XYLENE POWER LTD.

SPHEROMAK APPROXIMATION:

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

SPHEROMAK DESCRIPTION:
A spheromak quasi-toroid has a major axis and a minor axis. The spheromak's circulting current follows a closed spiral path which includes Np turns around the major axis and within the spheromak wall and by Nt turns around the minor axis and within the spheromk wall. This current path defines the quasi-toroidal shape of the sheet referred to as the spheromak wall.

In the space enclosed by the spheromak wall the magnetic field is toroidal and the electric field is zero. In the space outside the spheromak wall the magnetic field is poloidal and the electric field is quasi-radial. Outside the spheromak wall the electric field is normal to the spheromak surface. Within the core of the spheromak the net electric field is zero near R = 0. The cylindrically radial components of the electric field cancel out in the spheromak core. In the far field the electric field is nearly spherically radial. The spacial magnetic field distribution outside the spheromak wall caused by the circulating charge is modelled by a ring current Np I at R = Ro, Z = 0,and a ring charge Q at R = Ro, Z = 0. Hence the electric and magnetic field geometry is constant over time. At the spheromak wall both the toroidal and poloidal magnetic fields are tangent to the spheromak wall.

The spheromak field structure enables the existence of semi-stable plasma spheromaks and discrete stable atomic charged particles and acts as a store of energy.
 

ATOMIC PARTICLE SPHEROMAKS:
Atomic particle spheromaks have a quantized charge that superficially appears to be at rest with respect to an inertial observer. Isolated stable atomic particles such as electrons and protons hold specific amounts of energy (rest mass). When these particles aggregate with opposite charged particles the assembly emits photons. This photon emission decreases the total amount of energy in the assembly creating a mutual potential energy well.

In an atomic particle spheromak current moves uniformly around a closed spiral path at the speed of light. The spheromak net charge is uniformly distributed along the current path. The uniform charge distribution along the current path and uniform current cause constant electric and magnetic fields. The time until the current retraces its previous path is (1 / Fh) where Fh is the characteristic frequency of the spheromak.

An isolated spheromak in free space has a distorted circular cross section. However, the fields of an atomic particle spheromak may be further distorted by external electric and magnetic fields caused by adjcent spheromaks.
 

SPHEROMAK CONCEPT:
Conceptually a spheromak wall is a quasi-toroidal surface formed from the current path of a spheromak. The current path within the spheromak wall conforms to the quasi-toroid surface curvature.

The magnetic field of a spheromak has both toroidal and poloidal components. The current path gradually changes direction over the surface of the quasi-toroid while following a closed spiral.

In a spheromak quantized net positive and/or net negative charge moves along a closed spiral path embedded in the spheromak wall. A spheromak is cylindrically symmetric about the spheromak major axis and is mirror symmetric about the spheromak's equatorial plane. The net charge Qs is uniformly distributed over the current filament length Lh.

For an isolated spheromak in a vacuum, at the center of the spheromak the net electric field is zero. Inside the spheromak wall the magnetic field is purely toroidal and the electric field is zero. In the region outside the spheromak wall the magnetic field is purely poloidal. The spheromak has surface charge. Outside the spheromak wall for [Rw > Rs] the electric field is normal to the spheromak wall and is spherically radial in the far field. The current circulates within the spheromak wall which forms the interface between the toroidal and poloidal magnetic fields.

For [Rw < Ro] the sphermak has surface charge but the radial electric field components cancel due to the internal electric field from one side of the spheromak affecting the internal electric field on the other side of the spheromak.
 

SPHEROMAK ANALYSIS:
Accurate spheromak analysis requires precise knowledge of the function Zw(Rw) which function specifies the position of the spheromak wall. This function is required in order to accurately calculate energy integrals both inside and outside the spheromak wall. On this web page for Rw > Ro we develop an accurate equation for the spheromak wall position of the form:
1 / (Zw^2 + Rw^2)^2 = (A / Rw^2) - B^2
= B^2 [(A / B^2 Rw^2) -1].

On this web page we further show that the fundamental radius Ro of the spheromak is:
Ro^2 = [A / 2 B^2]
so that the equation for the spheromak wall position in the range:
Rc < Rw < Rs
becomes:
1 / (Zw^2 + Rw^2)^2 = B^2 [(2 Ro^2 / Rw^2) - 1]

The challenge is to accurately evaluate the constant B^2.

Finding B^2 involves finding the winding ratio:
(Np / Nt)
which involves finding the turn length ratio:
(Lp / Lt).
The poloidal turn length Lp is accurately known as:
Lp = 2 Pi Ro
but calculating the toroidal turn length Lt involves a line integral of such complexity that its evaluation needs a dedicated web page.
To obtain a rough idea as to the size of B^2, on this web page we use the approximation that:
[(Np Lp) / (Nt Lt)] ~ [1 / 2].
For accurate calculation of B^2 it is necessary to use the relationship:
[(Np Lp) / (Nt Lt)] =[Mp / Mt]
where, as shown on the web page titled (A HREF="GF Spheromak Winding Constraints.htm"> Spheromak Winding Constraints
[Mp / Mt] involves multiple possible pairs of prime numbers. Then:
[Np / Nt] =[Mp / Mt][Lt / Lp]

In summary, finding B^2 requirs finding [Np / Nt] which in turn requires finding [Lt / Lp], which means that Lt must be accurately calculated. That calculation is the subject of a dedicated web page.
 

SOLUTION APPROXIMATION:
Note that a neutral spheromak cannot exist because a spheromak relies on the distributed charge on the current filament to balance the attractive magnetic foces between adjacent current filaments. A net electriclly neutral particle such as a neutron must be composed of at least two spheromaks.

For Family A spheromaks to a good approximation:
[(Np Lp) / (Nt Lt)] ~ [1 / 2]

As shown on the web page titled: Spheromak Winding Constraints this approximation is accurate to about 1% but is not good enough for precision calculations. However, this approximation does indicate general spheromak behaviour not dependent on the winding ratio.

Note that Np and Nt are both integers which ultimately leads to a unique solution.

A real charged spheromak in a vacuum has an external radial electric field which modifies the field energy distribution.
A real spheromak case might be an electron spheromak around a positive nucleus. At large distances the electric fields cancel. Inside the spheromak walls the electric field is zero. Everywhere on the spheromak walls the poloidal magnetic energy density outside the wall plus the external electric field energy density normal to the wall equals the toroidal magnetic field energy density inside the wall.

In a plasma or a crystal the external electric fields almost cancel so spheromak energy is largely magnetic. Hence it is informative to calculate the theoretical Planck constant for the purely magnetic case.
 

SPHEROMAK GEOMETRY:
Assume that the spheromak shape is a symmetrical distorted toroid.
Rs = maximum spheromak wall radius in the equatorial plane;
Rc = minimum spheromak wall radius in the equatorial plane:
Ro = Spheromak characteristic radius where:
Rs^2 + Rc^2 = 2 Ro^2;
H = Z value of a point on the spheromak wall;
Ho = Z value at R = Ro;
Outside zone nearly radial electric field set by an imaginary charged ring of radius Ro with charge Qs located at (R = Ro, Z = 0);
Outside zone poloidal magnetic field set by an imaginary ring of radius Ro carring current Np I located at (R = Ro, Z = 0);
Inside zone toroidal magnetic field Bt = Muo Nt I / 2 Pi R

The outside region has three zones. they are:
R > Rs
R < Rc
and
Rc < R < Rs

The inside region has only one zone which applies for Rc < R < Rs
In each zone the total energy is found by integrating over cylindrical elements.
 

