StationaryMagnetic Field
and Coil Electric Generators

Table
of
Contents
Introduction
Stationary Magnetic Field and Coil Electric
Generator
Solid State Generator Y Test
Electric Motor Type Y
Principle Of Magnetic Amplification
Magnetic Field Direction Controlling
Gate
Conclusion
References.

Introduction
Attempts were made to try to
control the flow directions of magnetic fields from magnets using
electromagnetic fields. This magnetic field direction control may perhaps be
part of the operation or behavior of a free energy, or over unity device. The
electromagnetic fields such as in and emitted by  iron rods were to attract
magnetic fields of nearby magnets 2 and at the same time (the electromagnet)
magnetically repel magnetic fields of magnets 4 to control the flow of magnetic
fields of these magnets in iron rods or bars. The electromagnetic field from a
coil produced by an electrical current through the coil wire would then draw and
guide the magnetic field from a magnet 2 at the pole of this electromagnet. The
electromagnetic field in the iron (of electromagnet) would then be reinforced in
strength with the magnet's (magnet 2) magnetic field. The electromagnetic field
and magnetic field both would follow the same path through the solid iron as a
summed electromagnetic and magnetic fields. A change in motion of a mass like a
magnet 4 is an indication of electrical energy input. By combining the magnetic
and electromagnetic field magnitudes without mechanical motion, some small
amount of electrical energy was to be produced from the local space-time of the
magnetic fields of the device and due to the increased magnetic field strength
in the iron. Stationary magnetic field generators may perhaps be free energy or
over unity devices. It is called a stationary magnetic field generator, because
there are no moving generator parts in the stationary magnetic field motor to
produce changing magnetic field amplitudes. Some energy from generator's local
four dimensional space-time may be transfered into three dimensional space-time
by and in its magnetic field B. This page may perhaps show some magnetic and
electrical devices for doing this.

Generator
In the
stationary or motionless magnetic field and coil electric generator abreviated
s.m.f.c. generator, strips of steel or iron sheets pass between the stationary
magnet magnetic poles and the poles of an iron core of an output coil. The metal
strips redirect the magnetic
field of the magnet away from the iron core of
the output coil. The redirected magnetic field then flows away from the output
coil core which produces a changing magnetic field intensity in the coil core.
This changing magnetic field intensity in the output coil core induces and
electric voltage and current from the coil. Figure 1 shows the basic design of
the electric generator. It has a stationary magnet 2, the iron strips 3, the
output coil iron core 4 and the output coil 5. The stationary magnet 2 provides
the magnetic field intensity B. The iron strips 3 can absorb and redirect the
magnetic field with intensity B from the output coil core 4 as shown in figure
1(a). When the strips are not
present between the poles of magnet 2 and core
3, the magnetic field intensity B follows the path through the iron core 4 of
the output coil 5 as shown in figure 1(b). The output coil 5 consists of a
number of windings of copper magnet wire.

S.m.f.c. Electric
Generator

Figure 1.

The iron strips 3 are designed in the
form of a rotor as shown in figure 1(c).

Solid State Generator Y
Test
This page may perhaps show
a method for generating electricity with magnets and electromagnets without
moving mechanical parts.
The solid state
generator is designed to try to control the flow direction of magnetic fields by
electromagnetic fields such that the electromagnetic field amplitude is
reinforced by the magnetic field amplitude. In the solid state electric
generator, there is not relative motion between the stator magnet and the output
coil. The flow of the stator magnet magnetic field is controlled by an
electromagnetic field from a relatively stationary input coil.
Direct
electrical current may produce an electromagnetic field that is stronger with
the magnet.
Figures 2, 3 and 4 shows the basic design of an experiment for
controlling the flow direction of a magnetic field using an electromagnet. It
has stator cylindrical magnet 2 and stator electromagnet 3. It has movable disk
magnet 4.

Simple Magnet and Electromagnet Stator Actuator Test
Design

Figure 2.

The disk magnet 4 is placed over the rounded iron magnetic pole
6 of electric input electromagnet 3. Stator magnet 2 can be moved closer or
further from the other magnetic pole of electromagnet 3. The magnet poles N and
S directions are as shown in figures 2, 3 and 4. Stator magnet 2 has length
Ls and diameter Ds. Electromagnet 3 has soft iron
core diameter Dc and length
Le. Electromagnet 3 has copper coil 5
which receives electric current I from
an electric current supply 8.
Demonstration video 2 shows an experiment of
possible magnetic force direction control and amplification using an
electromagnet of figure 2 design. In the video 2, the electromagnet 3 without
magnet 2 is activated with current I,
and magnet 4 does not move much. The current meter in the background shows the
current I magnitudes. Then magnet 2 is moved closer to electromagnet 3 and
magnet 4 moves with much larger magnitudes with current I changes. Next magnet 2 is moved, without the
electromagnet 3 electromagnetic fields. Movement of magnet 4 were less. The
magnet 4 movements are larger. The electromagnetic field from electromagnet 3 is
not sufficient to move the magnet 4 much, but when stator magnet 2 is moved
closer to electromagnet 3, the movable magnet 4 can be made to move. This
indicates that the electromagnetic field of electromagnet 3 can control the flow
direction of magnetic field of magnet 2 in such as way as to strengthen the
magnet force from the electromagnet 3 that moves magnet 4.

Demonstration Video 2: Magnet and Electromagnetic Actuator
Test:
solidStateGenerator.WMV, file size: 438
kilobytes.

The differences in current I between
without and with magnet 2 are not very noticeable in the video 1. Adding magnet
2 to electromagnet 3 does decrease the magnetic attraction force between magnet
4 and electromagnet 3. When magnet 4 is replaced by a piece of iron that is not
initially magnetized, the magnetic field intensity of magnet 2 and
electromagnetic field intensity of electromagnet does add by
about:

B=Be+Bs.
(5)

