Major Mathematical Solution of Newman's Theory!

[fritz]

If you would have two moving electro magnets -
the mutual inductivity related to the magnetic coupling of
both magnets would change(oscillate) - which would result
in an oscillating inductivity component for each coil.
The same would apply if you switch to magnets.

[kmarinas86]

As stated earlier:

force=energy/distance
force=power/velocity
force=(change in power/change in time)/acceleration

So power=force*velocity, but what else does equal?

First we must understand that:

angle=distance/radius

and equivalently:

distance=radius*angle

Thus:

force=(energy/radius)/angle
force=(power/radius)/angular velocity
force=((change of power/change in time)/radius)/angular acceleration

Now we know that:

power=force*radius*angular velocity
[force*radius*angular velocity]/[reactive power] = -1*applied voltage/current^2
reactive power=[force*radius*angular velocity]/[-1*applied voltage/current^2]
reactive power=-1*[force*radius*angular velocity*current^2]/[applied voltage]

We remember that:
reactive power=voltage drop due to inductance * current

So:
voltage drop due to inductance * current=-1*[force*radius*angular velocity*current^2]/[applied voltage]
voltage drop due to inductance=-1*[force*radius*angular velocity*current]/[applied voltage]

We recall that:
voltage drop due to inductance = inductance * (change of current/change of time)

So:
inductance * (change of current/change of time)=-1*[force*radius*angular velocity*current]/[applied voltage]
inductance * (change of current/change of time)/current=-1*[force*radius*angular velocity]/[applied voltage]

We already know that:
acceleration/velocity=(change of current/change of time)/current

inductance * acceleration/velocity=-1*[force*radius*angular velocity]/[applied voltage]

And since:
angular velocity=velocity/radius

inductance * acceleration=-1*[force*velocity^2]/[applied voltage]
inductance * acceleration * applied voltage=-1*[force*velocity^2]
inductance * (acceleration/force) * applied voltage=-1*[velocity^2]
(inductance/mass) * applied voltage=-1*[velocity^2]
-1*(inductance/mass) * applied voltage=[velocity^2]
sqrt(-1)*(inductance/mass) * applied voltage=velocity

The last line here is an error.

Preferably, we shouldn't have to take the square root.

i*(inductance/mass) * applied voltage=velocity

I have a question for you.

Do you think the velocity in this equation is imaginary?

If so, what does it represent?

Can you imagine an imaginary acceleration?

If v^2/r was the kinematic (non-dynamic) definition of acceleration, would the acceleration be negative? Against what mass "m" would this acceleration act upon?

Skip that portion.

Next we have:

First let's setup the equation for mass:

i*(inductance/velocity) * applied voltage=mass

In this equation, we should replace (i/velocity), with (-1/velocity^2) to undo the error made. This will have little effect on the results however.

First let's setup the equation for mass:
-(inductance/velocity^2) * applied voltage=mass

The mass in question must be tied to the particles' (charges) forces and acceleration. Thus it is fair to assume that the masses in question will gain kinetic energy relative to objects of abritary mass upon which it acts. In doing so, it applies forces against these objects causing them to accelerate according to their mass. So what is acceleration?

[i*(inductance/velocity) * applied voltage]/force=mass/force
force/[i*(inductance/velocity) * applied voltage]=acceleration

The fix:
-(inductance/velocity^2) * applied voltage]/force=mass/force
-force/[(inductance/velocity^2) * applied voltage]=acceleration

The mere existence of:
1) force
2) inductance
3) velocity
4) applied voltage
5) acceleration

Make no reference to mass.

What is implied however is that, for a given force, mass increases with:
2) inductance
4) applied voltage

And decreases with:
3) velocity
4) acceleration

Since:

force = power / velocity
force = (change in power/change in time) / acceleration

This implies that such masses dissapate over time at they accelerate (which implies motion).

Which variables above can be considered imaginary?

