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### Author Topic: Joseph Newman on the concept of "Power per second"  (Read 5667 times)

#### kmarinas86

• Full Member
• Posts: 156
##### Joseph Newman on the concept of "Power per second"
« on: August 31, 2008, 03:22:52 PM »
Quote from: JosephNewman.com
January 2006

***************

To the World:

I, Joseph Newman, have repeatedly stated that mass means NOTHING if it does not have a MECHANICAL UNDERSTANDING . The equation for horsepower of rotating motor shaft so PROVES - "ft. lbs X rpm divided by 5252 = HP". Yet that equation does not designate the TIME of said horsepower. Constant of mass - rpm implies horsepower per minute.

James Watt in the late 1700s who established the HP as a unit of measurement specifically stated that such a measurement was dependent upon TIME: "33,000 lbs 1 ft. per MINUTE."

Accordingly, the HP of a rotating shaft must be measured in HP per minute of rotating shaft or, in shorthand, HP of rotating shaft per SECOND. ["550 lbs 1 ft. per second."]

I show you now a simple MECHANICAL explanation of HP per second of TRUTH: Take a shaft and attach it to the center of 1 ft radius [2 ft. diameter] pulley. Now imagine that you rotate the shaft of one full rotation of 360 degrees in one SECOND. Then "you" have lifted 550 lbs 6.283 ft in one second! ["1 ft. radius gives 6.283 ft in circumference"] That is, 6.283 HP per second! Per rotation of shaft! Multiply by [60 rpm] 60 SECONDS that is 376.98 HP per MINUTE!

Said above is WRONG! Vague conventional HP equation gives only 6.283 HP per MINUTE!

The TRUTH of MECHANICAL explanation given by me has been applied to rotating shaft of 400 lb magnetic rotary of my revolutionary motor that will shock the world in the immediate future. IT WILL PROVE WHO IS THE MASTER OF MY REVOLUTIONARY ENERGY INVENTION!

Fact: Conventional equation must be changed to the following:

HP of rotating shaft of motor:

HP = ft lbs X RPM divided by 5252 X 60 = HP per MINUTE.

Happy New Year to the people of the Earth.

Love,

Joseph Westley Newman

Newman's explanation of this is not quite perfect. Actually HP is power, not energy. However, change in power per change in time is actually something we ought to consider.

Now consider that:

Force = Energy / Distance

Force = Power / Velocity

Thus, it naturally follows that:

Force = (change in power / change in time) / Acceleration

For a given force, energy is proportional to the distance driven by that force, power is proportional the velocity driven by that force, and change in power / change in time is proportional to the acceleration driven by that force.

The distance driven by a force is dependent on its velocity, which in turn is dependent on its acceleration. But not only that, as seen in the equations above, it is also dependent on how much the power changes with respect to time. It seems that our equations requires power to change in order for force to exist in the first place!

Newman says the fundamental aspect of the universe consists of particles that travel in endless spirals. Therefore, these particles would have a radius at each moment in time around which they would pivot and accelerate.

The condition we now have is:

Force*Acceleration = (change in power / change in time)

Since velocities rise in proportion to acceleration and distances in proportion to velocity, it is fair to say that they are all proportional to each other. This is justified by the fact that time itself is independent of the position, velocity, and acceleration, yet position, velocity, and acceleration are all dependent on time. Likewise, power should rise in proportion to (change in power / change in time), just as energy rises in proportion to power. These notions are supported by the (http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus)

Thus, by designing a system where the rate of power fluctuation can be increased without a proportional increase in rate usage of electrostatic potential energy, one can increase the power and energy output of a system whose source of energy is otherwise based only on that electrostatic potential energy. In this sense, relying on POWER per SECOND (i.e. ENERGY per SECOND SQUARED) would purportedly be why this machine is a over-unity device. But let's investigate more before jumping to conclusions.