POLOIDAL MAGNETIC FIELD
Consider a thin closed ring coil with Np turns and radius Ro. The center of this coil is at Z = 0, R = 0. The coil lies in the plane Z = 0. Assume that current I is circulating through this coil.
The magnetic field Bpor at the center of this coil is given by:
Bpor = Muo Np I / 2 Ro

The magnetic flux Phi through the coil is given by:
Phi = Bp A
~ Bor Pi Ro^2
 

POLOIDAL MAGNETIC FIELD ENERGY:
Self inductance L of the coil at R = Ro, Z = 0 is:
L = Np Phi / I
= Np Bpor Pi Ro^2 / I
= Np [Muo Np I / 2 Ro] Pi Ro^2 / I
= Np^2 Muo Pi Ro / 2

The poloidal magnetic energy E stored around this coil is:
Ep = L I^2 / 2
= Np^2 Muo Pi Ro I^2 / 4

dEp = L I dI

I = Q C / Lh

dI = Q C d[1 / Lh]

The self inductance

dI = Q C d(1 / Lh)

dEp = L I Q C d(1 / Lh)

The coil length Lh is given by:
Lh^2 = (Np Lp)^2 + (Nt Lt)^2
= [(Np)^2 + (Nt^2 (Lt / Lp)^2] Lp^2
or
Lh / Lp = [(Np)^2 + (Nt^2 (Lt / Lp)^2]^0.5

Lp = 2 Pi Ro

Lh = [(Np)^2 + (Nt^2 (Lt / Lp)^2]^0.5 (2 Pi Ro)

(1 / Lh) = 1 / {[(Np)^2 + (Nt^2 (Lt / Lp)^2]^0.5 (2 Pi Ro)}

d(1 / Lh) = [1 / {[(Np)^2 + (Nt^2 (Lt / Lp)^2]^0.5 (2 Pi)}] d[1 / Ro]
= [Lp / {Lh 2 Pi}] d[1 / Ro]

Hence:
dEp = L I Q C d(1 / Lh)
= L I Q C [Lp / Lh 2 Pi] d[1 / Ro] = [Np^2 Muo Pi Ro / 2] I Q C [Lp / Lh 2 Pi] d[1 / Ro]

Assume that in conformity with spheromak existence requirements:
I = Q C / Lh
so that:
dE = [Np^2 Muo Ro / 4] Q^2 C^2 [Lp / Lh^2] d[1 / Ro]

Recall that:
Lp = 2 Pi Ro
Hence:
dEp = [Np^2 Muo (Lp / 2 Pi)(1 / 4)] Q^2 C^2 [Lp / Lh^2] d[1 / Ro]
= [Np^2 Muo (Lp^2 / Lh^2 )][1 / (8 Pi)] Q^2 C^2 d[1 / Ro]
= Muo [Np^2 Lp^2 / Lh^2](1 / 8 Pi) Q^2 C^2 d[1 / Ro]

Recall that: Lh = [(Np)^2 + (Nt^2 (Lt / Lp)^2]^0.5 Lp

Hence:
[Lp^2 / Lh^2] = 1 / [(Np)^2 + (Nt^2 (Lt / Lp)^2]

Thus:
dEp = {Muo Np^2 / [(Np)^2 + (Nt^2 (Lt / Lp)^2]}(1 / 8 Pi) Q^2 C^2 d[1 / Ro]

F = C / Lh

dF = C d(1 / Lh)
= C [1 / {[(Np)^2 + (Nt^2) (Lt / Lp)^2]^0.5 (2 Pi)}] d[1 / Ro]

Hence:
d[1 / Ro] = {[(Np)^2 + (Nt^2) (Lt / Lp)^2]^0.5 (2 Pi)}dF / C

Thus:
dEp = {Muo Np^2 / [(Np)^2 + (Nt^2) (Lt / Lp)^2]}(1 / 8 Pi) Q^2 C^2 {[(Np)^2 + (Nt^2 (Lt / Lp)^2]^0.5 (2 Pi)}dF / C
= {Muo Np^2 / [(Np)^2 + (Nt^2) (Lt / Lp)^2]^0.5}(1 / 4) Q^2 C}dF
= {Muo Np^2 Q^2 C / 4[(Np)^2 + (Nt^2) (Lt / Lp)^2]^0.5}} dF
= {Muo Np^2 Q^2 C Lp / 4 Lh} dF
=

The poloidal magnetic field contribution to the Planck constant is:
hp = dEp / dF
= [Muo Q^2 C][1 / 4][Np^2] [Lp / Lh]
 

TOROIDALIC FIELD ENERGY CONTRIBUTION TO THE PLANCK CONSTANT:
Changing the dimension Ro changes the amount of toroidal magnetic field energy whichmakes a further contribution to the Planck Constant.
 

RADIAL ELECTRIC FIELD ENERGY CONTRIBUTION TO THE PLANCK CONSTANT:
Changing the dimension Ro changes the electric field energy along the surface of the sphermak which also contribures to thePlanck constant.
 

DETERMINATION OF Rc AND Rs:
At R = Rc, Z = 0 there are no net electric fields at this location. Hence at this location:
Btc = Bpc.
Btc = [Muo Nt I / 2 Pi Rc]
Bpor = Muo Np I / 2 Ro
Refer to web pages titled:Theoretical Spheromak to find Bpc.

At R = Rs, Z = 0
Bts = Muo Nt I / 2 Pi Rs
Both Bps and Ers are important. The average surface charge density increases with increasing Rw.

The spheromak walls must intersect the equatorial plane at Rc and Rs.
 

SPHEROMAK FILAMENT
From the web page titled: Spheromak Structure the current path winding length Lh is given by:
[Lh]^2 = Nt^2 [Lt]^2 + [Np^2] [Lp]^2

For any spheromak:
F = C / Lh
and
I = Qs C / Lh
or
I^2 = Qs^2 C^2 / Lh^2
= Qs^2 C^2 / [Nt^2 Lt^2 + Np^2 Lpo^2]

Thus the magnetic field energy density inside the spheromak wall takes the form:
Bt^2 / 2 Muo = [Muo Nt I / 2 Pi R]^2 / 2 Muo
= [Muo / 8] [Nt / Pi R]^2 [I^2]
= [Muo / 8] [Nt / Pi R]^2 Qs^2 C^2 / [Nt^2 Lt^2 + Np^2 Lpo^2}

Recall from basic Gauss Law theory that:
Closed line integral [Bt dS] = Muo I Nt
or
Bto 2 Pi Ro = Muo Nt I
or
Bto = [(Muo Nt I) / (2 Pi Ro)
which is a well known result.

Recall from basic Gauss Law theory that:
Closed line integral [Bp dS] = Muo I Np
or at the spheromak wall:
Line Integral [Bp dLt] = Muo I Np
but along this path Bp varies over a wide range.

At R = Rc, Z = 0 there is no electric field. Hence Bpc = Btc = (Muo Nt I) / (2 Pi Rc).

At R = Rs, Z = 0 the external electric field is:
~[Q / (4 Pi Rx^2 Epsilono)]
giving an external electric field energy density:
[Epsilono / 2][Q / (4 Pi Rx^2 Epsilono)]^2
= [1 / 2 Epsilono] [Q / (4 Pi Rx^2)]^2

Hence:
[Bps^2 / 2 Muo] + [1 / 2 Epsilono] [Q Ro / (4 Pi Rx^2)]^2 = [Bto^2 / 2 Muo][Ro / Rs]^2
or
= [Bps^2 / 2 Muo] + [Muo C^2 / 2][Q / 4 Pi Rs^2]^2 = [Bto^2 / 2 Muo][Ro / Rs]^2

Thus:
Bps^2 + [Muo C]^2[Q / 4 Pi Rx^2]^2 = Bto^2 [Ro / Rs]^2

Recall that:
Bto = Muo Nt I / 2 Pi Ro
= Muo Nt Q C / Lh 2 Pi Ro
= Nt Q Muo C /Lh Lp

Hence for field energy density balance at R = Rs:
Bps^2 + [Muo C]^2[Q / 4 Pi Rx^2]^2} = [Nt Q Muo C / Lh Lp]^2 [Ro / Rs]^2
or
Nt Ro/ Lh Lp Rs > 1 / 4 Pi Rx^2

This condition is barely met by a spheromak.

However, at the spheromak outer wall at Z = 0:
(Bts^2) / 2 Muo = Bps^2 /2 Muo - [Epsilon / 2] [Ers]^2
and at the spheromak inner wall at Z = 0:
(Btc^2] / 2 Muo = [Bpc^2 / 2 Muo] -[Epsilon / 2][Erc]^2

For the quasi-toroid:
Lp = 2 Pi Ro
giving:
Bto = Muo I Nt / (2 Pi Ro)

Bt = Bto (Ro / R)

Hence:
Bt^2 = [Muo / 8] [Nt / Pi R]^2 Qs^2 C^2 / [Nt^2 Lt^2 + Np^2 Lp^2}
or
[Bt^2 / 2 Muo] = [1 / 16][Nt / Pi R]^2 Qs^2 C^2 / [Nt^2 Lt^2 + Np^2 Lp^2}
= Ut

Hence:
Uto = [Bto^2 / 2 Muo]= [1 / 16] [Nt / Pi Ro]^2 Qs^2 C^2 / [Nt^2 Lt^2 + Np^2 Lp^2}

Thus:
Ut / Uto = (Ro / R)^2

Note that inside the spheromak wall the field energy density Ut is proportional to
(Ro^2 / R^2).

