XYLENE POWER LTD.

SPHEROMAK APPROXIMATION:

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

ATOMIC PARTICLE SPHEROMAKS:
Atomic particle spheromaks have a quantized charge that superficially appears to be at rest with respect to an external 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.

The object of this web page is to develop relatively simple expressions that qualitiatively describe spheromak behavior. Spheromak geometry is complex, so it is unrealistic to expect these expresions to accurately predict the Planck Constant or the Fine Structure Constant.

In an atomic particle spheromak a quantum net charge Q is uniformly distributed along a closed spiral filament of length Lh. This charge moves axially along the filament path at the speed of light C causing a current:
I = Q C / Lh.
The uniform charge distribution along the current path and hence the uniform current cause constant electric and magnetic fields.

The time until an element of charge retraces its previous path is:
(1 / Fh) = Lh / C
where Fh is the spheromak's characteristic frequency.

An isolated spheromak in free space has an ellipsoidal cross section. In a solid the spheromak cross section may be distorted by external electric and magnetic fields from other nearby particle and atomic spheromaks.
 

SPHEROMAK CONCEPT:
Conceptually a spheromak wall is an ellipsoidal-toroidal shaped surface formed from the current path of a spheromak.

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

In a spheromak quantized net charge moves along a closed spiral path which defines 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. In the region 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 and the electric field is radial normal to the outer surface of the spheromak wall. Due to the distributed charge along the path length Lh the spheromak has a surface charge S(R). As shown on the web page titled: Spheromak Wall:
S(R) = So (Ro / R)
where:
So = Q / (Lp Lt) where: Lp = 2 Pi Ro
and
Lt = ellipse perimeter length.

In the far field the electric field is spherically radial. The current:
I = Q C / Lh
circulates along the filament on the surface of the spheromak wall which forms the interface between the toroidal magnetic field region and the mixed electric and poloidal magnetic field region.

Within the spheromak core the electric field components cancel each other out.
 

SPHEROMAK DIAGRAM:

On the above diagram:
Ho = 2.156
Rs = 2.156
Rc = 0.4802
 

SPHEROMAK GEOMETRIC PARAMETERS:
Assume that the spheromak shape is a symmetrical ellipsoidal toroid.
R = a radius from the spheromak major axis, herein termed theZ axis;
Rw = a value of R on the spheromak wall;
Rs = maximum spheromak wall radius on the equatorial plane;
Rc = minimum spheromak wall radius on the equatorial plane:
Ro = Spheromak characteristic radius where the surface electric field is A = the total surface area of one side of the spheromak wall.

SPHEROMAK DESCRIPTION:
A spheromak quasi-toroid has a major axis and a minor axis. The spheromak's circulting current follows a single layer closed spiral path which has Np turns around the quasi-toroid major axis and Nt turns around the quasi-toroid minor axis. This single layer of turns forms the spheromk 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, locally normal to the spheromak wall's outer surface. At the center of the spheromak at R = 0, Z = 0 the net electric field is zero. The cylindrically radial components of the electric field cancel out in the spheromak core.

In the far field the electric field is spherically radial and themagnetic field is zero,. The spacial magnetic field distribution outside the spheromak wall caused by the circulating charge can be modelled by a ring current Np I at R = Ro, Z = 0, and a ring charge Q at R = Ro, Z = 0. 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 electric and magnetic field energy.
 

SPHEROMAK STRUCTURAL ASSUMPTIONS:
1) A spheromak consists of a quantum charge Q circulating along a complex closed path at the speed of light C. 2) The closed path defines a quasi-toroid shaped wall;
3) The spheromak wall is radially symmetric about the Z axis;
4) The spheromak wall is mirror symmetric about the Z = 0 plane;
5) The spheromak wall minimum radius on the Z = 0 plane is Rc;
6) The spheromak wall maximum radius on the Z = 0 plan is Rs;
7) The closed filament path has Np poloidal turns where Np is a positive integer;
8) The closed filament path has Nt toroidal turns where Nt is a positive integer.
9) The integers Np and Nt have no common factor other than unity.
10) The length of one purely toroidal turn is Lt;
11) The length of one purely poloidal turn is Lp;
12) The charge circulation path length Lh is given by:
Lh^2 = (Np Lp)^2 + (Nt Lt)^2;
13) The spheromak frequency Fh is given by:
Fh = (C / Lh);
 

SPHEROMAK WINDING CONSTANTS:
1) For resonance [(Np Lp) / (Nt Lt)] is a simple rational number constant independent of spheromak energy;
2) For a stable spheromak Np and Nt are integer constants,
(Np / Nt) = constant;
3) Lp and Lt are each proportional to the linear size of the spheromk wall.
4) Hence:
(Lp / Lt) = constant independent of spheromak linear size or energy
 

FIELD ENERGY DENSITY BALANCE AT R = Rc, Z = 0
1) At R = Rc, Z = 0: Bpc = Muo Np I / 2 Ho;
2)At R = Rc, Z = 0: Btc = Muo Nt I / 2 Pi Rc;
3) Due to field energy density balance at R = Rc, Z = 0:
Bpc = Btc
19) Hence:
Np / Ho = Nt / Pi Rc;
or
(Ho / Rc) = Pi (Np / Nt)
Kb = 2 Ho /(Rs - Rc) (Rs - Rc) = Ka Ro  

