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Author Topic: Rosemary Ainslie COP>17 Circuit / A First Application on a Hot Water Cylinder  (Read 271452 times)

Offline Rosemary Ainslie

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Thanks Fritz.  That's a really comprehensive reply.  You're last para ?  That's depressing.  But I'm not sure that things aren't changing.  I'm already onto some electric car manufacturers and they just want to 'lighten' the battery load.  So there's some interest. But I get it that the 'art' of this is going to be related to the optimised 'rate of recharge'.  But that's a learning curve.  I still see zero emissions if this works - possibly again in conjunction with some solar cells to supply a trickle charge as required. 

And if we can get our hot water cylinder to cook by applying energy this way - then... the sky's the limit - or rather - it's NOT the limit.  LOL. 

Kindest regards,
Rosemary

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Offline fritz

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Hi Rosemary,

Traditional batteries are like smiling buddha;-)))
They perform perfect if they are full of charge and without load.  ;D ;D ;D ;D

This is why they don´t perform in electrical vehicle.
In my opinion - a well defined "compound" battery which fits the needs of an electrical verhicle has not been invented so far.
Its just adding up lots of general purpose cells. By having an electrolyte reservoir - it would be easy to have the proper trade off in terms of conversion capability and storage density. But this would not work with dry cells prefered for laptops and mobile use....
A tesla roadster is a nice bomb if it comes to the amount of spontaneus convertible energy available. Even the roadster would operate with only a fraction of that available conversion performance.
But there is a change and it will happen ;-)))


rgds.

Fritz

Offline nul-points

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    • Doc Ringwood's Free Energy blog
hi Rosemary

thanks for the feedback

have i understood correctly your thesis to predict that the heat generated by your 'RA Heater Circuit' will be anomalous compared to the 'conventional'  (I^2 x R x time) value as measured across the load, using standard equipment (eg. scopes, voltmeters, current probes, etc)?

if so, then presumably any testing for COP>1 needs to be done using calorimetry?

your own original accredited results showing  COP >> 1  appear to have been successfully done this way

the flip side of this requirement would be that conventional measurements are only going to show conventional energy conversion and gradual depletion of source battery

so in any future testing, there is little point in measuring volts and amps on the output side - although obviously it is still necessary (and fairly straightforward) to obtain the energy supplied from volts & amps input

we can then compare total (V*I*t Joules In) with total (Heat Joules Out)


(ASIDE - all this makes me wonder: why did the previous replicators use such 'heavy-duty' scopes and collect so much electrical data, if the excess output energy is thermal rather than conventional volts & amps?)


i hope these ramblings makes sense!

thanks
sandy
« Last Edit: July 27, 2010, 09:42:57 PM by nul-points »

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Offline Rosemary Ainslie

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Hello Sandy.  I saw your readcasts on Scribd.  Many thanks  ;D

The measure of energy dissipated at the load is impossibly complex.  We're inducing a large CEMF on an inductive/resistive load.  Energy measured here is traditionally computed as v^2/r - except on these switching circuits.  But v is changing and changing at a really fast rate.  I know Harvey attempted an analysis based on the resonance - but it was laughably inadequate and in any event intended to prove a zero gain.  I'm sure there are ways of doing this.  But - in as much as we're dissipating heat - then analysis according to temperature is empirical.  No one will argue the results. 

We needed that 'heavy duty' scope TDS3054C (my thanks to Tektronix) because we need to 'find' the required waveform across the resistor.  It holds the characteristic signature harmonic which is referred to as a 'shadow cycle' or 'preferred oscillation mode'.  Once this is evident we can confidently say that the system is in a kind of 'self resonating' frequency that invariably gives those gains.

Also, the assessment of current flow across the shunt needs to be acurately gauged.  Unlike the resistor, the shunt is non-inductive and therefore any measurements here are also correctly evaluated based on the ohmic values of the shunt.  The better the machine the less arguable the results.  So.  In short it's needed for the evaluation of the waveform across the load to determine the required resonance - and then for the voltage measurements acorss the shunt to evaluate the energy delivered by the battery.  Hope that makes it clearer.

But you're right that thesis predicts the level of 'heat' that is evident on the load.  The thinking is that when those 'binding' magnetic fields are broken - when they can no longer orbit - then they unravel.  They cascade out of the fast orbit and slow down to become apparent in our own measurable dimensions.  We see this as fire or we measure it as heat.  Effectively those little one dimensional fields lose velocity and gain mass (and temperature rise) in an inverse proportional relationship.  Effectively in the field orbit they're cold.  Outside of that orbit they're hot.  Which effectively means that the magnetic fields themselves hold a calorific value.  And that also means that 'fire' 'heat' 'light' are all a form of 'plasma'.  Not popular thinking as it is widely dismissed by mainstream.  My justification for this proposal is two fold.  In the first we know that fire does nothing to change the properties of the atoms that are associated with a fire (nuclear fire excepted).  It simply unbinds previously bound amalgams.  In the second - this would account for the total conservation of energy - which the thesis requires.  It has the added merit of explaining how fire can 'spread' accessing as it would - those binding fields in adjacent flammable materials.  In essense I'm proposing that fire is a 'plasma' - not just that 'chemical reaction which is well understood by mainstream'.  That's how this is usually dismissed.  LOL.

