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## Solid States Devices => solid state devices => Topic started by: mr2 on May 30, 2015, 12:48:36 AM

Title: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on May 30, 2015, 12:48:36 AM
Well.. does it?

The simplest way to explain:

When searching for my user name, I have been here for many years. Not actually active, but in the background..

I have patented my work, and tried to patent it in the rest of the word for the last 7 years.
But my invention is declined because my motor controller is actually a "radio" because it seems so.. by paper.

After seeing the Swedish program Patent 986, and my own experience, I will give my 12 years of experience away.
I am not the type "Look at me, I have the overunity machine".

And.. I might have wrong. But the theory is simple enough.

"Everyone" know of the capacitor paradox. Where did the half of the energy go?

If you place a resistor before the capacitor while charging, you don't loose energy, you loose time.
A loss of energy should show a waste of electrons, but in a closed circuit, the electrons have no way to go except filling the capacitor.
Then; What if you use the loss of time to spend on movement on a motor?

The capacitor have a simularity to a water tap. You have a bucket. You fill the bucket.. you walk to the flowers and watering them.
While you are watering, the next bucket is filled up. But if your walk takes 30 seconds, you don't care if the next bucket is filled in 28 seconds or 1 second.
But while filling the bucket in 28 seconds, the water tap has driven a generator, giving light for 28 seconds.
Hence the 28 seconds of light is free.

My invention is using a capacitor running a motor that have a charging time thru coil to capacitor less than passing time of magnets.
The discharge is the same time.
Hence I am using the same energy to push 2 magnets, while charging the capacitor just once. You have 2 pulses on one charge of a capacitor.

Sorry, it's in Norwegian.
Battery is 1 F. 11V. 0,7W storage. Pulsing the motor with 1500uF.
Done in normal way, the pulses is about 660 pulses á 1500uF.
In my way, the pulses is 980 pulses. Same 11V, 1F in both tests.

A more industrial version shown in Italy (destroying the best product they had in stock)

Notice the sound from original motor to my design:
0:30 compared to 1:10

My patent:
https://patentscope.wipo.int/search/en/detail.jsf?docId=WO2011005103

You are free to try it. Please give me an report if done..

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: MarkE on May 30, 2015, 01:06:05 AM
There is no paradox involved in what happens to the stored energy when connecting two capacitors together of the same capacity, where one is initially charged and the other is not charged.  Anyone who understands RC charging of a capacitor understands where half the energy goes.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on May 30, 2015, 01:13:31 AM
There is no paradox involved in what happens to the stored energy when connecting two capacitors together of the same capacity, where one is initially charged and the other is not charged.  Anyone who understands RC charging of a capacitor understands where half the energy goes.

Yes, please explain where the electrons go...
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: Void on May 30, 2015, 02:44:34 AM
Yes, please explain where the electrons go...

Hi mr2. I have pondered this as well. The total charge from one capacitor when divided between another capacitor
of the same value of capacitance as the first capacitor doesn't actually disappear anywhere. The total charge divided between the two
same valued capacitors remains the same. It is only the total energy that drops by half. I think it is a valid question to ask where
half the energy went when the charge from one capacitor is divided over to a second capacitor of the same capacitance value.

If I pour water into an empty cylinder, the water pours in the same all the way to the top, but charge is not like water
in that charge wants to repel like charge with a relatively strong force. So, the more charge we cram into a capacitor, the more
energy it takes to cram that charge into the capacitor, as the stronger the repelling force of the charge that is already crammed in
there becomes the more the capacitor is charged. When we discharge this crammed in charge to another capacitor of the same capacitance value,
the charge on the first capacitor then divides evenly between the two capacitors. There is now half the charge on each capacitor, giving a total charge
the same as before the discharge, but the total energy of the two capacitors is only half the energy that was originally on the first capacitor before we
discharged it to the second capacitor. Where does half the original energy go? Presumably it is dissipated in resistive losses as heat in the wire connecting
the two capacitors together and in the capacitor plates, although if you think about it the total energy should drop by about the same for different
wire resistance values and for different capacitor ESR values, should it not? This then makes me wonder where did half of the energy really all go?
I have done experiments similar to what I believe you are describing where I can end up with more energy from such a discharge than
just half the original energy before the discharge, using three capacitors and a transformer primary between two main capacitors, so it is still a
mystery to me where the energy really all does go when doing such a capacitor discharge.

Where does half the energy really go? Can it all be accounted for as heat gain due to resistive losses? If I place a transformer primary winding
between the two capacitors, I have a higher total wire resistance, but I can collect some of the normal energy losses on a third capacitor on the transformer
secondary (through a rectifier diode) and end up with somewhat more than one half of the original energy after the discharge. There is more wire resistance between
the two capacitors, but I end up with more than half the energy after the discharge. Something doesn't quite seem to add up there. What might I be overlooking?
At any rate this does seem to support the notion that the energy that is normally lost in such a capacitor discharge can be captured at least to some
extent with the right setup... It seems it would be hard to capture most of it however.