FIELD ENERGY DENSITY BALANCE AT R = Rs, Z = 0
1) At R = Rs, Z = 0: Bts = [(Muo N I) / (2 Pi Rs)]
2) At R = Rs, Z = 0: Es = [(So / Epsilono)(Ro / Rs)^2]
Assumes that charged core does not affect electric field at outside surface;
3) So = Q / (Lp Lt)
4) Lp = 2 Pi Ro
5) Ro = [(Rs + Rc) / 2]
6) Due to field energy density balance at R = Rs, Z = 0:
[Bts^2 / 2 Muo] = [(Epsilono / 2) Es^2]
7)Hence:
[(Muo Nt I) / (2 Pi Rs)]^2 / 2 Muo = (Epsilono / 2)[(So / Epsilono)(Ro^2 / Rs^2)]^2
or
[(Muo Nt I C) / (2 Pi Rs)]^2 = [(So / Epsilono)(Ro^2 / Rs^2)]^2
or
[(Muo Nt I C) / (2 Pi Rs)] = [(So / Epsilono)(Ro^2 / Rs^2)]
or
[( Nt I) / (C 2 Pi Rs)] = [(So)(Ro^2 / Rs^2)]
8) I = (Q C) / Lh
9) So = Q / (2 Pi Ro Lt)
10)Hence:
[( Nt Q C) / ( Lh C 2 Pi Rs)] = [Q (Ro^2 / Rs^2) / (2 Pi Ro Lt)]
or
[( Nt) / (Lh)] = [Ro / (Rs Lt)]
or
(Nt Lt) = Lh(Ro / Rs)
or
(Nt Lt)^2 = Lh^2 (Ro / Rs)^2
11) Recall that:
Lh^2 = (Np Lp)^2 + (Nt Lt)^2 12)Hence:
(Np Lp)^2 = Lh^2[1 - (Ro / Rs)^2]
13) (Np Lp)^2 / (Nt Lt)^2 = [1 - (Ro / Rs)^2] / (Ro / Rs)^2
14) Ro = (Rs + Rc) / 2
15)Hence:
(Np Lp)^2 / (Nt Lt)^2 = {1 - [(Rs + Rc) / 2 Rs]^2} / [(Rs + Rc) / 2 Rs]^2
= {(2 Rs)^2 - (Rs + Rc)^2} / {(Rs + Rc)^2}
= {3 Rs^2 - 2 Rs Rc - Rc^2} /(Rs + Rc)^2
 

SPECIAL CASE OF (Np Lp) = (Nt Lt)
If (Np Lp) = (Nt Lt)
then:
{3 Rs^2 - 2 Rs Rc - Rc^2} = Rs^2 + 2 Rs Rc + Rc^2 or
2 Rs^2 - 4Rs Rc - 2 Rc^2 = 0 or Rs^2 - 2 Rs Rc - Rc^2 = 0 Rs = {2 Rc +/- [4 Rc^2 - 4(1)(-Rc^2)]^0.5} / 2
= Rc(1 +/- [2]^0.5) *************************************************

Ett = Et + Ep + Er

Assume that:
Et = Kt / Lh
Ep = Kp / Lh
Er = Kr / Lh
Then:
Ke = Kt + Kp + Kr
giving:
Ett = Ke / Lh,
Hence:
dEtt = -Ke dLh / Lh^2
19) Fh = C / Lh
Hence:
dFh = - C dLh / Lh^2
20) Hence:
dEtt / dFh = Ke / C
= h
= Planck Constant
 

The above derivation rests on use of electromagnetic theory to prove that:
Ett = Ke / Lh

A constant relative spheromak geometry implies that for stable spheromaks:
Rs / Rc = constant #1

A constant relativespheromakgeometry impliesthat (Ho / Rc) = Constant #2.

SPHEROMAK FIELDS
Eo = Q / [Epsilono A],
where A is the total surface area of one side of the spheromak wall.

At the spheromak high point where Z = Ho:
Es = Q / [4 Pi Epsilono (Rw^2 + Ho^2)]
?????

At the spheromak high point where Z = Ho:
Es = Q Ro / (Epsilono A Rw)

If the poloidal magnetic field is negligible at the spheromak high point then at this point:
Q / [4 Pi Epsilono (Rw^2 + Ho^2)] = Q Ro / (Epsilono A Rw)
or
1 / 4 Pi(Rw^2 + Ho^2) = Ro / A Rw
or
A / 4 Pi (Rw^2 + Ho^2) = Ro / Rw
which indicates that the surface area A of a spheromak is smaller than the surface area of a sphere enclosing the spheromak.

This equation is important.

At the same high point:
(Epsilono / 2) Es^2 = Bt^2 / 2 Muo
Es = Q Ro / Epsilono A Rw
and
Bt = Bto Ro / Rw = (Muo Nt I) / (2 Pi Rw)
giving:
(Epsilono / 2) [Q Ro / Epsilono A Rw]^2 = [(Muo Nt I) / (2 Pi Rw)]^2 / 2 Muo
or
C^2 [[Q Ro / A Rw]^2 = [(Nt I) / (2 Pi Rw)]^2
or
C [[Q Ro / A Rw] = [(Nt I) / (2 Pi Rw)]

I = Q C / Lh

Hence:
Ro / A = Nt / 2 Pi Lh
or
Lh 2 Pi Ro / A = Nt

Lp = 2 Pi Ro

Lh Lp / A = Nt

Lh^2 Lp^2 / A^2 = Nt^2

A^2 = Lh^2 Lp^2 / Nt^2
= [Np^2 Lp^2 + Nt^2 Lt^2] Lp^2 / Nt^2
= (Np / Nt)^2 Lp^4 + Lt^2 Lp^2

This is a reoccuring expression, that is the result of the outside of the spheromak.

