http://uk.youtube.com/watch?v=1g3DwpshYxU
Omega thankyou pulsed energy changes the rules Chet
No ... :D
Its obviously not OU, but can anyone explain the increased voltage ?
hi all
he may not have achieved OU energy - yet - but he's going along the right path!
i - and others - (including NerzhDishual on this site) have also found the same charge anomaly when switching charge from one capacitor to another
a charge imbalance by definition will involve an energy imbalance
an interesting follow-on question is: "is it possible to cause sufficient charge imbalance, due to this anomaly, that the associated energy imbalance becomes overunity"
and the answer is...
...yes it is!
i've uploaded details of my experiments with some switched capacitor circuits which produced initial charge gains similar to those shown in this video - also added details of further experiments which went on to achieve around 120% energy converted by the circuit compared to total energy supplied (all energy input from charged capacitor)
http://ringcomps.co.uk/doc
some results & general circuit, copied from my site, are posted on this site (by Overunity.com member Feynman) at:
http://www.overunity.com/index.php/topic,4419.msg96175.html#msg96175
my parts list & schematic are posted on this site at:
http://www.overunity.com/index.php/topic,4419.msg97603.html#msg97603
(replies #25 & #26)
the 'extra step' needed to achieve OU is in making sure that the energy used to charge the 'output' capacitor doesn't get dissipated as heat or other radiation without doing sufficient 'useful' work first
if you don't 'use' the charging energy you're losing half of what you supply as input
even if you can arrange your circuit to make use of this other 50% of your input energy, you could only expect to approach the 'theoretical' maximum of 100% of input
the neat thing is that by switching the charge from input to output, as several people have found, there is a net gain of charge in the circuit and this enables the 'efficiency' to exceed 100% - even after the usual losses!
unexplainable by conventional Electrodynamics - totally permissible in a universe which is permeated by an energetic 'lattice' of primordial particles - the 'aether' - the vacuum medium - zero-point energy - as understood and accepted by both Maxwell and Einstein
have a nice day!
sandy
Doc Ringwood's Free Energy site http://ringcomps.co.uk/doc
Hi all,
There is no anomaly in what it was described.
Everyone can do some simple math for himself/herself, based on the energy stored in a capacitor: E=C*U*U/2
6V final into 2 identical capacitors starting with 1 capacitor at 12V is not unity at all but 50% efficiency and besides that there is maybe just not quite enough education/experience for the claimer.
Fact: every time one charges a capacitor from a constant voltage source, half of the energy is lost (as heat). If one calls it unity, then when later discovering the correct (isentropic) charging process, it will be amazed by an apparent 200% efficiency, which is obviously not the case.
Have a nice day,
Tinu
@Tinu
i think you'll find the 12V initial charge making two charges of 6+V was just showing you what happens when you connect the two capacitors together manually without the 'charge-switching' circuit
the title text before the vid clearly states "12V + 0V = 15V ?"
it is not necessary to lose the initial 50% energy, as i mentioned in my previous post, if you modify the circuit to make use of the charging energy - my circuit does this and converts at least 20% more energy than the total energy used from the input capacitor
this is not being 'amazed by an apparent 200% efficiency' just by preventing energy loss - this is real energy gained over the total supplied
let's look at the charge: in one of my first experiments i started with a total initial charge of 2 Coulombs - and finished with a total charge of 2.9 Coulombs - this is not an imaginary gain! ;)
ok, now let's look at the energy: the total initial energy stored in the input capacitor is 9.57 Joules, the final energy stored in the input capacitor is 7.33 Joules
therefore, total energy input to the circuit: 2.24 Joules
energy converted by the circuit:
0.25 Joules used by the switching oscillator
1.22 Joules passed thro' load resistor to charge up output capacitor
1.22 Joules discharged thro load-resistor from output capacitor at end
Total energy converted by circuit: 2.69 Joules
Energy Quotient of circuit: 2.69/2.24 = 1.2
and the charge anomaly that accompanied this result:
start charge: 2.39 Coulombs
final charge: 3.0 Coulombs
...this anomaly can provide very real overunity!
have a nice day
sandy
Doc Ringwood's Free Energy site http://ringcomps.co.uk/doc
Quote from: nul-points on August 07, 2008, 02:16:47 PM
let's look at the charge: in one of my first experiments i started with a total initial charge of 2 Coulombs - and finished with a total charge of 2.9 Coulombs - this is not an imaginary gain! ;)
Total charge is zero. It always was and it will remain so. Charge conserves until otherwise proved.
If you want to talk about charge on ONE plate of a capacitor, that?s a different story and it is not relevant as long as you decouple it from potential in respect to the other plate. So, you have to decide if you want to connect the final capacitors in series (in which case you?d have indeed 15V but only half ?charge?) or in parallel (in which case you obviously don?t have 15V any longer).