Newman says the fundamental aspect of the universe consists of particles that travel in endless spirals. Therefore, these particles would have a radius at each moment in time around which they would pivot. Each of these particles would have an angular momentum. The angular momentum of a particle is equal to:

Newman also emphasizes Einstein's equation of mass-energy equivalence, or E=MC^2. And thus:

angular momentum = massenergy*velocity*radius*sin(theta) / speed of light^2

By isolating massenergy, we have:

massenergy=angular momentum * speed of light^2 / [velocity*radius*sin(theta)]

Newman says matter in motion consists of innumerable particles are simultaneously mass and energy. Each particle, having its own gyroscopic action, is conserved, or otherwise, its constituent particles remain conserved, in terms of mass and energy. If so, the ratio of mass energy to angular momentum for such particles should be a constant, and we would have the following:

massenergy/angular momentum = speed of light^2 / [velocity*radius*sin(theta)]

massenergy/angular momentum = (speed of light^2/velocity^2) * velocity^2 / [velocity*radius*sin(theta)]

massenergy/angular momentum = (electric force / magnetic force) * angular velocity / [sin(theta)]

Where:

"electric force / magnetic force" (which is proportional to the charge velocity) is inversely proportional to angular velocity / [sin(theta)].

When charge velocity is a constant, angular velocity is proportional to with sin(theta).

When sin(theta) approaches zero, phase velocity approaches zero.

But what does this mean for the Newman Motor? It means slower velocity charge and bigger magnetic field coupling between permanent and temporary magnets with more force and larger radius.
« Last Edit: August 31, 2008, 03:43:07 PM by kmarinas86 »

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

##### Joseph Newman on the concept of "Power per second"
« on: August 31, 2008, 03:22:52 PM »

#### TinselKoala

• Hero Member
• Posts: 13968
##### Re: Joseph Newman on the concept of "Power per second"
« Reply #1 on: September 02, 2008, 05:19:09 PM »
The quote from Newman illustrates nothing more than his misconceptions and his contempt for real mathematical analysis. His motor is nothing more than a big flywheel, and is quite inefficient.

He doesn't think he's talking about d(HP)/dt.
He thinks he's talking about the power of his motor, not the way it changes over time.

The only way that "horsepower per time" makes any sense is as a change in power over a time period.
All other things held constant, a change in HP over time would indicate an acceleration (or deceleration). Acceleration requires force applied over time. This represents energy.
So, energy can be "extracted" by decelerating the Newman wheel. To accelerate the wheel, energy must be added over time.
To maintain the wheel at a constant rpm, the input energy over time (the power) must replace the losses (friction, etc.). If you increase the load, the wheel will slow down, unless you input more power.

It's really as simple as that. The Newman machine in its various incarnations is an inefficient electric motor, nothing more or less.

#### kmarinas86

• Full Member
• Posts: 156
##### Re: Joseph Newman on the concept of "Power per second"
« Reply #2 on: September 03, 2008, 03:50:05 PM »
The quote from Newman illustrates nothing more than his misconceptions and his contempt for real mathematical analysis. His motor is nothing more than a big flywheel, and is quite inefficient.

He doesn't think he's talking about d(HP)/dt.
He thinks he's talking about the power of his motor, not the way it changes over time.

The only way that "horsepower per time" makes any sense is as a change in power over a time period.
All other things held constant, a change in HP over time would indicate an acceleration (or deceleration). Acceleration requires force applied over time. This represents energy.
So, energy can be "extracted" by decelerating the Newman wheel. To accelerate the wheel, energy must be added over time.
To maintain the wheel at a constant rpm, the input energy over time (the power) must replace the losses (friction, etc.). If you increase the load, the wheel will slow down, unless you input more power.

It's really as simple as that. The Newman machine in its various incarnations is an inefficient electric motor, nothing more or less.

The point about my previous post above was that the magnetic field takes a constant power to drive a constant magnetic field energy depending on the parameters of the coil. It takes a change of electrical power to give a power to the magnetic field, in that its energy can actually change.

Via the calculus method of integration, we find that:

Power=initial power+integration of [(âˆ†Power/âˆ†time)] with respect to time
Energy=initial energy+integration of Power with respect to time

The part of voltage we are interested is the part that can actually produce the magnetic field. This is equal to Inductance*(âˆ†Current/sec). The part that cannot change the energy of a magnetic field without itself changing is equal to Current*Resistance. As a result, we have:

Power=initial power+Voltage on Inductance*Current
Power=initial power+Inductance*[âˆ†Current/âˆ†time]*Current
Energy=initial energy+integration of Power with respect to time

There is an exponential formula that relates the rate of current rise to the voltage, inductance, and resistance.

rate of current rise=[âˆ†Current/âˆ†time]=(Voltage/Inductance)*e^(-time*(Resistance/Inductance))

Where "(-time*(Resistance/Inductance))" is the exponent of e.

