Language:
To browser these website, it's necessary to store cookies on your computer.
The cookies contain no personal information, they are required for program control.
the storage of cookies while browsing this website, on Login and Register.

### GDPR and DSGVO law

Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding.
Amazon Warehouse Deals ! Now even more Deep Discounts ! Check out these great prices on slightly used or just opened once only items.I always buy my gadgets via these great Warehouse deals ! Highly recommended ! Many thanks for supporting OverUnity.com this way.

Many thanks.

# New Book

Products

WaterMotor kit

### Statistics

• Total Posts: 894447
• Total Topics: 15730
• Online Today: 44
• Most Online: 103
(December 19, 2006, 11:27:19 PM)
• Users: 4
• Guests: 10
• Total: 14

### Author Topic: Major Mathematical Solution of Newman's Theory!  (Read 6869 times)

#### kmarinas86

• Full Member
• Posts: 156
##### Major Mathematical Solution of Newman's Theory!
« on: September 10, 2008, 04:09:04 AM »
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.

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

change of power/change of time = applied voltage * reactive power/magnetic flux

Thus:

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

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

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

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

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

power/[reactive power] = -1*impedance/current

power/[reactive power] = -1*applied voltage/current^2

Now what exactly is this power?

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:

and equivalently:

Thus:

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

Now we know that:

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

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]

acceleration/velocity=(change of current/change of time)/current

And since:

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
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?

First let's setup the equation for mass:

i*(inductance/velocity) * 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 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

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)

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.

#### Free Energy | searching for free energy and discussing free energy

##### Major Mathematical Solution of Newman's Theory!
« on: September 10, 2008, 04:09:04 AM »

#### kmarinas86

• Full Member
• Posts: 156
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #1 on: September 10, 2008, 10:01:32 AM »
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

This is the result of Kirchhoff's voltage laws. The battery is a capacitor:
1) It does not let current through
2) It stores charge at a voltage
3) Voltages can pass through and so can the change of current.

The voltage drop across the battery also consists of a voltage drop due to resistance, which limits the maximum current that can be drained from them. The voltage drop due to the capacitance in a battery is as simple as the battery voltage (applied voltage) itself.

Quote
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.

This is an often overlooked result. As you will you see, it changes everything about our notion of what "power" is.

Quote
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. It results in acceleration.

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

change of power/change of time = applied voltage * reactive power/magnetic flux

Thus:

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

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

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

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

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

power/[reactive power] = -1*impedance/current

power/[reactive power] = -1*applied voltage/current^2

Now what exactly is this power?

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:

and equivalently:

Thus:

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

Now we know that:

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

reactive power=-1*[torque*angular velocity*current^2]/[applied voltage]
reactive power=-1*[power*current^2]/[applied voltage]
reactive power*[applied voltage/current^2]=-1*power
reactive power*impedance=-1*power*current
reactive power*impedance/linear charge density=-1*power*velocity

See next.

Quote
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]

acceleration/velocity=(change of current/change of time)/current

And since:

inductance * acceleration=-1*[force*velocity^2]/[applied voltage]
inductance * acceleration * applied voltage=-1*[force*velocity^2]

inductance * acceleration * applied voltage=-1*[force*(current/linear current density)^2]
inductance * linear current density * acceleration * applied voltage=-1*[force*(current)^2/linear current density]
inductance * (change of current/change of time) * applied voltage=-1*[force*(current)^2/linear current density]

It was given earlier that:
reactive power = magnetic flux * (change of current/change of time)

So:
inductance * reactive power/magnetic flux * applied voltage=-1*[force*(current)^2/linear current density]
reactive power/current * applied voltage=-1*[force*(current)^2/linear current density]
reactive power * applied voltage=-1*[force*(current)^3/linear current density]
reactive power * applied voltage=-1*[force*(current)^3*/[current/velocity]]
reactive power * applied voltage=-1*[force*velocity*(current)^2]
reactive power * applied voltage=-1*[power*(current)^2]
[applied voltage/(current)^2]=-1*[power/reactive power]
[impedance/current]=-1*[power/reactive power]
reactive power*[impedance/current]=-1*power

From last:
reactive power*impedance/linear charge density=-1*power*velocity
reactive power*[impedance/current]=-1*power

Same result.

Quote
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
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?

