MY PATENTED DEVICE DOES NOT WORK!

[mr2]

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 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)


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..

[MarkE]

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.

[mr2]

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...

[Void]

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...

[MarkE]

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.