When an electromagnet like 3 is magnetized by a magnetic field from
an external magnet like 2, the impedance Z of the electromagnet is expected to
decrease. This decrease in impedance Z
is assumed to be due to the magnetic particles of the iron core of the
electromagnet being able to rotate less due to the incoming magnetic field of
magnet 2. The impedance of the electromagnet 3 copper coil is Z=(X2+R2)1/2, where R is the resistance of the copper wire of the
coil 9 of electromagnet 3. R=47 ohms in
this design. The X=2×π×f×L is
the reactive impedance of the coil 9 of electromagnet 3. The f is the electrical frequency of the input
current I into coil 9, L is the inductance of the coil 9 of
electromagnet 3, and π=3.141592654. Coil 9 has N number of turns of copper wire that has
enamel coating. Ammeter 11 displays the electric current I=V÷Z
from battery 8 to coil 9, where V is the
electric supply 8 voltage. V=9.0 volts.
The electromagnetic field intensity of electromagnet 3 electromagnetic field is
Be, and magnetic field
intensity of magnet 2 magnetic field is Bs. The electromagnetic field
intensity be of electromagnet draws in the magnetic field intensity Bs of stator magnet 2 in such a way
as to increase the magnetic field intensity B of the electromagnet 3 by B=Be+Bs. The B can be the combined magnetic field intensity
of Be and Bs at the pole 6 of electromagnet
3. Ls is 0.1 to 0.12 metre.
Figures 3 and 4 shows the assumed and approximate magnetic field flux line
patterns in and around magnet 2 and electromagnet 3. In figure 3(a) the
electromagnet has no electromagnetic field intensity Be when there is no electric
current I in coil 9. Magnet 4 attracts
with force Fm to the iron
core 10. In figures 3(a) and 3(c), the coil 9 are not drawn in figures. In
figure 3(b), electric current I goes
through coil 9 and as a result magnetic field intensity Be is at iron core 10 pole 6.
Magnet 4 is magnetically repelled by magnetic force F. In figure 3(c) the magnetic field lines
(blue colored dashed lines) from magnet 2 goes through some of the iron core 10
of electromagnet 3 and then some of the lines 14 may come from the iron core 10.
Very little magnetic lines of force reach movable magnet 4 and magnetic field
intensity from electromagnet 3 is B<Bs. Only some of the magnetic lines
of force from magnet 2 reaches the magnetic field of movable magnet 4 in figure
3(c). The magnetic field from magnet 2 that goes into iron core 10 flows mainly
out the sides of iron core 10 in figure 3(c). In figure 4(a) when there is an
electromagnetic field intensity Be in electromagnet 3 due to
electric current I, more of the magnetic
lines of force 15 from magnet 2 stays in iron core 10 and reaches movable magnet
4. The magnetic field intensity B working against the magnetic field intensity
Bm of movable magnet 4 is
now: B=Be+Bs which produced magnetic repulsion force F near pole 6 in figure 4(a). The
electromagnetic field with intensity Be from electromagnet 3 and Bs of stator magnet 2 sum. The
magnetic field intensity Bs
is likely to reduce the magnetic attraction forces between iron core 10 and
Bm which allows the larger
motion of magnet 4. The energy efficiency e is assumed to be:

e=(F×s)÷(Fe×se).
(6)

cE=F×s=Fe×se.
(7)

Where Fe is axial
magnetic force between electromagnet 3 and magnet 4 alone and dFe is change in electromagnetic
force by electromagnetic field from electromagnet 3 alone. Axial magnetic force
is in the direction of the magnetic field direction. The se is motion of magnet 4 while
receiving change in force dFe. The s is movement of magnet 4 while under the
summed change in force dF. Fe=A4×Bm×k4÷r2. With A4 the surface area of magnet 4 and
length k4 of magnet 4, and
distance r between the
magnets.

Magnetic Flux Line Patterns Without and With Be

Figure
3.

Magnetic Flux Line Pattern With Intensities Bs and Be and Flux Line Vectors

Figure 4.

Figures 3 and 4 shows magnetic flux line direction vectors Xsi and Xei.  Figure 4(b) shows a two
dimensional Cartesian coordinate graph of the resultant vector Xi produced by summing vectors Xsi and Xei. The lengths of the arrows
represent the magnetic flux density amount at a point i in magnetic field, while
the direction of the arrow in the graph represents the direction of the magnetic
flux line at the same point i. Magnetic flux line density and direction vectors
Xsi and Xei are summed in figure 4(b) graph,
but the resultant vector Xi is
controlled by vector Xei. Perhaps
Xi=b×Xsi×Xsei. This may be magnetic
field vector amplification of Xsi
to Xi. Magnetic field directions
and intensities are manipulated by other electromagnetic or magnetic fields like
Xei at one location. The Xi vector (arrow) is more horizontal
than Xsi which produces larger
magnetic repulsion forces F between
B and Bm at pole 6. The phenomenon of a
magnetic field direction being controlled by a weaker electromagnetic or
magnetic field can be called magnetic field vector amplification.
Iron core
10 of electromagnet 3 has a large electromagnetic permeability uc.
The magnetic amplification
A of the device is assumed to be:

A=uc×Ds2÷Dc2=B÷Be∝dF÷dFe.
(8)
A=uc×N×Ds2÷(L10×Dc2)=B÷Be∝dF÷dFe.

With length
L10 of iron core 10, and N
the number of copper wire turns of coil 9. Perhaps in part:

B=(Be+Bs)×Bei
(9)

at magnetic pole 6. Exponent i is : 0<i<1. Then magnetic force
F against magnet 4 from magnets 2 and electromagnet 3 is:

F=B×Bm÷r2.
(10)

The r is the distance between pole 6 surface and magnet 4
surface.
The change in force:

dF=dB×Bm÷r2
(11)

appears to be larger with the inclusion of magnet 2 magnetic field
intensity Bs. This may be due
to the larger magnetic attraction force between Be and Bm. From video 2 there can be seen
that the magnetic repulsion forces between Be and Bm alone (without stator magnet 2)
are small, and that the difference between magnetic and electromagnetic field
intensities B and Be (with stator magnet 2) is
relatively large. Magnetic field intensity Be produced by the electric current
I in the electromagnet 3 core 10 appears
to draw some of the magnetic field intensity Bs of the stator magnet 2 through
iron core 10 to the magnetic pole 6 of electromagnet 3 where magnet 4 is closely
at.
Magnetic force F between magnet 4 and iron core 10 may be: F2=uc×Bm÷r2, where u is electromagnetic permeability of iron core
10. The internal energy or potential between iron core 10 and magnet 4 may
perhaps be: Ei=F2×se2 where Ei is a constant which is different
than the energy equation (6). The se2=se.
Demonstration video 3
shows an experiment using the same magnetic and electromagnetic actuator design
as in figure 2. In video 3 the magnet on weight scale is first pushed
magnetically in verticle directions by electromagnet 3 (beige colored
part). Then the cylindrical rod magnet ( black colored disks) was
added over electromagnet 3  pole producing larger swings of weight
indication on scale.  In this case, the magnetic repulsion forces between
magnet 4 and electromagnet 3, and electromagnet 3 with magnet 2 are measured by
a weight scale. The dFe is a
small change in magnetic force of electromagnet 3 alone on magnet 4. In video 3
there can see the small change in force dFe with electromagnet 3 alone. Then
a larger change in force dF with magnet
2 at other pole of electromagnet 3 can be seen. There can be seen that the
change in magnetic force dF with magnet
2 is larger than dFe with
only electromagnet 3. Electromagnet 3 does seem to control the magnetic flux
line flow of stator magnet 2. The small cylindrical black colored object is
ferrite magnet 4 with electromagnet 3 about 0.025 metre above it. Demonstration
video 3b is similar as video 3 except there is a smaller distance between the
pole 6 and magnet 4.