ENERGY DENSITY BALANCE NEAR R = Rs, Z = 0
Along the outside spheromak wall near Z = 0, R = Rs:
Bt balances Es giving:
Bt^2 / 2 Muo = Epsilono Es^2 / 2 or
[(Muo Nt I / 2 Pi Ro) (Ro / Rw)]^2 / 2 Muo = [Epsilono / 2] [Q Ro / (Epsilono A Rw)]^2
or
(1 / C^2)Nt^2 I^2 / (2 Pi Rw)^2 = Q^2 Ro^2/ ( A Rw)^2
or
Nt I / C 2 Pi = Q Ro / A
but I = Q C / Lh giving: Nt Q C / Lh C 2 Pi = Q Ro / A
or
Nt / Lh 2 Pi = Ro / A
For Lp = 2 Pi Ro:
Nt /Lh = Lp / A
or
Nt = Lp Lh / A or
Nt^2 A^2 = Lp^2 Lh^2
= Lp^2 [ Np^2 Lp^2 + Nt^2 Lt^2] A^2 = (Np / Nt)^2 Lp^4 + Lp^2 Lt^2
Note that A > Lp Lt
 

ENERGY DENSITY BALANCE NEAR R = Rc, Z = 0

At theinside wall near R = Rc, Z = 0:
Bpc = Btc or
Muo Np I / 2 Ho = Muo Nt I / 2 Pi Rc
or
Np / Ho = Nt / Pi Rc
or
Np / Nt = Ho / Pi Rc
 

ENERGY DENSITY BALANCE NEAR R = Rx, Z = Ho

Now consider what happens at Rw = Rx, Z = Ho
At this point the radial electric field becomes spherical so the electric field becomes:
Eo = Q / [Epsilono 4 Pi (Ro^2 + Ho^2)]

This field is balanced by the toroidal magnetic field:
Bto = Muo Nt I / 2 Pi Rx

Hence at R = Rx, Z = Ho:
(1 / 2 Muo)[Muo Nt I / 2 Pi Rx]^2 = (Epsilono / 2) [Q / (4 Pi Epsilono (Rx^2 + Ho^2))]^2
or
(1 / C^2)[Nt^2 I^2 / (2 Pi Rx)^2 = [Q / 4 Pi (Rx^2 + Ho^2)]^2
or
Nt I / 2 Pi Rx C = Q / (Rx2 + Ho^2)
or
Nt Q C / Lh 2 Pi Rx C = Q / 4 Pi (Rx^2 + Ho^2)
or
Nt / (Lh 2 Pi Rx) = 1 / 4 Pi (Rx^2 + Ho^2)

PLANCK CONSTANT:
The spheromak structure leads to the Planck constant and the related Fine Structure constant.

Spheromak analysis provides insight into the mechanism by which nature stores energy in rest mass and the reasons for quantum mechanical behavior.

A spheromak's electric and magnetic field structure allows quantized charges to act as stable packets of electro-magnetic energy. The behavior of these spheromak based energy packets is governed by the laws of electricity and magnetism. This web page shows the mathematical relationship between these laws and quantum mechanics.

Any physical measurement involves emission or absorption of radiant energy quanta by the system being examined. Hence there is always a potential uncertainty the equivalent of one energy quantum in the measure of any physical parameter. This issue is known as quantum uncertainty.

On Earth a fraction of the incident energy carried by solar radiation is absorbed by matter. That energy is later re-emitted from Earth via lower frequency infrared radiation which is characteristic of the surface temperature of the matter. This absorption of solar radiation and subsequent emission of infrared radiation determines the direction of evolution of many chemical reactions. Absorption and emissionof radiation by matter are governed by the Planck constant.

The Planck Constant is actually a composite of other physical constants. On this web site spheromak theory is used to derive the form of the Planck Constant from first principles. It is shown that the Planck constant h is in part a geometrical constant known as the Fine Structure constant and is in part a function of an electron charge quantum Q and the ratio:
[Muo / Epsilono]^0.5
where:
Muo = permiability of free space;
and
Epsilono = permittivity of free space
 

ENERGY EXCHANGE BETWEEN MATTER AND RADIATION

MATTER AT REST:
Matter at rest stores energy in electromagnetic structures known as spheromaks.
 

PHOTON ENERGY:
A photon is a quantum of radiant energy that is either emitted by or absorbed by a spheromak. To change energy a spheromak absorbs or emits photons of energy dEtt and frequency dFh.
 

SPHEROMAK OPERATION:
A spheromak's electric and magnetic field structure allows a quantized charge to act as a stable packet of electro-magnetic energy. The behavior of these spheromak based energy packets is governed by the laws of electricity and magnetism. This web page shows the mathematical relationship between these laws and quantum mechanics.
 

