XYLENE POWER LTD.

SPHEROMAK WINDING CONSTRAINTS

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

CURRENT FILAMENTS:
Our universe is composed of a large number of charged particles, each containing a closed filament with circulating charge. Each closed charge filament contains one quantum of net electric charge:
Q = 1.60217663 X 10^-19 coulombs
which flows along the filament at the speed of light:
C = 2.99792458 X 10^8 m / s.
The net charge Q is uniformly distributed along the filament. In a stable charged particle at every point along the filament the electric and magnetic forces are in balance. For an isolated charged particle in a vacuum that filament geometry forms the wall of a spheromak. Hence isolated electrons and isolated protons exhibit spheromak geometry.
 

The spheromak inner wall and the spheromak outer wall have different horizontal radii with respect to the spheromak main axis of symmetry. Hence, at the inner wall the filament to filament distance is less than at the outer wall. Since the charge motion along the filament is constant at the speed of light and because in a spheromak the frequencies are constant the filament radius of curvature with respect to the spheromak minor axis varies so that the net charge motion along the filament remains constant. This issue causes the spheromak height (2 H) as a function of radial position R to be a complex function.

Symmetry indicates that at the center of the spheromak core the net electric field is zero and the net toroidal magnetic field is zero.
 

SPHEROMAK WINDING:
The spheromak winding has Np poloidal turns and Nt toroidal turns. The numbers Np and Nt are positive integers with various special mathematical properties.

The filament winding turns are in a single layer that contains no cross overs. The spheromak turns must not overlap or intersect. The filament path only intersects itself at the filament path closure point. A spheromak will collapse if the number of poloidal turns Np and the number of toroidal turns Nt share a common integer factor other than one. For the same reason Np can not be an integer multiple of Nt or vice versa.
 

FILMENT LENGTH Lh:
The spheromak wall consists of a long filament of charge that forms a complex closed spiral current path with length Lh. This filament contains Nt toroidal turns each of length Lt and Np poloidal turns each of length Lp. The numbers Np and Nt have no common factors so the filament path never crosses itself. The current flows along this filament path at the speed of light.

From the web page titled: Spheromak Structure the filament length Lh is given by:
[Lh]^2 = Nt^2 Lt^2 + Np^2 Lp^2
where:
The term (Nt Lt) results in a toroidal magnetic field and
the term (Np Lp) results in a poloidal magnetic field.
 

Define:
Lh = Mh (Lamda) = C / F
Np Lp = Mp (Lamda)
Nt Lt = Mt (Lamda)
where Mh, Mp and Mt are positive integers that have no common integer factors other than one.

Thus:
[Lh]^2 = Nt^2 Lt^2 + Np^2 Lp^2
becomes:
Mh^2 (Lamda)^2 = Mp^2 (Lamda)^2 + Mt^2 (Lamda)^2
or
Mh^2 = Mp^2 + Mt^2
where:
Mh (Lamda) = C / F

The key spheromak resonance assumption is that:
(Lamda) = (Np Lp / Mp)
and
(Lamda)= (Nt Lt / Mt)
giving:
[Np Lp / Mp] = [Nt Lt / Mt]
or
[Np Lp Mt = Nt Lt Mp
giving:
[(Np Lp) /(Nt Lt) = (Mp / Mt)

The spheromak filament length is:
Mh^2 dL^2 = Mp^2 dL^2 + Mt^2 dL^2
which gives:
Mh^2 = Mp^2 + Mt^2
or
2 Mh dMh = 2 Mp dMp + 2 Mt dMt

At times when the spheromak filament length is stable:
dMh = 0
giving:
2 Mp dMp + 2 Mt dMt= 0
or
[dMt / dMp] ~ - [Mp / Mt]

Define:
X = Lh / Np Lp
or
X^2 = [Lh / Np Lp]^2

Recall that:
[Lh]^2 = Nt^2 Lt^2 + Np^2 Lp^2
or
[Lh / Np Lp]^2 = [Nt^2 Lt^2 / Np^2 Lp^2] + 1
giving:
[X^2 - 1] = [Nt Lt / Np Lp]^2

Let Pa and Pb be prime numbers. The two circustances under which Mp and Mt have no common integer factors are:
Family A:
Pa = Mpa + 2 Mta
giving:
dMpa = -2 dMta
or
(Mta / Mpa) ~ 2

Family B: Pb = Mtb + 2 Mp
giving:
dMtb = - 2 dMp.
Mtb / Mpb ~ (1 / 2)
Pb = Mtb + 2 Mpb
~(Mtb) + 2 (2 Mtb)
~ 5 Mtb

Remember that Mpa, Mta, Mpb and Mtb are integers.
 