Demonstration Video 3: Magnet and Electromagnetic Actuator Force
Test:
MagneticGenerators/dFe2.WMV, File size:
530 kilobytes.
Demonstration Video 3b: Magnet and Electromagnetic Actuator Force
Test:
dFe4.WMV, File size: 518
kilobytes.

The weight scale meter near centre of
the video 3 can be made to change through a larger range by electromagnet 3 with
magnet 2 in use. Moving electromagnet 3 iron 10 alone over magnet 4 on weight
scale did not show a noticeable weight or force Fe change.
This would indicate that the dF>dFe or increased motion of magnet 4
was not due to reduced magnetic attraction between iron 10 of electromagnet 3
and magnet 4. This may indicate that F×s
and Fe×se
from above equation (7) are not equal. A question is: will the larger weight or
force swings increase the electrical efficiency of the generator Y
design? Larger weight or force F
variations may indicate that:

F×s>Fe×se.
(12)

The current meter on the left side is ammeter 11 and shows input
current I into coil 9. Sometimes the
weight scale meter does not return to its orginal positions indicating that the
iron 10 has some magnetic memory.
Using calculus integration mathematics, the
potential energy difference of equation (12) becomes:

E=∫ F×ds=∫
k×s×ds
=(k÷2)×(sf2-si2)
Ee=∫ Fe×dSe=∫ k×se×dSe=(k÷2)×(sef2-sei2).
(13)

The symbol ∫ is the calculus integration sign. The k is the spring constant of the spring 17 of
the weight scale 16 that works against the magnetic forces F and Fe. The tension force of the spring
17 of the weight scale 16 on which magnet 4 rests on is made mathematically
equal to F or Fe. Variable sei is initial position of magnet 4
which is also the stretch distance of the spring 17 with spring constant k. This is initial displacement with magnetic
field intensity Be alone.
Fe=Be×Bm÷r2. Variable Sef is the final position of magnet
4 with magnetic field intensity be alone from electromagnet 3 alone. Variable
Si is position of magnet 4
and stretch distance of metal spring of weight scale with magnetic field
intensity Bs only of stator
magnet 2. Variable sf is position of magnet 4 due to magnetic force
F caused by magnetic field intensity
B=(Be+Bs)×Bei when stator magnet 2
and electromagnet 3 are both used to produced a magnetic repulsion force F on magnet 4. Examples: k=3×101 newton/metre, sei=0.0 metre, sef=0.0017 metre, si=0.0025 metre, sf=0.0055 metre of video 3b
magnetic actuator design. Copper coil 9 outer diameter Do=0.040 metre, uc=300 to 500, and core 10 diameter
Dc=0.016 metre. The E and Ei are expected to be equal, but
seem not to be equal with Ei=Ee. Possibly E=Ei+Ee, but where does Ei go to? Perhaps the energy E is also E= , A4=2.8×10-4
metre2, k4=0.0096
metre.
∫ F×ds=(m÷2)×v2, where m is the mass of magnet 4 and v is the average vertical velocity of the
magnet 4. The moving speed v may also be
an indication of energy E as well as its
motion magnitude s. The speed v of motion may be measured by counting the
amount of times distances s and se are done within a certain time
period t. The time period for 20 s span motions is t= , and for 20 se motions is te= . The m=0.0131 kilogram. The oscillation frequency
is f=1÷T, and distance s moved should be proportional to average
velocity v. The T=t÷20
cycles, and Te=te÷20 cycles. The device may be
working as an electromagnetic field vector or intensity and direction amplifier.
The magnetic field flux line directions from a magnet like 2 may be redirected
by another electromagnetic or magnet field like Be in such a manner as to sum the
magnetic field strengths of both magnetic fields. There seem to be little
required electromagnetic field intensities to re-direct a magnetic field
polarities or flux lines from a magnet like 2. Figure 5 may show the assumed and
inaccurate magnet 4 displacements se and s, and input currents I versus time t. Figure 5(a) shows the displacements se of magnet 4 and input current
I without stator magnet 2. Figure 5(b)
shows the displacements s and input
current I versus time t with magnet
2.

Magnet
4 Displacements s And Input Currents
I Versus Time t

Figure 5.

The currents I
amplitudes in figures 5(a) and 5(b) are similar, but current I in figure 5(b) exists a little longer in
time t. There is a minimum input current
Imin level to move movable
magnet 4 from the iron core 10 of electromagnet 3, which is: Imin=Bm÷(Bs×l), where l is coil 9 copper wire length that carries
the input current I. Figure 6 shows a
similar design to the design of figures 2, 3 and 4. Figure 6 design has a stator
magnet 2 and and input electromagnet 3. The coil 9 of electromagnet 3 is
slightly smaller in size relative to the iron core 10 diameter Dc than the design of figures 2 and
3. A movable magnet 4 operates as a projectile. The input coil receives a larger
current I relative to the size of the
coil 3.

Electromagnet Actuator For Larger relative Input Current
I
Figure
6.

Electromagnet 3 coil 9 has about N=400 turns of copper wire, with coil 9 copper
wire resistance R=4 ohms, and Dc=0.006 metre, l=45 metres. The length L4 of the movable cylindrical
shaped magnet 4 must be large enough. Demonstration video 4 shows the magnetic
actuator of figure 6 in operation. This magnetic actuator is also a magnet
accelerator. The movable magnet 4 travels vertically further with the magnet 2
than without magnet 2 with electromagnet 3 is energized with input current I. This and the vertical speed of magnet 4
indicates that the magnetic actuator will work for smaller time periods T and Te of figure 5 graph. The ammeter
at lower left on the screen shows the input current I. In video 4, the current I is initially applied to coil 9 (the copper
colored wire), and little movement of magnet 4 is seen. Then when magnet 2 is
placed underneath electromagnet 3 of coil 9, magnet 4 is quickly
electromagnetically repelled by electromagnet 3 iron core 10. The lower magnet 2
reduces the magnetic attraction force between magnet 4 and iron core 10 which
permits magnet 4 to be pushed away by electromagnetic field of electromagnet 3.
Maximum current I in both tests seems to
be about I=  ampere.

