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# New Book

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### Author Topic: Free Energy from Electromagnetic Wave Fields  (Read 7898 times)

#### ZL

• Jr. Member
• Posts: 67
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #60 on: February 27, 2017, 11:03:55 AM »

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

##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #60 on: February 27, 2017, 11:03:55 AM »

#### shylo

• Sr. Member
• Posts: 448
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #61 on: February 27, 2017, 06:19:57 PM »
ZL your quote: "A hint that will give away the working principle: gravity has nothing to do with it. It would work the same (or nearly the same) in zero gravity in space as well."

Gravity is what makes the machine work. He lifts the arm to create the wave. Or he pushes the balanced arm down, which lifts the opposite end. Gravity then takes over when he releases it.

Also his math is incomplete, you need to calculate the lengths of the arms in reference to the center axis.
Interesting video ,but the more waves, the more unpredictable the outcome.
artv

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

##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #61 on: February 27, 2017, 06:19:57 PM »

#### ZL

• Jr. Member
• Posts: 67
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #62 on: February 27, 2017, 08:45:31 PM »
Gravity is what makes the machine work. He lifts the arm to create the wave. Or he pushes the balanced arm down, which lifts the opposite end. Gravity then takes over when he releases it.

OK, if you think you know this better, then please educate us, and explain in detail how the machine works, and why it works that way. Does it support then gravity waves? How would it behave in zero gravity on a space station, and why? Can you write down the equations that govern the wave propagation in the device? We need specifics, detailed explanation, and real understanding. Without that we can not proceed with wave energetics.

Quote
Also his math is incomplete, you need to calculate the lengths of the arms in reference to the center axis.

Exactly what is wrong with his math? Again, please be specific. Show us where the error is, and how you would calculate the same in a correct manner.

#### shylo

• Sr. Member
• Posts: 448
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #63 on: February 28, 2017, 02:19:34 AM »
It's a bunch of balanced weights.
I disrupt that balance by, either lifting or pushing down, then release, the work I put in is translated.
To me the real observation is, what he didn't do was ,interrupt the wave.
Or let the wave interact.Not the same as when he added the second machine. It just had shorter tines.
artv

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

##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #63 on: February 28, 2017, 02:19:34 AM »

#### ZL

• Jr. Member
• Posts: 67
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #64 on: February 28, 2017, 11:41:01 AM »
It's a bunch of balanced weights. I disrupt that balance by, either lifting or pushing down, then release, the work I put in is translated.

This sounds like Bessler Wheel logic. But just saying this will not bring it home to you. Therefore let us pick apart your explanation, and examine the validity of each claim and idea.

If the wave machine is only a bunch of balanced weights for you, then why don’t you start checking your theory by examining one single balanced weight? Imagine a modified special seesaw, which is not close to the ground, but it is at a height which allows the arms to make complete rotations around the axis. Now, you do the same as Prof. did in the video. You grab one end of the balanced seesaw, move it down a little and immediately move it up again into its starting position. Before you release the seesaw, make sure that the arm is stationary (v=0) just like it was before the start of our experiment. Now will this seesaw perform any oscillatory movement like it was observed in the wave machine? If gravity is the restoring force of any potential oscillation (like in the case of a simple pendulum), then your seesaw must oscillate.

An alternative experiment is to just rise one end of the seesaw, make sure that it is stationary, and release it. Will it start to oscillate like a pendulum?

I don’t respond to the second part of your post, because it makes no sense at all, and first we have to get you understand what is wrong with your explanation.

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

##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #64 on: February 28, 2017, 11:41:01 AM »

#### telecom

• Sr. Member
• Posts: 370
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #65 on: February 28, 2017, 05:05:56 PM »
This sounds like Bessler Wheel logic. But just saying this will not bring it home to you. Therefore let us pick apart your explanation, and examine the validity of each claim and idea.