Hope that clarifies things.
Kindest as ever,
Rosemary
http://www.scribd.com/doc/33988924/DARK-MATTER-MFM

edited.  'except on these switching circuits'
« Last Edit: July 28, 2010, 02:01:20 AM by Rosemary Ainslie »

Offline Rosemary Ainslie

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Hi Rosemary,

Traditional batteries are like smiling buddha;-)))
They perform perfect if they are full of charge and without load.  ;D ;D ;D ;D

A tesla roadster is a nice bomb if it comes to the amount of spontaneus convertible energy available. Even the roadster would operate with only a fraction of that available conversion performance.
But there is a change and it will happen ;-)))
rgds.

Fritz

 ;D

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Offline SkyWatcher123

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Hi folks, these videos by gotoluc still have merit in my opinion and I think is similar in effect to Rosemary's circuit. All I can say is that I think gotoluc was convinced he had nothing special, when there is something special going on here.
http://www.youtube.com/profile?user=gotoluc#p/u/70/WsmPyUzZtgQ
http://www.youtube.com/profile?user=gotoluc#p/u/66/xvE7IGCra14
peace love light
Tyson

Offline Rosemary Ainslie

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Hi folks, these videos by gotoluc still have merit in my opinion and I think is similar in effect to Rosemary's circuit. All I can say is that I think gotoluc was convinced he had nothing special, when there is something special going on here.
http://www.youtube.com/profile?user=gotoluc#p/u/70/WsmPyUzZtgQ
http://www.youtube.com/profile?user=gotoluc#p/u/66/xvE7IGCra14
peace love light

Hello Tyson.  I love goto's work.  But that's work that was done over a year ago and we're not discussing apples and apples in this test.  He has not got inductive resistors - or not highly inductive ones as we require - and there's no self oscillating resonance - which is very much required.  But I do NOT want to detract from gotoluc's work.  I'm a very big fan of his.  Just not sure that this reference is in any way appropriate.  Not sure why you favour this over those videos that show this very clearly. ??

Regards,
Rosemary

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Offline Rosemary Ainslie

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Hi Guys,

I need to point out a quandary that I've been battling with.  I was rather hoping that someone would point it out.  Another question.  In that battery confirguration (post 75).  Take away the resistor and assume that during the 'on' period of any cycle the discharge battery is 24 volt to the recharge battery 12.  No impedance in the circuit other than a nominal voltage drop across the diodes.  All things being equal then the 24V source will discharge plus minus 2 amps = vi 48 watts.

Here's the kicker.  2 amps is supplied to a 12 volt battery.  Two amps * 12 volts (recharging battery) = vi 24 watts.  Net loss to the system is therefore 24 watts and in terms of conventional measurement protocols there's no useful point to this circuit or this excercise.  Every time that the 24 volt supply discharges it recharges the supply with 24 watts - it marginally heats up two diodes - and the remaining plus/minus 24 watts are lost?  Where?  It's not in heat at the supply batteries - because discharging batteries are NOT known to heat up.  The only impedance/resistance in the circuit is the recharge battery and IT only has 12 volts.  The assumption is that this will restrict the supply current to plus/minus 2 amps.  SO.  Where does that surplus wattage get wasted?  Technically it never reaches the recharge battery.  And the only thing connecting the two of them are some circuit wires. 

Hopefully someone can enlighten me.  At this stage I'm wondering if this circuit is simply pointing to a measurement paradox.  Either that or - as is more likely - I've made a blunder.  The one thing I would point out is that the 24 watts is not in the wires.  Those batteries are effectively in parallel.  And the maximum current flow discharge from the source batteries would be 2 amps. 

regards,
Rosemary

Offline gyulasun

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Hi Rosemary,

Perhaps the first thing to examine is the 2A current (true RMS, average, peak) value and how dependable the measurement was.

It may start from the switch's duty cycle that eventually controls the current value able to flow due to the 12V voltage difference between the batteries. Pulsed current measurements are not easy you surely know.

And in case it turns out the true current is maybe less than the 2A value, then it can only warm up slowly the inside body volumes of the batteries, making it hardly noticeable, especially in case the measurements lasted for only say 10-15 minutes. (I guess this time duration of course.)

rgds,  Gyula

PS:  Sorry for 'chiming' in.  I am convinced that your switched heating circuit with the inductive resistor can only be evaluated by calorofic measurements (Sandy nicely described this above.)

So this means an enclosure of the inductive resistor into an isolated oil volume, (into a thermos bottle filled with vegetable oil you use for cooking) and using a thermometer inside to note the temperature changes etc then compare the input DC energy consumption to the calculated heat energy.