THE ISSUE OF CONSTANT RELATIVE SHAPE

THE ISSUE OF SPHEROMAK DISTOTRTED CROSS_SECTION:

We need to find a comparable equation giving (Rs / Rc) in terms of Np, Nt, Lp, Lt

The spheromak surface area is: A

The spheromak surface charge density is proportional to (1 / R)

Assume that the equivalent surface charge density at R = Ro is;
[Q / A]
Then the surface charge density at R = Rc is:
(Ro / Rc)[Q / A]
and the surface charge density at R = Rs is:
(Ro / Rs)[Q /A]

The outside radial electric field at R = Rs is:
(1 / Epsilono)(Ro / Rs)[Q / A]

The outside radial electric field energy density at R = Rs is:
(Epsilono / 2){(1 / Epsilono)(Ro / Rs)[Q /(Lp Lt)]}^2
= (1 /2 Epsilono){( Ro / Rs)[Q / A]}^2

The toroidal magnetic field at R = Rs is: Bts = [Muo Nt I / 2 Pi Rs]

The toroidal magnetic field energy density at R = Rs is:
(1 / 2 Muo)[Muo Nt I / 2 Pi Rs]^2

At Z = 0, R = Rs these two field energy densities are equal giving:
(1 / 2 Epsilono){(Ro / Rs)[Q /A]}^2 = (1 / 2 Muo)[Muo Nt I / 2 Pi Rs]^2
or
C^2{(Ro / Rs)[Q / A]}^2 = [Nt I / 2 Pi Rs]^2
or
C{(Ro / Rs)[Q / A]} = [ Nt I / 2 Pi Rs]

Recall that I = Q C / Lh
Hence:
C {(Ro / Rs)[Q / A]} = [Nt Q C / 2 Lh Pi Rs]
or
{(Ro)/ A} = [Nt / 2 Lh Pi]
or
Lh / A = Nt / 2 Pi Ro

However:
Lp = 2 Pi Ro
giving:
Lh / A = Nt / Lp

Lh = A Nt / Lp
or
Lh^2 = A^2 Nt^2 / Lp^2
but
Lh^2 = (Nt Lt)^2 + (Np Lp)^2

Hence:
A^2 Nt^2 / Lp^2 = (Nt Lt)^2 + (Np Lp)^2
or
A^2 = Lp^2 Lt^2 + Lp^4 (Np / Nt)^2
This is a key spheromak relationship. This equationindicates that one condition for spheromak existence is:
A > Lp Lt

If (Np / Nt) = constant
then a spheromak retains a constant relative geometry independent of spheromak energy.

Note that for a theoretical round cross section spheromak,
A = Lp Lt and Np goes to zero.

ROUNDED DEE SPHEROMAK CROSS SECTION:
In order for a spheromak to exist its cross sectional shape must be like a D with the rounded portion of the D facing the spheromak center and the straight portion of the D facing out.

Assume that the rounded portion of the spheromak wall is described by:
(Zw / Ho)^2 + [(Rs - Rw) / (Rs - Rc)]^2 = 1
for Rc < Rw < Rs and for this special case:
Ho = (Rs - Rc)

The straight portion of the spheromak wall is described by:
R = Rs for -Ho < Z < + Ho

Then:
Ro = (Rs - Rc) / 2
giving:
Lp = 2 Pi Ro
= Pi (Rs - Rc)

Lt = 2 Ho + Pi (Rs - Rc)
= (2 + Pi)(Rs - Rc)

Hence:<
Lp Lt = Pi (Rs - Rc)(2 + Pi)(Rs - Rc)
= Pi (2 + Pi)(Rs - Rc)^2

Area of spheromak perimeter strip:
= Ap = 2 Pi Rs (2 Ho)
which for Ho = (Rs - Rc) gives:
Ap = 2 Pi Rs (2 (Rs - Rc))
= 4 Pi Rs (Rs - Rc)

the surface area of the curved part of the spheromak is:
Ac = Integral from Theta = 0 to Theta = (P / 2) of:
2 (Rs-Rc) d(Theta)[2 Rs - 2(Rs - Rc) cos(Theta)] Pi
 
= 4 Rs Pi (Rs - Rc) (Pi / 2)
- 4 Pi (Rs - Rc)^2 sin(Pi / 2)
 
= 4 Pi (Rs- Rc)[Rs (Pi / 2) - (Rs - Rc)]

The total spheromak surface area is:
A = Ap + Ac
= 4 Pi Rs (Rs - Rc) + 4 Pi (Rs- Rc)[Rs(Pi / 2) - (Rs - Rc)]
= 4 Pi(Rs - Rc){ Rs + (Rs Pi / 2) - (Rs - Rc)}
= 4 Pi (Rs - Rc){(Rs Pi / 2) + Rc)}

Hence:
A / Lp Lt
= 4 Pi (Rs - Rc){( Rs Pi / 2) + Rc} / [Pi (2 + Pi)(Rs - Rc)^2]
= {2 Rs Pi + 4 Rc} / (2 + Pi)(Rs - Rc)}
= {2 Pi (Rs / Rc) + 4} / [(2 + Pi)((Rs / Rc) - 1]

This term is always greater than unity indicating spheromak existence for a rounded Dee cross section
 P> *************************** Insert elliptical calculation here *********************************************

Consider the toroidal magnetic field at R = Rc:
Btc = Muo Nt I / 2 Pi Rc
and
Btc = Bpc = Muo Np I/ 2 Ho

Hence:
Muo Nt I / 2 Pi Rc = Muo Np I/ 2 Ho
or
Nt / Pi Rc = Np / Ho
or Ho / Rc = (Np / Nt)(Pi)

Thi is a spheromak characteristic shape parameter.