Recall that:
Lh^2 = [Nt^2 Lt^2 + Np^2 Lp^2}

Recall that:
F = C / Lh
 

Bpw SOLUTION:
At R = Rw, Z = Zw the electric field is:
Ers = Qs / (4 Pi Epsilono (Rw^2 + Zw^2)

The electric field energy density at radius R = Rw, Z = Zw is:
(Epsilono / 2) Ers^2
= (Epsilono / 2)[Qs / (4 Pi Epsilono (Rw^2 + Zw^2)]^2
= [1 / Epsilono] [Qs^2 / (32 Pi^2 (Rw^2 + Zw^2)^2)]

Recall that:
C^2 = 1 / (Epsilono Muo)
or
1 / Epsilono = Muo C^2
which gives the energy density in the electric field for R = Rw, Z = Zw as:
[Qs^2 / 32 Pi^2][Muo C^2 / (Rw^2 + Zw^2)2]

Outside the spheromak core energy density balance at the spheromak wall gives:
[Bpw^2 / 2 Muo] + [Qs^2 / 32 Pi^2][Muo C^2 / (Rw^2 + Zw^2)^2] = [Bto^2 / 2 Muo] [Ro / Rs]^2
or
[Bpw^2 / 2 Muo] = [Bto^2 / 2 Muo] [Ro / Rs]^2 - [Qs^2 / 32 Pi^2][Muo C^2 / (Rw^2 + Zw^2)^2]
or
[Bpw^2] = [Bto^2] [Ro / Rs]^2 - [Qs^2 / 16 Pi^2][Muo^2 C^2 / (Rw^2 + Zw^2)^2]
or
Bpw = {[Bto^2] [Ro / Rs]^2 - [Qs^2 / 16 Pi^2][Muo^2 C^2 / (Rw^2 + Zw^2)^2]}^0.5

In the spheromak core partial electric field cancellation causes:
Bpw to vary from:
Bpw = {[Bto^2] [Ro / Rs]^2 - [Qs^2 / 16 Pi^2][Muo^2 C^2 / (Rw^2 + Zw^2)^2]}^0.5
at R = Ro
to
Bpw = {[Bto^2] [Ro / Rs]^2
at R = Rc.

Lh^2 = (Np Lp)^2 + (Nt Lt)^2
where:
Lp = 2 Pi Ro

Thus:
Lh^2 = [Np Lp]^2 + [Nt Lt / Lp]^2 Lp^2
or
Lh^2 = Lp^2 [ Np^2 + Nt^2 (Lt / Lp)^2]

Recall that Lh = C / Fh. Hence:
C / Fh = (2 Pi Ro){Np^2 + Nt^2 (Lt / Lp)^2}^0.5
or
F / C = 1 /{2 Pi Ro [Np^2 + Nt^2 (Lt / Lp)^2]^0.5}
or
[1 / Ro] = 2 Pi [Np ^2 + Nt^2 (Lt / Lp)^2]^0.5 [F / C]

For any spheromak:

I = (Qs C / Lh)

Line Integral over Theta of [Bpw dLt] = Muo Np I

Due to mirror symmetry about Z = 0:
Line integral from R = Rc to R = Rs of:
[Btw dLt
= Muo Np I / 2

Recall that:
Bto = Muo I Nt / Lp
or
Bto^2 = [Muo I Nt / Lp]^2

Recall that:
I = Qs C / Lh

Hence outside the spheromak core:
Bto^2 = [(Muo Qs C Nt) / (Lh Lp)]^2

Hence:
Bpw = {[Bto^2] [Ro / Rw]^2 - [Qs^2 / 16 Pi^2][Muo^2 C^2 / (Rw^2 + Zw^2)^2]}^0.5
= {[(Muo Qs C Nt) / (Lh Lp)]^2 [Ro / Rw]^2 - [Qs^2 / 16 Pi^2][Muo^2 C^2 / (Rw^2 + Zw^2)^2]}^0.5
= {[(Muo Qs C Nt) / (Lh Lp)]^2 [Ro / Rw]^2 - [Qs^2 / 16 Pi^2][Muo^2 C^2 Ro^4 / Ro^4 (Rw^2 + Zw^2)^2]}^0.5
= {[(Muo Qs C Nt) / (Lh Lp)]^2 [Ro / Rw]^2 - [Qs^2 / 16 Pi^2][Muo^2 C^2 / Ro^4][Ro^4 /(Rw^2 + Zw^2)^2]}^0.5

Lp = 2 Pi Ro

Bpw = = {[(Muo Qs C Nt) / (Lh Lp)]^2 [Ro / Rw]^2 - [Qs^2 / 4][Muo^2 C^2 / Ro^2 Lp^2][Ro^4 /(Rw^2 + Zw^2)^2]}^0.5

dLt = [(dZw / dRw)^2 + 1]^0.5 dRw

Due to mirror symmetry about Z = 0:
Line Integral from Rw = Rc to Rw = Rs of:
Bpw dLt
= [Muo Np Qs C / 2 Lh]

Hence:
Line integral from Rw = Rc to Rw = Rs of:
[1/ Lp] {[(Nt / Lh)]^2 [Ro / Rw]^2 - [1/ 4][Ro^2 / (Rw^2 + Zw^2)^2]}^0.5 [(dZw / dRw)^2 + 1]^0.5 dRw

This integration needs two parts, one for Rc < Rw < Ro and one for Ro < Rw < Rs.

Hence:
Line integral from Rw = Rc to Rw = Rs of:
[2 Lh / Np] {[(Nt / Lh)]^2 [Ro / Rw]^2 - [1/ 4][Ro^2 / (Rw^2 + Zw^2)^2]}0.5 [(dZw/ dRw)^2 + 1]^0.5 dRw
= Lp
= 2 Pi Ro
 

USE THE LINE INTEGAL TO AROUND (Np I) TO DEVELOP THE SPHEROMAK WALL FUNCTI0N:
Hence:
Line integral from Rw = Rc to Rw = Rs of:
[Lh / 2 Np] {[(Nt / Lh)]^2 [1 / Rw]^2 - [1/ 4][1 / (Rw^2 + Zw^2)^2]}0.5 [(dZw / dRw)^2 + 1]^0.5 dRw
= [Pi / 2]

This equation is of the form:
Line integral from Rw = Rc to Rw = Rs of:
[Lh / 2 Np] {[C / Rw^2] - [D /(Rw^2 + Zw^2)^2]}^0.5 [(dZw / dRw)^2 + 1]^0.5 dRw
= [Pi / 2]

Now try the potential spheromak wall function:
(1 / (Rw^2 + Zw^2)^2 = [A / Rw^2] - B^2

Then:
(Zw^2 + Rw^2)^2 = Rw^2 /(A - B^2 Rw^2)
or
(Zw^2 + Rw^2) = +/- Rw /(A - B^2 Rw^2)^0.5
which is a distorted circle.

{[C / Rw^2] - [D / (Rw^2 + Zw^2)^2]}^0.5
= {[C / Rw^2] - D [[A / Rw^2]- B^2]}^0.5
= {[(C - (D A)) / Rw^2] + [D B^2]}^0.5
= {[D B^2]}^0.5
if:
A = C / D
= 4 [Nt / Lh]^2

Then:
Line integral from Rw = Rc to Rw = Rs of:
[Lh / 2 Np] {[C / Rw^2] - [D /(Rw^2 + Zw^2)^2]}^0.5 [(dZw / dRw)^2 + 1]^0.5 dRw
= [Lh /2 Np] {D B^2}^0.5 [(dZw / dRw)^2 + 1]^0.5 dRw
= [Pi / 2]
becomes:
Line integral from Rw = Rc to Rw = Rs of:
[Lh / 2 Np] {D B^2}^0.5 [(dZw / dRw)^2 + 1]^0.5 dRw
= [Pi / 2]
or
Line integral from Rw = Rc to Rw = Rs of:
[Lh / 4 Np] B [(dZw / dRw)^2 + 1]^0.5 dRw
= [Pi / 2]

Now recognize that:
Line integral from Rw = Rc to Rw = Rs of:
[(dZw / dRw)^2 + 1]^0.5 dRw = (Rs - Rc) Pi / 2

Thus:
Line integral from Rw = Rc to Rw = Rs of:
[Lh / 4 Np] B (Rs - Rc) Pi / 2
= Pi / 2
or
[Lh / 4 Np] B (Rs - Rc) = 1
or
B = 4 Np / [Lh (Rs - Rc)]
or
B^2 = 16 Np^2 / [Lh^2 (Rs - Rc)^2]
 

******************************************************************

DETERMINE THE FUNCTION THAT POSITIONS THE SPHEROMAK WALL:
Recall that the spheromak wall function is:
1 / (Zw^2 + Rw^2)^2 = (A / Rw^2) - B^2
= [4 (Nt / Lh)^2 (1 / Rw^2)] - [16 Np^2 / Lh^2(Rs - Rc)^2]
where:
A = 4 (Nt / Lh)^2
and
B = 4 Np / Lh (Rs - Rc)

This is the funamental spheromak wall equation. It describes a three dimensional spheromak wall in cylindrical co-ordinates. Note that a particular spheromak is identified by two constants, A and B.