Thus we have:

Power=initial power+Inductance*(Voltage/Inductance)*e^(-time*(Resistance/Inductance))*Current
Power=initial power+Voltage*Current*e^(-time*(Resistance/Inductance))
Energy=initial energy+integration of Power with respect to time

If you double time, e^(-2*time*(Resistance/Inductance)) drops to 0.135335283 times its original value. In other words, this power related to the magnetic field decreases with time.

Keep in mind, this refers to the power delivered to the magnetic field (i.e. the imaginary component power) as opposed to the power delivered to the resistance as net dissipation (i.e. real component of complex power). The point is that the longer you keep the circuit on per pulse, the less power you will be able to get from changes in magnetic field energy
« Last Edit: September 03, 2008, 04:54:08 PM by kmarinas86 »

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

##### Re: Joseph Newman on the concept of "Power per second"
« Reply #2 on: September 03, 2008, 03:50:05 PM »

#### kmarinas86

• Full Member
• Posts: 156
##### Re: Joseph Newman on the concept of "Power per second"
« Reply #3 on: September 03, 2008, 05:31:07 PM »
The point about my previous post above was that the magnetic field takes a constant power to drive a constant magnetic field energy depending on the parameters of the coil. It takes a change of electrical power to give a power to the magnetic field, in that its energy can actually change.

Via the calculus method of integration, we find that:

Power=initial power+integration of [(âˆ†Power/âˆ†time)] with respect to time
Energy=initial energy+integration of Power with respect to time

The part of voltage we are interested is the part that can actually produce the magnetic field. This is equal to Inductance*(âˆ†Current/sec). The part that cannot change the energy of a magnetic field without itself changing is equal to Current*Resistance. As a result, we have:

Power=initial power+Voltage on Inductance*Current
Power=initial power+Inductance*[âˆ†Current/âˆ†time]*Current
Energy=initial energy+integration of Power with respect to time

There is an exponential formula that relates the rate of current rise to the voltage, inductance, and resistance.

rate of current rise=[âˆ†Current/âˆ†time]=(Voltage/Inductance)*e^(-time*(Resistance/Inductance))

Where "(-time*(Resistance/Inductance))" is the exponent of e.

Thus we have:

Power=initial power+Inductance*(Voltage/Inductance)*e^(-time*(Resistance/Inductance))*Current
Power=initial power+Voltage*Current*e^(-time*(Resistance/Inductance))
Energy=initial energy+integration of Power with respect to time

If you double time, e^(-time*(Resistance/Inductance)) drops to 0.135335283 times its original value. In other words, this power related to the magnetic field decreases with time.

Keep in mind, this refers to the power delivered to the magnetic field (i.e. the imaginary component power) as opposed to the power delivered to the resistance as net dissipation (i.e. real component of complex power). The point is that the longer you keep the circuit on per pulse, the less power you will be able to get from changes in magnetic field energy

e^x=.5 when x=ln(.5)=-0.693147181, and this is where the power delivered to the magnetic field will match the power delivered to the resistance of the coil. However, for the energy delivered to the magnetic field to match the energy delivered to the resistance, the integration of e^x is equal to 0.5. Thus e^x+xe^x, the integration of e^x, must equal 1/2. So when the energy delivered to the magnetism matches the energy delivered to the resistance, x=-.314923=-time*(Resistance/Inductance), such that time=0.314923*Inductance/Resistance. e^x becomes 0.72984507 and current would be 1-e^x or 27.015493% of its maximum value. Any less than this will result in having the energy delivered to magnetic field energy to exceed the energy delivered to the resistance of the wire.  The only power utilized from the batteries in net is the heat dissapated in the resistor while all power delivered to the magnetic field will return to its origin (because the power delivered to the magnetic field is reactive power and reactive power is not consumed).