Demonstration Video 4: Magnet and Electromagnetic Actuator Force Test

Using Larger Relative Input Current I:

magnetLauncher.WMV, File size: 265 kilobytes.

Without magnet 2 there was less change in motion of magnet 4.
Smaller
time periods T and Te appear to require larger input
currents I. I=1÷T.
Adding stator magnet 2 underneath electromagnet 3 reduces the magnetic
attraction force between movable magnet 4 and iron core 10 of electromagnet.
This enables magnet 4 to leave core 10 easier, but the purpose of video 4
experiment is to see if smaller time periods Te between peaks of se
will also work. The vertical upwards acceleration ay of magnet 4 in video 4 at a
moment of time is assumed to be:

ay=(Fe-(Fm-F2)-Fw)÷m,

where F2 is the magnetic repulsion force
between magnets 2 and 4 at magnetic pole 6. The m is the mass of magnet 4, and Fw=m×g is
the weight of the magnet 4. The Fe can be the magnetic repulsion
force between magnet 4 and electromagnet 3 alone when input current I is applied is applied to electromagnet 3
coil 9. Force F2 reduces
Fm. Examples: m=3×10-2 kilogram, g=9.81 newtons/kilogram, F2=0.01 newton, ay=1
metre/second2.
When a smaller magnet 3 with a stronger
magnetic field intensity B3 is close to another larger magnet 2 with
a weaker magnetic field intensity B2, the smaller magnet's magnetic
field attraction force may overcome the weaker magnetic field repulsion force as
shown in figures 8(a), 8(b) and demonstration video 5. The smaller magnet 3 will
be magnetically attracted to magnet 2 ferrite material even if magnets are at
magnetic repulsion orientation (north to north poles N). Magnetic field
intensity B3 magnetic flux
lines (green colored lines) of smaller magnet 3 push aside the magnetic flux
line (blue colored dashed lines) of magnetic field intensity B2 of the larger magnet 2. The
magnetic lines of force shown as dashed lines are at an imaginary plane 4
intersecting magnet 2. B3>1.5×B2, B2=0.01 tesla.
Figures 8(a) and 8(b) may show a more correct model of magnetic field flow
patterns between repelling magnetic fields when one magnetic field intensity
(B3) is much stronger. the
stronger magnetic field intensity like B3 dominates over a weaker magnetic
field intensity like B2.
Figure 8(c) may show a less correct model of magnetic field flux line flow
directions for B3>1.2×B2. The repelling magnetic lines of
force are bend away from each other by each other. Magnet 2 diameter D2=0.04 metre with length L2=0.008 metre, magnet 3 diameter
D3=0.003 metre with length
L3=0.0015
metre.

Magnetic Flux Lines Between Magnetic Fields
Of Different Intensities

Figure 8.

Demonstration Video 5. Figure 8 Magnets
Experiment:
magneticgenerators/
.WMV,
file size: kilobytes, (not available
yet).

When the distances r between the magnets are further, there is some
magnetic repulsion forces. When the distance r is shorter, magnetic attraction
between magnet 3 and magnet 2 iron can dominate the magnetic repulsion force.
This type of magnetic lines of force behavior must be taken into consideration
when designing a magnetic actuator.

Figure 9 may show a superior
and similar design to figures 2, 3, and 4 design. In this similar the design,
both magnetic poles of stator magnet 2 are used. In this design the stator
magnet 2 has a u shaped design. Stator magnet 2 consists of two magnets 2a and
2b and an iron magnetic flux guide 2c. Movable magnet 4 also has its south S and
north n poles used. Movable magnet 4 has two magnets 4a , 4b and an iron
magnetic flux guide 4c. Electromagnet 3 now has two iron cores 10a and 10b and
two corresponding input coils 9a and 9b. Electromagnet 3 also has a third curved
shaped iron core 10c that is between cores 10a and 10b. Between cores 10c, and
10a is an air gap with span y.

Dual Pole Magnet and
Electromagnet Stator Actuator Design
Figure 9.

The
electromagnet 3 core section 10b needs an air gap span y of sufficient width,
because the electromagnet repulsion force at 20 are not easily redirected.
Magnetic repulsion force vectors may not be easily re-directed. as shown in
figure 10. In figure 10(a) the magnets are far apart with magnetic repulsion
force vectors as shown. When the magnets are brought closer together, the
magnetic repulsion forces or force vectors do not vary much.

Magnetic Repulsion Force Vectors
Figure
10.

Electric current Io induction by magnetic and
electromagnetic fields may be: Io=Lr×dH/dt,
where Io is varying direct
current. Where H=B×π×Dr2, Lr is inductance of output or
secondary coil 11, and Dr is
diameter of secondary iron core 12 of coil 11.
Magnetic
reluctance is the ability of a ferrous materials to resist the flow of magnetic
fields through it. Reluctance is the opposition produced by a magnetic substance
to magnetic flux; specifically the magnetic potential difference to the
corresponding magnetic flux in the same material. In this case the magnetic
reluctance ґ has units of
metre2 tesla/newton. Figure 10b may show a design of a magnetic
reluctance meter that uses these units of measure. It has a stator magnet 2 with
magnetic field intensity β and surface
area A2 at its magnetic pole.
A rotor 3 can be used to measure the magnetic force. Magnetic materials (rods) 4
and 5 under test which transfer the magnetic field from magnet 2 to rotor 3.

A Magnetic Reluctance meter.
Figure 10b.

The
magnetic reluctance per length Lm of material has units of
metre3 tesla/newton. Imaginary examples: β=0.02 tesla, A2=0.001 metre2, ґ =1 metre2 testa/newton, ґ /Lm=0.01 metre
testa/newton.