If the wave machine is only a bunch of balanced weights for you, then why don’t you start checking your theory by examining one single balanced weight? Imagine a modified special seesaw, which is not close to the ground, but it is at a height which allows the arms to make complete rotations around the axis. Now, you do the same as Prof. did in the video. You grab one end of the balanced seesaw, move it down a little and immediately move it up again into its starting position. Before you release the seesaw, make sure that the arm is stationary (v=0) just like it was before the start of our experiment. Now will this seesaw perform any oscillatory movement like it was observed in the wave machine? If gravity is the restoring force of any potential oscillation (like in the case of a simple pendulum), then your seesaw must oscillate.

An alternative experiment is to just rise one end of the seesaw, make sure that it is stationary, and release it. Will it start to oscillate like a pendulum?

I don’t respond to the second part of your post, because it makes no sense at all, and first we have to get you understand what is wrong with your explanation.
I think you are saying that the total potential + kinetic energy of the system is unchanged, like in a pendulum?

BTW, I wasn't able to watch the video - can't open that file.

#### ZL

• Jr. Member
• Posts: 67
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #66 on: February 28, 2017, 06:37:18 PM »
I think you are saying that the total potential + kinetic energy of the system is unchanged, like in a pendulum?

I was explaining that gravity can not be responsible for an oscillatory motion in a balanced seesaw; neither can it play any major role in the operating principle of the wave machine. This is not a pendulum. A simple pendulum is unbalanced when it is released at an off center point. The seesaw is balanced in all possible positions at all angles, therefore gravity can not develop an active unbalanced resultant force on it at any angle. Without such a force it can not accelerate or decelerate, thus it can not oscillate. Wherever you turn it, it will remain in that position as long as no other force acts upon it.

Quote
BTW, I wasn't able to watch the video - can't open that file.

Which video? This one?
AT&T Archives: Similiarities of Wave Behavior (Bonus Edition)

I suspect you are talking about the java applet, which is an interactive simulator, not a video:
Wave Machine Model
http://www.opensourcephysics.org/items/detail.cfm?ID=10481

Download the download 1694kb.jar double click on it, and you are ready to start playing with waves and observe their motion in real time, just like the Prof. did in the video.

If it does not work then you will have to install Java on your PC first:

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

##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #66 on: February 28, 2017, 06:37:18 PM »

#### telecom

• Sr. Member
• Posts: 370
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #67 on: March 01, 2017, 03:30:01 AM »
In the above video an original energy is supplied by  the torque
when stressing the supporting wire.
It can be equal to the force x arch of the bend.
This in turn creates a torsion stress in the above wire.
The rest of the system behaves as a multitude of the pendulums
oscillating around the point of the equilibrium.

#### telecom

• Sr. Member
• Posts: 370
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #68 on: March 01, 2017, 03:56:39 AM »
Trying to generalize,
it looks that the energy of the waves is greatly dependent on the properties of
the  medium. In this case, the elasticity of the wire material.
What is elasticity?
Its a property of the matter, in this case, the steel.
This matter consists of the atoms.
Perhaps by stressing it we can influence the atomic structure ?

#### ZL

• Jr. Member
• Posts: 67
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #69 on: March 01, 2017, 09:20:04 AM »
This in turn creates a torsion stress in the above wire. The rest of the system behaves as a multitude of the pendulums oscillating around the point of the equilibrium.

Nice one! Now you are getting close to really understanding how the wave machine works. As I have already started to point out in my previous post, the wave propagation is based on the oscillatory movement of mass particles in mechanics and acoustics (or on the oscillatory variation of EM fields in electromagnetics). A pendulum is also an oscillator; therefore it makes complete sense to imagine the wave machine to be a chain of coupled torsion pendulums.

There are several different pendulums:
https://en.wikipedia.org/wiki/Pendulum
But in this case we are interested in the torsion pendulum:
https://en.wikipedia.org/wiki/Torsion_spring
More specifically we are interested in torsional harmonic oscillators, which are also briefly discussed on the above page.

Quote
This matter consists of the atoms. Perhaps by stressing it we can influence the atomic structure?

