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Author Topic: MY PATENTED DEVICE DOES NOT WORK!  (Read 13110 times)

Offline mr2

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MY PATENTED DEVICE DOES NOT WORK!
« 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.

https://www.youtube.com/watch?v=54Bj_V0acAo

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 รก 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)


https://www.youtube.com/watch?v=YubFvO3NgGI

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..


Free Energy | searching for free energy and discussing free energy


Offline MarkE

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #1 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.


Offline mr2

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #2 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...

Offline Void

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #3 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...



Offline MarkE

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #4 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.

Free Energy | searching for free energy and discussing free energy

Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #4 on: May 30, 2015, 07:51:31 AM »
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Offline MarkE

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #5 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.

Offline sm0ky2

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #6 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.


Free Energy | searching for free energy and discussing free energy

Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #6 on: May 30, 2015, 01:20:47 PM »
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Offline MarkE

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #7 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.

Offline poynt99

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #8 on: May 30, 2015, 02:18:46 PM »
Indeed resistive losses, no matter how small the resistance between the caps.

See this document clearly illustrating the effect.

Offline tinman

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #9 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.

Free Energy | searching for free energy and discussing free energy

Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #9 on: May 30, 2015, 03:56:21 PM »
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Offline Void

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #10 on: May 30, 2015, 04:00:32 PM »
Indeed resistive losses, no matter how small the resistance between the caps.
See this document clearly illustrating the effect.

Very nice and clear explanations. Thanks poynt99.
All the best...

Offline Paul-R

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #11 on: May 30, 2015, 04:03:25 PM »

My patent:
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:
http://worldwide.espacenet.com/publicationDetails/biblio?DB=EPODOC&II=0&ND=3&adjacent=true&locale=en_EP&FT=D&date=20110113&CC=WO&NR=2011005103A1&KC=A1

but I cannot see what the point of the device is.


Offline sm0ky2

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #12 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.

Offline sm0ky2

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #13 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.






Offline tinman

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Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #14 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.

Free Energy | searching for free energy and discussing free energy

Re: MY PATENTED DEVICE DOES NOT WORK!
« Reply #14 on: May 30, 2015, 06:38:33 PM »

 

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