For some purposes it is helpful to express this equation as:
1 / (Zw^2 + Rw^2)^2 = (A / Rw^2) - B^2
which in unitless normalized form becomes:
1 / [(Zw / Ro)^2 + (Rw / Ro)^2]^2
= Ro^4 / (Zw^2 + Rw^2)^2
= (Ro^4 A / Rw^2) - (Ro^4 B^2)
= [4 Nt^2 (Ro / Lh)^2 (Ro / Rw)^2] - [4 Np^2 (Ro / Lh)^2[(2 Ro)/(Rs - Rc)]^2]

Note that in normalized form a particular spheromak is characterized by its dimension constant Ro and by its shape constant:
[(Rs - Rc) / (2 Ro).

Note that the shape of a spheromak wall is superficially shaped like the inner tube of an old automotive pneumatic tire.

Note that, as shown later herein, to obtain two real solutions it is essential that:
(A / Rw^2) > B^2
 

GROSS RANGE OF VALIDITY CONSTRAINT:
In the spheromak wall function the LHS is always positive. Hence the spheromak wall function is only valid for Rw values that make the RHS positive. That range of validity is restricted to:
Rw^2 < (A / B^2)
or
Rw^2 < [4 (Nt /Lh)^2] / [4 Np / Lh(Rs - Rc)]^2}
or
Rw^2 < [(Nt^2 (Rs - Rc)^2/ 4 Np^2)]
or
[Rw^2 / (Rs - Rc)^2] < (Nt / 2 Np)^2

This inequality constrains the range of the turns ratio (Nt / 2 Np)^2.

The spheromak wall radius Rw is further restricted by the requirement that:
Rc^2 < Rw^2 < Rs^2
 

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FINDING Zw(Rw):
Recall that the original spheromak wall function is:
1/ (Zw^2 + Rw^2)^2 = [(A / Rw^2) - B^2]
which normalized to radius Ro becomes:
Ro^4 / (Zw^2 + Rw^2)^2
= 1 / [(Zw / Ro)^2 +(Rw / Ro)^2]^2
= [(A Ro^2 (Ro / Rw)^2] - [B^2 Ro^4]

[(Zw / Ro)^2 +(Rw / Ro)^2]^2
= 1 / {[(A Ro^2 (Ro / Rw)^2] - [B^2 Ro^4]}
= Rw^2 / {[(A Ro^2 Ro^2] - [B^2 Rw^2 Ro^4]}
= (Rw / Ro)^2 / {[(A Ro^2] - [B^2 Rw^2 Ro^2]}
= (Rw / Ro)^2 / {B^2 Ro^2[(A / B^2) - (Rw^2)]}

(Zw / Ro)^2 + (Rw / Ro)^2 = +/-(Rw / Ro) / {B^2 Ro^2[(A / B^2)- Rw^2]}^0.5)

Hence:
(Zw / Ro)^2 = [+/- (Rw / Ro)/ {B^2 Ro^2 [(A / B^2)- Rw^2]}^0.5)] - [(Rw / Ro)^2]
= [+/- (Rw / Ro)/ {B^2 Ro^4 [(A / (B^2 Ro^2))- (Rw / Ro)^2]}^0.5)] - [(Rw / Ro)^2]

 

PRACTICAL ROUTE TO (Zw / Ro):
Recall that:
Zw^2 = [+(Rw / {B^2 [(A / B^2)- Rw^2]}^0.5)] - [Rw^2]

The range of validity is:
Rw^2 < (A / B^2)
and will be further restricted by:
Rc < Rw < Rs

This expression has Zw = 0 at:
[+(Rw / {B^2 [(A / B^2)- Rw^2]}^0.5)] = [Rw^2]

Hence Zw = 0 at:
1 = Rw^2 [A - B^2 Rw^2
or
B Rw^4 - A Rw^2 + 1 = 0

This is a quadratic equation with positive real solutions at:
Rw^2 = {A +/- [A^2 - 4 B^2 (1)]^0.5} / 2 B^2
= [A / 2 B^2][1 +/- [1 - (4 B^2 / A^2)]^0.5

Define:
Ro^2 = [A / 2 B^2]

Then:
(4 B^2 / A^2) = (2 B^2 / A)(2 / A) = (2 / A Ro^2)

Thus at Zw = 0:
Rw^2 = Ro^2 {1 +/- [1 - (4 B^2 / A^2)]^0.5}
= Ro^2 {1 +/- [1 - (2 / A Ro^2)]^0.5
which gives:
Rs^2 = Ro^2 {1 + [1 - (2 / A Ro^2)]^0.5}
and
Rc^2 = Ro^2 {1 - [1 - (2 / A Ro^2)]^0.5}

Note that for a real spheromak:
(2 / A Ro^2) < 1
or
0 < (4 B^2 / A^2) < 1
which implies that:
(2 / A) < Ro^2
or
A Ro^2 > 2
Conclusion: For (A Ro^2) > 2 there are two positive Rw values where (Zw / Ro) goes through zero.
 

EVALUATE (Zw / Ro)^2 AT Rw = Ro
Note that Rw = Ro is part way between Rw = Rc and Rw = Rs.

Recall that:
Zw^2 = [+(Rw / {B^2 [(A / B^2)- Rw^2]}^0.5)] - [Rw^2]

At Rw = Ro:
Zw^2 = [+(Ro / {B^2 [(A / B^2)- Ro^2]}^0.5)] - [Ro^2]

Recall that:
Ro^2 = [A / 2 B^2]
Hence:
Zw^2 = [+(Ro / {B^2 [2 Ro^2 - Ro^2]}^0.5)] - [Ro^2]
Zw^2 = [+(Ro / {B^2 [2 Ro^2 - Ro^2]}^0.5)] - [Ro^2]
= [+(Ro / {B^2 Ro^2}^0.5)] - [Ro^2]
= [1 / B] - [Ro^2]

Recall that:
B = 4 Np / Lh (Rs - Rc)

Hence to obtain positive real solutions for Zw^2:
[1 / B] > [Ro^2]
or
[Lh (Rs - Rc) / 4 Np] > Ro^2
which is an important condition both for spheromak existence and for plotting a spheromak profile.

Recall that:
Ro^2 = [A / 2 B^2]

Hence:
For real Zw^2 solutions:
[1 / B] > [A / 2 B^2]
or
1 > [A / 2 B]

However,for positive real values of Rc and Rs there is a requirement that:
[4 B^2 / A^2] < 1.

It appears that for spheromaks:
(Zw / Ro)^2 takes -ve values, implying that spheromaks exist in imaginary Zw space.
Nevertheless we can still plot - (Zw / Ro)^2 versus (Rw / Ro)^2 to find spheromak shape in the (Zw / Ro)^2 vs (Rw / Ro)^2 domain.

We can also plot +/- {[- (Zw / Ro)^2]^0.5} versus (Rw / Ro) to find the spheromak wall profile

FINDING PARAMETER "A" FROM (Rs / Rc):
The parameter (Rs / Rc) can easily be measured on plasma spheromak photographs and becomes the input data to the following equations.
Typically (Rs / Rc) ~ 4. The resulting values of (2 / A Ro^2) are an input to the spheromak profile equation.