Note that for i=sqrt(-1), when ai=bi, a=b. Also, i^2=1/i^2, and is verified by the fact that i^4=1. It now seems appropriate that we also define the equation in terms of i^2 and 1/i^2

The section here is less relevant now as we find that the square root was improperly performed.

force=acceleration*[i*(inductance/velocity) * applied voltage]

force/[acceleration*(inductance/velocity) * applied voltage]=i
[force/[acceleration*(inductance/velocity) * applied voltage]]^2=i^2

i*(inductance/mass) * applied voltage=velocity

i=velocity/[(inductance/mass) * applied voltage]
i^2=velocity/[(inductance/mass) * applied voltage]
i^2=[(inductance/mass) * applied voltage]/velocity

[i^2]*[1/i^2]=
[force/[acceleration*(inductance/velocity) * applied voltage]]*[(inductance/mass) * applied voltage]/velocity=
[force/[acceleration*mass]]=1

This means out of the variables:
1) force
2) acceleration
3) mass

None of them are imaginary, for:
i=sqrt(-1)
i^3=-1*sqrt(-1)

This section is has no errors.

Euler's formula:

e^(ix) = cos x + i sin x
ix = ln(cos x + i sin x)

The torque of the magnetic field depends on the cross product of the magnetic moment and the magnetic field. Torque varies with the sine of the angle between them.
The energy of the magnetic field depends on the cross product of the magnetic moment and the magnetic field. Energy varies with the cosine of the angle between them. Thus we can state that:

The derivative of energy is proportional to the derivative of cos x, which is -sin x * dx
If torque is the derivative of energy divided by the derivative of angle x, then torque = -sinx

[magnetic moment * magnetic field] TIMES Euler's formula

magnetic moment * magnetic field * e^(ix) = magnetic moment * magnetic field * (cos x + i * sin x)

magnetic moment * magnetic field * e^(ix) = (Energy - i * Torque)

If units of torque * -i = units of Energy, then angle x is in units of -i, and e^(ix) becomes real.

units of -i becomes synonomous with angle.

[kmarinas86]

BY KMARINAS86

September 9, 2008, 9:18 PM

voltage*charge is a given for a set of batteries with some voltage and charge capacity.

[(dm/dt)*q] = [voltage * charge] = Electrostatic Potential Energy

[(dq/dt)*m] = [current * magnetic flux] = Torque of Magnetic Field

While voltage*charge is normally referred to as the energy in the battery, we are instead interested in the energy of the magnetic field as derived from interactions which happen during circuit operation.

We'll see if the magnetic field's energy comes from battery chemistry alone.

force = energy / distance
force = power / velocity
force = (change in power/change in time) / acceleration

linear charge density = charge / distance
linear charge density = current / velocity
linear charge density = (change of current/change in time) / acceleration

applied voltage = voltage drop due to capacitance + voltage drop due to resistance + voltage drop due to inductance
applied voltage = charge/capacitance + current*resistance + (change in current/change in time)*inductance
applied voltage = linear charge density * (distance/capacitance + velocity*resistance + acceleration*inductance)

linear charge density = charge / distance = applied voltage * capacitance/distance

applied voltage = applied voltage * (1 + velocity*resistance*(capacitance/distance) + acceleration*inductance*(capacitance/distance))

velocity*resistance*(capacitance/distance) + acceleration*inductance*(capacitance/distance) = 0

velocity*resistance*(capacitance/distance) = -1*acceleration*inductance*(capacitance/distance)

velocity*resistance = -1*acceleration*inductance

-1*velocity/acceleration = inductance/resistance

-1*power/(change in power/change in time) = inductance/resistance

Therefore, as (change in power/change in time) decreases, so does power.

This kind of power is a type of power that cannot exist without changing. It is called reactive power.

In other words, the equation assumed that "power"="reactive power" (specifically "inductive reactance").

A short circuit will produce a low resistance/inductance. This will mean that (change in power/change in time) is high relative to power.