Sure, by twisting a metal rod you are distorting the crystal structure of the material. The bonds between the atoms get stretched or compressed. But in order to understand the working principle of the wave machine and calculate the energy content of a wave pulse, we don’t need to go down to atomic level. All we need is to understand the macroscopic behavior of elasticity and its quantitative analysis.

The next step for you is to study a bit the torsion pendulum and its mathematical analysis. The equation of motion of the torsional oscillator is already given on the wiki page. But there is a wealth of information online about this subject at all possible levels of understanding. We need university level of understanding here, because we must be able to derive the formula for the energy content of a wave pulse.

Google has offered many hits for torsion pendulum. Here are some:
http://vlab.amrita.edu/index.php?sub=1&brch=280&sim=1518&cnt=1
http://vlab.amrita.edu/index.php?sub=1&brch=280&sim=1518&cnt=4

When you are done with digesting the material, then please try to summarize and explain the working mechanism of the wave machine as a series of coupled torsional oscillators. You may also perform a google search about coupled oscillators, or coupled pendulums to give you an idea how to attack the problem.

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

##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #69 on: March 01, 2017, 09:20:04 AM »

#### telecom

• Sr. Member
• Posts: 370
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #70 on: March 02, 2017, 04:47:13 AM »

When you are done with digesting the material, then please try to summarize and explain the working mechanism of the wave machine as a series of coupled torsional oscillators. You may also perform a google search about coupled oscillators, or coupled pendulums to give you an idea how to attack the problem.

The biggest amplitude is at the point of the most twisted wire, and a vise versa.
So the energy can be determined from the amount of work which takes to
perform the twisting.
It equals F x S, where F is the twisting force, and S is the distance of the
arch of the applied force.
The total for the wave would be the  integral of these over the length of the wave.
This is how much I could extend myself to figure it out.
Zoltan, I think we all be greatly benefitting from your input at this point.
Regards

#### ZL

• Jr. Member
• Posts: 67
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #71 on: March 02, 2017, 09:27:54 PM »
The biggest amplitude is at the point of the most twisted wire, and a vise versa. So the energy can be determined from the amount of work which takes to perform the twisting.

We want to know the energy content of a half sine wave on the wave machine, independently of its source. We don’t need to know who created the wave with what amount of invested work. Knowing this wouldn’t be of much use anyway if the wave we observe has suffered significant attenuation, especially if it has been also reflected from an impedance mismatch (like that of the open end of the transmission line).

Here is what we know: the properties of the transmission line (given by the manufacturer), the amplitude, wave length, and shape of the wave (sinusoid). That’s all. From this we should be able to calculate the energy content of the wave.

Quote
It equals F x S, where F is the twisting force, and S is the distance of the arch of the applied force.

The formula for calculating the work W=F*s is correct, but when we are dealing with rotation then engineers don’t use such notation. In case of rotation the work (or energy) is calculated as W=M*theta where M is a torque (M=F x r, vector product - where r is the radial distance of the attacking force from the axis) and theta is the angle of rotation (in radians).

Quote
The total for the wave would be the integral of these over the length of the wave.

This is partially true. First of all, in case of the wave machine, for the calculation of energy content, we can break it up into small discrete segments, and treat each segment as a simple torsion pendulum. Each rod with its piece of central torsion wire is a single segment. Therefore, instead of integration, we can simply use a summation to add together the energy contents of all the segments.

But what you have calculated with W=F*s or better with W=M*theta is only the potential energy stored in the elastic distortion of the wire (like in a spring), which is only half of the story. The torsion pendulum has got kinetic energy as well, which has to be added to the potential energy in order to get the total energy content.

Let’s summarize what we have figured out so far. The wave machine demonstrates the propagation of torsional waves in an elastic rod or wire as a transmission line. In order to let us see the wave movement, it uses balanced rods periodically attached to the torsion wire. The rods are either soldered to the torsion wire, or fixed to it with other techniques in such a way that they don’t slip. The rods serve dual purpose; they slow the wave down, and they also convert the torsion into translation to make the amplitude more visible. The wave machine can be analyzed as a series of individual torsion pendulums all connected together.