Family A Precise Calculation:
Pa = Mpa + 2 Mta

Recall that due to resonance:
[(Np Lp) /(Nt Lt) = (Mp / Mt)

Hence:
Pa = Mpa + 2 Mp [(Nt Lt) / (Np Lp)]
= Mpa + 2 Mp [X^2 - 1]^0.5

Hence:
(Pa - Mpa)^2 = 4 Mp^2 [X^2 - 1]
or
Pa^2 - 2 Pa Mpa + Mpa^2 - 4 Mpa^2 [X^2 - 1] = 0
or
Pa^2 - 2 Pa Mpa + Mpa^2 [1 - 4 X^2 + 4] = 0
or
Pa^2 - 2 Pa Mpa + Mpa^2 (5 - 4 X^2) = 0

This quadratic equation gives:
Pa = {2 Mpa +/- [4 Mpa^2 - 4(1)Mpa^2 (5 - 4 X^2)]^0.5} / 2
= Mpa {1 +/- [1 - (5 - 4 X^2)]^0.5}
= Mpa{1 +/- [4X^2 - 4]^0.5}
= Mpa + 2 Mpa [X^2 - 1]^0.5
= Mpa + 2 Mta

Hence:
Mta = Mpa [X^2 - 1]^0.5
or
Mta / Mpa = [X^2 - 1]^0.5 = Nt Lt / Np Lp
as expected.
 

Family B Precise Calculation:
Pb = Mtb + 2 Mpb

Recall that:
Nt Lt / Mt = Np Lp / Mp

Hence:
Pb = (Nt Lt / Np Lp) Mp + 2 Mpb
= (X^2 - 1)^0.5 Mpb + 2 Mpb

(Pb - 2 Mpb) = (X^2 - 1)^0.5 Mpb

Pb^2 - 4 Mpb Pb + 4 Mpb^2 = (X^2 - 1)Mpb^2

Pb^2 - 4 Mpb Pb + 4 Mpb^2 - (X^2 - 1)Mpb^2 = 0

Pb^2 - 4 Mpb Pb + Mpb^2(4 - X^2 + 1) = 0

Pb = {4 Mpb +/- [16 Mpb^2 - 4 (1)Mpb^2(4 - X^2 + 1)]^0.5} / 2
= {2 Mpb +/- (1 / 2) [4 Mpb^2(X^2 - 1)]^0.5}
= 2 Mpb +/- Mpb [X^2 - 1]^0.5)

Pb and Mpb are both positive integers with Pb > 2 Mpb.

Hence:
Pb = 2 Mpb + Mpb [X^2 - 1]^0.5
= 2 Mpb + Mtb

Hence:
Mtb = Mpb [X^2 - 1]^0.5
or
(Mtb / Mpb) = (X^2 - 1)^0.5 = [Nt Lt / Np Lp]
as expected.

A key issue is determination of the prime numbers Pa and Pb.

FAMILY A:
Mta ~ 2 Mpa Nominal
2 Mpa +/- 1 = Mta Corrected by adding +/- 1 to the doubled amount

Pa = Mpa + 2 Mta
= Mpa + 2 [2 Mpa +/-1]
= 5 Mpa +/- 2

Mta ~ 2 Mpa
or
Mpa / Mta ~ (1 / 2)

Recall that:
Np Lp / Mpa = Nt Lt / Mta
or
Mpa / Mta = [Np Lp / Nt Lt]

Hence for Family A:
[Np Lp / Nt Lt] ~ (1 / 2)
 

FAMILY B
2 Mtb ~ Mpb Nominal, 2 Mtb +/- 1 = Mpb Corrected by adding +/- 1 to the doubled amount

Pb = Mtb + 2 Mpb
= Mtb + 2[2 Mtb +/- 1]
= 5 Mtb +/- 2

Mpb ~ 2 Mtb
or
[Mpb / Mtb] ~ 2

Recall that
Np Lp / Mp = Nt Lt / Mt
or
Mp / Mt = [Np Lp / Nt Lt]

Hence for Family B:
[Np Lp / Nt Lt] ~ 2
 

ARE BOTH FAMILY A AND FAMILY B REAL?
A spheromak identity is:
Bpor / Bto = [Muo Np I / 2 Ro] / [Muo Nt I / 2 Pi Ro] = (Np Pi / Nt)

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

Hence for Family A:
[Bpor / Bto] = Pi Lt / 2 Lp

Following a similar arguement for Family B gives:
Bpor / Bto = Pi 2 Lt / Lp

Note that the ratio [Bpor / Bto] with Family B is four times larger than for Family A. Hence Family B will dominate in real spheromaks.
 

CANDIDATE VALUES FOR Pa AND Pb:
Pa = 5 Mpa +/- 2, Pb = 5 Mtb +/- 2, Pa, Mpa,Pb, Mtb are all integers,
On a table of primes find pairs of prime numbers ending in 3 and 7.

A table of prime numbers up to 1997 reveals the following candidate prime number pairs. These primes are applicable to both Pa and Pb:
P = 43, 47; Mpa = 9, Mta = 17, 19; Mtb = 9, Mpb = 17, 19
P = 103, 107; Mpa = 21 Mta = 41, 43; Mtb = 21, Mpb = 41, 43
P = 163, 167
P = 193, 197
223, 227
313, 317
463, 467
613, 617
643, 647
673, 677
823, 827
853, 857
883, 887
P = 1093, 1097; Mpa = 219, Mta = 437, 439; Mtb = 219, Mpb = 437, 439
1213, 1217
1303, 1307
1423, 1427
1483, 1487
1663, 1667
1693, 1697
1783, 1787
1873, 1877
P = 1993, 1997; Mpa = 399, Mta = 797, 799; Mtb = 399, Mpb = 797, 799

We must use other criteria to reduce this potential candidate prime number list and to find if both Family A and Family B correspond to real solutions.