First let's setup the equation for mass:

i*(inductance/velocity) * 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 abritaryarbitrary 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

Since:Because:

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

This implies that such masses dissapatedissipate 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

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)

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 synonomoussynonymous with angle.

From earlier, we found that:
force/[i*(inductance/velocity) * applied voltage]=acceleration

So:
power/[i*(inductance) * applied voltage]=acceleration
power/[i*(inductance) * applied voltage]=(change in power/change in time)/force
power/(change in power/change in time)=[i*(inductance) * applied voltage]/force
velocity/acceleration=[i*(inductance) * applied voltage]/force
velocity=[i*(inductance) * applied voltage]/mass
mass*velocity=[i*(inductance) * applied voltage]
momentum=[i*(inductance) * applied voltage]

The momentum exchange is proportional to the inductance of the material and the voltage produced

Quote
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
5) acceleration

And decreases with:
3) velocity
4) acceleration

Since:
Because:

#### hartiberlin

• Hero Member
• Posts: 7975
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #2 on: September 10, 2008, 10:37:51 AM »
Hi,
surely the rotating magnet inside the aircoil changes its
inductance L all the time, so we have a factor of
dL/dt also, which is always forgotten, when a Newman
machine is calculated.

But to my experience, the main electrical OU effect in the machine
is due to oxidation of the graphite brushes and thus the release of free
electrons into the circuit.

Regards, Stefan.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #2 on: September 10, 2008, 10:37:51 AM »

#### gyulasun

• Hero Member
• Posts: 4068
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #3 on: September 10, 2008, 11:31:39 AM »

Hi,
surely the rotating magnet inside the aircoil changes its
inductance L all the time, so we have a factor of
dL/dt also, which is always forgotten, when a Newman
machine is calculated.
....

Hi Stefan,

Respectfully I disagree: when you insert a permanent magnet into an aircore coil and move or rotate it inside the coil, the coil's L inductance does not change but a very little, practically it is negligible.  The reason for this is that the permeability of any permanent magnet is close to 1 (usually 1.2 - 1.3 (or less) for ceramic and 1.03 - 1.1 for rare earth types).  This sounds suprising for sure. I mentioned this already on this forum too, see here:
http://www.overunity.com/index.php/topic,1988.msg25821.html#msg25821

regards,  Gyula

#### hartiberlin

• Hero Member
• Posts: 7975
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #4 on: September 10, 2008, 11:55:09 AM »
Hi,
yes, surely depends on the permanent magnet.
But surely you will also put there in some connection iron
between the magnets, so this might count then also .

Anyway, the dL/dt component is mostly forgotten,
when one calculates these types of motors..so watch out...

#### Free Energy | searching for free energy and discussing free energy

##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #4 on: September 10, 2008, 11:55:09 AM »

#### fritz

• Sr. Member
• Posts: 424
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #5 on: September 10, 2008, 11:04:35 PM »
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

• Full Member
• Posts: 156
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #6 on: September 10, 2008, 11:17:25 PM »
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:

and equivalently:

Thus:

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

Now we know that:

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

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]

acceleration/velocity=(change of current/change of time)/current

And since:

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.

Quote
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:

Quote
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

Quote
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.

Quote
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.

Quote
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.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #6 on: September 10, 2008, 11:17:25 PM »

#### kmarinas86

• Full Member
• Posts: 156
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #7 on: September 11, 2008, 07:54:16 PM »
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").

Quote
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.
« Last Edit: September 11, 2008, 08:55:19 PM by kmarinas86 »

#### kmarinas86

• Full Member
• Posts: 156
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #8 on: September 11, 2008, 08:58:02 PM »
Two possible sources of confusion in this thread are the following mathematical errors:

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

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

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

#### Free Energy | searching for free energy and discussing free energy

##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #8 on: September 11, 2008, 08:58:02 PM »

#### hartiberlin

• Hero Member
• Posts: 7975
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #9 on: September 11, 2008, 10:28:03 PM »
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.

#### BEP

• TPU-Elite
• Hero Member
• Posts: 1289
##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #10 on: September 12, 2008, 12:31:48 AM »
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)

All this work you are doing is very interesting but at this point I have my doubts about the above bold text.
In fact, if you are wrong then we are also looking at a positive time constant for capacitive reactance.

#### Free Energy | searching for free energy and discussing free energy

##### Re: Major Mathematical Solution of Newman's Theory!
« Reply #10 on: September 12, 2008, 12:31:48 AM »