All the best...

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: MarkE on May 30, 2015, 07:51:31 AM
Yes, please explain where the electrons go...
Unless so much voltage is applied that the capacitors break down, the electrons don't go anywhere.  They remain within the dielectric of the capacitors.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: MarkE on May 30, 2015, 07:53:57 AM
Hi mr2. I have pondered this as well. The total charge from one capacitor when divided between another capacitor
of the same value of capacitance as the first capacitor doesn't actually disappear anywhere. The total charge divided between the two
same valued capacitors remains the same. It is only the total energy that drops by half. I think it is a valid question to ask where
half the energy went when the charge from one capacitor is divided over to a second capacitor of the same capacitance value.

If I pour water into an empty cylinder, the water pours in the same all the way to the top, but charge is not like water
in that charge wants to repel like charge with a relatively strong force. So, the more charge we cram into a capacitor, the more
energy it takes to cram that charge into the capacitor, as the stronger the repelling force of the charge that is already crammed in
there becomes the more the capacitor is charged. When we discharge this crammed in charge to another capacitor of the same capacitance value,
the charge on the first capacitor then divides evenly between the two capacitors. There is now half the charge on each capacitor, giving a total charge
the same as before the discharge, but the total energy of the two capacitors is only half the energy that was originally on the first capacitor before we
discharged it to the second capacitor. Where does half the original energy go? Presumably it is dissipated in resistive losses as heat in the wire connecting
the two capacitors together and in the capacitor plates, although if you think about it the total energy should drop by about the same for different
wire resistance values and for different capacitor ESR values, should it not? This then makes me wonder where did half of the energy really all go?
I have done experiments similar to what I believe you are describing where I can end up with more energy from such a discharge than
just half the original energy before the discharge, using three capacitors and a transformer primary between two main capacitors, so it is still a
mystery to me where the energy really all does go when doing such a capacitor discharge.

Where does half the energy really go? Can it all be accounted for as heat gain due to resistive losses? If I place a transformer primary winding
between the two capacitors, I have a higher total wire resistance, but I can collect some of the normal energy losses on a third capacitor on the transformer
secondary (through a rectifier diode) and end up with somewhat more than one half of the original energy after the discharge. There is more wire resistance between
the two capacitors, but I end up with more than half the energy after the discharge. Something doesn't quite seem to add up there. What might I be overlooking?
At any rate this does seem to support the notion that the energy that is normally lost in such a capacitor discharge can be captured at least to some
extent with the right setup... It seems it would be hard to capture most of it however.

All the best...
If you want to understand this, then consider what happens when you charge an ideal 1uF capacitor from zero through a 1 Ohm resistor using a voltage source, and then after discharging the capacitor again,  when you charge that same capacitor through a 1K Ohm resistor using the same voltage source.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: sm0ky2 on May 30, 2015, 01:20:47 PM
To really understand this problem, you have to understand what the capacitor is actually doing.
The bucket analogy works, if you think of it as filling the bucket through a hole at the bottom of the bucket.
the more water you put in, the more the potential energy resists further filling of the bucket.
Follow the curve, and you see how this affects charging time.

Also:
under the standard view of the paradox, the "total energy" is viewed as only the at-rest potential energy of the charged capacitor(s). Potential energy is only part of the problem, we must account for all changes in energy throughout the system.
Q = CV ;   E = QV/2 or E = CV^2/2     hmmmm V^2/2... where have we seen this?
let's try it:  V = 2 ;  E= (2)^2 / 2 = 2
V = 1 ;  E= (1)^2 / 2 = 0.5 (x2) = 1
Why?

The answer is in the kinetic energy. When you move the charges, Pe is converted into Ke.
With the capacitor fully charged, Ke goes to Max. When the capacitor is half charged, Ke only goes to approx. 1/4 of max. Since both capacitors balance each other out at equilibrium, the final at-rest Pe of the system is half of the kinetic maximum. Half of the total E is stored as Pe (1/4th in each cap), This means that there is only 1/2 of the energy remaining as kinetic in the moving charge, which cannot further forces its' way into the bucket. It must go somewhere else.
What happens when we resist an electric charge? --- electromagnetic radiation.

Further evidence of this can be obtained by draining the second capacitor during the charging time, where-in the full Pe of the system can be recovered, as it is converted into kinetic energy.

It is my argument that the remainder of the Kinetic energy, that is not converted into potential energy in each of the half-charged capacitors, is radiated as EMF.
(somewhere in the upper range of RF scaling to microwaves with the value of C)
This will occur as a single pulse with simple plate capacitors, or a series of short, truncated bursts with layered caps.

There is really no paradox, it is simply a matter of perspective. If you only look at half of the problem, you only see half of the answer.

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: MarkE on May 30, 2015, 02:01:47 PM
The truth is far simpler.  It is almost all just resistive loss.  Charge any capacitor from a voltage source through a resistor and the resistor dissipates the same amount of energy as ends up stored in the capacitor.  The resistor value just contributes to how long the charging process takes.