Recall that:
[Rs^2 / Rc^2] = {1 + [1 - (2 / A Ro^2)]^0.5} / {1 - [1 - (2 / A Ro^2)]^0.5}
Hence:
Rc^2{1 + [1 - (2 / A Ro^2)]^0.5} = Rs^2{1 - [1 - (2 / A Ro^2)]^0.5}
Hence:
(Rs^2 + Rc^2)[1 - (2 / A Ro^2)]^0.5 = Rs^2 - Rc^2
Hence:
[1 - (2 / A Ro^2)]^0.5 = (Rs^2 - Rc^2 )/ (Rs^2 + Rc^2)
Hence:
1 - (2 / A Ro^2) = [(Rs^2 - Rc^2 )/ (Rs^2 + Rc^2)]^2
or
(2 / A Ro^2) = 1 - [(Rs^2 - Rc^2 )/ (Rs^2 + Rc^2)]^2
= [(Rs^2 + Rc^2)^2 - (Rs^2 - Rc^2)^2] / [Rs^2 + Rc^2]^2 = 4 Rs^2 Rc^2 / (Rs^2 + Rc^2)^2
=4 (Rs / Rc)^2 / [(Rs/ Rc)^2 + 1]^2
 

FOR (Rs / Rc) = 4:
(2 / A Ro^2) = = 4 (Rs / Rc)^2 / [(Rs/ Rc)^2 + 1]^2
= 64 / (17)^2

[1 - (2 / A Ro^2)]^0.5
= [1 - [64 / (17)^2]^0.5
= [(17^2 - 64 )/ 17^2]^0.5
= [(289 - 64) / 289]^0.5
= [225/ 289]^0.5
= (15 / 17)

A = (2 / Ro^2) (17^2 / 64)

Note that the requirement that (A Ro^2) > 2 that is needed for real values of Rc^2 and Rs^2 is met.

The practical way to plot the position of a spheromak wall is:
a)Choose value of:
(Rs / Rc):
Eg (Rs / Rc) = 4 observed in plasma experiments

b) Calculate:
[2 / (A Ro^2)]^0.5 = 2 (Rs / Rc) / [(Rs / Rc)^2 + 1]
For (Rs / Rc) = 4:
[2 / (A Ro^2)]^0.5 = 8 / 17

c) For (Rs / Rc) = 4 Calculate:
[2 / A Ro^2] = {[2 / (A Ro^2)]^0.5}^2
= 64 / 289

d) For (Rs / Rc) = 4 Calculate:
(Rc / Ro) = {1 - {1 - (2 / A Ro^2)}^0.5}^0.5
= {1 - {1 - (64 / 289)}^0.5}^0.5
={1 - {225 / 289}^0.5}^0.5
= {1 - (15 / 17)}^0.5
= {2 / 17}^0.5
= 0.3429971703

Note that (Rc / Ro)^2 = (2 / 17)

e) For (Rs / Rc) = 4 Calculate:
(Rs / Ro) = {1 + {1 - (2 / A Ro^2)}^0.5}^0.5
= {1 + {1 - (64 / 289)}^0.5}^0.5
= {1 + {225 / 289)}^0.5}^0.5
= {1 + (15 / 17)}^0.5
= {32 / 17)}^0.5
= 1.371988681

Note that (Rs / Ro)^2 = (32 / 17)

f)For (Rs / Rc) = 4: Form function:
(Zw / Ro)^2 = [((2 / A Ro^2)^0.5 ) / [{(2 (Ro/ Rw)^2 - 1}^0.5)]] - [(Rw / Ro)^2]
= [(8 / 17) / [{(2 (Ro / Rw)^2 - 1}^0.5)]] - [(Rw / Ro)^2]

Use this function to calculate (Zw / Ro)^2 values for (Rw / Ro) varying from (Rc / Ro) to (Rs / Ro).

g) Choose initial value of (Rw / Ro). Typically start at (Rw / Ro) = (Rc / Ro)

h) For (Rs / Rc) = 4 calculate:
(Zw / Ro) at (Rw / Ro)^2 = (Rc / Ro)^2 = (2 / 17)
The result should be zero.

i) (Zw / Ro)^2 = [((2 / A Ro^2)^0.5 ) / ({(2 (Ro / Rw)^2) - 1}^0.5)] - [(Rw / Ro)^2]

j) Calculate:
(Zw / Ro) = +/- [(Zw / Ro)^2]^0.5

k) Increment (Rw / Ro)

l) Repeat steps (g) to (k) above incrementing (Rw / Ro) until (Rw / Ro) = (Rs / Ro)

m) Plot [-(Zw / Ro)^2]^0.5 versus (Rw / Ro) for Rc < Rw < Rs

n) Plot same data for corresponding -ve values of Rw

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CARTESIAN GRAPH PLOTTING:
The objective of graph plotting is to plot:
Y = +/- {[- (Zw / Ro)^2]^0.5} versus X = (Rw / Ro) in four quadrents
Remember that Zw contains the factor i (square root of -ve one) so that (Zw / Ro)^2 is a negative number. Hence to plot (Zw / Ro) versus (Rw / Ro) on a normal apparatus it is necessary to calculate:
[- (Zw / Ro)^2]^0.5
During the plotting process for each argument point in the range:
(Rc / Ro)^2 < (Rw / Ro)^2 < (Rs / Ro)^2
it is necessary to:
a)Increment X = (Rw / Ro)
b) Calculate X^2 = (Rw / Ro)^2
c)Use the function provided to calculate (Zw / Ro)^2 which is a negative quantity;
d)Calculate Y = +/-{[- (Zw / Ro)^2]^0.5}
e)Plot Y versus X in four quadrants

For the chosen spheromak shape condition of (Rs / Rc) = 4:
The argument is: X = (Rw / Ro)
The argument range is:
(Rc / Ro) < (Rw / Ro) < (Rs / Ro)
or
(Rc / Ro)^2 < (Rw / Ro)^2 < (Rs / Ro)^2
depending on how the apparatus is configured.

(Rc / Ro)^2 = (2 / 17)

(Rs / Ro)^2 = (32 / 17)

The function is:
(Zw / Ro)^2 = [((2 / A Ro^2)^0.5 ) / ({(2 (Ro / Rw)^2) - 1}^0.5)] - [(Rw / Ro)^2]
where:
(2 / A Ro^2)^0.5 = [8 / 17]

Remember that we are trying to plot:
Y = +/- {[-(Zw / Ro)^2]^0.5} versus X = (Rw / Ro)

Don't try to plot (Zw / Ro) directly unless you are certain that your math package will correctly process the imaginary numbers. The section prior to this one contains my manual calculations of representative points. There are function zeros at (Rw / Ro) = (Rc / Ro) and at (Rw / Ro) = (Rs / Ro).

MANUAL POINT EVALUATION:
(Zw / Ro)^2 = [((2 / A Ro^2)^0.5 ) / ({(2 (Ro / Rw)^2) - 1}^0.5)] - [(Rw / Ro)^2]
= [(8 / 17) / ({(2/ X^2) - 1}^0.5)] - [X^2]

For X^2 = 2 / 17, (Zw / Ro)^2 = 0
Hence for X = 0.3429983364, (Zw / Ro) = 0

For X^2 = 5 / 17,
(Zw / Ro)^2 = [(8 / 17) / ({2 (17 / 5) -1}^0.5)] - [5 /17]
[(8 / 17) / ({29 / 5}^0.5)] - [5 /17]
= [0.4705882353 /2.408318916] - 0.2941176471
= -.098716518

For X^2 = 8 / 17,
(Zw / Ro)^2 = [(8 / 17) / ({2 (17 / 8) -1}^0.5)] - [8 /17]
= [(8 / 17) / ({26 / 8}^0.5)] - [8 / 17]
=[(0.4705882353 / (1.802775638)] - 0.4705882353
= -.2095528489

Hence at X = [8 / 17]^0.5 = 0.6859943406,
[-(Zw / Ro)^2]^0.5 = .4227745664

For X^2 = 17 / 17,
(Zw / Ro)^2 = [((2 / A Ro^2)^0.5 ) / ({(2 (Ro / Rw)^2) - 1}^0.5)] - [(Rw / Ro)^2]
= [((8 / 17 ) / 1] - [1]
= -(9 / 17)
Hence for X = 1, [-(Zw / Ro)^2]^0.5 = 0.7276068719

For X^2 = 25 / 17
(Zw / Ro)^2 = [(8 / 17) / ({2 (17 / 25) - 1}^0.5)] - [25 /17]
= [(8 / 17))] / (3 / 5)] - [25 /17]
= [40 / 3(17)] - [25 / 17]
= -.6862745098