Although the shape of the pulse on the attached photo superposition.png in reply #56 of this thread is not exactly sinusoid, for the sake of simplicity let’s calculate the energy content of one half of a sine wave. Let’s assume that the wavelength is λ, and there are 21 rods (20 torsion wire segments) within the half wavelength. The amplitude of vertical displacement of the wave is A, which corresponds to an angular rotation of the rod #10 in the center of the half sinusoid theta_max. Thus we have 20 complete mini torsion pendulums within this length λ/2, which contain the wave pulse and its energy. The displacement of the first and last bar is zero. Let us assume that the angular rotation of each rod in our half sinusoid can be calculated according to the equation attached below (0<=n<=20 is the number of the examined rod). The first rod is #0, the second rod is #1, the central rod is #10, and the last rod is #20.

Now all you have to do is calculate the total energy content of the wave, by adding together the energies of all 20 individual mini torsion pendulums that contain the pulse. Please give us the formula that contains both potential and kinetic energies, and can be used for this calculation. This should not be too difficult for you since you have said that you are a mechanical engineer, and I have also given a number of references that discuss torsion pendulums in detail. You can also dust off your old textbooks and refresh your memory about the subject. Then we can demonstrate its use on a specific example, to calculate the numerical value of a specific wave pulse.

Finally we will be able return to the original subject of analyzing whether we can gain excess energy from the superposition of two waves that propagate in opposite directions or not.

Finally here are two very useful documents for those who are seriously interested in the practical understanding and building a wave machine:

Wave Motion Demonstrator - Instruction Manual
ftp://ftp.pasco.com/support/Documents/English/SE/SE-9600/SE-9600%20Manual.pdf

Coupled Torsion Pendulum
http://physics.unipune.ernet.in/~phyed/24.3/24.3_Pathare.pdf

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

##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #71 on: March 02, 2017, 09:27:54 PM »

#### telecom

• Sr. Member
• Posts: 370
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #72 on: March 03, 2017, 03:34:57 AM »

Here is what we know: the properties of the transmission line (given by the manufacturer), the amplitude, wave length, and shape of the wave (sinusoid). That’s all. From this we should be able to calculate the energy content of the wave.

The formula for calculating the work W=F*s is correct, but when we are dealing with rotation then engineers don’t use such notation. In case of rotation the work (or energy) is calculated as W=M*theta where M is a torque (M=F x r, vector product - where r is the radial distance of the attacking force from the axis) and theta is the angle of rotation (in radians).

Ok, understood.

Quote
This is partially true. First of all, in case of the wave machine, for the calculation of energy content, we can break it up into small discrete segments, and treat each segment as a simple torsion pendulum. Each rod with its piece of central torsion wire is a single segment. Therefore, instead of integration, we can simply use a summation to add together the energy contents of all the segments.

Yes, for sure, this will simplify things!

Quote

But what you have calculated with W=F*s or better with W=M*theta is only the potential energy stored in the elastic distortion of the wire (like in a spring), which is only half of the story. The torsion pendulum has got kinetic energy as well, which has to be added to the potential energy in order to get the total energy content.

This part makes me puzzled - it probably involves the period of the above pendulum.

Quote
Let’s summarize what we have figured out so far. The wave machine demonstrates the propagation of torsional waves in an elastic rod or wire as a transmission line. In order to let us see the wave movement, it uses balanced rods periodically attached to the torsion wire. The rods are either soldered to the torsion wire, or fixed to it with other techniques in such a way that they don’t slip. The rods serve dual purpose; they slow the wave down, and they also convert the torsion into translation to make the amplitude more visible. The wave machine can be analyzed as a series of individual torsion pendulums all connected together.