Electric Motor Type
Y
The magnetic actuator effects
and design of figures 2 through 4 can be used and made into an electric motor
design called electric motor type Y. The movable magnet 4 of figures 2 through 4
can become part of a piston and crank shaft assembly 16 in figure 11(a) which
turns the reciprocating motion of magnet 4 into rotary motion. Magnet 4 is
attached to aluminum piston 18. The piston transfers the magnetic repulsion
force F causes by electromagnet 3 and
magnet 4 onto flywheel 20 via cranks shaft 19. The electric motor type Y of
figure 11(a) has similar parts to the design of figures 2 and 3. The motor has
stator electromagnet 2 and stator magnet 2. A commutator and switch assembly 17
produces the required current input I
timing T. Magnetic repulsion force F=Bm×(Bs+Be)÷r2. Magnetic field from stator 2
can produce magnetic repulsion forces with the piston magnets 4 through iron
core 10, and reduce magnetic attraction force between piston magnet 4 and
electromagnet 3 iron core 10. There may be a small air gap z between electromagnet 3 iron core 10 and
stator magnet 2 to help reduce these magnetic repulsion forces as the magnet 4
approach the electromagnet 3 pole 6. Figure 11(b) shows the electric circuit
schematic diagram of the coil 9 input circuit. Input voltage V into input coil 9 can be V=+18 to +20 volts.

Magnet
Motor Type Y
Figure 11.

Demonstration video 5 will likely show
an electric motor type Y test of figure 11.

Demonstration Video 6.
Electric Motor Type Y  Experiment:

http://www.machines-x.info/magneticgenerators/ .WMV,
file
size: kilobytes, (not available yet).

Principle of Magnetic
Amplification
Magnets of increasing size but with similar
magnetic field intensities B are permitted to rotate about a fixed axis each.
Figure 13 shows the design of magnets of increasing sizes as a linear array of
cylindrical magnets or linear array of magnet rotors of magnets of decreasing
size. The linear array of rotors has magnets 2, 3, 4, and 5 of increasing size
that are permitted to spin about their fixed axii 7a, 7b, 7c, and 7d
respectively. The magnet spin speed is w. The rotating magnet array should have
at least 4 magnets which have their magnet pole axii being able to be aligned as
these spin at speed w. The axii of rotation of the magnets are parallel to each
other and on the same imaginary horizontal line. The magnetic poles like N or S
of ferrite magnet 5 can attract magnetic poles of adjacent magnets like 4.
Rotating magnet 5 abouts is spin axis 7d can rotate magnet 4 due to attracting
magnetic forces between the two. The same occurs between magnets 4 and 2 and
then also between magnets 3 and 2. Rotating the largest magnet 5 will rotate
then rotate the smallest magnet 2 via magnets 4 and 3. Rotating magnet 5 by
applying forward torque Tsf
can rotate magnets 2, but rotating magnet 2 with forward torque Tof may not rotate large magnet
5.

Linear Array Of Magnet Rotors

Figure 13.

Rotating the
largest magnet 5 rotates the second magnet (second largest magnet 4). This
second magnet 4 in turn rotates the third larger magnet 3 beside it. Rotating
the smallest magnet 2 has very little affect on the largest magnet 5 rotation.
The entire assembly has aluminum support 6 in which the physical axis or axle 9
of magnet 5 can be allowed to rotate in. Frame 8 holds the magnet 5 onto axle 9.
The frame 6 is not shown in the figures on the left of the draft figure 13. If
Tof is small, then the rotor
magnets 2 through 5 rotate in unisen and the Tof=Tsf.
Magnetic field strength at
the pole of each cylindrical shaped rotor magnet like 2 and 3
is:

Hn=B×rn2×π÷4,

for
n=1 to i magnets. Where rn is
the radius of the nth magnet of the array. The larger magnet 5 has a
larger magnetic field strength H5 than H3 of magnet 3, and so magnet 5 is
more difficult to rotate with magnet 3 via magnet 4. Magnet 3 cannot rotate
magnet 5 via magnet 4, because the magnetic attraction is stronger between
relatively larger magnets 4 and 5 than between magnets 3 and 4. The difference
between magnetic field strengths H3 and H5 in the magnet array may perhaps
cause magnetic amplification. Examples: r5=0.013 metre, B=0.01 tesla, w=6 radians/second, π=3.141592654, i=4
magnets, r3=0.009 metre,
Tof=Tsf=0.0001 newton metre.

Can use an electrical transformer and replace the magnetic field with an
electromagnetic field. The affects are opposite to the magnets of figure 13.
Figure 14 shows and electrical transformer design. This transformer T is called
a variable or non-uniform core sectional area transformer. The iron core
sectional areas in the primary and secondary coils need not be the same. It has
a toroid shaped iron core 2 with core length or average circumference c. The core 2 has two different sectional area
dimensions. Primary coil 3 has coil core 4 sectional area Ap. Secondary coil 5 has iron core
6 sectional area Ao.

Electrical Transformer
With Two Sectional Area Dimensions
Figure
14.

Apþp=Aoþs,
þs/þp=R,

where þp is the electromagnetic flux line
density of primary coil electromagnetic field in primary coil core sectional
area Ap. The þs is the electromagnetic flux line
density in secondary coil core sectional area Ao of secondary coil. By placing
iron filings or dust on magnet poles, observations have shown that the magnetic
flux density þp or þs is  and is  of
core length c. The peak output voltage
Vs from the secondary coil
is:

Vs≈Vp×ls÷lp

for large Ii;

Vs≈Vp×Ns÷Np,

where Vp is the peak input voltage into
primary coil, lp is length of
coil wire of primary coil and ls is length of secondary coil
wire. Ii is the primary coil
3 input signal current. Np is
primary coil 3 number of wire turns. Ns is number of secondary coil 5
wire turns. Induced output voltage
Vs with no load RL current Ii appears to be: Vs∝Ls×Ii×dIi/dt. Magnetic field strength Hp in primary coil 3 iron core 4
is:

Hp=B×Ap.