Along this wall The toroidal magmetic field Btw increases as:
Btw =Bts (Rs / Rw). Thus the toroidal internal mgnetic field cancels the radial external electric field.
Along this curve Rw is slowly decreasing and Zw is increasing. Eventually a critical point is reached at Rw = Rx, Zw = Ho at which Bp starts to become significant in the outside field energy density and Ep starts to fall off due to opposite side field cancellation. At smaller values of Rw, Bp rather than Ew becomes the dominant source of outside field energy density.

In the spheromak core:
Bpw outside cancels Btw inside,

At the extreme outer point:
Epx = Q /[Epsilono 4 Pi (Rx^2 + Ho^2)]

At this point:
Btwx = Bts (Rs / Rw)
 

To a first approximation the spheromak cross section is half elliptical. The vertical center line of the ellipse is at:
R = Rs.

At R = Rs at the spheromak wall Z = Ho.

The equation for this ellipse is:
{(Rs - R) /[Rs - Rc)]}^2 + {Z / Ho}^2 = 1

At the point Z = Ho, R = Rs the electric field is about:
Q / {Epsilono A)

The corresponding electric field energy density is:
Ue = (Epsilono / 2) [Q / {Epsilono A)]^2}

At the same point inside the spheromak wall the toroidal magnetic field is:
Bth = Btc [(2 Rc) / (Rc + Rs)]
= [Muo Nt I / 2 Pi Rc][(2 Rc) / (Rc + Rs)]
= [Muo Nt I / Pi(Rc + Rs)]

The corresponding magnetic field energy density is:
Ut = (1 / 2 Muo)[Muo Nt I / Pi(Rc + Rs)]^2

I = Q C / Lh

Ut = (Muo / 2)[Nt Q C / Lh Pi (Rc + Rs)]^2

Ue = Ut
gives:
(1 / 2 Epsilono) Q^2 / {(H^2 + [(Rs + Rc) / 2]^2}^2 = (Muo / 2)[Nt Q C / Lh Pi (Rc + Rs)]^2
or
(1 / Muo Epsilono)Q^2 / {(H^2 + [(Rs + Rc) / 2]^2}^2 = [Nt Q C / Lh Pi (Rc + Rs)]^2
or
C Q / {(H^2 + [(Rs + Rc) / 2]^2} = [Nt Q C / Lh Pi (Rc + Rs)]
or
1 / {(H^2 + [(Rs + Rc) / 2]^2} = [Nt / Lh Pi (Rc + Rs)]

Recall that:
H / Rc = (Np / Nt) Pi

Lp = 2 Pi (Rc + Rs) / 2
= Pi (Rc + Rs)

Express Lt in terms of H and (Rs - Rc) / 2
= length of ellipse perimeter P

P = Pi {3 (a + b) - [(3 a + b)(a + 3 b)]^0.5}
Lt = Pi {3 (H + (Rs - Rc) / 2) - [(3 H + (Rs - Rc) / 2)(H + 3 (Rs - Rc) / 2)]^0.5}
= Pi Rc {3 [(H / Rc) + ((Rs / Rc) - 1) / 2] - [(3 H / Rc) + ((Rs / Rc) -1)/ 2)((H / Rc) + 3((Rs/ Rc) -1) / 2)]^0.5}

For a real spheromak the shape parameter (Rs / Rc) is likely set by the balance between Bt and the external electric field Ee on the outside wall where Bp is very small.

At Z = 0, R = Rs the external electric field is cylindrical. The external electric field can be evaluated From Q, Lh, Nt, Rs and should approximately equal Bt at R = Rs Likewise the spherical external electic field should balance Bt at the peaks of the spheromak.

The energy content of a spheromak is defined by Lh. Spheromaks also adopt a low energy state where d(Lh) = 0.

For optimization with respect to Np, Nt the parameters Lp and Lt are constant. Hence:
Np Lp^2 dNp + Nt Lt^2 dNt = 0

Hence:
dNp / dNt = [Nt Lt^2 / Np Lp^2] = -2, (-1 / 2)

These two criteria, together with the total spheromak energy Ett, uniquelydefine a particular spheromak. ********************************************************** Assume that the poloidal current is concentrated at:
R = (Rc + Rs) / 2

Lp = 2 Pi (Rc + Rs) / 2 = Pi (Rc + Rs) Lt = (Rs - Rc) Pi (Lp + Lt) / Pi = 2 Rc Rs = (Lp + Lt) /( 2 Pi) Rc = (Rb Pi - Lt) / Pi = Rs - (Lt / Pi) = (Lp + Lt) / 2 Pi - (Lt / Pi)
= (Lp - Lt) / 2 Pi

(Rc + Rbs = (Lp - Lt)/ 2 Pi + (Lp + Lt) / 2 Pi = Lp / Pi Hence: [Np / Nt] = (Rc + Rs) / (Ra 2 Pi)

PLANCK CONSTANT:
Note that spheromak energy Ett has three components:
a) Toroidal magnetic field energy;
b)Poloidal magnetic field energy; c)Quasi-radial electric field energy

The effect of spheromak surface charge, which causes quasi-radial electric field energy, is to increase the total field energy density at the spheromak wall's outer surface. This increase in outside field energy density reduces the linear size and volume of the spheromak which reduces the spheromak path length Lh and hence increases the spheromak natural frequency Fh.
 