Hooking two capacitors together does radiate a little bit.  The lion's share of energy loss goes directly to heat dissipated in the wiring and ESR of the two capacitors.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: poynt99 on May 30, 2015, 02:18:46 PM
Indeed resistive losses, no matter how small the resistance between the caps.

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: tinman on May 30, 2015, 03:56:21 PM
To really understand this problem, you have to understand what the capacitor is actually doing.
The bucket analogy works, if you think of it as filling the bucket through a hole at the bottom of the bucket.
the more water you put in, the more the potential energy resists further filling of the bucket.
Follow the curve, and you see how this affects charging time.

No-no good smoky2,as it takes just as much energy to fill the bucket from the top as it would from the bottom. The energy required to fill the bucket(regardless as to wether you fill from the bottom or the top)is the exact potential energy you have in the bucket at the end of the fill-as long as you dont change the hight of the bucket.

There is a big missbelief that you loose half of the net charge when you charge a capacitor,but this is not the case. The charge that apparently went missing,or was disipated as resistive losses is still actually in the cap-->well most of it anyway. You will loose a small amount to resistive losses as heat,but it is not half. the only way to see this is through experimenting with the cap it self. for this you need a long clean bench,and some carful soldering/wire fixing skills.

When a capacitor is charged,you are only moving the already existing charge from one plate to another. Electrons dont just disappear,and it is the flow of electrons that cause heat,and the more you restrict that flow,the more heat you create. What you see as a discharged capacitor is just a capacitor that has an equal charge on each plate,and thus the potential difference in charge is 0-but the charge is still there. It is the same with your bucket reference. If you take a second bucket of the same size,and you link a hose from the bottom of one bucket to the other,you now have your capacitor.,Then let the water flow from the full bucket to the empty bucket,and you can place any type of resistance in that hose that joins the two buckets,and it will not change the outcome-->the two buckets will still have the same amount of water in them as the amount of water you started with,but now you have no potential difference.

Lets explain on a bigger scale than a small bucket. Lets take two 1000ltr water tanks that have a diameter of X amount and are two meters tall-->this give us a decent head pressure to work with,and the head pressure is what we will call charge in the capacitor. We have our hose that joins the two tanks at the bottom of the tanks,with a valve at one end of the hose. Now,lets say that hose is 1 inch in diameter,and we place a small resistance in that hose-say a sleeve with a 20mm hole in it. We open the valve,and the water flows from the full tank to the empty tank,and it keeps flowing until the water level is equal in each tank-but we still have the same amount of water in both tanks that we started out with. Now we place a large resistance in the hose,say a sleeve with a 5mm hole in it. We run the test again,and although it took longer for the water to equalise in the tanks,we still have the same amount of water we started out with. That flowing water could have been driving a small turbine,and running a small light ,and it still wouldnt change the outcome-that being ,we still have the same amount of water at the end. When you charge a cap,you are simply pumping the water out of one tank,and putting it back into the other.

There are those here that will disagree,but i have done the test and experiments,and most of the charge is still in the cap-regardless of what type of load you place on it.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: Void on May 30, 2015, 04:00:32 PM
Indeed resistive losses, no matter how small the resistance between the caps.

Very nice and clear explanations. Thanks poynt99.
All the best...
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: Paul-R on May 30, 2015, 04:03:25 PM

My patent:
https://patentscope.wipo.int/search/en/detail.jsf?docId=WO2011005103 (https://patentscope.wipo.int/search/en/detail.jsf?docId=WO2011005103)

Small translation issue -

Abstract line 1. "comprehending" should be "comprising"

There may be more.

I find this site better:

but I cannot see what the point of the device is.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: sm0ky2 on May 30, 2015, 06:22:35 PM
The truth is far simpler.  It is almost all just resistive loss.  Charge any capacitor from a voltage source through a resistor and the resistor dissipates the same amount of energy as ends up stored in the capacitor.  The resistor value just contributes to how long the charging process takes.

Hooking two capacitors together does radiate a little bit.  The lion's share of energy loss goes directly to heat dissipated in the wiring and ESR of the two capacitors.

resistance doesn't account for but a fraction of the losses in the system.
we're talking about a small conducting wire, which can be substituted with a superconductor @ 0 Ohms.
E = 1/2 still
This effect was explained in several physics papers 25-30 yrs ago. College professors still like to use it to get their students to think.

If you look at both the kinetic and potential energies of the system, everything is accounted for.
We are only using 1/2 of the potential energy.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: sm0ky2 on May 30, 2015, 06:38:10 PM
@ Tinman,

Nice example with the buckets.

And yes, the Total Charge in the system remains about the same.
This is true. Q = CV, charge = capacitance x voltage

What changes, is the Total Energy.
The equation that defines energy of a capacitor: E = CV^2/2 = QV/2 = Q^2/(2C)
divide the charge in half, and you divide the energy by 4.
there's 2 of them, so total energy is 1/2.