For X^2 = 28 / 17
(Zw / Ro)^2 = [(8 / 17) / ({2 (17 / 28) - 1}^0.5)] - [28 /17]
= [(8 / 17) / ({6 / 28}^0.5)] - [28 /17]
= [0.4705882353 / 0.4629100499] - 1.647058824
= -.6304720478

Hence at X = 1.283377896, [-(Zw / Ro)^2]^0.5 = 0.7940226998

For X^2 = 32 / 17,(Zw / Ro)^2 = 0
Hence for X = 1.371988681, (Zw / Ro) = 0

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SPHEROMAK SIZE AND SHAPE:
The size of a spheromak is set by Ro where:
Ro^2 = (A / 2 B^2)

The parameter that sets the shape of a spheromak is:
[A / 2 B]
Note that as shown below, this important parameter is a function of (Rs / Rc). If we specify (Rs / Rc) in effect we set (A / 2 B).
Typically in plasma spheromaks:
(Rs / Rc) ~ 4
 

HELPFUL IDENTITY:
Rs^2 / Ro^2 = [1 + [1 - (4 B^2 / A^2)]^0.5}
and
Rc^2 / Ro^2 = [1 - [1 - (4 B^2 / A^2)]^0.5}

Hence:
(Rs^2 / Ro^2) - 1 = - [(Rc^2 / Ro^2) - 1]
or
Rs^2 + Rc^2 = 2 Ro^2
 

FIND (Rs / Ro) AND (Rc / Ro) IN TERMS OF (Rs / Rc):
Rs^2 + Rc^2 = 2 Ro^2
(Rs / Rc)^2 Rc^2 + Rc^2 = 2 Ro^2
[(Rs / Rc)^2 + 1] Rc^2 = 2 Ro^2
Hence:
Rc^2 / Ro^2 = 2 / [(Rs / Rc)^2 + 1]
or
(Rc / Ro) = {2 / [(Rs / Rc)^2 + 1]}^0.5

Similarly:
Rs^2 + Rc^2 = 2 Ro^2
Rs^2 + (Rc / Rs)^2 Rs^2 = 2 Ro^2
Rs^2 [(1 + (Rc / Rs)^2] = 2 Ro^2
Rs^2 / Ro^2 = 2 / [(1 + (Rc / Rs)^2] (Rs / Ro) = {2 / [(1 + (Rc / Rs)^2]}^0.5

These are important relationships used for plotting the spheromak wall function.
 

DETERMINE THE [Rs^2 Rc^2] PRODUCT:
Rs^2 Rc^2 = Ro^2 [1 - [1 - (4 B^2 / A^2)]^0.5] Ro^2 [1 + [1 - (4 B^2 / A^2)]^0.5]
= Ro^4 [1 - [1 - (4 B^2 / A^2)]]
= Ro^4 [4 B^2 / A^2]
= Ro^4 [1 / Ro^2 B^2]
= Ro^2 / B^2

Hence:
B^2 = Ro^2 / Rs^2 Rc^2
or
B = Ro / Rs Rc
= [4 Np / Lh (Rs - Rc)]
 

FIND (Rs - Rc) IN TERMS OF Ro:
Recall that:
Rs^2 = [A / 2 B^2]{1 + [1 - (4 B^2 / A^2)]^0.5}
= Ro^2 {1 +[1 + [1 - (1 / (Ro^2 B^2)]^0.5

Rc^2 = [A / 2 B^2]{1 - [1 - (4 B^2 / A^2)]^0.5}
= Ro^2 {1 - [1 - (1 /(Ro^2 B^2)]^0.5

Rs = [A / 2 B^2]^0.5 {1 + [1 - (4 B^2 / A^2)]^0.5}^0.5
and
Rc = [A / 2 B^2]^0.5 {1 - [1 - (4 B^2 / A^2)]^0.5}^0.5
 

FIND RELATIONSHIIP BETWEEN [Rs / Rc] AND: [2 B / A]:
Rs^2 / Rc^2 = {1 + [1 - (4 B^2 / A^2)]^0.5} /{1 - [1 - (4 B^2 / A^2)]^0.5}
or
[Rs / Rc]^2 {1 - [1 - (4 B^2 / A^2)]^0.5} = {1 + [1 - (4 B^2 / A^2)]^0.5}
or
(Rs / Rc)^2 - 1 = [(Rs^2 / Rc^2) + 1][1 - (4 B^2 /A^2)]^0.5
or
[(Rs^2 - Rc^2) / (Rs^2 + Rc^2)]^2 = 1 - (4 B^2 / A^2
or
(4 B^2 / A^2) = 1 - [(Rs^2 - Rc^2) / (Rs^2 + Rc^2)]^2
= [Rs^4 + 2 Rs^2 Rc^2 + Rc^2 + Rs^4 - 2 Rs^2 Rc^2 + Rc^4] / [Rs^2 +Rc^2)]^2
= 2 [Rs^4 + Rc^4] /[Rs^2 + Rc^2]^2

Hence:
[2 B / A] = [2 (Rs^4 + Rc^4)]^0.5 / [Rs^2 + Rc^2]

This is an important relationship in spheromak analysis because for plasma spheromaks the ratio [Rs/ Rc] is readily available from plasma photographs.
 

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PLOTTING EXAMPLES:

CHARACTERIZE A TYPICAL PLASMA SPHEROMAK:
For a typical plasma spheromak:
FIND [2 B / A] = [1 / (B Ro^2)]:
If:(Rs / Rc)= 4, then:
[1 / (B Ro^2)] = [2 B / A] = [2 (Rs^4 + Rc^4)]^0.5 / [Rs^2 + Rc^2]

FIND (Rc / Ro):
If: (Rs / Rc)= 4, then:
(Rc / Ro) = {2 / [(Rs / Rc)^2 + 1]}^0.5
= {2 / 17}^0.5 = 0.3429971703

FIND (Rs / Ro):
If: (Rs / Rc) = 4, then:
(Rs / Ro) = {2 / [(1 + (Rc / Rs)^2]}^0.5
= {2 / [(1 + (1 / 16)]}^0.5
= {32 / 17}^0.5
= 1.371988681

***********************************************************

Experimental plasma spheromaks tend to be stable at:
Rs ~ 4 Rc

(Rs - Rc) = Ro ({1 + [1 - (1 / Ro^2 B^2)]^0.5}^0.5 - {1 - [1 - (1 / Ro^2 B^2)]^0.5}^0.5)
which is critically dependent on the(B Ro) product.

Recall that:
B = [4 Np / Lh (Rs - Rc)]

Clearly a spheromak is characterized by two constants. They are:
[1 / B] = [Lh (Rs - Rc) / 4 Np]
and
[A / 2 B] = [4 Nt / Lh]^2 [1 / 2] = B Ro
where:
[2 B / A]^2 < 1

Note that as an alternative the spheromak wall can be specified via A and Ro, where Ro sets the spheromak size and A sets the spheromak shape.

Recall that:
Ro^2 = (A / 2 B^2)

Hence:
[A / 2 B] = B Ro^2

Hence the normalized spheromak wall profile becomes:
[Zw / Ro] = +/- [Rw / Ro] {-1 + [1 / [(A - B^2 Rw^2) Rw^2]]^0.5}^0.5

This normalized (Zw / Ro) profile is valid for:
Rw > 0
and
Rc < Rw < Rs

For the special case of Rs = 4 Rc:
(Rc / Ro) = 0.342997
and
(Rs / Ro) = 1.371988681
 

At X = 1:
[Zw / Ro] = +/- {-1 + [1 / (B Ro^2)]}^0.5

For the special case of Rs = 4 Rc:
[1 / (B Ro^2)] = 1.3336
giving:
[Zw / Ro] = + / - {-1 + 1.3336}^0.5 = +/- 0.57758

Then:
2 Ho / Ro = 1.15516
 

We can plot intermediate computed points at X = (Rw / Ro) values of:
0.35, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3

Hence for Rs / Rc = 4:
[Zw / Ro] = +/- [X]{-1 + [[1.3336] /[[(2 - X^2)(X^2)]^0.5]]}^0.5
 

Function Evaluation:
For X = 0.4:
[Zw / Ro] = +/- [X]{-1 + [[1.3336] /[[(2 - X^2)(X^2)]^0.5]]}^0.5
= +/- [0.4]{-1 + [[1.3336] /[[(2 - 0.16)(0.16)]^0.5]]}^0.5
= +/- [0.4]{-1 + [[1.3336] /[0.542586]]}^0.5
= +/- [0.4]{1.45785}^0.5
= +/- 0.48296

FINDING (Np / Nt)^2:
Note that the ratio:
(Nt / Np) is tied to the ratio (Lp / Lt) through the stability requirement that:
(Np Lp / Nt Lt)= [Mp / Mt] ~ (1 / 2).
The ratio:
(Lp / Lt)
comes from the spheromak geometry.
 

