1.version 15.12.96, 2. version 20.2.97, revised 3.3.97

**3. Computer simulation of the Brown generator**

** **Our model bases on one magnetic circuit, see fig.4 + 5.
The source of the magnetism is a permanent magnet of constant magnetic
voltage
. The magnetic flux due to his voltage is limited by the network of magnetic
resistances in the circuit consisting of the inner magnetic resistance
of the permanent magnet ,
of the magnetic resistance of the stray fields ,
and of the resistance
of the airgap plus the rotor which consists of a rotating piece of iron
in the air gap controling the magnetic flux .
We describe it as variable resistance
or dependent
from position of the angle
during rotation. A coil in the magnetic circuit connected with a impedance
of constant resistance R gives off the magnetic field energy excited by
the rotating iron in the form of a time dependent electric current I(t).
By this current I the coil with n windings has a magnetic voltage =-n.I
. (The sign of
has to be minus because - as we will see later- the differential equation
of the generator reduces to an usual LR-circuit under stationary conditions
with =0 only
if this is the case.) In order to derive the differential equation of this
generator we have to find the flux
at the rotor and through the coil. Therefore we analyse the network at
the black dot in fig.5. At this node Kirchhoff's law holds

For the three magnetic resistors involved Ohms law applies

From the last two equations we obtain the magnetic voltage

whereby we used the abbreviations
and . As it will
be shown in the appendix
can be identified with the flux through the frontal surface of the permanent
magnet and b is a constant whose value depends from geometry and distance
l of the airgap, if the magnetic circuit is closed by the rotor. Therefore,
using (2c) and (3)
becomes

Using , U=R.I, =-n.I
and the definition
the differential equation for the generator is derived to

If we add a capacitor C parallel to the resistor R (see dotted capacitor
in fig.5) we would have to insert I=U/R + C.dU/dt in =-n.I
and in (4) and we would obtain an differential equation of second order
for the generator

where we abbreviated .

Before we solve these equations we have to specify
and all other model parameters.

contains two
components which are added on in series:
is the magnetic resistance of the rotor which is neglected because
-the time dependent magnetic resistance of the airgap- is much bigger.
We split the last resistance again in in two parts: the constant resistance
of the remaining airgap if the rotor closes the magnetic cycle nearby at
in fig.4 and resistance
of rest of airgap left open additionally at other angels. If we use the
formula
for the magnetic resistance of a volume of length l and area A which is
filled with matter of the ferromagnetic constant ,
then we can estimate all values numerically. The values used in our calculation
are compiled in tab.3.

Because is
small against
it is neglected and we get at frequency f

(In order to facilitate iteration in our calculation the edges at
of the |cos|-function were smoothed in the intervals
by a biquadratic spline .
The coefficient c and d can be found assuming the function and its first
derivative to be continuous.)

In order to estimate the efficiency we have to calculate the electrical
and the mechanical work done after half a cycle. The electrical work output
at half a cycle of frequency f=1/T can be calculated to

The mechanical work input is

whereby the torque for a ferromagnetic
stick has been calculated to

In the third line we used the approximation ,
because
for ferromagnetic materials, where :=magnetic
permeability. Deriving the last line we used
and =B.A, i.e. the definitions
for the magnetic resistance and magnetic flux.

Starting condition of our cycle was t=0, I=0 at 50 Hz normally. A standard
Runge Kutta library routine has been used which can be found in [7].

Results:

Our numeric results are periodic.The initial phase starting the generator
plays a role only for low load, see upper trace in fig.6. The lower trace
there shows current I versus time t for a rotation frequency of 50Hz and
R= 240 Ohm. The diagram is periodic for half a cycle. We see that the generator
produces a pair of current spikes of opposite polarity especially at angles
where the rotor closes the magnetic circuit.The first spike is higher than
the second. Fig. 7 show the electric and mechanic work
versus time t at every moment measured at the resistor R at 50Hz and 240
Ohm. After half a cycle (0.01 sec) ~0.5 J are consumed by the resistor.
Therefrom, we can calculate the mean electric output power of 50 W at 50
Hz delivered by the generator estimated by our calculation. The mechanic
work done at this time is 2.3D-05 J. Therefrom we can estimate an efficiency
2.17D4. Fig. 8 shows
versus .
It is the magnetic work diagram of the coil for half a cycle and is identical
to the electrical work delivered after half a cycle acc. to equ.8. It has
an area of ~0.5 J. Fig.9 shows
versus resistance R, fig. 10 shows
versus frequency f.