It takes an increasing amount of energy to put more charge into a capacitor.
During discharge, this comes out as momentum, velocity, kinetic energy of a moving charge.

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: tinman on May 30, 2015, 06:38:33 PM
resistance doesn't account for but a fraction of the losses in the system.
we're talking about a small conducting wire, which can be substituted with a superconductor @ 0 Ohms.
E = 1/2 still
This effect was explained in several physics papers 25-30 yrs ago. College professors still like to use it to get their students to think.

If you look at both the kinetic and potential energies of the system, everything is accounted for.
We are only using 1/2 of the potential energy.

That is correct,the other half is still in the cap.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: forest on May 30, 2015, 06:59:29 PM
Would it be too hard to experimental prove if there is EM radiation in this case ? Do you know any scientific papers of such experiment ?
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: tinman on May 30, 2015, 07:33:49 PM
Would it be too hard to experimental prove if there is EM radiation in this case ? Do you know any scientific papers of such experiment ?
There is no EM losses or any other losses-->the !/2 charge that seems to be missing is still in the cap. The electrons dont just go zipping of into the sky,or get radiated as heat-the electron flow is what creates the heat. The higher the restriction to this flow,the more heat is generated-electrical flow friction.

When you charge a cap or a battery,your charging source is nothing more than a pump. This pump is what pumps the electrons from the positive plate,back to the negative plate. When there is no voltage across the cap,you havnt used the charge,you have simply provided a path for the electrons to flow to the positive plate. This is the very same as the water in the two tanks wanting to reach an equilibrium-->the charge is still in the caps. The problem is,there is now no potential difference between the two charged plates.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: MarkE on May 31, 2015, 07:47:43 AM
Would it be too hard to experimental prove if there is EM radiation in this case ? Do you know any scientific papers of such experiment ?
Maxwell's equations tell us that any rate of change of current through a circuit radiates E/M energy.  Experiments go back to the 1800's.  Everyday thousands of engineers conduct these experiments in EMC labs doing what they need to to reduce the radiation to acceptable levels.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: pomodoro on May 31, 2015, 02:46:32 PM
Beginners should also note that the capacitor holds no extra electrons when charged than it did when discharged.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: tinman on May 31, 2015, 03:30:51 PM
Beginners should also note that the capacitor hold no extra electrons when charged than it did when discharged.
Spot on ;)
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on June 12, 2015, 11:44:19 PM
This was fun.

People who knows what they are talking about.

Have you tested it live? Nope, you have formulas!
Hence the truth.. Stupid...

This explains the theory. But sorry, in Norwegian..

1F, 11V. 1000uF capacitor as drive-capacitor.
First tests: 11V, each pulse 1000uF: about 660 pulses.

With my change of same components, but charging the capacitor via the electromagnet;
The time charging the capacitor is used to run the motor. The loss of time is adjusted to the time to charge and discharge the capacitor.
980 pulses and a much higher voltage and therefore highter restvalue of energy.

Same energy when starting, same energy per pulse.. and a closed circuit..

They say: Where did the half of the energy go? And all proof is math. But electrons have no other way to go, but via copper wires.

Yet another test comparing a "state of the art" motor from any shop:
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on June 12, 2015, 11:50:12 PM
... AND it works on any motor the method is connected to..

So; if it have been explained as "radiation" or all other stupid explanations, it would not work, since half of the energy is wasted when charging and discharge a capacitor.
It is prooved by several tests; both homemade motor, and commercial motors that it actually work..
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on June 12, 2015, 11:54:14 PM
Maxwell's equations tell us that any rate of change of current through a circuit radiates E/M energy.  Experiments go back to the 1800's.  Everyday thousands of engineers conduct these experiments in EMC labs doing what they need to to reduce the radiation to acceptable levels.

Fun... Protecting the statements by talking about old experiments with components from that time.. And have the knowledge just by reading it..
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on June 13, 2015, 12:04:16 AM
There is no EM losses or any other losses-->the !/2 charge that seems to be missing is still in the cap. The electrons dont just go zipping of into the sky,or get radiated as heat-the electron flow is what creates the heat. The higher the restriction to this flow,the more heat is generated-electrical flow friction.

When you charge a cap or a battery,your charging source is nothing more than a pump. This pump is what pumps the electrons from the positive plate,back to the negative plate. When there is no voltage across the cap,you havnt used the charge,you have simply provided a path for the electrons to flow to the positive plate. This is the very same as the water in the two tanks wanting to reach an equilibrium-->the charge is still in the caps. The problem is,there is now no potential difference between the two charged plates.

You have understood it!

When charging a capacitor you have only the loss of time, a time represented by R.
If you then have a motor running at 1000 RPM, there is x cm between the magnets, you have a Y ms time between the magnets.
You then have to calculate the size of capacitor that have at least 87% the time of charge and discharge.