SPHEROMAK SHAPE AND ITS CONSTANT B^2:
Recall that:
1 / (Zw^2 + Rw^2)^2 = (A / Rw^2) - B^2

Since via [Mp / Mt] there are multiple possible solutions for B. The real solution is selected by minimization of the total system energy. The other issue is that [Np / Nt] is a ratio of integers which is tied to the integer ratio in [Mp / Mt].

Note that in the region Rc < Rw < Rs
2 Ro^2 > Rw^2
 

FIND SPHEROMAK RELATIVE HEIGHT(2 Ho / Ro) AT Rw = Ro:
Recall that:
= Zw / Ro = +/- [Rw / Ro] {-1 + [1 / (B Ro^2)][1 /[(2 - (Rw / Ro)^2) (Rw / Ro)^2]]^0.5}^0.5

At Rw = Ro:
= (Ho / Ro) = +/- (1){[- 1 + [1 / (B Ro2)]][1 /(2 - 1)(1)]]^0.5}^0.5
= {(2 B / A) - 1}^0.5

Thus:
2 (Ho / Ro) = 2 {(2 B / A) - 1}^0.5
 

DETERMINE THE SPHEROMAK'S Lt VALUE:
This calculation has been moved to a dedicated web page titled:
Spheromak Lt Calculation
 

FINDING (Np / Nt):
Recall that from the definitions of A and B:
A = 4 (Nt / Lh)^2
B = 4 Np / Lh (Rs - Rc)

Hence:
(2 B / A) = 2 [4 Np / Lh (Rs - Rc)] / [4 (Nt / Lh)^2]
= 2 (Np / Nt^2) [Lh / (Rs - Rc)]

From the web page titled: Spheromak Winding ConstraintsrRecall that for Family A spheromaks:
[(Nt Lt) / (Np Lp)] = 2 +/-(1 / Mpa)
or
(Nt Lt)^2 = (Np Lp)^2 [2 +/- (1 / Mpa)]^2
= (Np Lp)^2 [4 +/- 2 / Mpa]

Lh^2 = (Nt Lt)^2 + (Np Lp)^2
= (Np Lp)^2 [4 +/- 2 / Mpa] + (Np Lp)^2
= 5 (Np Lp)^2 +/- [(2 / Mpa)(Np Lp)^2

Hence:
Lh = {5 (Np Lp)^2 +/-[2 / Mpa)(Np Lp)^2}^0.5
= 5^0.5 Np Lp [1 +/- (2 / Mpa) / 5]^0.5
= 5^0.5 Np Lp [1 +/- (1 / 5 Mpa)]

Hence:
(2 B / A) = 2 (Np / Nt^2)[Lh / (Rs - Rc)]
= 2 (Np / Nt)^2 (5^0.5) Lp [1 +/- (1 / 5 Mpa) / (Rs - Rc)

Recall that Lp = 2 Pi Ro giving:
(2 B / A) = 2 (Np / Nt)^2 (5^0.5) 2 Pi Ro [1 +/- (1 / 5 Mpa)] / (Rs - Rc)]

Recall that:
[2 B / A] = [2 (Rs^4 + Rc^4)]^0.5 / [Rs^2 + Rc^2]

Solve these two equations to find (Np / Nt)^2

Examine the term:
(Rs - Rc) / Ro

Recall that: (Rs - Rc) / Ro = ({1 + [1 - (1 / Ro^2 B^2)]^0.5}^0.5 - {1 - [1 - (1 / Ro^2 B^2)]^0.5}^0.5)
We need the Lt line integral to quantitatively determine the parameter (Ro B)^2.

RATIO OF Rs / Rc
Note that (Np / Nt)^2 is only very weakly dependent on the ratio: X = (Rs / Rc)
 

PLOTTING (Zw / Ro) VERSUS X:

For a typical plasma spheromak:
(Rs / Rc) = 4:
implying that:
(2 B / A) = (1 / B Ro^2) = (1.3336)

At X = 1:
[Zw / Ro] = +/- {-1 + [1 / (B Ro^2)]}^0.5

For the special case of Rs = 4 Rc:
Then:
[Zw / Ro] = + / - {-1 +(4 /3)}^0.5 = (1 / 3)^0.5 = 0.57735

Hence for a typical plasma spheromak with (Rs / Rc) = 4:
(1.3336) = 2 (Np / Nt)^2 (5^0.5 Lp) / (Rs - Rc)

Recall that:
Lp = 2 Pi Ro
giving:
(5 / 13)= 2 (Np / Nt)^2 (5^0.5) 2 [Pi Ro / (Rs - Rc)]
or
(5 / 52) = (Np / Nt)^2 (5^0.5)[Pi Ro / (Rs - Rc)]
or
(5^0.5) / 52 = (Np / Nt)^2 ---------

Note that:
Nt^2 ~> 4 Ro^2 Np^2 / (Rs - Rc)^2
or
[Nt^2 / Np^2] ~> [4 Ro^2 / (Rs - Rc)^2]
 

WINDING FAMILY SELECTION:
In the Family B winding stability analysis it is shown that:
(Np Lp / Nt Lt) ~ 2
giving:
(Nt Lt)^2 = (Np Lp)^2 / 4

Hence:
(Nt^2 / Np^2) = [Lp^2 / 4 Lt^2]

Hence for Family B:
[Lp^2 / 4 Lt^2] ~> [4 Ro^2 / (Rs - Rc)^2]

However:
Lp = 2 Pi Ro
giving:
4 Pi^2 Ro^2 / 4 Lt^2 ~> [4 Ro^2 / (Rs - Rc)^2
or
(Pi^2 / Lt^2)~> 4 / (Rs - Rc)^2
or
(Pi / Lt ) ~> 2 / (Rs - Rc) or
(1 / Lt) ~`> 2/ Pi (Rs - Rc) or
Lt < [Pi (Rs - Rc) / 2 ] However, geometry requires that Lt > 2 (Rs - Rc). Hence Family B does not describe real spheromaks.
 

In the Family A winding stability analysis it is shown that:
(Np Lp)/ (Nt Lt) = (1 / 2)
giving:
[(Nt Lt) / (Np Lp)]^2 = 4

Hence:
[Nt / Np]^2 = 4 {Lp / Lt}^2

Recall that:
[Nt^2 / Np^2] ~> [4 Ro^2 / (Rs - Rc)^2]

Recall that:
Lp = 2 Pi Ro

Thus:
4 [2 Pi Ro / Lt]^2 ~> [4 Ro^2 / (Rs - Rc)^2]
or
[4 Pi^2 / Lt ^2] ~> [1 / (Rs- Rc)^2]
or
[1 / Lt^2] ~> [1 / 4 Pi^2 (Rs - Rc)^2]
or
[1 / Lt] ~> [1 / 2 Pi (Rs - Rc)]

This inequality describes a real Lt path that is a squished circle. This inequality also limits the range of the constant B^2.

1 / (Zw^2 + Rw^2)^2 = [4 (Nt / Lh)^2 (1 / Rw^2)] - [16 Np^2 / Lh^2(Rs - Rc)^2]
where:
A = 4 (Nt / Lh)^2
and
B^2 = 16 Np^2 / Lh^2 (Rs - Rc)^2
or
B = [4 Np / Lh (Rs - Rc)]

The spheromak characterizing constants are:
[A / 2 B] = [4 Nt^2 / Lh^2] / [8 Np / Lh (Rs - Rc)]
= [Nt^2 / 2 Lh Np] = B Ro
and
B = [4 Np / Lh (Rs - Rc)]

An alternative way of expressing the spheromak wall position formula is:

[Lh^2 / 4] / (Zw^2 + Rw^2)^2 = [(Nt / Rw)^2 - 4 (Np^2 / (Rs - Rc)^2)]
where:
Lh^2 = (Np Lp)^2 + (Nt Lt)^2

For real (Family A) spheromaks:
[(Np Lp) / (Nt Lt)] ~ (1 / 2)

Lp = 2 Pi Ro

Ro = [1 / 8]^0.5 [Nt (Rs - Rc) / Np]
Rc^2 = Ro^2 - [1 / 8][Lh (Rs - Rc) / Np] {(Nt^4 (Rs - Rc)^2)/ (Lh^2 Np^2))- 4}^0.5
Rs^2 = Ro^2 + [1 / 8][Lh (Rs - Rc) / Np] {(Nt^4 (Rs - Rc)^2)/ (Lh^2 Np^2))- 4}^0.5

The spheromak will likely adopt a (2 B/ A) value that minimizes the spheromak total energy content.

