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.
 

NATURAL FREQUENCY F:
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 and Np poloidal turns. 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. The length Lh of this filament and the constant speed of light C together give the spheromak a characteristic natural frequency F where:
F = C / Lh

From the web page titled: Spheromak Structure the winding length Lh is given by:
[Lh / 2 Pi]^2 = Nt^2 [(Rs - Rc) / 2]^2 + [Np^2 / 2] {(Rs^2 + Rc^2) / 2}
= Nt^2 [(Rs - Rc) / 2]^2 + Np^2 [Ro^2]
where:
Ro^2 = [(Rs^^2 + Rc^2) / 2]

Each spheromak has a filament length Lh and a characteristic radius Ro.
The ratio:
(Lh / 2 Pi Ro)
is the same for all isolated charged particle spheromaks.
Thus:
Lh = (Lh / 2 Pi Ro) (2 Pi Ro)
and:
Lh F = C
or
F = C / Lh
= [C /(Lh / 2 Pi Ro)][1 / 2 Pi Ro]

Hence:
F = [C / (Lh / 2 Pi Ro)][1 / 2 Pi] [1 / Ro]
and
dF = [C / (Lh / 2 Pi Ro)][1 / 2 Pi] d[1 / Ro]
which expression is important in evaluating the Planck Constant and the Fine Structure constant. The parameter:
(Lh / 2 Pi Ro)
is a function of constant spheromak relative geometry.
 

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

At the spheromak's stable operating point:
d[Lh / 2 Pi]^2 = 0 = 2 Nt dNt [(Rs - Rc) / 2]^2 + [2 Np dNp / 2] {Rs^2 + Rc^2}
or
Nt dNt [(Rs - Rc)^2 / 2] = - [Np dNp] {Rs^2 + Rc^2}
or
Np / Nt = - (dNt / dNp)[(Rs - Rc)^2 / 2] / [Rs^2 + Rc^2]

Prime number theory shows that in a spheromak either:
(dNp / dNt) = (- 2)
or
(dNp / dNt) = (- 1 / 2)

FAMILY B:
P = 2 Np + Nt
where:
P = prime number.
Note that for this family of winding turn numbers Nt is always odd and that:
dNp / dNt = - (1 / 2)

A spheromak has two potentially stable low energy operating states:
CASE A
P = Np + 2 Nt
where:
P = prime number.
Note that for this family Np is always odd and that there is an Lh relative maximum at:
d[Lh / 2 Pi]^2 / dNt = 0
which gives:
dNp = - 2 dNt.
or
[Np / Nt] = - (dNt / dNp)[(Rs - Rc)^2 / 2] / [Rs^2 + Rc^2]
= 2 [(Rs - Rc)^2 / 2] / [2 Ro^2]
= [(Rs - Rc)^2] / [2 Ro^2]

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

CASE B:
P = Nt + 2 Np
where:
P = prime number.
Note that for this family Nt is always odd and that there is an Lh relative maximum at:
d[Lh / 2 Pi]^2 / dNp = 0
which gives:
dNp = - (1 / 2) dNt.

[Np / Nt] = (dNt / dNp)[(Rs - Rc)^2 / 2] / [Rs^2 + Rc^2]
= (1 / 2)[(Rs - Rc)^2 / 2]/ [2 Ro^2]
=[(Rs - Rc) / Ro]^2 [1 / 8]

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

Generally only one of thee two Np, Nt winding turn families is physically valid.

The spheromak must remain stable in the presence of small geometric disturbances that can cause either Np or Nt to integer increment or decrement away from their nominal Np = Npo and Nt = Nto values. In a stable spheromak, Npo and Nto do not share integer factors nor do the new values of Np and Nt share common integer factors after the aforementioned integer incrementation or decrementation.

Note that both Case A and Case B arise from the same expression for the spheromak filament length Lh.
 

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
%nbsp;

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 DIMENSIONS:
Rs^2 + Rc^2 = 2 Ro^2

In a spheromak the values of [(Rs - Rc) / 2 Ro], (Hm / 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
 

It is anticipated that the prime number selection process will make only one of these solutions real.

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 August 27, 2024.