Magnetic field strength
Hs in secondary coil 5 core 6
is:

Hs=B×Ao,

where B is electromagnetic field intensity in iron
core 2 of transformer.
Ao>Ap. This design works on the idea
that the magnetic field intensity B from
a magnet's pole does decrease when the opposite magnetic pole is attracted to a
large piece of iron. A larger Ao does not diminish B in primary coil core sectional area Ap. Primary coil 3 has input
impedance:

Zp∝uo×u×(Ap/Pp)×Np2÷hp,

where Pp=2wp+2dp is the perimeter of area
Ap.
secondary coil 5
impedance:

Zs∝uo×u×(Ao/Ps)×Ns2÷hs,

for iron core
electromagnetic permeability u>100. Where hp is length of primary coil 3, and
hs is length of secondary
coil 5. Ps=2ws+2ds is the perimeter of area
Ao=ws×ds. ws=ds.
Pseudo energy
efficiency constant may be:

sE=(Hs×∂Hs/∂t)1/2÷(Hp×∂Hp/∂t)1/2,

with change
∂þs of flux line
density þs during time ∂t. With change ∂þp of flux line density
þp during time ∂t.
Figure 15 shows one design. It is similar to the design of figure 14 except the
iron core is an H-core shape. The primary coils 3a and 3b working as one coil 3
have sectional areas Ap<Ao. Ap=wp×dp. The area of a rectangle like
Ap is largest when the side:
wp=dp. The secondary coil 5 is in the
middle in the middle section of the transformer T.

Variable
Sectional Area Transformer With An H-Core Shape
Figure
15.

Vp and Vs are forward voltages when the
primary coil 3 is the input coil, and secondary coil 5 is the output coil as
normal. Vrp is the reverse
voltage from the primary coil when the primary coil 3 becomes the output voltage
and Vrp is the reverse
voltage into the secondary coil when secondary coil 5 becomes the input coil of
the transformer. Vrp can be less than Vs even when Ns=Np=200 turns. Ns=Np=200, 300, and 350 turns.
Demonstration video 8 shows this using the design of figure 15. The digital
multimeter on the left back in video 8 measures input voltage Vp=5.5 and 6.3 volts-peak. The
green screen in the middle shows the input currents Ii=1.2 ampere-peak and then Iri=0.48 ampere-peak. The meter on
the right shows Vs=9.5
volts-peak and then Vrp=2
volts-peak. The H-core transformer T is shown last in video 8.

Demonstation Video 8: Forward Voltage Vs, Reverse Voltage Vrp
and Input
Voltage: reverseVoltage.WMV, file size: 260 kilobytes.

Voltage Vrp should be equivalent to counter
voltage in primary coil on forward voltage Vp. Vp=5.6 volts-peak as sine wave
frequency f=60.0 hertz. Vrp=2 volts-peak, Vs=9.5 volts-peak. Ii=4×0.15 v-p÷0.5 ohm=1.2
ampere-peak. The electric input current when transformer T is reverse driven
(with secondary coil 5 as electric input) is: Iri=1.6×0.15 V-p÷0.5 ohm=0.48
ampere. Zs÷Zp=Iri÷Ii. hs=hp=0.022 metre.

Vp÷Vrp=Gv×(Zp÷Zs).

Ii=(Vp-Vrp)÷Zp. Examples: Ap=0.001 metre2, ∂þp=þp=1000 magnetic lines of
force/metre2, Ao=0.002 metre2, ∂t=0.01 second, þs=∂þs, Vp=9 volts-peak, lp=ls=5 metres, B=0.01 tesla-peak, Hs=∂Hs, Hp=∂Hp, Ii=dIi=1 ampere, ∂t=dt=1÷(4×F), f=60.0 hertz, secondary coil inductance Ls=0.3 henry, Np=Ns=200 turns, u=500.
Can try to see the change
ΔIi in input current
Ii as the output current
Is is changed by
ΔIs of the design of
figure 15 and video 8. Demonstration video 9 shows a sample test. The upper
oscilloscope trace shows the input current Ii and the lower trace shows output
current Is from secondary
coil 5. The alternating current meter on the right shows input voltage Vp in the primary coil
3.

Demonstration Video 9: Testing Change in Input Current
ΔIi:

changeInCurrents.WMV, file size: 158 kilobytes.

The test results
from video 9 experiment were: ΔIs=(0.19-0.06)V-p/0.35 ohm =0.37
ampere-peak, ΔIi=(0.27-0.28)V-p/0.5 ohm=-0.02
ampere-peak,
Np=410
turns, Ns=350 turns, ΔVp=10 V-p - 10.5 V-p=-0.5
volts-peak, load RL=0.7 ohm,
bi=0.0 to 5 degrees, 0.03
metre>Ap÷hp>0.021 metre. V-p means
volts-peak root mean square.
The ΔIi is the change in input current
Ii produced by a change of
output current ΔIs of
output current Is.
Transformer T current gain:

Gi=(ΔIs÷ΔIi)×(Np÷Ns).

Then for the design of
video 9, the Gi=22
amperes/ampere. Then the pseudo power gain of the variable core area transformer
T is:

Gw=Gv×Gi.

The real electric input
power Wp into primary coil 3
is:

Wp=Vp×Ii×cos bi,

where bi is the phase angle between the
peaks of Vp and Ii. The real electric output power
Ws from the secondary coil 5
is:

Ws=Vs×Is×cos bs,

where bs is the phase angle between the
peaks of Vs and Is which should be measured by an
induction type alternating current watt-hour meter.
If there are n
number of transformers Tn, then the electrical current input into
each transformer is Iin+1,
and electrical input voltage to each primary coil is Vpn+1. Ii1 is electrical current input
without an electrical transformer T when load RL is directly driven by Ii1.
Ii2=Ii1/2. The Nsn+1 is the
number of coil wire turns of the primary coil of  nth non-uniform sectional
area transformer Tn. Examples: bi=bs=20 degrees, Ii1=0.2 ampere-peak, Vp1=9 volts-peak. Figure 16 shows a
different electrical transformer T shape. The primary coil 3 is the central
coil, and secondary coil 5 is divided into two equal coils 5a and 5b. The
secondary coils 5a and 5b have a core sectional areas that sum into Ao>Ap. The larger primary coil 3 core
sectional area Ap is allowed
to be larger in this design like in conventional electrical transformer designs.
When the magnetic field strength Hp leaves the primary coil 3 iron
core, it is divided into two. One half goes to coil 5a core and the other half
goes into coil 5b core. Electrical power transfers less easily from the primary
coil giving a magnetic amplifier.