Quote from: nul-points on August 07, 2008, 02:16:47 PM
ok, now let's look at the energy: the total initial energy stored in the input capacitor is 9.57 Joules, the final energy stored in the input capacitor is 7.33 Joules
therefore, total energy input to the circuit: 2.24 Joules
energy converted by the circuit:
0.25 Joules used by the switching oscillator
1.22 Joules passed thro' load resistor to charge up output capacitor
1.22 Joules discharged thro load-resistor from output capacitor at end
Total energy converted by circuit: 2.69 Joules
Energy Quotient of circuit: 2.69/2.24 = 1.2
I can?t comment on energy estimations but how can you be sure that figures are correct?
Quote from: nul-points on August 07, 2008, 02:16:47 PM
and the charge anomaly that accompanied this result:
start charge: 2.39 Coulombs
final charge: 3.0 Coulombs
...this anomaly can provide very real overunity!
I wish it also.
Electrolytics have memory and recover a large amount (several good percents and often much more!) of initial energy when discharged. In fact, I think both apparent surplus charge as well as energy as in the above quotes are due to this memory effect. It?s not free energy neither OU: they reside in your power supply (or battery).
Cheers,
Tinu
Quote from: nul-points on August 07, 2008, 02:16:47 PM
...this anomaly can provide very real overunity!
Isn't it ironical ?
You know the secret of free energy and you are still paying electricity bills and running your car on oil.
Quote from: tinu on August 08, 2008, 09:48:16 AM
Total charge is zero. It always was and it will remain so. Charge conserves until otherwise proved.
If you want to talk about charge on ONE plate of a capacitor, that's a different story and it is not relevant as long as you decouple it from potential in respect to the other plate
i think most forum members will be comfortable with the concept of 'charging' a capacitor by applying a voltage across its terminals
so when i report that the total initial 'charge' in the capacitors was 2 Coulombs, and the total final charge was 2.9 Coulombs, people can come to their own conclusions about the implication of that
my observation - from actually doing the experiment, not just talking about it - was that the total amount of Coulombs stored in the capacitors had increased without any further application of voltage across the capacitors from external circuitry
Quote from: tinu on August 08, 2008, 09:48:16 AM
I can't comment on energy estimations but how can you be sure that figures are correct?
anyone who is interested enough can see the measurement readings on my website which i linked above (and which is also given at the end of this post) - there are also some FAQs on the site answering questions i've received about possible ways in which errors could have been introduced into the results
Quote from: tinu on August 08, 2008, 09:48:16 AM
Electrolytics have memory and recover a large amount (several good percents and often much more!) of initial energy when discharged. In fact, I think both apparent surplus charge as well as energy as in the above quotes are due to this memory effect. It's not free energy neither OU: they reside in your power supply (or battery).
i agree, there is a well known 'memory' effect easily seen with larger valued capacitors where, after a capacitor has been discharged, it self-recharges to some extent
this effect is not confined to modern Electrolytics - there are reports dating back to the 1920s recording self-recharge on capacitors using Rochelle salt crystal as a dielectric
i've noted in my description of the experiments that self-recharge was observed - however, this effect takes place after the discharge of the input & output capacitors - and by this time the experiment has finished and the measurements already taken
i am interested to know what is the difference in energy between that in a capacitor that i have 'charged' to voltage 'V' from a battery and the same capacitor which has self-recharged to voltage 'V' after a full discharge
i ask this because if there is NO difference then even 'self-recharge' is free energy!
if there IS a difference between the two cases then how can we ever hope to calculate the energy stored in a capacitor?
i'm not aware of ever having seen TWO equations for calculating the Coulombs stored in a capacitor - one for battery/PSU applied voltages and a different one for self-recharged voltages
Quote from: tinu on August 08, 2008, 09:48:16 AM
It's not free energy neither OU: they reside in your power supply (or battery).
i have to disagree - the additional energy does not "reside in my power supply (or battery)" - it enters the circuit from some external source (energetic medium of space? aether? zero-point energy?) during the operation of the circuit which is not connected to a power supply or battery
all i can say is that once my experiment starts running the only energy which i supplied is contained in the input capacitor - and when the experiment stops more energy has been converted than is drawn from that capacitor
it's not rocket-science to confirm this anomaly - i'm aware of several people (Introvertebrate included) who have put together a handful of simple components and seen the charge anomaly effect for themselves
Quote from: Omega_0 on August 08, 2008, 11:22:32 AM
Isn't it ironical ?