PLANCK CONSTANT DERIVATION:
Energy is quantized because the structure of a stable spheromak consists of stable integer numbers of poloidal and toroidal current path filament turns that define the spheromak wall. The parameter h is constant because the spheromak energy Ett and the spheromak frequency Fh are both exactly inversely proportional to the nominal spheromak radius Ro, so the ratio:
(spheromak energy) / (spheromak frequency)
is constant.

Theoretical evaluation of h involves theoretical evaluation of Ke.
 

Photographs of plasma spheromaks indicate that in general the cross section of a spheromak is not round. However. to find a first approximation of spheromak operation it is mathematically convenient to assume a round cros section.

FIRST APPROXIMATION:
Consider a round toroid. Let minor radius = Ra Let major radius = Rb The toroidal magnetic field at R = Ra, Z = 0 is:
Bt = Muo Nt I / 2 Pi Ra

If the poloidal windings were concenrated at:
R = (Ra + Rb) / 2
Then the poloidal magnetic field at R = 0, Z = 0
would be:
Bp = (Muo Np I) / (Ra + Rb)

These windingsare located both above and below Z = 0 and have R values in the range:
Ra < R < Rb.

In order to find Bp at Z = 0, R = 0 it is necesary to consider the three dimensional nature of the poloidal windings. The poloidal magnetic field along the Z axis can be calculated fron thesum of many purely poloidal current rings along the spheromak wall. This calculation can be done for a round spheromak. Total poloidal current is Np I.
Each current ring has current Np I d(theta) / 2 Pi

The centerline of the space inside the spheromak wall is at:
Z = 0, R = [(Ra + Rb) / 2]

The height of the current ring above the Z = 0 plane is {[(Rb - Ra) / 2] sin(Theta)}^2

The horizontal distance of the current ring from the Z axis is:
{[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}

The diagonal distance D of the current ring from the point Z = 0, R= 0 is:
D = [{[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}^2 + {[(Rb - Ra) / 2] sin(Theta)}^2]^0.5

The vertical fraction of the element of magnetic field induced at Z = 0, R = 0 by this curent ring is:
{[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)} / [{[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}^2 + {[(Rb - Ra) / 2] sin(Theta)}^2]^0.5

From basic magnetic theory:
dB = Muo i dl X r / 4 Pi r^2

In this formula:
Muo i = Muo Np I d(Theta) / 2 Pi l X r = 2 Pi {[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)} {[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}
/ [{[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}^2 + {[(Rb - Ra) / 2] sin(Theta)}^2]^0.5

Collecting terms gives:
dB = {Muo Np I d(Theta) / 2 Pi} 2 Pi {[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)} {[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}
/ [{[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}^2 + {[(Rb - Ra) / 2] sin(Theta)}^2]^0.5
4 Pi [{[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}^2 + {[(Rb - Ra) / 2] sin(Theta)}^2]
 
= {Muo Np I d(Theta) / 2 Pi} {[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)} {[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}
/ 2 [{[(Ra + Rb) / 2] - [(Rb - Ra) / 2] cos(Theta)}^2 + {[(Rb - Ra) / 2] sin(Theta)}^2]^1.5
The result for Bp|Z = 0, R = 0 is: 2 X integralfrom Theta = 0 to Theta = Pi **************************************************************************************************************************

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.
 

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 Lp^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 Lp^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 just outside 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 at R = Rc:
Bpc = Btc = (Muo Nt I) / (2 Pi Rc).

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

FIND Rs:
At R = Rs:
[Bps^2 / 2 Muo] + [1 / 2 Epsilono] [Q / (4 Pi Rs^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 Rs^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 Rs^2]^2} = [Nt Q Muo C / Lh Lp]^2 [Ro / Rs]^2

Recall that:
Lp = 2 Pi Ro
giving:
Bps^2 + [Muo C]^2[Q / 4 Pi Rs^2]^2 = [Nt Q Muo C / Lh 2 Pi Ro]^2 [Ro / Rs]^2
= [Nt Q Muo C / Lh 2 Pi Rs]^2

Hence:
Bps^2 = [Nt Q Muo C / Lh 2 Pi Rs]^2 - [Muo C]^2[Q / 4 Pi Rs^2]^2
= [Muo Q C / 2 Pi Rs]^2 [(Nt / Lh)^2 - (1 / 4 Rs^2)]
= [Muo Q C / 2 Pi Rs]^2 [{(Nt^2) / [(Np Lp)^2 + (Nt Lt)^2]} - (1 / 4 Rs^2)]

Bps^2 Rs^4 = [Muo Q C / 2 Pi]^2 [{(Nt^2 Rs^2) / [(Np Lp)^2 + (Nt Lt)^2]} - (1 / 4)]

Bps^2 Rs^4 - [Muo Q C / 2 Pi]^2 [{(Nt^2 Rs^2) / [(Np Lp)^2 + (Nt Lt)^2]} + [Muo Q C / 2 Pi]^2(1 / 4)] = 0

This is a quadratic equation that in principle can be solved to find Rs^2 if Bps is known.