Note that in essence in Family A: Mpa ~ (Mta / 2)
whereas in Family B: Mpb ~ (2 Mtb)

SUMMARY:
In a real spheromak:
[C / F]^2 = Lh^2 = (Np Lp)^2 + (Nt Lt)^^2
or
[Mh^2 (Lamda)^2] = Mp^2 (Lamda)^2 + Mt^2 (Lamda)^2
or
Mh^2 = Mp^2 + Mt^2

The Family B winding condition requires that:
Pb = Mtb + 2 Mpb
where:
(Mpb / Mtb) ~ 2

In the context of a spheromak where:
Mpb = 2 Mtb +/- 1 :
Pb = Mtb + 2 Mpb
= Mtb + 2 [2 Mtb +/- 1]
= 5 Mtb +/- 2

Recall that:
Np Lp / Mp = Nt Lt / Mt
or
Mp / Mt = (Np Lp) / (Nt Lt)

Lh^2 = (Np^2 Lp^2 + Nt^2 Lt^2)

[Lh / Np Lp]^2
= [1 + (Nt Lt / Np Lp)^2]
= [1 + (Mt / Mp)^2]
= [(Mp^2 + Mt^2 ) / Mp^2]
=[(2 Mtb +/- 1)^2 + Mtb^2] / (2 Mtb +/- 1)^2

For large Mtb values:
[Lh / Np Lp]^2 ~ (5 / 4)

An important exact quantity is:
{(Npb Lpb) / (Ntb Ltb)]
= (Mpb / Mtb
= 2 +/- (1 / Mtb)

Thus:
Mh^2 = [2 Mtb +/- 1]^2 + Mtb^2
=[4 Mtb^2 + 1 +/- 4 Mtb + Mtb^2]
= [5 Mtb^2 + 1 +/- 4 Mtb]

Hence Mh^2 has two values. The larger value Mhh is:
Mhh^2 = 4 Mtb^2 + 4 Mtb + 1 + Mtb^2
= 5 Mtb^2 + 4 Mtb + 1

The smaller value Mhl is:
Mhl^2 = 4 Mtb^2 - 4 Mtb + 1 + Mtb^2
= 5 Mtb^2 - 4 Mtb + 1

A transition from Mhh to Mhl causes a 4 integer change in P.

Pb = Mtb + 2 Mpb
= Mtb + 2 [2 Mtb +/-1]
= 5 Mtb +/- 2

Since Mtb is an integer:
Pb = 5 Mtb +/- 2

If Mtb is even there a no Pb primes. Hence:
Mtb is odd. Then the formula: Pb = 5 Mtb +/-2
yields potential prime pairs terminating in 3 and 7.

These primes are listed herein.
 

UNIQUE SOLUTION:

(Lh / Lp)^2 has a unique solition.

Comparing the two different family equations for (Lh /Lp)^2 gives:
[(Pa - Mpa)/ 2 Mpa]^2 = [(Pb - 2 Mpb) / Mpb]^2
or
[(Pa - Mpa)/ 2 Mpa] = [(Pb - 2 Mpb) / Mpb]
or
[(Pa - Mpa) / (Pb - 2 Mpb) = 2 Mpa / Mpb

If Pa = Pb and if Mpa = 2 Mpb
then there is a conflict. Hence Pa an Pb are two different prime numbers. In theory each spheromak might have two different solutions, one for Family A and one for Family B.

[(Pa - Mpa) / 2 Mpa] = [(Pb - 2 Mpb) / Mpb]
or
Mpb (Pa - Mpa) = 2 Mpa (Pb - 2 Mpb)

Summary:
For Family A:
(Lh / Lp)^2 = Np^2 {1 + [(Pa - Mpa) / (2 Mpa)]^2}

Mpb (Pa - Mpa) = 2 Mpa (Pb - 2 Mpb)

For Family B:
[X^2 - 1] = [Nt Lt / Np Lp]^2
= [Mtb / Mpb]^2

Pb = Mtb + 2 Mpb

Pb - 2 Mpb = Mtb

Pb^2 - 4 Pb Mpb + 4 Mpb^2
= Mtb^2
= Mpb^2 {X^2 - 1]
or
Pb^2 - 4 Pb Mpb + Mpb^2 [4 - X^2 + 1] = 0

Hence:
Pb^2 - 4 Mpb Pb - Mpb^2 [(X^2 - 5] = 0 Pb = (4 Mpb +/- {16 Mpb^2 + 4 Mpb^2 [X^2 - 5]}^0.5) / 2
or
Pb = (2 Mpb +/- {4 Mpb^2 + Mpb^2[X^2 - 5]}^0.5)
= Mpb (2 +/- {4 + [X^2 - 5]}^0.5)
= Mpb (2 + {X^2 - 1}^0.5)

Recall that:
[X^2 - 1]^0.5 =(Mtb / Mpb)
where (Mtb / Mpb) ~ (1 / 2)
but Mtb and Mpb share no common integer factors.