You know the secret of free energy and you are still paying electricity bills and running your car on oil.
you can make fun of me all you like Omega_0, that doesn't alter the fact that this anomaly exists and people are starting to pick up on it
i started these experiments with the switched capacitors in March of this year - i think i've been very fortunate to get to a point where i can record Energy Quotients greater than 1 in that time
i'm following it through now with further investigations of the effect with a view to scaling up the power, looking for practical confirmation of excess energy and developing my basic test circuit to make it a continuous process
but then, that's me - i'm just a 'get on and do it, kinda guy', rather than a 'tell other people it's impossible' kinda guy ;)
all the best
sandy
Doc Ringwood's Free Energy site http://ringcomps.co.uk/doc
Sorry no OU here. Voltage alone is not a measure of an increase in energy, you must consider the capacitance which is now one half since your capacitors are now in series.
(remember the formula for two equal value capacitors in series ? Look it up or just divide one of them by two.)
Your original charged cap started out with more energy (not voltage) than the final two capacitors in series. Even though their final combined voltage was higher their effective capacitance was halved by the act of putting them in series. Energy (Joules) =1/2 CV^2.
Many starters make this fundamental error.
Quote from: ORION on August 08, 2008, 10:18:18 PM
Sorry no OU here.
the only thing
not happening 'here'... is a considered inspection of other people's experimental results! ;)
in the case of Introvertebrate's test, he clearly states that he doesn't have a net increase in energy - but the increase in voltage is in violation of conservation of charge in his closed circuit
in the case of my experiments, i measure a clear excess (about 50%, in one test) of total Coulombs stored in the circuit after a test run
if anybody bothered to do the Coulomb calcs for this anomaly they would find that the total Coulombs stored on the two caps at the end of the test are greater than the total Coulombs stored at the start of the test
charge conservation has been violated
Quote from: ORION on August 08, 2008, 10:18:18 PM
Voltage alone is not a measure of an increase in energy, you must consider the capacitance which is now one half since your capacitors are now in series.
i agree that it's of no value to connect the two capacitors together at the end and say, 'look, more voltage' - that's why i don't do it
it's the increased total charge in the
system which is important (i'm sure Introvertebrate understands that)
this is not about capacitors in series doubling the voltage & halving the capacitance - the capacitors are connected by a switched circuit in these tests - not a piece of wire
in order to store Coulombs of charge into a capacitor, work must be done - if work is done, then energy is expended
if you've stored more Coulombs than you expected, you must have used more energy than you expected
a charge gain anomaly will always be accompanied by an energy gain anomaly
this has happened! on a workbench near you! to Introvertebrate - to NerzhDishual - to me - to people in other OU forums
these are real measured events
- we could deny they ever happened, and hope they go away
- or we could investigate further ...because we've found something unexpected ...because it could be useful
so - the big question is... is it possible for this 'unexpected' charge gain to be sufficiently high to enable the total energy use of the circuit to go overunity?
the results indicate: yes, it is!
in the case of my test circuit (which uses inductance and a resistive load to capture the usual 50% energy loss - a technique sometimes used in switched-mode power supply design) i'm measuring similar gains of Coulombs stored - plus excess energy converted by the circuit over total energy supplied
this is not theory - these are measured results
if anyone has any issue with this it will have to be on the basis of the experimental setup and measurements
** health warning: this will involve some reading, thinking and calculating ** :)
Quote from: ORION on August 08, 2008, 10:18:18 PM
Your original charged cap started out with more energy (not voltage) than the final two capacitors in series. Even though their final combined voltage was higher their effective capacitance was halved by the act of putting them in series
Many starters make this fundamental error.
i'm sure they do
i don't connect my capacitors in series - i use a switched-charge circuit ;)
this is not a time for the textbook - this is a time for the workbench!
...we can amend the textbook later
all the best
sandy
Doc Ringwood's Free Energy site http://ringcomps.co.uk/doc
Quote from: nul-points on August 09, 2008, 03:42:42 AM
in the case of my experiments, i measure a clear excess (about 50%, in one test) of total Coulombs stored in the circuit after a test run
if anybody bothered to do the Coulomb calcs for this anomaly they would find that the total Coulombs stored on the two caps at the end of the test are greater than the total Coulombs stored at the start of the test
charge conservation has been violated
Everything is normal in your description exactly the way it should be less the last line, which is false. In a general setup if you step down the voltage, you?ll certainly increase the current, hence the ?charge?.
No offense, but you do not understand what charge conservation is. Please read more about it, otherwise you may create a lot of confusion.
Quote from: nul-points on August 09, 2008, 03:42:42 AM
if you've stored more Coulombs than you expected, you must have used more energy than you expected
Not necessarily at all!
See also the above explanation.
Cheers,
Tinu
QuoteI am interested to know what is the difference in energy between that in a capacitor that i have 'charged' to voltage 'V' from a battery and the same capacitor which has self-recharged to voltage 'V' after a full discharge
i ask this because if there is NO difference then even 'self-recharge' is free energy!