SPECIAL CASE:
If Bps = 0. Then:
(1 / 4) = [{(Nt^2 Rs^2) / [(Np Lp)^2 + (Nt Lt)^2]}
or
[(Np Lp)^2 + (Nt Lt)^2] = 4 Nt^2 Rs^2
or
4 Rs^2 = [(Np Lp)^2 + (Nt Lt)^2 / Nt^2<]BR> or
Rs^2 = Lh^2 / 4 Nt^2
or
Rs = Lh / 2 Nt
 

FIND Rc:
At the spheromak inner wall Rw = Rc at Z = 0:
(Btc^2] / 2 Muo = [Bpc^2 / 2 Muo] -[Epsilon / 2][Erc]^2

However, at the inner wall at Z = 0: Erc = 0.
Hence:
Bpc^2 = Btc^2 = [Muo Nt I / 2 Pi Rc]^2

However:
Bpc = Muo Np I / 2 Ro

As a test point moves along the spheromak wall along Lt from Zw = 0, Rw = Rc going towards Z = 0, Rw = Rs the radial electric field gradually increases and the poloidal magnetic field gradually decreases. Over this path the toroidal magnetic field decreases from its initial value of (Muo Nt I / 2 Pi Rc) to its final value of (Muo Nt I / 2 Pi Rs). Over this same path the poloidal magnetic field decreases from its intial value of Bpc = [Muo Nt I / 2 Pi Rc] to almost zero.

The Zw(Rw) function has zeros at Rw = Rc and at Rw = Rs.

There is a further requirement that:
Line integral from Rw = Rc to Rw = Rs of:
Bp dLt = Muo Np I

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

Thus:
Line integral from Rw = Rc to Rw = Rs of:
Bp dLt = Muo Np I
Line integral from Rw = Rc to Rw = Rs of:
Bp [1 + (dZw / dRw)^2]^0.5 dRw = Muo Np I

At the spheromak wall in the central core of the spheromak:
Bp^2 = Bt^2 = [Bto (Ro / R)]^2

Outside the central core Bp rapidly falls to zero. Hence in effect this line integral limits the spheromak central core length.

For an ellipse described by:
Z^2 = K^2 (Rs - R)(R - Rc)

The peak value of this ellipse is at:
R = Ro = (Rs + Rc) / 2

At that peak:
Zo^2 = K^2 (Rs - [(Rs + Rc) / 2])([(Rs+ Rc) / 2] - Rc)
= K^2 [(Rs - Rc) / 2]^2

Hence:
Zo = K (Rs - Rc) / 2

Now consider the spheromak wall near Rw = Rs

In this region Bp ~ 0 so that at the spneromak wall Bt is balanced by Er. Hence in this region:
[Bto (Ro / R)]^2 / 2 Muo = (Epsilono / 2)Er^2

At R= Rs:
Er^2 = Ers^2 [Rs^4 / (R^2 + Z^2)^2]

Hence at R = Rs:
[Bto (Ro / Rw)]^2 / 2 Muo = (Epsilono / 2)Ers^2 [Rs^4 / (Rw^2 + Zw^2)^2]

This electric field representation is only precisely true at Rw = Rs. We need a better representation for the surface electric field decrease with decreasing R.

At Rw= Rs, Z = 0:
[Bto (Ro / Rs)]^2 / 2 Muo = (Epsilono / 2)Ers^2

Hence:
[Bto (Ro / Rw)]^2 / 2 Muo = {[Bto (Ro / Rs)]^2 / 2 Muo} K^2 {[Rs^2 (Rs + Rc - (Rc Rs / Rw))^2] / (Rw^2 + Z^2)^2}
or
(Ro / Rw)^2 = {Ro^2} K^2 (Rs + Rc - (Rc Rs / Rw))^2 / (Rw^2 + Zw^2)^2
(Zw^2 + Rw^2)^2 = Ro^2 K^2 (Rs Rw + Rc Rw - Rc Rs)^2 / Ro^2
= K^2 (Rs Rw + Rc Rw - Rc Rs)^2
or
Zw^2 + Rw^2 = K[Rs Rw + (Rc Rw - Rc Rs)]
or
Zw^2 = Rs Rw - Rw^2 + Rc Rw - Rc Rs
= (Rs - Rw)(Rw - Rc) where K = 1

Thus an ellipse should be OK for the spheromak outer wall if the electric field representation is correct. Note that at Rw = Rs:
[Rs^2 (Rs + Rc - (Rc Rs / Rw))^2 = Rs^4
so the electric field seems correct at Rw = Rs.

For K = 1 this is a circle of radius Ro and perimeter (2 Pi Ro),
where:
Ro = (Rs - Rc) / 2

Assume an ellipse shaped spheromak wall cross section.