Note that via:
[(Np Lp) / (Nt Lt)] ~ (1 / 2)
the (Np / Nt) ratio is set by the (Lp / Lt) ratio, which is set by the line integral for length Lt.

This line integral for Lt must be accurately calculated.

Recall that:
Lh^2 = (Np Lp)^2 + (Nt Lt)^2
and (Np Lp / Nt Lt) = (Mp / Mt) ~ (1 / 2)
[(Np Lp) / (Nt Lt)]^2 = (Mp / Mt)^2 ~ (1 / 4)
Lh^2 = (Np Lp)^2 + (Mt / Mp)^2 (Np Lp)^2
= [1 + (Mt / Mp)^2](Np Lp)^2
~ 5 (Np Lp)^2
This calculation is only approximate. For exact rsults it is necessary to replace the 5 with the quantity:
[1 + (Mt / Mp)^2].

Thus in exact calculations:
Lh^2 = [1 + (Mt / Mp)^2][Np Lp]^2
 

MAGNETIC FIELD ENERGY CONTAINED INSIDE THE SPHEROMAK WALL:
E = Integral from Rw = Rc to Rw = Rs of:
2 Zw 2 Pi Rw dRw [Bto^2 / 2 Muo](Ro / Rw)^2

Recall that:
(Zw^2 + Rw^2) = Rw /(A - B^2 Rw^2)^0.5
or
Zw^2 = [Rw /(A - B^2 Rw^2)^0.5] - Rw^2
or
Zw = {[Rw /(A - B^2 Rw^2)^0.5] - Rw^2}^0.5
= Rw {1 / Rw (A - B^2 Rw^2)^0.5 - 1}^0.5

Hence:
E = Integral from Rw = Rc to Rw = Rs of:
{[1 / (Rw (A - B^2 Rw^2)^0.5)] - 1}^0.5 dRw] [Bto^2 / 2 Muo] 4 Pi Ro^2

Recall that:
A = 4 [Nt / Lh]^2
and
B^2 = 16 Np^2 / [Lh^2 (Rs - Rc)^2]

Note that the integrand has a sharp peak at:
Nt^2 = [4 Np^2 Rw^2 / (Rs-Rc)^2
or
Nt = 2 Np Rw / (Rs - Rc)
or
Rw = [(Rs - Rc) / 2] [Nt / Np]

Note that the small difference between Nt^2 and [4 Np^2 Rw^2 / (Rs - Rc)^2 causes Nt to have many turns that increase the value of Bto and hence the spheromak energy and the Planck constant.

Recall that for the case of F = 4:
Lh (Rs - Rc)= Zo^2 4 Np
or
Np / (Rs - Rc) = [Lh / 4 Zo^2]
giving:
Rw (A -B^2 Rw^2)^0.5 = (2 Rw / Lh){Nt^2 - 4 Np^2 Rw^2 / (Rs - Rc)^2}^0.5
= (2 Rw / Lh){Nt^2 - 4 Lh^2 Rw^2 / 16 Zo^4}^0.5
= {[4 Rw^2 Nt^2 / Lh^2] - [Rw^4 / Zo^4]}^0.5

Then:
Integral from Rw = Rc to Rw = Rs of:
{[1 / (Rw (A - B^2 Rw^2)^0.5)] - 1}^0.5 dRw]
= Integral from Rw = Rc to Rw = Rs of:
= {[1 / {[4 Rw^2 Nt^2 / Lh^2] - [Rw^4 / Zo^4]}^0.5] - 1}^0.5 dRw

This integration may need to be numerical.
 

EFFECT OF EXTERNAL MAGNETIC FIELD:
Recall that: Bpor = Muo Np I / 2 Ro
and
Bto = Muo Nt I / 2 Pi Ro

Hence:
Bpor / Bto = Np Pi / Nt
or
(Bpor / Bto)^2 = Np^2 Pi^2 / Nt^2

Hence:
Bto^2 = Bpor^2 Nt^2 / Np^2 Pi^2

Similarly if the spheromak poloidal magnetic field and the external magnetic field are not aligned: Bpor^2 changes to Bpor^2 - 2 Bpor Be

Change in toroidal field energy density is = 4 Bpor Be / 2 Muo

The change in spheromak contained toroidal magnetic field energy is:
Delta E = Integral from R = Rc to R = Rs of:
{[1 / (Rw (A - B^2 Rw^2)^0.5)] - 1}^0.5 dRw] [4 Bpor Be / 2 Muo] [4 Ro^2 Nt^2 / (Np^2 Pi)]
CONTINUE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX Q = 1.602 X 10^-19 coul
Muo = 4 Pi X 10^-7 web / amp -m
C = 2.997 X 10^8 m / s

Q^2 Muo C / 8 = [2.5664 X 10^-38 / 8][4 Pi X 10^-7][2.997 X 10^8]
= 12.082 X 10^-37

 

Assume that based on plasma spheromak photographs:
Rs = 4 Rc

To a first approximation spheromak core magnetic field Bpo is given by: Bpo = Muo Np I / 2 Ro

Spheromak core energy density is:
Upor = Bpor^2 / 2 Muo
= [Muo Np I / 2 Ro]^2 [ 1 / 2 Muo]
= Muo Np^2 I^2 / 8 Ro^2

Inside a central sphere of radius Rc the field energy of a spheromak is approximately:
Upor [4 Rc^3 / 3]
= [Muo Np^2 I^2 / 8 Ro^2][4 Rc^3 / 3]
= Muo Np^2 I^2 Rc^3 / 6 Ro^2

Spheromak existence requirement is:
I = Qs C / Lh

Hence:
I^2 = Qs^2 C^2 / Lh^2

Muo = 4 Pi X 10^-7 ______
Qs = 1.602 X 10^-19 coul
C = 2.997 X 10^8 m / s
giving:
h = Np [(Muo Qs^2 C) / 2^1.5 Pi]
= Np X 4 Pi X 10^-7____ X (1.602)^2 X 10^-38 coul X 2.997 X 10^8 m / s X (1 / 2^1.5 Pi) x [(4 / 3) + (Pi / 4)]
= Np X 2 X 5.429 X 10^-37 joule second X [(4 / 3) + (Pi / 4)]
~ Np X 23 X 10^-37 joule second

The experimental value of h is:
h = 6.62607015 × 10−34 joule second
suggesting that:
Np ~ Nt
~ [6.62607015 × 10−34 joule second] / [23 X 10^-37 joule second]
~ 288

Thus we have developed a crude expression for the Planck constant:
h = (E / F)

However, to get the lead coeeficient right we cannot use the approximation that (Np Lp) / (Nt Lt) = (1 / 2). Instead exact expressions must be used. that are mathematically much more complex.

We need a more accurate calculation of the field energy distribution outside the spheromak wall.

The spheromak wall will tend to adopt the fixed geometric parameter values (Rc / Ro), (Rs / Ro) and (Ho / Ro) that result in a stable spheromak wall position. The spheromak wall geometry will determine K. The ratios (Rc / Ro) and Rs / Ro) are set by boundary conditions at (Rc, 0) and (Rs, 0). The integers Np and Nt arise from application of prime number theory to the aforementioned parameters.

The boundary condition at (Rc, 0) generates a factor of about Pi in the ratio of (Np / Nt) which in large measure determines the integers Np and Nt. At any particular controlling prime number P the Np and Nt values increment and decrement together to find the best Np, Nt number pair for meeting this boundary condition. The mechanism by which a spheromak finds its controlling prime number P is not yet fully understood.
The likey source is 2^0.5 relationship which tends to produce a P = 577. 8^0.5 P = integer
8^0.5 (577) = 1632.002451
 

This web page last updated Aug.17, 2025.