OverUnity Demonstration ... by Introvertebrate

[nul-points]

it's in progress, Orion

[tinu]

it appears that no-one can provide any difference...

...in that case, i suggest that capacitor 'self-recharge' (aka Dielectric absorption) can be considered as free energy also

I missed that part. Probably you should?ve found the answer through experimenting already, all by yourself. Nevertheless, the answer is: energy taken from power supply/battery is much greater during the charge of an electrolytic than that theoretically required. Anybody can check it using an oscilloscope or, even without one, by taking enough I(t),U(t) values, by plotting them and by integrating the area. So, the electrolytic is taking self-recharging energy in advance from your power source, as long posted above.
In addition, one may measure the leakage current for an electrolytic, which is quite large also. For that, enough to keep the capacitor connected to the power supply and measure I(t) while U=ct (voltage is that of the source, t>>5*R*C). Where do you think this large energy goes?
All in all, electrolytics function somewhat similar with a rechargeable battery but the energy given back (the memory effect) is significant mostly after an abrupt discharge and, in general, the effect decreases with time (i.e. the capacitor does not keep the ability to give back its initially stored energy for a very long time, pretty much like an old/bad rechargeable).

Is a rechargeable battery OU? 

more voltage on the capacitor -> more Coulombs charge stored

Correct. But the voltage part was initially completely missing there. You may read again your original statement.

All in all, for an isentropic transfer of energy from one capacitor to the other identical one and constant voltage for both at the end, starting with 12V one may get close to 8.48V on BOTH capacitors. This is unity. Did you actually get OU? I can?t see it from the movie?

Cheers,
Tinu

[tinu]

@Sandy,

I couldn?t read the page until now because I thought it?s dead (the original link doesn?t work). Finally, I added the www and it worked: http://www.ringcomps.co.uk/doc/

I?ll read it all, but at a first glance I can tell you really have to work on the charge conservation issue. Here it is an example: take a small battery and charge one capacitor. Charge another one. Keep going. I bet you don?t have enough capacitors around to deplete the battery. Now: what is the original charge of the battery? And what?s the final charge stored by capacitors? See?! I know you said no offense taken although immediately afterwards you show a bit offended but I have to say it again: think more on charge conservation. Look at simple examples until it is crystal clear to you. Right now you are way off base and there are only two cases imho: those who know will dismiss your entire work at once because of it and those who don?t know are terribly confused.

Cheers,
Tinu

[nul-points]

Quote from: nul-points on August 09, 2008, 10:30:00 PM

it appears that no-one can provide any difference...

...in that case, i suggest that capacitor 'self-recharge' (aka Dielectric absorption) can be considered as free energy also

I missed that part. Probably you should've found the answer through experimenting already, all by yourself

you're right - i've already come to a conclusion about that answer through my tests

i was inviting other people to consider the question for themselves - people who currently believe that the electrical energy imparted into a capacitor by self-recharge (dielectric absorption) is any way a different kind of energy to that which we can store in the capacitor ourselves, from a battery or other DC source

ie. does it have the same Joules per Coulomb stored? can it power the same loads? etc

Nevertheless, the answer is: energy taken from power supply/battery is much greater during the charge of an electrolytic than that theoretically required. Anybody can check it using...

the 'greater .. than .. theoretically required' energy taken from a supply to charge a capacitor is actually the work required to charge the capacitor - as mentioned in my earlier posts above:-

> "if you don't 'use' the charging energy you're losing half of what you supply as input"
> "to store Coulombs into a capacitor, work must be done"

i also mentioned that it can be avoided - as sometimes done in switched-mode power supplies - and my test circuit uses inductance and a resistive load to reclaim that loss

if you can avoid the loss up front then it can't be available for self-recharge later, after a discharge - therefore, self-recharge has to be a different phenomenon

So, the electrolytic is taking self-recharging energy in advance from your power source, as long posted above

nope - see my previous answer

In addition, one may measure the leakage current for an electrolytic, which is quite large also. For that, enough to keep the capacitor connected to the power supply and measure I(t) while U=ct (voltage is that of the source, t>>5*R*C). Where do you think this large energy goes?

i think the leakage current goes into heating my room 

Is a rechargeable battery OU? 

if a rechargeable battery can provide energy due to dielectric absorption (and i've no idea if it can) - which is extra to the initially-stored chemical energy and any subsequent personally-applied re-charge electrical energy - then that would be free energy in my book 

Quote from: nul-points on August 09, 2008, 10:30:00 PM

more voltage on the capacitor -> more Coulombs charge stored

Correct. But the voltage part was initially completely missing there. You may read again your original statement.

my original statement said: "if you've stored more Coulombs than you expected,you must have used more energy than you expected"

i'm sure most attentive forum members will understand that the Coulombs stored (Q) are proportional to the Voltage (V) on the capacitor (Q = V * C)

i gave the process in more detail the second time (showing the full 3 step relationship between increased volts and increased work done) because your response to the statement above was: "Not necessarily at all!"

All in all, for an isentropic transfer of energy from one capacitor to the other identical one and constant voltage for both at the end, starting with 12V one may get close to 8.48V on BOTH capacitors. This is unity. Did you actually get OU? I can?t see it from the movie

i didn't get OU in the movie - it's not my movie, it's Introvertebrate's

you remember, we were discussing his experiment way back in the thread 

i mentioned things like:-

> "he may not have achieved OU energy - yet - but he's going along the right path!"

and

> "in the case of Introvertebrate's test, he clearly states that he doesn't have a net increase in energy"

...i'm just teasing - i realise your mistake with the movie

i think you've looked at my site now - so you'll have seen that the results of my experiments indicate overunity efficiency

I couldn't read the page until now because I thought it's dead (the original link does't work). Finally, I added the www and it worked: http://www.ringcomps.co.uk/doc/

http://ringcomps.co.uk/doc/

http://www.ringcomps.co.uk/doc/

both work for me


...did you pay your last electric bill?

I know you said no offense taken although immediately afterwards you show a bit offended but I have to say it again: think more on charge conservation

really, tinu - no offense taken - if only you could see my smile!


well, we both seem to have presented our cases sufficiently - i think we can now finish this discussion and let the jury decide

...all they need is a couple of capacitors, scope/DVM, and a switching circuit - oops, and i nearly forgot that rechargeable battery !


i'll let you have the last word because  ...well, just because that's the kind of guy i am 

all the best
sandy
Doc Ringwood's Free Energy site  http://ringcomps.co.uk/doc

[ORION]

The real pleasure here is seeing a couple of experimenters present their cases and disagree without having to flame each other. This is the mark of gentlemanly scientific attitude.

The final proof, when all is said, will be in the pudding, the harvesting of the free energy.

Can you supply some details or proposed  schematic on this work in progress? Very curious?

Thanks in advance.