Although the shape of the pulse on the attached photo superposition.png in reply #56 of this thread is not exactly sinusoid, for the sake of simplicity let’s calculate the energy content of one half of a sine wave. Let’s assume that the wavelength is λ, and there are 21 rods (20 torsion wire segments) within the half wavelength. The amplitude of vertical displacement of the wave is A, which corresponds to an angular rotation of the rod #10 in the center of the half sinusoid theta_max. Thus we have 20 complete mini torsion pendulums within this length λ/2, which contain the wave pulse and its energy. The displacement of the first and last bar is zero. Let us assume that the angular rotation of each rod in our half sinusoid can be calculated according to the equation attached below (0<=n<=20 is the number of the examined rod). The first rod is #0, the second rod is #1, the central rod is #10, and the last rod is #20.

Now all you have to do is calculate the total energy content of the wave, by adding together the energies of all 20 individual mini torsion pendulums that contain the pulse. Please give us the formula that contains both potential and kinetic energies, and can be used for this calculation. This should not be too difficult for you since you have said that you are a mechanical engineer, and I have also given a number of references that discuss torsion pendulums in detail. You can also dust off your old textbooks and refresh your memory about the subject. Then we can demonstrate its use on a specific example, to calculate the numerical value of a specific wave pulse.

Honestly, Zoltan, you are overestimating my limited mental faculties.
I would rather prefer to go the route of the lesser resistance, and have it
already done for us by some superior mind.
In any case, it will take some time for me to digest all this not very
intuitive info.

Quote

Finally we will be able return to the original subject of analyzing whether we can gain excess energy from the superposition of two waves that propagate in opposite directions or not.
But I'm very interested in this subject anyway, it really makes me think hard.

Finally here are two very useful documents for those who are seriously interested in the practical understanding and building a wave machine:

Wave Motion Demonstrator - Instruction Manual
ftp://ftp.pasco.com/support/Documents/English/SE/SE-9600/SE-9600%20Manual.pdf

Coupled Torsion Pendulum
http://physics.unipune.ernet.in/~phyed/24.3/24.3_Pathare.pdf

#### telecom

• Sr. Member
• Posts: 370
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #73 on: March 03, 2017, 03:58:50 AM »
Hi Zoltan,
just want to expand on your remarkable formula.
According to what you've described,
The potential energy of the wave should be the
sum from 1 to 20 for the each torsion bar element..
Ptotal =SUM [1-20]( T x Q )
where Q is calculated according to your formula.
Regards

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

##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #73 on: March 03, 2017, 03:58:50 AM »

#### ZL

• Jr. Member
• Posts: 67
##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #74 on: March 04, 2017, 07:38:41 PM »
Honestly, Zoltan, you are overestimating my limited mental faculties.

Overestimating? I was expecting form you to be able to calculate these things, because you wrote me that you are a mechanical engineer who has got his diploma from a university. Calculating the energy content of a sine wave on the wave machine supposed to be a simple routine task for a mechanical engineer. Now it seems obvious that your claim about your qualification was a porky, which I don’t appreciate.

Quote
I would rather prefer to go the route of the lesser resistance, and have it already done for us by some superior mind.

That is fine, but in that case you are not doing any FE research. You are expecting from the “superior mind” to do the research for you, find the solution, and give you the blueprint of a developed FE machine, and all this for free. Ahem… don’t you think that your expectations are unrealistic and selfish? If you want to do your own research in this subject then you have to learn the required basics first. If you think it is too difficult for you, then you can still contribute to the cause by supporting the research of those who can do it. If you are not willing to contribute even this little, then you will have to wait until Santa brings you a present that you can replicate.

Perhaps this is a good point to hibernate this thread again for a while. The rest of the explanations about the line of thought we have discussed here will be shared with those who actively support my research. If anybody is interested, you can contact me via the contact form on my website:

https://feprinciples.wordpress.com/contact/

I respond to everybody (at least to their first email). If you would not get any response within a week, then something is wrong. Either try again, or let me know in a post here on this thread.

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

##### Re: Free Energy from Electromagnetic Wave Fields
« Reply #74 on: March 04, 2017, 07:38:41 PM »