Then:
Pb = Mpb (2 + {X^2 - 1}^0.5)
= Mpb (2 + (Mtb / Mpb)
= 2 Mpb + Mtb
as expected.

FIND Pb AND Sb

For X = 3 apply this test to the below listed table of prime numbers,

For prime Pb = 577:
8^0.5 (577) = 1632.002451

Hence Pb = 577, Sb = 1632

Recall that:
Pb = Mtb + 2 Mpb

FIND Mtb, Mpb:
Test a table of primes.
If 8^0.5 Pb = integer, then adopt that Pb value.
Pb = 577,
Sb = 8^0.5 (577) = 1632.00251 ~ 1632
 

Prospective Pb Results that are prime numbers.

Then:
Sb^2 = Pb^2 (N^2 - 1) Mtb = (2 Pb^2 + 4 Pb)/ 4 (Pb + 1)

Mpb = Pb^2 / 4(Pb + 1)

Mtb + 2 Mpb
= (2 Pb^2 + 4 Pb)/ 4 (Pb + 1) + 2 Pb^2 / 4(Pb + 1)
= (4 Pb^2 + 4 Pb) / 4 (Pb +1)
= Pb
as expected.

For Pb = 577:
Mtb = ((2 Pb^2 + 4 Pb)/ 4 (Pb + 1)
= ((665,858 + 2308) / 2312 = 288.9991349 = 289
Mpb = Pb^2 / 4(Pb + 1) = 144.0004325 = 144

Pb = Mtb + 2 (Mpb)
= 289 + 2(144) = 577
as expected.
HOWEVER, 289 IS NOT PRIME p  

CONSISTENCY CHECK:
Mpb = [ -2 Pb + Sb] / (N^2 - 5)
(N^2 - 5) Mpb + 2 Pb = Sb
[(N^2 - 5) Mpb + 2 Pb]^2 = Sb^2
(N^2 - 5)^2 Mpb^2 + 4 Pb (N^2 - 5) Mpb + 4 Pb^2] = Pb^2 (N^2 - 1)
(N^2 - 5)^2 Mpb^2 + 4 Pb (N^2 - 5) Mpb] = Pb^2 (N^2 - 5)
(N^2 - 5) Mpb^2 + 4 Pb Mpb = Pb^2 Mpb[(N^2 - 5) + 4 Pb] = Pb^2

Try N = 3, then:
4 Mpb(1 + Pb) = Pb^2
then:
Mpb = Pb^2 / 4(Pb + 1)

Pb = Mtb + 2 Mpb

Mtb = Pb - 2 Mpb
= Pb - 2 Pb^2 /4(Pb + 1)
= [Pb 4(Pb + 1) - 2 Pb^2] / 4(Pb + 1)
= [4 Pb^2 + 4 Pb - 2 Pb^2)/4(Pb + 1)
= (2 Pb^2 + 4 Pb)/ 4 (Pb + 1)

Pb = Mtb + 2 Mpb
= (2 Pb^2 + 4 Pb)/ 4 (Pb + 1) + 2 Pb^2 / 4(Pb + 1)
= (4 Pb^2 + 4 Pb) / 4(Pb + 1
= Pb
as expected. Results seem consistent. Hence N = 3.

Pb = Sb / (N^2 - 1)^0.5
Pb^2(N^2 - 1) = Sb^2
Hence: 8 Pb^2 = Sb^2

Recall that:
(Lh / Lp)^2 = Np^2 {1 + [(Pb - 2 Mpb) / Mpb]^2}
= Np^2 {1 + [Mtb / Mpb]^2}
= Np^2 {1 + [(2 Pb^2 + 4 Pb)/ Pb^2]^2}
= Np^2 {1 + [2 + (4 / Pb)]^2}
= Np^2 {5 + (16 / Pb) + (16 /Pb^2)}

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Hence in effect a spheromak is characterized by two prime numbers Pa and Pb where:
Mpb (Pa - Mpa) = 2 Mpa (Pb - 2 Mpb)

RHS = always even
(Pb - 2 Mpb) is always odd
Mpa either odd or even

If Mpa = odd then Mpb = odd or even
If Mpa = even then Mpb = even
 

FAMILY A:

In Family A:
P = Mpa + 2 Mta
a) Mpa is always odd,
b) [(1 / 2)(Mp / Mt)^2]Nt = odd integer,
c) [(1 / 2)(Mp / Mt)^2] an integer
d) (Mp / Mt)^2 is an even integer
e) (Mp / Mt)^2 = 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484, 576,.....
f)Pa = Nt {2 + (2, 8, 18, 2, 50, 72, 98, 128, 162, 200, 242, 288, ....
Note that none of these are prime.
 

FAMILY B:

In Family B:
Pb = Mtb + 2 Mpb:
a) Mt is always odd,
b) 4 (Mp / Mt)^2 Nt = 2 Np is always even integer,
c) [2(Mp / Mt)^2] is always an even integer,
e) (Mp / Mt)^2 can be either odd or even.
f) (Mp / Mt) = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, ....
g) Pb = Nt [5, 17, 37, 65, 145, ....]