Discussion:

Our model shows that overunity efficiency is possible principally from
theory. We think that the tendency of our numerical results are correct
surely, because a first glance at equ.6 for
shows that the mechanical input work can be reduced arbitrarily choosing
as high as possible without influencing the electrical output
(equ.4) at all. This indicates the possibility of overunity efficiency
without any numeric calculation. Furthermore, our calculation reproduces
a curious behaviour described by Kromrey[9]: The generator protects itself
against being short circuited and produces less power at very low load,
see fig.9. We believe that the output values are calculated in right order
of magnitude. A calculation with 4 x 4 cm^2 cross section iron core resulted
in 70 W output. If we realize that Brown used electromagnets for excitation
whose iron cores are probably in the magnetic saturated stated to reduce
back emf then it is very probable that these magnets have more than 1 T
field strenght at their surface. And because output power goes with ~ I^2
or~^2 then it seem
possible that his (probably RMS-) values can be correct. However, Brown's
experimental input values can not be calculated correctly. This could have
following causes:

1) friction:

Any friction is neglected in our model. For a real machine a input
power loss P due to friction has to be added to the input. Its dependence
is quadratic proportional to the rotation frequency, i.e. .
The proportional constant has to be measured or estimated according experience.

2) saturation of iron:

Magnetic material shows a saturation. This can be accounted for in
the equations by making
dependent from
in our equations according to the material chosen.

3) hysteretic losses and finite transmission time of magnetism
through iron:

This problem is inherent and can be reduced only by appropriated choice
of magnetic material.

4) eddy currents:

Although Brown took care to avoid this problem using laminated transformer
cores his setup had metallic ground plates which were electrically excited
by the rotation of the machine as he proved himself. Therefore, the use
of non- metallic material in the neighborhood of the machine is recommended.
If possible the permanent magnets should laminated as well - or should
be made out of non conducting magnetic ceramic.

5) finite time of transmission of the magnetic flux

Browns generator used laminated transformer cores, which are probably
to slow for the faster harmonics of the spikes. The problem can be overcame
using appropriate iron core materials.

6) bad description of the magnetic resistance
versus

We made the approximative assumption that the field across the surface
of the magnet is constant. The assumption for the dependence of the magnetic
resistance from angle is simple
but arbitrary.

The sense of our model was to show qualitativ dependences with present
data. With complete data the calculations can be improved by exact field
calculation of the magnetic resistance at every angle -or-
by using empirically measured values of the magnetic resistance for every .
The measurement can be done by determing L at each angle
and calculating the magnetic resistance acc. to the equation
which can be found by comparing coefficients of the generator equation
under stationary conditions and =0
with the ordinary L/R-circuit.

**Appendix:** Derivation of
and b from magnet data

1.) According to [11] the the magnetic field along the axis of an permanent
magnet obeys the formula

2.) If two identic cylindric magnets face one another with poles opposed, Br must be doubled at the point P.

3.) Using the same geometry like 2.) but connecting the magnets with soft iron on the back L has to be replaced by 2L and Br has to be doubled.

If we short the permanent magnet(see fig.5) at the surface (by soft iron) then we can calculate the flux at the surface to acc. to equ.(2c). Using the last formula we can estimate the flux at the surface for a cylindical neodymium magnet (R=1cm, L=2cm, Br=1,33T, l=2mm, geometry see point 2.) above ) from (A.1) and obtain =3.73D-4 Tm2. Then, the constant b can be determined using equ.(3) under stationary conditions (=0).

we get (2mm)=5D6/(Ohm.s) and b(2mm)=4.96D-8 Vs/A .

**Bibliography:**

Fig.4: Our model of the Brown-Ecklin generator

Fig.5: The Brown-Ecklin generator represented as closed magnetic circuit of magnetic flux current

Fig.6: Current I versus time of the Brown generator at 50 Hz and

lower trace 2 Ohm. upper trace 240 Ohm

Fig7: Work W vs. time t of the Brown generator at 50Hz and 240 Ohm load

Fig.8: The magnetic work diagram versus of half a cycle of 50 Hz at 240 Ohm load

Fig.9: Power P versus load R of the Brown generator at 50 Hz

Fig.10: Power P versus frequency f of the Brown generator at 240 Ohm load