Then you have like a "AC signal" on the oscilloscope because of LC.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: Void on June 13, 2015, 12:40:54 AM
mr2, what sort of overall efficiency have you measured?
All the best...

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: ATOM1 on June 14, 2015, 04:29:03 AM
The quantum effects of electron coherence in an alignment of high harmonic resistance over a point of charge and discharge can fall out off the ohms law and into a new law and one where time it self creates energy  ! To many time we use the water bucket to measure with ! Our minds are moulded to an old way of thinking . We forget the universe was made from nothing and time these two components hardly ever are included ????????

I have had many experiments over the years that show a time delay as the cause of an OU result but not with normal resisters or standard set ups for cap banks ! I all ways use harmonics in the formulas and low to high resistance and now square laws to convert micro farads into electron capacitance .. To me if the point of the beginning of the universe was smaller than an electron than even a capacitor can store many times its normal power.

quantum events of strangeness is what I look for and totally avoid all other research ! Christ feed 500 people from one small basket and I have to say it does seem possible to do the same with a capacitor ....

An electron is a strange thing it is never in the same place its always hoping in and out of time and space , This it self is something beyond the normal rules for electrical mathematics ,  to many times push and pull the electron around but when we let it jump in the way it likes to move than sometimes we get far more than we expected .....

From nothing came everything , but what if everything is a lot less than what we expected and there is only one electron in the universe and being shared , it would fit its quantum profile and is in more than one way possible.... We  also tend to think we know it all done it all and expect the same out come every time we do something yet there is other avenues to explore where there are rewards beyond our imagination..........

Look again

ATOM1

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on January 16, 2016, 12:14:44 AM

ATOM1 is wasted. Quantum effects of what?

You have a "battery" bucket with 100 L water.
Your "resistor" is a hose with diameter 2 cm.
Your capacitor is a bucket with capacity 2 L.

You will always have the 100 liters as the start... the hose will restrict the flow. The "capacitor" will ONLY contain 2L at max. Then the flow will stop.
If you have a impeller in the hose, you produce energy. ALWAYS the same amount of water to fill the "capacitor".

When you then use the water in the "capacitor" to flow thru the same hose, you make the same energy again, with the same water, the same volume, and the same energy as when the "capacitor" is charged.
Hence; 2L water have passed a generator, making double energy.

Some say; You loose energy when charging a capacitor.
No.
You have a closed circuit. Charging a 2Watt capacitor, is using the 2W. Not 3... not 4. Because; How would the extra watts pass the capacitor max capacity?

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on January 16, 2016, 03:09:24 AM
About capacitors and the loss of energy charging it (the capacitor paradox)
You charge a 1W capacitor. The electrons goes from the negative of battery, to negative of capacitor.
The capacitor send 1W of electrons from the positive side of the capacitor to the battery.

Battery have lost 1W. The capacitor have 1W at the positive side, and -1W at the negative side making the difference 2W.
That will be the rated level of this capacitor of 2W.
Half the energy would be "lost".
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: TinselKoala on January 16, 2016, 08:31:47 AM
Capacitors are not rated in Watts and neither are batteries. The Watt is NOT a unit of energy, it is a unit of POWER. You must understand the difference. A Watt is one Joule of energy per second. That is, one Joule of ENERGY passing your point of measurement in one second. If you have that same one Joule of energy passing your measurement point in 0.1 second, you have 10 Watts of power. If it passes in 0.01 second, you have 100 Watts of power. One Joule of energy can represent megawatts of power if it happens in a short enough time interval (a microsecond or less).

A Joule, on the other hand, is one Watt-second. One Watt of power, sustained for a period of one second, requires one Joule of energy. And so on.

Voltage is not energy, current is not energy, power is not energy. Only _energy_ is the conserved quantity among these variables.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: Hoppy on January 16, 2016, 12:28:01 PM
The quantum effects of electron coherence in an alignment of high harmonic resistance over a point of charge and discharge can fall out off the ohms law and into a new law and one where time it self creates energy  ! To many time we use the water bucket to measure with ! Our minds are moulded to an old way of thinking . We forget the universe was made from nothing and time these two components hardly ever are included ??? ??? ??

I have had many experiments over the years that show a time delay as the cause of an OU result but not with normal resisters or standard set ups for cap banks ! I all ways use harmonics in the formulas and low to high resistance and now square laws to convert micro farads into electron capacitance .. To me if the point of the beginning of the universe was smaller than an electron than even a capacitor can store many times its normal power.

quantum events of strangeness is what I look for and totally avoid all other research ! Christ feed 500 people from one small basket and I have to say it does seem possible to do the same with a capacitor ....

An electron is a strange thing it is never in the same place its always hoping in and out of time and space , This it self is something beyond the normal rules for electrical mathematics ,  to many times push and pull the electron around but when we let it jump in the way it likes to move than sometimes we get far more than we expected .....