Note that Mp and Mt are positive integers that share no common factors, Mta and Mtb are both always odd.
 

RATIO (Mtb / Mpb):
Mtb = (2 Pb^2 + 4 Pb)/ 4 (Pb + 1)

Mpb = Pb^2 / 4(Pb + 1)

[Mtb / Mpb] = (2 Pb^2 + 4 Pb)/ Pb^2
= 2 + (4 / Pb)
= 2 + (4 / 577)
= 1158 / 577
= [2 (579) / 577] = [(2(3)(193) / 577]

SUMMARY:
The requirement that (Np Lp) and (Nt Lt) have a stable ratio with no common factors leads to Family B solution:
(Mtb / Mpb) = [(2(3)(193) / 577]
= [2 (579) / 577]

Hence for Family B the toroidal winding is slightly more than twice the length of the poloidal winding. Note that the ratio [Mtb / Mpb] involves two prime numbers 193 and 577.
 

FIND Lp /Lt:
Recall that:
[(Np Lp)/ Mp] = [(Nt Lt) / Mt]
or
[Lt / Lp] = (Np / Nt)(Mt / Mp)

Pb = Mt + 2 Mp
577 = 193 + 2(192)
Mt = 193
Mp = 192

Np, Nt are integers,
Np not equal to Nt,
Mt = prime integer,
giving:
Mt = (Nt / Np) (Lt / Lp) Mp 193 = (193/192)(1 / 1)(192)

Solution is:
Nt = 193
Np = 192

The number Np = 192 is a key part of the Planck constant h and the Fine Structure constant [Alpha].

FIND [Lh / Lp]^2:
Recall that:
[Np Lp / Mp] = [Nt Lt / Mt]

Lh^2 = (Np Lp)^2 + (Nt Lt)^2
= (Np Lp)2 [1 + (Nt Lt / Np Lp)^2]
= (Np Lp)^2 [1 + (Mtb / Mpb)^2]
= (Np Lp)^2 {1 + [2 + (4 / Pb)]^2}

[Lh / Lp]^2 = Np^2 {1 + [2 + (4 / Pb)]^2}
= Np^2 {1 + [2 +(4 / 577)]^2}
= Np^2 {1 + [2 (3)(193) / 577]^2}
= Np^2 {1 + [1158 /577]^2}
= Np^2 {[(577)^2 + (1158)^2]/ (577)^2}
= Np^2{[332929 + 1340964] / 332929}
= Np^2 {1673893 / 332929}

The ratio: [Lh / Lp}^2 is a key part of the Planck Constant h and the Fine Structure Constant [Alpha].

Hence:
The parameter:
{Lh / Lp]^2 = Np^2 {1673893 / 332929}
= Np^2 [5.027777694]
should be be used to determine the theoretical value of the Planck constant.
 

PRIME NUMBER THEORY:

THE NO COMMON INTEGER FACTOR CONSTRAINT:
An important constraint on the existence of a spheromak is that Np and Nt have no common integer factors and both dNp and dNt are small integers. A mathematical expression of this statement of no common factors is:
[Np / Nt] = [(N + 1) / ((prime) - 2(N + 1))]
 
for
0 <= N <= [(Prime - 1) / 2]
This mathematical function was devised by Heather Rhodes. The quantity (prime) can be any prime number.
 

Prove that the function:
(Np / Nt) = {[(N + 1)] / [prime - 2(N + 1)]}
generates (Np / Nt) values with no common factors.
If (N + 1) = (Fo M)
where Fo is a factor of (N + 1) then the denominator is
(prime - 2 Fo M) which can only be reduced if Fo is a factor of prime, which it is not due to the definition of a prime number.

However, the prime number can potentially adopt a range of values. The value that the prime number adopts should minimize the error in (Np / Nt) where:
Np = N + 1
and
Nt = [prime - 2(N + 1)]

[Np / Nt] = [(N + 1) / [(prime) - 2 (N + 1)]

This prime number theory together with the requirement for no common integer factors in Np and Nt gives two families of number pairs Np, Nt that share no common integer factors other than one. Those families are:
FAMILY A:
P = Np + 2 Nt
and

FAMILY B:
P = 2 Np + Nt

Due to prime number theory, at the spheromak stable operating point where:
d[(L / 2 Pi Ro)^2] = 0
either:
dNp / dNt = -2
as in Family A or
dNp / dNt = -(1 / 2)
as in Family B
 

FINDING dH / dR:
In order to accurately calculate the field energy content of a spheromak it is necessary to first calculate the spheromak height (2 H) as a function of radial position R. In order to do so it is necessary to determine the function (dH / dR) and then integrate this function from R = Rc to R = Rs.