A Newman motor has a very high inductance/resistance (i.e. L/R time constant). That means for a given change of power, the power is already larger. When power changes from the off state, a given change in power in a given time would mean a larger power output. This means that there is power already built into the device before the system even turns on.

For a given voltage, the change of power per change of time is:
change of power/change of time = voltage * (change of current/change of time)

The reactive power is voltage drop due to inductance, which is equal to:
reactive power = voltage drop due to inductance * current
reactive power = inductance * (change of current/change of time) * current
reactive power = magnetic flux * (change of current/change of time)

Which applies even if inductance is not constant. The formula "inductance = change of magnetic flux / change of current" does not apply if inductance itself changes.

Inductance can be thought of as two ways:
1) The relation between a change in applied magnetic flux and the current it induces in some random material.
2) The relation between the current in a material, and the magnetic flux that it produces as a result.

Therefore:
(change of current/change of time) = reactive power/magnetic flux

This was given by the previous equation, and based on the idea that reactive power is the "power" in question.

In other words:

(change of current/change of time) = power/magnetic flux

Why?

In order for a force to do work, not only must a displacement occur, it must result in an acceleration, since a force times a relative velocity does no work, say, an object sitting on top of another object is seen from car traveling at 55 mph. Therefore the very basis for power is reactive power, since it can only exist when charges accelerate. Charges which accelerate lose kinetic energy (Larmor radiation), resulting in a loss of heat.

Since the inductive voltage drop here is the same as -1 * resistive voltage drop. Either this means a negative time constant for the inductor (which is not the case), or it means an acceleration and velocity of charges opposite polarity. This would mean the forces would have to push on the charges in the opposite direction they are moving.

That mathematical description can be explained in better detail by a mechanical description of what happens.

Applicant [Joseph Newman] used a permanent magnet D.C. motor as a generator. He demonstrated that the resistance of the copper windings of the generator was only 3 ohms. Therefore, he demonstrated the so-called work load would be nil if the two leads from the generator were connected, and the generator shaft were then rotated by hand (pulling a cord wrapped around a 1.5 inch diameter pulley attached to shaft of generator). I was asked to then pull the cord lightly once and then briskly. I immediately experienced noticeable resistive force the harder I pulled the cord, although there was no conventional work load hooked in the system. Applicant mechanically explained these results by his teachings of gyroscopic particles; that when the atoms of the rotating coils of the generator hit the gyroscopic particles (at some degree of a right angle) which were being emitted from the atoms of the permanent magnets in the generator, that the gyroscopic particles then went down the length of copper wire coils (but that their gyroscopic spin would then be at some degree of a right angle to the balance of the spin of the gyroscopic particles still moving in the magnetic field from the permanent magnets), therefore when the leads were hooked together this then allowed the gyroscopic particles to then try to re-enter the influences of said gyroscopic particles of said permanent magnets, but that their spin would be at some degree of a right angle to one another, therefore they try to push away from each other, resulting in the coils of [the] generator then having resistance to rotation. And that this effect was multiplied the faster you turned the coils, because then the more gyro-particles you would cause to be released from said permanent magnetic field, resulting in an ACCUMULATIVE EFFECT of gyro-particles in the closed system (coils), then trying to re-enter the influence of gyro-particles moving in said magnetic field of said permanent magnets and therefore would always more vigorously resist your acceleration of the coil and its shaft of the generator, although there was no conventional work load placed in the system.

[kmarinas86]

Two possible sources of confusion in this thread are the following mathematical errors:

power*magnetic flux/[reactive power] = -1*inductance*[applied voltage/resistance]

power*magnetic flux/[reactive power] = -1*inductance*impedance

-1*(inductance/mass) * applied voltage=[velocity^2]
sqrt(-1)*(inductance/mass) * applied voltage=velocity

[hartiberlin]

So how can you optimize the mechanical power output due to your calculations
with the lowest electrical input power ?

(sorry am very busy right now, so could not study it deeper)

Many thanks.