From nothing came everything , but what if everything is a lot less than what we expected and there is only one electron in the universe and being shared , it would fit its quantum profile and is in more than one way possible.... We  also tend to think we know it all done it all and expect the same out come every time we do something yet there is other avenues to explore where there are rewards beyond our imagination..........

Look again

ATOM1

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on January 30, 2016, 02:17:33 AM
The world is loaded with explainers of the obviously...
But not able to knowing why...

Capacitors are not rated in Watts and neither are batteries. The Watt is NOT a unit of energy, it is a unit of POWER. You must understand the difference. A Watt is one Joule of energy per second. That is, one Joule of ENERGY passing your point of measurement in one second. If you have that same one Joule of energy passing your measurement point in 0.1 second, you have 10 Watts of power. If it passes in 0.01 second, you have 100 Watts of power. One Joule of energy can represent megawatts of power if it happens in a short enough time interval (a microsecond or less).

A Joule, on the other hand, is one Watt-second. One Watt of power, sustained for a period of one second, requires one Joule of energy. And so on.

Voltage is not energy, current is not energy, power is not energy. Only _energy_ is the conserved quantity among these variables.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: TinselKoala on January 30, 2016, 05:33:30 AM
The world is loaded with explainers of the obviously...
But not able to knowing why...

We heal, we reflect, we are reborn. Consciousness consists of ultra-sentient particles of quantum energy. “Quantum” means an evolving of the non-local. This life is nothing short of a summoning revolution of unified wisdom. Without truth, one cannot vibrate. How should you navigate this perennial nexus? The solar system is calling to you via bio-electricity. Can you hear it? Child, look within and unify yourself.

We are at a crossroads of hope and discontinuity. We are in the midst of an unified blossoming of transformation that will enable us to access the quantum matrix itself. Throughout history, humans have been interacting with the totality via ultrasonic energy.

Soon there will be a blossoming of truth the likes of which the quantum soup has never seen. It is a sign of things to come. The future will be an authentic evolving of insight.
It is in deepening that we are guided.

It can be difficult to know where to begin. Although you may not realize it, you are sublime. If you have never experienced this lightning bolt at the speed of light, it can be difficult to grow. Kabala may be the solution to what’s holding you back from an unimaginable harmonizing of faith. As you reflect, you will enter into infinite potential that transcends understanding. You will soon be recreated by a power deep within yourself — a power that is mythic, internal.

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: sm0ky2 on January 30, 2016, 08:10:29 AM
@ MR 2

I think you are missing the conceptual premise behind what you are doing.

we can shorten the duty cycle of power going into a device. this is for certain.
and most of our devices, the way we use them, are very inefficient by design.
things like storing the energy in capacitors, or a flywheel, or shortening the pulse time with a Joule Thief, etc.
these tricks can make our use more efficient.

But you are not creating any sort of "overunity" when you do this.

I will try to explain.

A motor, when given a steady power input, of maximum potential, has a given torque potential over time, vs the RPM.
meaning, under a heavy load, all of the energy sent into the motor can be converted into motive force.

When you operate a motor at a lower duty cycle, meaning, you put less energy over time through the coils.....
you lose some % of this torque over time.
it is a precisely inversely proportional relationship.

the same motor, when run through your device, under heavy load, you will see the loss of power over time.

just running the motor, at a certain rpm, or turning a small fan, this does not load the motor. you cannot see the power you lose.
what you see, is a more efficient method of turning the motor to a given rpm. basically, uselessly fooling yourself.

if all you do is run a fan, or some low-power use of the motor,.
sure, do what you are doing, and you can save "energy" cost of running that motor.
this is because we waste a lot of energy to turn that fan. you make it more efficient by shortening the duty cycle.
a joule thief will do the same thing when connected to a motor, or visible lighting.

it is only under constant heavy load, that the motor is strained enough to observe how the total energy over time is lower.

test this, and you will see what I am talking about.

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: conradelektro on January 30, 2016, 08:51:14 AM
What I say will be very disturbing for mr2, but I have to say it, because mr2 is the perfect example:

mr2 wasted 12 or at least 7 years because he did not discuss his "invention" publicly with the people who would have told him 12 years ago where his error is (making a motor more efficient at low power, for which the motor was not designed).

mr2 got lost in ill-designed tests because he always feared that someone would take away his "invention" or "discovery". He avoided to consult the right experts, because he wanted to hold up his dream.

I saw this behaviour in many "inventors", in all fields of technology. Specially the "field of OU contraptions" attracts people who think they have something others will take away from them. They forget, that if it were such a miracle, it would be taken away no matter what they do. You can never protect a miracle machine, it would just be too important for everybody not to take it by force or at least by deception or fraud.

It is also typical with what ferocity and pertinence the "inventors" defend their brain child, no matter what experts tell them. The experts are always accused of being part of a grand conspiracy against humankind and the inventor in particular. Legions of researchers have been wrong, only the "inventor" finally has seen the light (although he is no expert in the field of his "invention").