From the web page titled: Spheromak Structure:
[dH / d(Theta)]^2 = [(Rs - Rc) / 2]^2 cos^2(Theta) + [(Rs - Rc) /2]^2 sin^2(Theta)] [Np / Nt]^2
- 2 Rx [(Rs - Rc) / 2] cos(Theta)[Np / Nt]^2

[(dH / dTheta)]^2 = [dH / dR]^2 [dR / d(Theta)]^2
or
[dH / dR]^2 = (dH / dTheta)]^2 / [dR / d(Theta)]^2

Recall that:
R = Rx + [(Rs - Rc) / 2] cos(Theta)
or
dR / d(Theta) = - [(Rs - Rc)/ 2] sin(Theta)
giving:
[dR / d(Theta)]^2 = [(Rs - Rc)/ 2]^2 sin^2(Theta)

Then:
[dH /dR]^2 = { [(Rs - Rc) / 2]^2 cos^2(Theta) + [(Rs - Rc) /2]^2 sin^2(Theta)] [Np / Nt]^2
- 2 Rx [(Rs - Rc) / 2] cos(Theta)[Np / Nt]^2 }
/ {[(Rs - Rc)/ 2]^2 sin^2(Theta)}

= {cos^2(Theta) + sin^2(Theta)] [Np / Nt]^2
- [4 Rx / (Rs - Rc)] cos(Theta)[Np / Nt]^2} / { sin^2(Theta)}

Then:
[dH / dR] = +/- {cos^2(Theta) + sin^2(Theta)] [Np / Nt]^2
- [4 Rx / (Rs - Rc)] cos(Theta)[Np / Nt]^2} ^0.5 / {sin(Theta)}
 

Express cos(Theta) and sin(Theta) as functions of R

Recall that:
R = Rx + [(Rs - Rc) / 2] cos(Theta) or
cos(Theta)= 2 (R - Rx) / (Rs - Rc)
and
cos^2(Theta) = [2 (R - Rx) / (Rs - Rc)]^2
= [4 (R^2 - 2 R Rx + Rx^2) / (Rs - Rc)^2]
= [(4 R^2 - 4 R (Rs + Rc) + Rs^2 + 2 Rs Rc + Rc^2) / (Rs - Rc)^2
= [(2 R - (Rs+ Rc)]^2 / (Rs - Rc)^2

sin^2(Theta) = {1 - [2 (R - Rx) / (Rs - Rc)]^2}
= {1 - [(4 R^2 - 4 R (Rs + Rc) + Rs^2 + 2 Rs Rc + Rc^2) / (Rs - Rc)^2]}
= {(Rs^2 - 2 Rs Rc + Rc^2) - (4 R^2 - 4 R (Rs + Rc) + Rs^2 + 2 Rs Rc + Rc^2) / (Rs - Rc)^2}
= {(- 4 Rs Rc) - (4 R^2 - 4 R (Rs + Rc)) / (Rs - Rc)^2}
= {( - 4 Rs Rc - 4 R^2 + 4 R Rs + 4 R Rc) / (Rs - Rc)^2}
= {4 [R (Rs - R) + Rc (R - Rs)] / (Rs - Rc)^2}
= {4 [(Rs - R)(R - Rc)] / (Rs -Rc)^2}

sin(Theta) = [2 / (Rs - Rc)] [(Rs - R)(R - Rc)]^0.5

These terms can now be substituted into the expressions for [dH / dR] with the objective of determining H(R).

Knowledge of H(R) is needed to make an exact calculation of spheromak energy.
 

Recall that:

[dH / dR] = +/- {cos^2(Theta) + sin^2(Theta) [Np / Nt]^2
- [4 Rx / (Rs - Rc)] cos(Theta)[Np / Nt]^2}^0.5
/ sin(Theta)
 

EVALUATING TERMS OF dH / dR:

cos^2(Theta) = [2 R - (Rs+ Rc)]^2 / (Rs - Rc)^2]

sin^2(Theta)[Np / Nt]^2 = {4 [(Rs - R)(R - Rc)] / (Rs -Rc)^2}[Np / Nt]^2

- [4 Rx / (Rs - Rc)] [cos(Theta)][Np / Nt]^2
= - [4 Rx / (Rs - Rc)] [2 (R - Rx) / (Rs - Rc)][Np / Nt]^2
= - [8 Rx] [(R - Rx) / (Rs - Rc)^2][Np / Nt]^2

- sin(Theta) = [2 / (Rs - Rc)] [(Rs - R)(R - Rc)]^0.5

Thus:
dH / dR = +/- {cos^2(Theta) + sin^2(Theta) [Np / Nt]^2
- [4 Rx / (Rs - Rc)] cos(Theta)[Np / Nt]^2}^0.5
/ sin(Theta)

= +/- { [[2 R - (Rs+ Rc)]^2 / (Rs - Rc)^2]] + {4 [(Rs - R)(R - Rc)] / (Rs -Rc)^2}[Np / Nt]^2
- [8 Rx] [(R - Rx) / (Rs - Rc)^2][Np / Nt]^2}^0.5
/ [2 / (Rs - Rc)] [(Rs - R)(R - Rc)]^0.5

= +/- { [[2 R - (Rs+ Rc)]^2][1 / 4]] + {[(Rs - R)(R - Rc)]}[Np / Nt]^2
- [2 Rx] [(R - Rx)][Np / Nt]^2}^0.5
/ [(Rs - R)(R - Rc)]^0.5

Case A:
[Np / Nt]^2 = [(Rs - Rc) / Ro]^4 [1 / 4]

At R = Rm, dH / dR = 0. Hence:
[4 [Rm - Rx]^2]] + {[(Rs - Rm)(Rm- Rc)]}[(Rs - Rc) / Ro]^4
- [2 Rx] [(Rm - Rx)][(Rs - Rc) / Ro]^4}^0.5
This equation is quadratic in Rm and hence is readily solvable for Rm
 .