Partial Toroid Transformer
This section may help
explain some magnetic field flows in toroidal transformers or
partial toroid transformer that has an air gap  in the toroid core.
Figure 16 shows the basic toroidal transformer design with an air gap. The
toroid transformer has a ferrous toroid core 2  with a small air gap 3 in
figure 16(a). On the other side of the air gap 3 is the primary coil 4 for
electrical input Vi4×Ii4 . The primary coil
4 receives electrical input current Ii4. There is a movable
secondary coil 5 that can have its angular position ase
relocated during the test. The secondary coil collected electromagnetic power
from the electromagnetic field in toroid 2 and turns it into electrical output
to drive  a light bulb 6. The air gap 3 then has at its centre a magnetic
neutral zone 9. This magnetic neutral zone is from the electromagnetic field
emitted by primary coil 4. Experiments shows that the electrical
output Pout5 from secondary coil 5 is largest with secondary
coil 5 is wound directly over the primary coil 4 in figure 16(a). The electrical
power from the secondary coil 5 is least when the secondary coil 5 is furthest
from primary coil 4 and is (secondary coil 5) over the air gap 3 as shows
in figure 16(b). The electrical power output  is medium when the secondary
coil 5 is half between the air gap 3 and primary coil 4 centre as shown in
figure 16(c). Demonstration video 2c shows this. The light bulb 6 was brightest
when secondary coil 5 is just next to primary coil 4, and then the light
bulb is dimmest when close to the air gap 3 in video 2c.

Toroidal
Transformer With An Air Gap.
Figure 16.

Graph 7 in figure 16(c) shows the output powers Pout5
through light bulb 6 from secondary coil 5 versus angular
location ase  of centre of secondrary coil 5 relative to
centre of primary coil 4.

Demonstration Video 2c: Toroidal Transformer With An Air Gap
Test;
file size: 503 kilobytes
.

The output power  as can be seen in the
video seems to be:

Pout5=
Vi4×Ii4×cos
(ase/2)+z.
(20)

The output
wattage Pout5  in light bulb does not seem to change
with the inverse of the square root of the distance between secondary coil 4
centre and primary coil 4 centre so that this equation (20) may perhaps be more
accurate. At ase= 180 degrees, the secondary coil 5 centre is
furthest from the primary coil and is over (or within) the air gap 3. This
would indicate that electromagnetic field amplitude changes in one coil
4 can have reduced affects in another coil 5 electrical
current at ase=180 degrees angular location. May
perhap be able to use air gaps in toroidal cores in stationary magnetic
field generator designs. Variable Bi4 can
be electromagnetic field intensity vector at centre of primary coil 4. The
electrical power Pout5  may perhaps be:

Pout5∝A2 Bi4[(N÷(r2×ase
))+ (S÷(r2 ×(360º-ase
)))]+z,

where N=+1 when secondary coil 5 is at the magnetic
north pole, and S=-1 when seondary coil 5 is at the south magetic
pole. This equation says that the electromagnetic field is also determined by
the magnetic polarities N and S. The r2 is the
radius of the toroid core 2.  The A2=W2
×d2 is the surface area of the core 2 magnetic pole in figure
16(a). Variable y3 is the width of the air gap 3, and the
output power reduced as air gap y3 increases as shown
near the completion of video 2c such that:

Pout5∝A2 Bi4[(N÷(r2×ase
))+ (S÷(r2 ×(360º-ase
)))]÷y3 +z.

Test values: Vi4=9
volts-peak, Ii4= ampere-peak at f=60 hertz, 10
degrees<ase<180 degrees, Pout5=0.2
watt, W2= , d2= ,  y3=0.0003
metre. Imaginary examples:
Bi4=0.005 tesla-peak ↑,
z=Pout5 /7.
The magnetic field
direction controlling gate may perhaps show the principle of magnetic
amplification also.

Magnetic Field Direction Controlling
Gate
The magnetic force Fx is weakest at
the magnetic neutral zone 6 in figure 2T. Figure 2T shows the design of the
experiment. The design has two similar ferrite magnets 2 and 3, that are
spaced apart by a plastic or wood spacer 4. An iron sheet 5 is allowed to move
freely in the x-axis direction in linear path 7 between the magnetic poles of
magnets 2 and 3. The linear path 7 allows iron sheet 5 only to move in the
magnetic field nuetral zone 6. Iron sheet 5 would be magnetically attracted two
magnet 2 or 3. Demonstration video 2T shows the experiment of figure 2T. Can see
from video 2T that there is generally very little magnetic force Fx in x-axis
direction and towards the central axis 8 of magnetic field of By for thin
sheets 5. Magnets 2 and 3 being similar have width W and depth d.

Magnetic Force Fx At Magnetic
Nuetral Zone Demonstration Device
Figure 2T.

Demonstration Video 2T: Force Fx
Demonstration: forceFX.WMV,
file size: 436
kilobytes.

Demonstration video 3T (not available
yet) experiment is similar to video 2T experiment except the iron
sheet 5 thickness Tp in video 3T is
thicker.

Demonstration Video 3T:

Magnetic force
Fx in the
direction of movement of iron sheet 4 between magnetically attracting magnetic
fields is defined mathematically as:

Fx∝By×A×(cos a)×u ×uo×Tp,
Fx∝A×(Σ( By×cos
ai)÷i)×u×uo×Tp÷dp,
(2)

if the general direction of By is
perpendicular to Fx direction in
sheet 5. This is for small distances between iron sheet 5, and magnets
2 and 3 surfaces. Where By
is the magnetic field intensity of magnets 2 and 3 in the y axis direction and
dp is the depth of iron
sheet 5 that is parallel to direction d. The reason why variable dp is used in equation
(2) is that reducing iron sheet 5 depth length dp bends magnetic flux
lines of force toward iron sheet 5 and increases the magnetic flux line
concentration in iron sheet 5 which increases force Fx
. Variable a or ai is angle between the
directions of Fx and By. The magnetic
field lines from magnets 2 and 3 poles should bend towards the iron sheet 5
changing the angles ai. Variable
u is the electromagnetic permeability of iron
sheet 5 that has thickness Tp. Symbol A represents the surface area of magnet
2 or magnet 3 that is perpendicular to By direction; A=W×d in
this case. The direction By is
tangent to magnetic line of force at the particular starting  point of
vector By
. The W  in this case is in the
direction of x-axis. Some magnetic field lines of force do escape toward
the sides of magnets 2 and 3 and this lines of force have magnetic field
angles ai that do
not equal to 90 degrees. Magnetic force F between a magnet and
a sheet of iron 5 in general in figure 3T is:

F=ABy2(cos a)÷r82,
F∝dBy2(cos a)÷r82,

when
magnetic field strengths Hv=Hv2, Hv=Hv3 of stator magnets
2 and 3 respectively are identical or nearly indentical. Where a is the angle between directions
of F and By, and
r8 is the distance
between the iron sheet 5 and magnet.