Z^2 = K (Rs - R)(R - Rc)
2 Z dZ =K [Rs - 2 R + Rc] dR
[dZ / dR] = K [Rs - 2 R + Rc] / 2 Z
[dZ / dR]^2 = K^2 [Rs + Rc - 2 R]^2 / 4 Z^2

Line integral from R = Rc to R = Rs of:
[(dZ / dR)^2 + 1]^0.5 dR
= Pi Ro Z^2 + R^2 = Ro^2 Integral from R = Rc to R = Rs of = [1 + (dZw / dRw)^2]^0.5 dRw = Ro Pi
where:
Ro = (Rs - Rc) / 2

CORE ANALYSIS
From the web page titled: Theoretical Spheromak
along the Z axis the field energy density is given by:
Ue(R = 0, Z) + Up(R = 0, Z) = = [Muo C^2 Qs^2 / (Ro^2 + Z^2)^3] {(Z^2 / 32 Pi^2) + (Ro^4 Np^2 / [8 Lh^2])}

Then at the origin:
Up(R = 0, Z = 0) = [Mu C^2 Qs^2 / (Ro^2)^3] {Ro^4 Np^2 / [8 Lh^2]}
= [Mu C^2 Qs^2 / (Ro^2)]{ Np^2 / [8 Lh^2]}

At the inside of the spheromak core wall:
Bt = Bto (Ro / R)
or
Ut = Bt^2 / 2 Muo = (1 / 2 Muo) Bto^2 (Ro / R)^2
= (1 / 2 Muo)(Ro / R)^2 (Muo Nt I / 2 Pi Ro)^2
= (Muo / 8 R^2)(Nt Qs C / Lh Pi)^2

Equate Ue + Up to Ut to find the path of the spheromak core wall:
(Muo / 8 R^2)(Nt Qs C / Lh Pi)^2 = [Muo C^2 Qs^2 / (Ro^2 + Z^2)^3] {(Z^2 / 32 Pi^2) + (Ro^4 Np^2 / [8 Lh^2])}
or
(1 / 8 R^2)(Nt / Lh Pi)^2 = [1 / (Ro^2 + Z^2)^3] {(Z^2 / 32 Pi^2) + (Ro^4 Np^2 / [8 Lh^2])}
or
(Ro^2 + Z^2)^3 (1 / 8 R^2)(Nt / Lh Pi)^2 = {(Z^2 / 32 Pi^2) + (Ro^4 Np^2 / [8 Lh^2])}
OR
(Ro^2 + Z^2)^3 = [8 R^2 Lh^2 Pi^2 / Nt^2]{(Z^2 / 32 Pi^2) + (Ro^4 Np^2 / 8 Lh^2)}
= [R^2 Lh^2 Z^2 / 4 Nt^2] + [R^2 Pi^2 Ro^4 Np^2 /Nt^2]
= [R^2 / Nt^2] [(Lh^2 Z^2 / 4) + (Pi^2 Ro^4 Np^2)]

Note that:
Bts / Btc = Rc / Rs
Bto / Bts = Rs / Ro
Bto / Btc = Rc / Ro

At Z = 0, R = Rc:
Ro^6 = Rc^2 Pi^2 Ro^4 Np^2 /Nt^2
or
Rc^2 = (Ro Nt / Pi Np)^2
or
Rc = (Ro Nt / Pi Np)

This is the same value for Rc as previously calculated.

Recall that:
(Ro^2 + Z^2)^3 = [R^2 / Nt^2] [(Lh^2 Z^2 / 4) + (Pi^2 Ro^4 Np^2)]
= [R^2 / Nt^2] [(Lh^2 Z^2 / 4) + (Ro^4 Ro^2 Nt^2 / Rc^2]
= R^2 [(Lh^2 Z^2 / 4 Nt^2) + (Ro^6 / Rc^2)]

This equation should be valid in the spheromak core for Ro >> R > Rc

Rearranging gives:
R^2 = (Ro^2 + Z^2)^3 /{[(Z^2 Lh^2) / (4 Nt^2)] + [Ro^6 / Rc^2]}

Identify the circumstances under which [d(R^2) / d(Z^2)] goes to infinity at Z = Zo

[d(R^2) / d(Z^2)] = {[(Zo^2 Lh^2) / (4 Nt^2)] + [Ro^6 / Rc^2]} 3 (Ro^2 + Zo^2)^2 - (Ro^2+ Zo^2)^3 [Lh^2 / 4 Nt^2] = 0
or
{[(Zo^2 Lh^2) / (4 Nt^2)] + [(Ro^6 / Rc^2]} 3 - (Ro^2+ Zo^2) [Lh^2 / 4 Nt^2] = 0
or
2 Zo^2 Lh^2 / 4 Nt^2 = (Ro^2 Lh^2 / 4 Nt^2) - (3 Ro^6 / Rc^2)
or
Zo^2 = [2 Nt^2 / Lh^2][Ro^2 Lh^2 / 4 Nt^2 - 3 Ro^6 / Rc^2]
= Ro^2 [ (1 / 2) - ((2 Nt^2 Ro^4)/ (3 Lh^2 Rc^2))]

Recall that:
Rc = [Nt Ro / Np Pi]

Hence:
Zo^2 = Ro^2 [(1 / 2) - (2 Nt^2 Ro^4 Pi^2 Np^2 / 3 Lh^2 Nt^2 Ro^2)]
= Ro^2 [(1 / 2) - (2 Ro^2 Pi^2 Np^2 / 3 Lh^2)]
= [Ro^2 / 4] [1 - ((Lp Np)^2 / (3 Lh^2))]

We need a different equation for the outer surface of the spheromak where the expression for axial field energy density is no longer valid.

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]
 

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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

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

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

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

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

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

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.
 

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

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.