CASE B:
[Np / Nt]^2 = [(Rs - Rc)/ Ro]^4 [1 / 64]

Recall that:
Ro^2 = [(Rs^2 + Rc^2) / 2]

At R = Rm, dH / dR = 0. Hence,
[[4 [Rm - Rx]^2]] + {[(Rs - Rm)(Rm - Rc)]}[(Rs - Rc)/ Ro]^4 [1 / 16]
- [2 Rx] [(Rm - Rx)][(Rs - Rc)/ Ro]^4 [1 / 16] = 0

This eqution is quadratic in Rm and hence is readily solvable for Rm.

Knowledge of Rm allows integration of the winding analysis with the field analysis which potentially gives Rm as the R value on the spheromak wall where Bz = 0.

Knowledge of H(R) is needed to make an exact calculation of spheromak energy.

These expressions for (dH / dR) needs to be integrated from R = Rc to R = Rs in order to determine H(R) in the region:
Rc < R < Rs
In order to facilitate this integration it is helpful to graph both (dH / dR) and H(R) as functions of R.
 

COMMENTS
The denominator of [dH / dR] approaches zero as R approaches either Rc or Rs. Hence the side walls of the spheromak wall are vertical at R = Rc, Z = 0 and at R = Rs, Z = 0.

dH / dR must be positive near R = Rc, must go to near R ~ (Rs + Rc) / 2 and then must become negative for larger R values. Note that in the range Rc < R < Rs the denominator is always positive.

In the range:
Rx < R < Rs
(dH / dR) changes sign from positive to negative. The point where:
(dH / dR) = 0
is the peak in the spheromak wall.
 

SPHEROMAK PARAMETERS:
Rs^2 + Rc^2 = 2 Ro^2 _______

In a spheromak the values of [(Rs - Rc) / 2 Ro], (Ho / Ro), Nt and Np are all constants independent of Ro.

The width of the spheromak wall enclosed region is:
(Rs - Rc)

The center line of the spheromak wall enclosed region is at:
Rx = (Rs + Rc) / 2

The outside equatorial diameter of the spheromak is 2 Rs and the inside equatorial diameter is 2 Rc.

The cross section of the spheromak wall is a distorted circle with circumference:
2 Pi (Rs - Rc)

From the web page titled Spheromak Structure dH / dR = +/- {cos^2(Theta) + [sin^2(Theta)] [Np / Nt]^2
- [4 Rx / (Rs - Rc)] cos(Theta)[Np / Nt]^2}^0.5 / {- sin(Theta)]}

Note that the field structure positions Rc and Rs with respect to Ro.
Rx = [(Rs + Rc) / 2]
is slightly different than
Ro = [(Rs^2 + Rc^2) / 2]^0.5

An issue of concern is whether the peak in the spheromak wall profile as derived from the field configurtion is coincident with the peak in the above derived spheromak wall profile. We know from the magnetic field configuration that this peak is not at Ro. We also know that it is not at Rx.
 

PRIME NUMBERS
Prime numbers less than 2003 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193 , 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003
 

PRIME NUMBER CLUSTERS:

Each prime number P yields a family of (Np / Nt) integer pairs. The spheromak steps along its Family A or Family B existence line:
P = Np + 2 Nt
or
P = 2 Np + Nt
to find its most stable state.

A stable spheromak needs to be tolerant of outside disturbances that cause fluctuations in Np and Nt.

The issue is that the prime number P will normally not significantly change but an external disturbance can potentially cause Np and Nt to temporarily deviate from their normal values Npo = Nto + 1. A stable spheromak should not collapse due to such a disturbance.

This tolerance to disturbance is improved if P is located within a cluster of prime numbers.

There are prime number clusters at:
223, 227, 229, 233;
and at
821, 823, 827, 829;
and at
877, 881, 883, 887;
and at
1087, 1091, 1093, 1097
and at
1297,1301, 1303, 1307;
and at:
1483, 1487, 1489, 1493;
and at
1867, 1871, 1873, 1877
and at
1993, 1997, 1999, 2003
 

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Note that spheromak field energy components are each proportional to [1 / Ro].

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

TURN WINDING:
Imagine that a single layer of thin wire is wound on a toroidal form. At its maximum equatorial perimeter the toroidal form has a horizontal equator line. Imagine starting the winding by wraping the thin wire around the form such that it makes M complete toroidal turns in one complete poloidal turn. As the next toroidal turn is added a decision must be made as to how to prevent successive turns on the form intersecting previous turns. The solution is to lay the next turn slightly ahead of or slightly behind the the previous turn. However, as the next turn initially crosses the equatorial line exactly M toroidal turns will be complete but the corresonding poloidal turn count will be one turn plus (1 / N) turns. At completion of 2 M toroidal turns the correponding poloidal turn count will be:
[2 + (2 / N)].
After completion of (M N) toroidal turns the poloidal turn count will be:
[N + N (1 / N)] = (N + 1).
Thus in this example:
Nt = (M N)
and
Np = (N + 1).
In this eample the turns ratio is:
(Np / Nt) = (N + 1) / (M N)
which for large N is approximately equal to:
(1 / M).