Angle a Between Force F and Magnetic Field Vector
By
Directions

Figure 3T.

The mechanical energy Ux to move sheet 5 along
distance sx in the
x-axis direction is:

Ux=Fx×sx.

The dy is the distance between the
facing pole surfaces of magnets 2 and 3 and is slightly larger
than Tp .
The magnetic
flux lines tend to attach to iron shield 5 and cause the magnetic flux lines to
bend when iron sheet 5 is moved. The larger the distance dy between magnets 2 and 3,
the less easier the magnetic flux lines detach from iron sheet 5.This makes the
angle a less than 90 degrees.

Imaginary examples: Fx=0.00010 newton → =F ,
By=0.02
telsa ↑ , Byi=0.02
telsa ↑,  a= 90
degrees,  u=200, Tp=0.0005 metre, sx= 0.0200 metre, A=0.0013 metre2=0.047 m.×0.022
m.,  d= 0.022 m.=dp ,  uo=4π×10-7 tesla
metre/ampere, Ux=0.000001 N.
m., π= 3.141592654, 50 degrees<ai<90 degrees for i= 1 to 150
angles, dy=1.01×Tp  =  0.0091
metre. This design is based on John Ecklin's "Stationary Armature
Generator" machine idea.

Figure 4T.

Figure 5T.

Can use the transductor to generated a force
Fy
that has a direction perpendicular to the direction of Fx. Figure 6T
shows the basic design. It has two stator magnets 2 and 3 with magnet lengths
L 2 and L 3 respectively. Magnets 2 and 3
may have similar magnetic field intensites By at
points at the centre of their surfaces A which (By) are
independent on size of the magnets. It has a movable magnet 8 that is free
to move up and down on a weight scale section 9 along the y-axis direction. The
weight scale can be used to measure verticle force Fy. An
iron plate 5 with dimensions a little larger than W and d can be inserted in the magnetic field
nuetral zone between stator magnets 2 and 3 to vary the magnetic field strength

H v from
the same magnets 2 and 3. Where Hv exists between magnets
3 and 8. Iron sheet 5 would be a magnetic field direction controlling gate. It
send much of the magnetic field from magnet 2 into magnet 3. Variable
ym is the distance between the surfaces of magnets 2 and 4
poles.

Transductor As A Varying Force
Generator.
Figure 6T.

Demonstration Video
4T: Force Fy
Demonstration: forceFY.WMV,
file size:
kilobytes (not avaliable yet).

The verticle magnetic force Fy should
be:

Fy=Hv8 Hv÷r82,

Fy=co
Hv8
Hv÷r82,

Hv=A By(L2+L3),

where Hv8 is the magnetic
field strength of magnet 8 and r8 is the distance between
magnets 3 and 8 centres. The co is the ability of a
magnet like magnet 3 to conduct another magnetic field through itself.
The co=(Fv-Fv3)÷Fv2;  the force
Fv produced by two
magnets 2 and 3 together, force Fv2 is force of
magnet 2 alone and force Fv3 is force of magnet 3
alone. Magnet 2 length L2  is added to the
length L 3 of magnet 3 with
iron sheet 5 making magnets 2 and 3 somewhat operate as one magnet against
movable magnet 8. The Hv8 would change the
Hv of
magnet 3, so that equation (2) may not apply. The Fx
magnitude may be weaker than the Fy magnitude. The
mechanical energy efficiency would then be:

e=∫Fy dsy ÷ ∫Fx dsx,

with the x-axis origin
on By axis. Where sy is the verticle
displacement of magnet 8. Demonstration videos 2T and 3T shows that iron
sheet 5 thicknes
Tp produces little difference in
forcesFx, and
that most of the angles of ai are close to 90 degrees.
This may perhaps indicate that mechanical efficiency e can be a little larger than 1
joule/joule. This means that if force
Fx in direction a=90 degrees does not exist in above
conditions, then changes in the force Fy is nearly a free
magnetic force. Each Hv of both magnets 2
and 3 may not remain identical with changes in Hv8, so equations
(2) may not apply. Fx amplitudes
are much smaller than magnetic attraction force in direction of
By even if magnetic field strengths of magnets 2 and 3 are not
identical.  When displacement y is going downwards in this case the
y is positive and for reverse displacement y, the
y is negative (minus signed). The magnitudes Fx seem smaller
than Fymagnitudes with
iron sheet 5 in between and close  to magnets 2 and 3.
Imaginary
examples: Fy=0.0007 newton ↓,
Hv8=8×10-7
tesla metre3 , Hv=1×10-6 tesla
metre3 ↑, r8=0.20
metre, L2=L3=0.008 metre, sy=0.006 metre ↓, e=0.9 N. m./N. m., dsy=0.006 metre, dsx=sx,
ym≈Tp,

∆ym≈Tp/2, Fv=0.2 newton , Fv2= 0.1
newton, Fv3=0.1 newton.

There can be magnetic field intensity Bx
from a second magnet that has a direction perpendicular to the direction
of By in
an iron core. The resultant vector when magnetic memory in iron
core is neutralized is:

Bo=o
Bo=Bx+By

which also the orientation vector of an iron atom in the iron, where
magnetic field intensity magnitudes:

Bx=||Bx ||,
By=||By||,
Bo=||Bo||.

Angle
of resultant Bo from the
y-axis:

b= tan (Bx/By).

Output:

Boy=Bo cos
b.

Unit vector o can also represent
the magnetic field orientation of an iron atom in the iron. Increasing the angle
b would cause less magnetic field of By to travel through the iron by:

Byy=By cos b.

This
Byy  is the magnetic field intensity
magnitude output through the iron and out of the iron along the
y-axis

Imaginary examples: 1
degree< b< 60 degrees, Bx=0.02 tesla →.
Similarly, this
effect seems to be able to be done also by inserting a magnet
between the poles of a C shaped iron core whose pole width is longer than
the magnet's width W
.

Conclusion

References
1. The Motionless Electromagnetic
Generator:
http://www.cheniere.org/megstatus.htm .
2.
The Manual of Free Energy Devices and Systems; by Don
Kelly;
from Cadake Industries

Incorporated, Box:1866, Clayton, Georgia 30525.
2. Stationary
Armature Generator; By: Jown W. Ecklin;

from: Rex Research,
http://www.rexresearch.com.

November 21, 2003.
Updated on: 05-11-
2007.

By: Leonard Belfroy,

Canada.