Conservation of Charge violated?

[Sephiroth]

I am currently uploading a video to youtube demonstrating a simple experiment that I think defies the "law" of conservation of charge but I thought I would clear it with you wise people first!

First, I took two identical 10,000uf capacitors and shorted one, and charged the other to 12.45v.

I then connected the capacitors in parellel and, no suprise, they each read about half the original voltage in the charged capacitor.

Then I took the two identical capacitors and charged one to 12.45 and left the other one shorted.

I then connected them in parrellel THROUGH the solid state oscillator in a way that would allow the flyback voltage to return to the empty capacitor.

the voltage on the two caps was then 8.47v and 7.24v.... Connected in series this makes 15.71v???

Of course I thought this was GREAT!!!! But then I did the joule calculations and it turns out there is less joules then what I started with... bugger.

So it isn't overunity (at least in the way it applies to conservation of energy) but hasn't this just gone against the law of conservation of charge?

I remember Elias doing a similar experiment a while back but I cant find the thread now.

[Sephiroth]

Anyone care to comment?

[Sephiroth]

This is basically how I have it set up though it isn't the exact schematic.

[peper10]

Hi Seph !!

I have tried something verry similar!
I have charged a 25volts 10000uf cap i have it pulsed through a transformer
and have colected the back end in two 400 volts 3300uf caps.
After douing this,i mesure the volts in the caps and i read 19.2 volts in both!
So i taugh,by having a switch with a magnet i can put them in seri and
come with 38.4 volts!!!!
I think it could be great for a Newman motor and put the colected volts
on the second phase of the rotation to have more torque.
Anyway!It's just an idea..
Great to see you brought this up!!!!

Maybe somebody else could comment this path..

Alain D.

[Aaron]

cap to cap

Yes, there is something interesting someone sent me in regards to this concept...almost, but it might put some of it in perspective. I'll ask if they don't mind me posting it.

[Sephiroth]

ok.... well here is the vidoe of the demonstration.

Please let me know if there is anything inaccurate in the vid so I can edit it!

Cheers!

YouTube - OverUnity Demonstration ... 12v + 0v = 15v???

[elias]

Yeah, I did this experiment too:
http://www.energeticforum.com/renewa...html#post13434

Elias

[Peter Lindemann]

No Violation Here!

ok.... well here is the vidoe of the demonstration.

Please let me know if there is anything inaccurate in the vid so I can edit it!

Cheers!

YouTube - OverUnity Demonstration ... 12v + 0v = 15v???

Sephiroth,

Great vid. It shows exactly what you are doing. I ran these experiments over 10 years ago and was quite confused by the results for a while. The problem of "conservation of charge" is not related to the voltage in the caps, but the total quantity of energy stored.

In the first case, you start with 12.4 volts in a 10,000uf cap. By the equation J=1/2CV2 (Energy in Joules equals one half times C (capacitance in Farads) times V2 (voltage squared). In this case, the stored energy is equal to 0.768 Joules. This is the starting point of the experiment.

When you connect this capacitor to an equal sized capacitor, the VOLTAGE looks conserved, as each capacitor ends up with half. Unfortunately, the total CHARGE stored is much less. Now we have the equivalent of a 20,000uf capacitor charged to 6.23 volts. By the same formula, we see that the total charge stored in your caps has DROPPED to 0.388 Joules.

By the same method, you can calculate the charge stored at the end of the second experiment, and you will find, it is still LESS than what you started with.

Its maddening, I know, but that's the way it is.

Peter

[Sephiroth]

Hi Peter,

Thanks for the feedback!

Yes, I understand it doesn't go over COP1 in joules

At the start of the experiment I had 0.7250125 joules in the capacitor

Afterwards I had 0.381938 + 0.262088 = 0.644026 joules

However, doesn't Charge Conservation relate to the coulombs and not the joules?

I started with 0.1245c

and ended up with 0.0874 + 0.0724 = 0.1598c

Which is a gain in charge, no?

If we look at the first example where I simply connect the two caps together I start with

0.1245c and 0.7250125 joules

then I had (approx)

0.1245c and 0.3969 joules

So I lost half my joules but conserved my charge.