Let mr2's story be a teaching. If you think to have discovered something extraordinary, discuss it openly with the real experts and forget patents and secrecy. The last 200 years of technology have taught us, that it is most likely you have made a conceptual error which many have made before you.

Sorry mr2, you are giving a perfect example which should not be overlooked by future "inventors of miracles".

@TinselKoala: I finally can master my life by nipping from the Quantum soup. It was almost too late for me, but you saved me. Why have you not told me sooner? You probably tried to patent this wisdom? And now you want to save your dark soul by doing good.

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on August 26, 2017, 12:47:31 AM
What I say will be very disturbing for mr2, but I have to say it, because mr2 is the perfect example:

mr2 wasted 12 or at least 7 years because he did not discuss his "invention" publicly with the people who would have told him 12 years ago where his error is (making a motor more efficient at low power, for which the motor was not designed).

mr2 got lost in ill-designed tests because he always feared that someone would take away his "invention" or "discovery". He avoided to consult the right experts, because he wanted to hold up his dream.

I saw this behaviour in many "inventors", in all fields of technology. Specially the "field of OU contraptions" attracts people who think they have something others will take away from them. They forget, that if it were such a miracle, it would be taken away no matter what they do. You can never protect a miracle machine, it would just be too important for everybody not to take it by force or at least by deception or fraud.

It is also typical with what ferocity and pertinence the "inventors" defend their brain child, no matter what experts tell them. The experts are always accused of being part of a grand conspiracy against humankind and the inventor in particular. Legions of researchers have been wrong, only the "inventor" finally has seen the light (although he is no expert in the field of his "invention").

Let mr2's story be a teaching. If you think to have discovered something extraordinary, discuss it openly with the real experts and forget patents and secrecy. The last 200 years of technology have taught us, that it is most likely you have made a conceptual error which many have made before you.

Sorry mr2, you are giving a perfect example which should not be overlooked by future "inventors of miracles".

@TinselKoala: I finally can master my life by nipping from the Quantum soup. It was almost too late for me, but you saved me. Why have you not told me sooner? You probably tried to patent this wisdom? And now you want to save your dark soul by doing good.

Said to Sheldon Cooper: "You can't say people are stupid!! Say you're sorry!!"
Sheldon Cooper: "I'm sorry you're stupid!!"

Sorry to say, it works. And universities versified it. I just didn't pay any attention to this forum.
Was looking for another email alert, and this one paid my attention.

The title when writing this was only a click-bait.
And @conradelektro has no clue whatsoever..
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on August 26, 2017, 12:58:37 AM
And just to clearify, I just don't care about ignorants with theoretical arguments.
Overunity does not need to be explained. Just do it. If it runs forever, you do NOT need to explain that it runs forever.
No maths, no graphs.. no nothing. Put a energy supply without mains and let it run.. if it never stops.. you have it.. if it stops.. sorry..
And there's thousands of messages with "how to"...
The Earth is flat.. and I can proove it. See to the horizont.. there is the end of the world and you will fall...
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: Cadman on August 28, 2017, 10:53:10 PM
Thanks mr2. I just caught this thread a few days ago. Your patent works.

You never claimed it was OU, did you.

.

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: webby1 on August 29, 2017, 08:36:28 PM
These are 2 methods I have used for decades,, just the basic parts without the contact protections and recovery stuff.
IIRC I got these from an Electronics Pocket Reference book.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: webby1 on August 29, 2017, 08:37:47 PM
This is a slight variation that I found useful,, again without the protections and recovery parts.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: forest on August 30, 2017, 12:51:23 AM
magnetic field is overunity
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: antijon on August 31, 2017, 02:54:12 PM
Webby, I was thinking about your first simple circuit, what is it used for? To make an inverter? I thought the same way as the OP, at first. If you put a light bulb and capacitor in series, you light the bulb and charge the cap, then if you break the supply and short the bulb/cap you use the same energy twice. It seems like the same power could be used twice, but that's not the case. As the voltage is divided between the bulb and cap, the total output power is going to equal the input power. So what's the purpose of the inductor in the circuit? Just to smooth the output?

As for the demo videos, the excess noise in the first fan is caused by underloading the motor, as smoky said. Because the RPM is based on line frequency, underloading causes the motor to try to spin faster, and as the polarity changes the rotor has to snap back to its proper location. This causes over amping and noise. It's common in PSC motors with a capacitor on the start winding. If the capacitor is too large, it pulls too much amps and makes a loud hum. Also leads to the motor overheating.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: webby1 on August 31, 2017, 03:48:00 PM
As a controller for a motor.

The coil is either a coil,, or it is a motor.

A thing I noticed a long time ago,, when you use a core once you have the core polarized you do not need as much input to hold it that way, and if the magnetic load on that coil changes, well things change.

The double cap setup is the main one I used to use all the time,, that one allowed me to turn my batteries into alternating pulsed DC.

