kc - R)

At Kc = 1:
So^2 = (1 + (2^0.5 / Pi)) / (1 - (2^0.5 / Pi)
= (Pi + 2^0.5) / (Pi - 2^0.5)
= (3.14159265 + 1.414213562) / (3.14159265 - 1.414213562)
= (4.555806212 / 1.727379088)
= 2.637409613
 

STABLE STATES WITH RESPECT TO WINDINGS:
When a spheromak is at one if its potentially stable states with respect to the number of windings:
d{Np^2 (So^2 + 1)^2 + Nt^2 (So^2 - 1)^2 Kc^2} = 0
or
Np dNp (So^2 + 1)^2 + Nt dNt (So^2 - 1)^2 Kc^2 = 0
 

AT THE POTENTIALLY STABLE STATE dNt = - dNp:
At the stable state:
dNp = - dNt
giving:
Np (dNp) (So^2 + 1)^2 + Nt(- dNp)(So^2 - 1)^2 Kc^2 = 0
or
Nt (So^2 - 1)^2 Kc^2 = Np (So^2 + 1)^2
or
Nr = (Np / Nt)
= (So^2 - 1)^2 Kc^2 / (So^2 + 1)^2
= R^2

Recall that the boundary condition gives:
Nr^2 + R^2 = (2 / Pi)^2

Hence:
[R^2]^2 + R^2 = (2 / Pi)^2
or
R^4 + R^2 - (2 / Pi)^2 = 0
or
R^2 = {- 1 +/- [1 + 4(1)(2 / Pi)^2]^0.5} / 2
= {- 1 +/- [1 + (4 / Pi)^2]^0.5} / 2
= {- 1 +/- [1.618993188]} / 2
= 0.3094965939

R = 0.5563241806

Recall that:
R = [((So^2 - 1) Kc) / (So^2 + 1)]

or (So^2 - 1) Kc = R (So^2 + 1)
or So^2 {Kc - R} = {Kc + R}
or
So^2 = (Kc + R) / (Kc - R)

= (Kc + 0.5563241806) / (Kc - 0.5563241806)

For Kc = 1:
So^2 = 1.5563241806 / 0.4436758194
= 3.507795811
 

AT THE POTENTIALLY STABLE STATE dNt = - 2 dNp:
At the stable state:
dNp = - 2 dNt
giving:
Np (dNp) (So^2 + 1)^2 + Nt(- 2 dNp)(So^2 - 1)^2 Kc^2 = 0
or
2 Nt (So^2 - 1)^2 Kc^2 = Np (So^2 + 1)^2
or
Nr = (Np / Nt)
= 2 (So^2 - 1)^2 Kc^2 / (So^2 + 1)^2
= 2 R^2

Recall that the boundary condition gives:
Nr^2 + R^2 = (2 / Pi)^2

Hence:
[2 R^2]^2 + R^2 = (2 / Pi)^2
or
4 R^4 + R^2 - (2 / Pi)^2 = 0
or
R^2 = {- 1 +/- [1 + 4(4)(2 / Pi)^2]^0.5} / 8
= {- 1 +/- [1 + (8 / Pi)^2]^0.5} / 8
= {- 1 +/- [2.735791616]} / 8
= 0.216973952

R = 0.4658046286

Recall that:
R = [((So^2 - 1) Kc) / (So^2 + 1)]

or (So^2 - 1) Kc = R (So^2 + 1)
or So^2 {Kc - R} = {Kc + R}
or
So^2 = (Kc + R) / (Kc - R)

= (Kc + 0.4658046286) / (Kc - 0.4658046286)

For Kc = 1:
So^2 = 1.4658046286 / 0.5341953714
= 2.743948576
 

AT THE POTENTIALLY STABLE STATE dNt = - 3 dNp:
At the stable state:
3 dNp = - dNt
giving:
Np (dNp) (So^2 + 1)^2 + Nt dNt (So^2 - 1)^2 Kc^2 = 0
or
Np (dNp) (So^2 + 1)^2 + Nt (- 3 dNp) (So^2 - 1)^2 Kc^2 = 0
or
3 Nt (So^2 - 1)^2 Kc^2 = Np (So^2 + 1)^2
or
Nr = (Np / Nt)
= 3 (So^2 - 1)^2 Kc^2 / (So^2 + 1)^2
= 3 R^2

Recall that the boundary condition gives:
Nr^2 + R^2 = (2 / Pi)^2

Hence:
[3 R^2]^2 + R^2 = (2 / Pi)^2
or
9 R^4 + R^2 - (2 / Pi)^2 = 0
or
R^2 = {- 1 +/- [1 + 4(9)(2 / Pi)^2]^0.5} / 18
= {- 1 +/- [1 + (8 / Pi)^2]^0.5} / 18
= {- 1 +/- 3.652514326} / 18
= 0.147361907

R = 0.3838774635

Recall that:
R = [((So^2 - 1) Kc) / (So^2 + 1)]

or (So^2 - 1) Kc = R (So^2 + 1)
or So^2 {Kc - R} = {Kc + R}
or
So^2 = (Kc + R) / (Kc - R)

= (Kc + 0.3838774635) / (Kc - 0.3838774635)

For Kc = 1:
So^2 = 1.3838774635 / 0.6161225365
= 2.246107522
 

SUMMARY:
The potential R and So^2 values at Kc = 1 are:
R = 0.4501581586 corresponding to equal magnetic vectors
So^2 = 2.637409613

R = 0.5563241806 corresponding to dNp = - dNt
So^2 = 3.507795811

R = 0.4658046286 corresponding to 2 dNp = - dNt
So^2 = 2.743948576

R = 0.3838774635 corresponding to 3 dNp = - dNt
So^2 = 2.246107522
 

Thus the largest stable So^2 potential value appears to be 3.507795811. Note that if Kc > 1 then So^2 will decrease.
 

This web page last updated April 19, 2019.

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