But in the second experiment I start with the same

0.1245c and 0.7250125 joules

but end up with

0.1598c and 0.644026 joules

So I have ended up with a GAIN in charge and lost only about 11% of the total joules... I think that's pretty good

[Sephiroth]

could someone please correct me if I am wrong! Thank you!

[Peter Lindemann]

Wild Stuff.....

could someone please correct me if I am wrong! Thank you!

Sephiroth,

I am not an expert in these matters, so I am not saying you are wrong. We are just having a conversation.

This is what I see. In the first situation, you have a capacitor of 10,000uf. When you transfer energy to the second capacitor, it is obvious that transferring it THROUGH the inductor is more efficient. But I do not see a gain in charge. When you measure the voltage by putting the capacitors in series, yes, the voltage is higher. But the capacitors in series also means that voltage is in a capacitor of HALF of the original value, since two 10,000uf capacitors in series equals a capacitor of 5,000uf.

If you take two graduated cylinders, the first "two inches" in diameter and the second "one inch" in diameter, and place a cup of water in the first cylinder. The water will rise a certain height up the column. Now, if you transfer the water to the narrower cylinder, (and only spill a little bit of it...LOL) it obviously rises higher up the column (more voltage). It doesn't mean you have more water, it means you have changed the shape of your container.

I think the math is playing tricks on you because there is a change in reference that is not being taken into account. At no time during the experiment do you have the ability to do more work that you had at the start.

Peter

[Sephiroth]

sorry, peter, I wasnt trying to be argumentative... I'm just trying to understand!

So is my error in thinking that charge is cumulative?

so if I have a capacitor with 0.0874c charge, and another capacitor with 0.0724c charge, it doesn't mean that I have a total of 0.1598c charge?

That's pretty bad... since that would mean in my first experiment I lost half my charge AND half my joules...

[peper10]

Question for Peter.

Hi again Peter!!
I was just thinking about what you mention earlier about water transfer and
coulomb mesurements,but,something might hit me in the face later!!!!
Isn't truth that with the majority of motors(involve in Renewable Energy)
mostly run by low current (amps) and have better efficiency by the voltage raised?????
I don't want to be a pain in the ass but,just revised Ren window motor
and by reducing the amps seems to have no affect on the motor,almost same
speed,still have decent torque and seems to charge too..

So my question is: Can we use this finding for the motors (electricaly drive)
mostly mention in this forum,or ,i'm away in the WOODS.....

If you can comment it would be appreciate!!!

Alain D.

[thaelin]

Hi Seph:
I played around with this idea some time ago and found basicly the same results. Using relays to switch things, I tried to charge a battery up. I lost in the end.
I am in the process of setting up an experiment that will not only switch caps in and out but will be dischargeing them through a blocking oscillator to recapture with. I at one time had a 3 trifilar coil system running that initially drew 350ma and when I recirculated the bemf to the front, the draw from the feed batt went down to 14ma! This was using regulation on the caps to the next stage and then hooked into the input after a pair of diodes to keep it from feeding into the batt. I was very pleased with the outcome. I am looking to add this idea to a motor/charger system to see if I can get it to help.

I can post a diagram of the circuit if you would like. Very simple really but fun to play with.

thaelin

[Peter Lindemann]

sorry, peter, I wasnt trying to be argumentative... I'm just trying to understand!

So is my error in thinking that charge is cumulative?

so if I have a capacitor with 0.0874c charge, and another capacitor with 0.0724c charge, it doesn't mean that I have a total of 0.1598c charge?

That's pretty bad... since that would mean in my first experiment I lost half my charge AND half my joules...

Sephiroth,

Yes, the math says if you just connect the capacitors, like in the first example, HALF of the charge is lost and HALF of the Joules are lost! I have always wondered where they go, but they are definitely gone. That is why I said that making the transfer through the inductor was WAY BETTER! But still, there is a loss. Just not as much.

Your experiment, showing an increase in voltage, only happens when you put the capacitors in series. But the capacitors in series creates a SMALLER capacitor. If your charge was completely conserved, then the voltage in the capacitors, together in series, would have had to be 17.53 volts. You show that your voltage is only 15.71 volts. This shows that your total charge is less than when you started.

Peter