Equally possible are the turns ratios:
(Np / Nt) = (M + 1) / (M N)
or
Np / Nt = (N M) / (N + 1)
or
Np / Nt = (N M) / (M + 1)

Summary:
If Np < Nt then the permitted turns ratios are of the form:
[Np / Nt] = [(M + 1) / (N M)]
or
[Np / Nt] = [(N + 1) / (N M)]

and if Np > Nt then the permitted turns ratios are of the form:
[Np / Nt] = [(N M) / (M + 1)]
or
[Np / Nt] = [(N M) / (N + 1)]

The Family A spheromak constraint is:
P = Np + 2 Nt

Application of this additional constraint gives:
P = (N + 1) + 2 M N = N (2 M + 1) + 1
P = (M + 1) + 2 M N = M (2 N + 1) + 1
P = N M + 2 (N + 1) = N (M + 2) + 2
P = N M + 2 (M + 1) = M ( N + 2) + 2

The Family B spheromak constraint is:
P = 2 Np + Nt

Application of this additional constraint gives:
P = 2 (N - 1) + N M = N (M + 2) - 2
P = 2 (N + 1) + N M = N(M + 2) + 2
P = 2 N M + (N - 1) = N (2 M + 1) - 1
P = 2 N M + (N + 1) = N (2 M + 1) + 1

These two sets of family constraints are redundant.

In general the objective is to first determine the M value and then to slighty modify that value using a:
[(N - 1) / N]
correction factor or a
[(N + 1) / N]
correction factor.

The spheromak will spontaneously step through Np and Nt values liberating energy until the lowest stable state is reached for the prevailing value of P.

A stable spheromak needs to be tolerant of outside disturbances that cause fluctuations in Np and Nt.

The issue is that the prime number P will not significatly change but an external disturbance can potentially cause Np and Nt to temporarily deviate from their normal values Npo and Nto. A stable spheromak should not collapse under such a condition.

This tolerance to disturbance is improved if P is located within a cluster of prime numbers.

Recall that for stability:
(dNt / dNp) =-(1/ 2) or - 2

We need to use the field energy distribution to find Rc and Rs. XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

SPHEROMAK SOLUTION:
The method of finding the mathematical model for an isolated spheromak is:
1) Find equations for the field energy density both inside and outside the spheromak wall;

2) Solve the field energy density equations simultaniouly to find the equation for the spheromak wall;

3) Find equations for the length Lp of a poloidal turn and for the length Lt of a toroidal turn;

4) Find the equation for the length Lh of the current filament closed loop.

5) Since Lh is inversely proportional to the spheromak energy, at the spheromak stable energy:
dLh = 0

6) At the spheromak stable energy apply the condition that:
dNp = - 2 dNt or 2 dNp = - dNt

7) The resulting equations describe stable spheromak behavior.

The contained energy of an isolated spheromak is constant. Hence, for an isolated spheromak:
d{[Lh / 2 Pi Ro]^2} = 0.

During these tradeoffs the spheromak must survive, meaning that at all times Np and Nt must have no shared integer factors other than one, which forces either:
dNp / dNt = - 2
in which case:
Np + 2 Nt = P
as in Case A above or
dNp / dNt = - (1 / 2)
in which case:
2 Np + Nt = P
as in Case B above.
 

Natural spheromaks are very stable, which suggests that an external influence which causes minor incrementation of P should not induce spheromak instability. Hence it is likely that the naturally stable value of P occurs in the middle of a sequence of closely spaced prime numbers. Similarly, the value of Npo or Nto that increments or decrements by 1 likely occurs between two prime numbers and the value of Npo or Nto that increments or decrements by 2 likely occurs within a sequence of closely spaced prime numbers.

It is anticipated that in a stable spheromak Lt may slightly adjust to conform with the constraints on Npo and Nto. Thus an electromagnetic solution of a spheromak only gives an approximate result. The precise result is a function of the actual governing prime number P.
 

Note that the turn numbers Np and Nt can only take discrete integer values that are further limited by the requirement that Np and Nt share no common integer factors other than unity. This requirement indicates that the Np and Nt values of spheromaks normally are governed by a single prime number P.

SPHEROMAK FORMATION:
At the commencement of spheromak formation there is an initial packet of energy. That packet of energy is consistent with a prime number P value. That choice of P value enables a spheromak existence line of constant P of the form:
P = Np + 2 Nt or P = 2 Np + Nt.
Then over time Np and Nt can step along this constant P line in accordance with:
dNp = - 2 dNt
or
dNt = -2 dNp
until the spheromak reaches its stable state. The probability of spheromak deep stability is much higher if the desired value of P is in the middle of a cluster of primes.
 

On this web site it is shown that:
[113 / 355] ~ [1 / Pi]
2(113) + 355 = 581
which is not prime
but:
2(355) + 113 = 823
which is prime and is in a dense group of primes 821, 823, 827, 829
 

This web page last updated July 28, 2025.