#### kmarinas86

• Full Member
• Posts: 156
##### Re: Joseph Newman on the concept of "Power per second"
« Reply #4 on: September 03, 2008, 08:42:49 PM »
The part of voltage we are interested is the part that can actually produce the magnetic field. This is equal to Inductance*(âˆ†Current/sec). The part that cannot change the energy of a magnetic field without itself changing is equal to Current*Resistance. As a result, we have:

Power=initial power+Voltage on Inductance*Current

Power=initial power+Inductance*[âˆ†Current/âˆ†time]*Current
Energy=initial energy+integration of Power with respect to time

There is an exponential formula that relates the rate of current rise to the voltage, inductance, and resistance.

rate of current rise=[âˆ†Current/âˆ†time]=(Voltage/Inductance)*e^(-time*(Resistance/Inductance))

Where "(-time*(Resistance/Inductance))" is the exponent of e.

Thus we have:

Power=initial power+Inductance*(Voltage/Inductance)*e^(-time*(Resistance/Inductance))*Current
Power=initial power+Voltage*Current*e^(-time*(Resistance/Inductance))

The following should be considered as well:
Current=(Voltage/Resistance)*(1-e^(-time*(Resistance/Inductance))

Thus we end up having:
Reactive Power=initial power+(Voltage^2/Resistance)*(e^(-time*(Resistance/Inductance)))*(1-e^(-time*(Resistance/Inductance)))
Reactive Power=initial power+(Voltage^2/Resistance)*(e^x)*(1-e^x)

While
Real Power=initial power+(Voltage^2/Resistance)*(1-e^x)*(1-e^x)

The power overall is:
Apparent Power=initial power+(Voltage^2/Resistance)*(1-e^x)
And is maximized when e^x=0, x<<0 (i.e. x is a very large negative number).

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

##### Re: Joseph Newman on the concept of "Power per second"
« Reply #4 on: September 03, 2008, 08:42:49 PM »

#### kmarinas86

• Full Member
• Posts: 156
##### Re: Joseph Newman on the concept of "Power per second"
« Reply #5 on: September 03, 2008, 10:31:34 PM »
[FOR] e^x=.5 when x=ln(.5)=-0.693147181, and this is where the power delivered to the magnetic field will match the power delivered to the resistance of the coil.

This is correct.

Quote
However, for the energy delivered to the magnetic field to match the energy delivered to the resistance, the integration of e^x is equal to 0.5.

This is false.

Quote
Thus e^x+xe^x, the integration of e^x, must equal 1/2. So when the energy delivered to the magnetism matches the energy delivered to the resistance, x=-.314923=-time*(Resistance/Inductance), such that time=0.314923*Inductance/Resistance. e^x becomes 0.72984507 and current would be 1-e^x or 27.015493% of its maximum value. Any less than this will result in having the energy delivered to magnetic field energy to exceed the energy delivered to the resistance of the wire.  The only power utilized from the batteries in net is the heat dissapated in the resistor while all power delivered to the magnetic field will return to its origin (because the power delivered to the magnetic field is reactive power and reactive power is not consumed).

So this is all wrong.

Now for the correction:

both
e^x*(1-e^x)
and
(1-e^x)^2
must be integrated from x to 0, for x<0

Only by doing so can it be arrived that when the energy delivered to the energy of the magnetic field=energy delivered as heating in the coil:

real power/apparent power = apparent power/final apparent power

Where
real power = 0.467584558 * final apparent power
apparent power = 0.683801548 * final apparent power
final apparent power = 100% of final apparent power, when current reaches a maximum for a given voltage
x=-1.151385251
time=1.151385251*Inductance/Resistance
time=1.151385251*(L/R time constant)
current=68.3% of final current

So when all the LHS of the equations above are less than their corresponding RHS would the energy delivered to the magnetic field exceed that which was delivered as ohmic heating.

#### Michelinho

• Sr. Member
• Posts: 452
##### Re: Joseph Newman on the concept of "Power per second"
« Reply #6 on: September 03, 2008, 10:39:21 PM »

Hi kmarinas86,

Quote
So when all the LHS of the equations above are less than their corresponding RHS would the energy delivered to the magnetic field exceed that which was delivered as ohmic heating.

Don't forget that the Newman motor coil is colder than the ambient temperature while doing work. So you have to use a negative ohmic heating factor.

Take care,

Michel

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

##### Re: Joseph Newman on the concept of "Power per second"
« Reply #6 on: September 03, 2008, 10:39:21 PM »