Dielectric absorption, the electret effect and PVDF piezo films are probably worthy of further investigation.
Quote from: tinu on August 09, 2008, 08:08:24 AM
No offense, but you do not understand what charge conservation is.
no offense taken :)
charge conservation can be viewed as an empirical law by considering the 'continuity equation' in electromagnetic theory
the continuity equation provides a relationship between the electric current density (Amps / square metre) and the charge density (Coulombs / cubic metre)
the equation basically states that if charge is leaving (or entering) a certain enclosed space then the rate of change of the remaining charge density varies with the opposite sign - ie. remaining charge density decreases with a removal of charge, and it increases with an addition of charge
- if charge density has increased in a closed system it must be as a result of charge entering the system from 'outside';
- if charge density has decreased in a closed system it must be as a result of charge leaving the system and flowing to something 'outside'
hence charge conservation: if no external flow of charge (in or out) occurs then the total charge density in a closed system must remain constant
what's not to understand? ;)
Quote from: tinu on August 09, 2008, 08:08:24 AM
Please read more about it, otherwise you may create a lot of confusion.
i believe that other forum members, who are sufficiently interested in these anomalies, are perfectly capable of verifying for themselves any statements which i - or you - make
Quote from: tinu on August 09, 2008, 08:08:24 AM
Quote from: nul-points on Today at 07:42:42 AM
if you've stored more Coulombs than you expected, you must have used more energy than you expected
Not necessarily at all!
absolutely at all! :)
Q (Coulombs charge) = V (volts) * C (capacitance)
more voltage on the capacitor -> more Coulombs charge stored
E (Joules energy) = 1/2 * C * V * V = 1/2 * Q * V
more Coulombs charge stored -> more Joules energy stored
E (Joules energy) = W (Watts work) * t (time)
more Joules energy stored -> more work done
so - as i said above, if you've stored more Coulombs than you expected, you've also used more energy than you expected (more stored plus more as work done)
a charge gain -> an energy gain
a charge gain anomaly -> an energy gain anomaly
Quote from: ORION on August 09, 2008, 09:38:21 AM
Dielectric absorption, the electret effect and PVDF piezo films are probably worthy of further investigation.
i agree
Quote from: nul-points on August 08, 2008, 07:57:20 PM
Quote
i am interested to know what is the difference in energy between that in a capacitor that i have 'charged' to voltage 'V' from a battery and the same capacitor which has self-recharged to voltage 'V' after a full discharge
i ask this because if there is NO difference then even 'self-recharge' is free energy!
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
all the best
sandy
Doc Ringwood's Free Energy site http://ringcomps.co.uk/doc
Assuming this is for real, what is the difficulty in arranging a series of switches, mechanical or solid state, that can make the anomaly into a continuous process rather than a one shot affair?
Has anyone gotten down to building such a device to harvest the energy on a continuous basis?
Shuttling the capacitors around so that the energy or charge keeps building?
What is the roadblock here?
it's in progress, Orion
Quote from: nul-points on August 09, 2008, 06:30:00 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. 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? ;)
Quote from: nul-points on August 09, 2008, 06:30:00 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.
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
@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
Quote from: tinu on August 10, 2008, 02:37:20 AM
Quote from: nul-points on August 09, 2008, 10:30:00 PM
Quote
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
Quote from: tinu on August 10, 2008, 02:37:20 AM
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
Quote from: tinu on August 10, 2008, 02:37:20 AM
So, the electrolytic is taking self-recharging energy in advance from your power source, as long posted above
nope - see my previous answer
Quote from: tinu on August 10, 2008, 02:37:20 AM
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 :)
Quote from: tinu on August 10, 2008, 02:37:20 AM
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: tinu on August 10, 2008, 02:37:20 AM
Quote from: nul-points on August 09, 2008, 10:30:00 PM
Quote
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!"
Quote from: tinu on August 10, 2008, 02:37:20 AM
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
Quote from: tinu on August 10, 2008, 02:37:20 AM
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? ;)
Quote from: tinu on August 10, 2008, 02:37:20 AM
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
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.
hi Orion, thanks for the interest - PM on its way
Hmm hes connecting x devise in series with other capacitor. Effect is : current is passing device x.
By the way devise x got a coil, what can increase voltage. Wheres trick?
I was wondering about Introvertebrate's video - if maybe there was some charge (pre-stored) in the oscillater circuit that could have affected the test results. If not then....
As I understand it.....(during the test results) both caps showed about 12 V when adding their seperate voltages together (not in series) after discharging to each other...
Next both caps showed about 15 V when adding their seperate voltages together after going through the bendin circuit... also the same results when connected in a series. Please help me understand!!!!
If not OU, how is the bedini circuit changing the voltages of the caps?
Bill