As far as re-using the charge,,  it is the unused portion of the first use of the charge, if you will, that you can recycle.  I sometimes used 2 different duty cycles and reduced the consumption of my motors,, no OU but things that can increase the efficiency are good things to play with as well.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: conradelektro on August 31, 2017, 04:35:40 PM

Overunity does not need to be explained. Just do it. If it runs forever, you do NOT need to explain that it runs forever.
No maths, no graphs.. no nothing. Put a energy supply without mains and let it run.. if it never stops.. you have it.. if it stops.. sorry..

I agree completely with your argument (show that it does not stop and no further theoretical explanations will be necessary).

So, does your contraption (motor) never stop?

This is the only thing you have to show. Can you demonstrate that your contraption never stops? If not, sorry!

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on September 08, 2017, 08:45:51 PM

I agree completely with your argument (show that it does not stop and no further theoretical explanations will be necessary).

So, does your contraption (motor) never stop?

This is the only thing you have to show. Can you demonstrate that your contraption never stops? If not, sorry!

Actually, I do not know.. I can't. There's no-one wanting to do that. You can do that, Conrad? If not, you're sorry!

But what I know: I got a notice of being charged by the authorities of false references if I used the name of a professor of NTNU (www.ntnu.no (https://www.ntnu.no/)) if I used his name as a reference.

The theory is easy;
A capacitor is like a bucket. You move energy / water to the capacitor / bucket.
You can move the energy / water to wherever you want.

A bucket / capacitor can be filled at any speed. You use a funnel to restrict speed to fill the bucket. You use a resistor to restrict the speed to fill the capacitor.

The funnel make you waste time. The resistor can be changed to a motor making work. You "waste" time by running a motor.

And the capacitor paradox is wrong. There is no paradox. The formula is wrong.

Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on September 08, 2017, 09:30:03 PM
Webby, I was thinking about your first simple circuit, what is it used for? To make an inverter? I thought the same way as the OP, at first. If you put a light bulb and capacitor in series, you light the bulb and charge the cap, then if you break the supply and short the bulb/cap you use the same energy twice. It seems like the same power could be used twice, but that's not the case. As the voltage is divided between the bulb and cap, the total output power is going to equal the input power. So what's the purpose of the inductor in the circuit? Just to smooth the output?

As for the demo videos, the excess noise in the first fan is caused by underloading the motor, as smoky said. Because the RPM is based on line frequency, underloading causes the motor to try to spin faster, and as the polarity changes the rotor has to snap back to its proper location. This causes over amping and noise. It's common in PSC motors with a capacitor on the start winding. If the capacitor is too large, it pulls too much amps and makes a loud hum. Also leads to the motor overheating.

You can't use a bulb as an direct comparison as you need a step-up voltager. A motor works as it's best as when a magnetic pole is passed. And you need the pulse at it's best push. That a capacitor can do. Not a constant voltage. Or current. You waste energy when the magnetic pole is passed.
Title: Re: MY PATENTED DEVICE DOES NOT WORK!
Post by: mr2 on September 08, 2017, 10:02:48 PM
@ MR 2

I think you are missing the conceptual premise behind what you are doing.

we can shorten the duty cycle of power going into a device. this is for certain.
and most of our devices, the way we use them, are very inefficient by design.
things like storing the energy in capacitors, or a flywheel, or shortening the pulse time with a Joule Thief, etc.
these tricks can make our use more efficient.

But you are not creating any sort of "overunity" when you do this.

I will try to explain.

A motor, when given a steady power input, of maximum potential, has a given torque potential over time, vs the RPM.
meaning, under a heavy load, all of the energy sent into the motor can be converted into motive force.

When you operate a motor at a lower duty cycle, meaning, you put less energy over time through the coils.....
you lose some % of this torque over time.
it is a precisely inversely proportional relationship.

the same motor, when run through your device, under heavy load, you will see the loss of power over time.

just running the motor, at a certain rpm, or turning a small fan, this does not load the motor. you cannot see the power you lose.
what you see, is a more efficient method of turning the motor to a given rpm. basically, uselessly fooling yourself.

if all you do is run a fan, or some low-power use of the motor,.
sure, do what you are doing, and you can save "energy" cost of running that motor.
this is because we waste a lot of energy to turn that fan. you make it more efficient by shortening the duty cycle.
a joule thief will do the same thing when connected to a motor, or visible lighting.

it is only under constant heavy load, that the motor is strained enough to observe how the total energy over time is lower.

test this, and you will see what I am talking about.

I have to agree, to a point.
When loading a motor 100% there is no savings. For now. There is no interest of saving energy. 90% of my country have seen my invention and the politics say: Why do we need to save energy?
We can't tax that..
Even Bellona that many of you might know about said: Good luck! Who really care about "free energy".. ?

But my invention saved energy on the BEST motors from any manufacturers, even 1 BILL USD investments done of ELCO in Italy.

http://www.elco-spa.com/ (http://www.elco-spa.com/)
Even the papst motors.

Been there, didn't get any response.. "not invented here..."