This was already suggested by Dr Stiffler too last year, Peter Lindemann agreed with him then, but still nobody did it from the energetic forum decent team, instead they dealt with day long scope measurements and you know what.
Yes, I know that measurements, evaluating the waveforms are also important but where is a down to earth proof, a practical 'gadget' that utilizes the extra heat coming from somewhere?
I mean I put on the table two heat sources, I run both simultenously, I measure both input power, suppose they are about the same and then I find one of the heat sources gives out a definitely more, well sensible heat 'stream' than the other.  (Of course, the the heatsource performing better than the other would include your circuit.)
Sorry if such tests have already been done, I surely have missed them.

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Offline fritz

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Take away the resistor and assume that during the 'on' period of any cycle the discharge battery is 24 volt to the recharge battery 12.  No impedance in the circuit other than a nominal voltage drop across the diodes.  All things being equal then the 24V source will discharge plus minus 2 amps = vi 48 watts.

If we assume 3 identical 12V batteries  20Ah capacity, 25mOhm internal Resistance, identical charge condition (1+1 == 24V, 1 == 12V), I would expect the following:
Battery Voltage each 12.6V, 3times internal resistance = 75mOhm, Voltage loss diode+switch 2V-> voltage difference==10.6V, resistance 75mOhm
Well, that would give a theoretical value of 140Amps. The major impact will be the wiring and contacts involved (which is missing but would play the dominant role). I would expect something around 30 Amps.
The problem with 140Amps diodes is that they are not the fastest ones.

I would expect the transfered charge as u(diff)(10.6)*i(==u(diff)/(internal resistances+wiring)) *t(on)
->@ 320W (@30Amps) - adding some losses, heating up - we transfer maybe @250W

The diodes would heat up (@30Amps, 1V drop) with 30W each.

So I would assume that you blow any diodes and switches using 20Ah lead-acid with fast diodes at the first pulse.


Here's the kicker.  2 amps is supplied to a 12 volt battery.  Two amps * 12 volts (recharging battery) = vi 24 watts.  Net loss to the system is therefore 24 watts and in terms of conventional measurement protocols there's no useful point to this circuit or this excercise.  Every time that the 24 volt supply discharges it recharges the supply with 24 watts - it marginally heats up two diodes - and the remaining plus/minus 24 watts are lost?  Where?  It's not in heat at the supply batteries - because discharging batteries are NOT known to heat up.  The only impedance/resistance in the circuit is the recharge battery and IT only has 12 volts.  The assumption is that this will restrict the supply current to plus/minus 2 amps.  SO.  Where does that surplus wattage get wasted?  Technically it never reaches the recharge battery.  And the only thing connecting the two of them are some circuit wires. 

Do you use batteries with 2Amps short circuit current ?
I would expect that the heating up of batteries is the same on charging and discharging - only dependent from the current. On top of that you would have extra heating if the battery is full and you try to overcharge it - than heating up will increase dramatically. Same if you charge with too much current.

rgds, fritz.

BTW: You would achieve the maximum transfer if the internal resistance of the battery which is derived from the charge condition is identical. If a full battery charges an empty one - the current would be limited by the internal resistance of the empty one which would be definitely higher. If you charge the full battery with the empty one - again the high internal resistance of the empty one will limit the maximum current.


« Last Edit: July 28, 2010, 12:28:55 PM by fritz »

Offline gyulasun

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Hi Fritz,

You wrote a very good post, I agree with it in every respect and unfortunately I surely misunderstood Rosemary's earlier post by assuming the batteries are connected by a controlled switch. This is why I answered that way in my previous post.

Thanks,  Gyula
« Last Edit: July 28, 2010, 12:47:45 PM by gyulasun »

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Offline fritz

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a controlled switch would make sense  ;D
especially with an inductive load.

Offline Rosemary Ainslie

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Gyula and Fritz,  yet again thanks for the explanations.  I see where I was wrong.  There's far less resistance in the battery than I realised.  And however hard one switches - it's going to be increasingly difficult to keep that potential difference between the batteries as those switching cycles develop.  Therefore we would need to use that inductance on the resistor as per the usual circuit.  I've also had it on good authority that the two battery arrangement may not necessarily be any better than one.

That leaves us with the same problem.  How do we use a MOSFET which is known to be brittle at higher wattages.  One can put them in series - as explained by just about everyone.  But my own experience of this is disasterous.  One blown and the others follow in an expensive cascading series where we're left with nothing but that acrid smell of burnt metal (very distinctive). 

By the way Gyula - the preferred 'oil bath' calorimetric test is acknowledged as preferred.  It just was not really required as we were not measuring 'marginal' over unity results.  Had we been doing so then - of course - it would have been required to immerse the resistor in some sort of oil or somesuch liquid.  Those protocols that we applied to all our tests were designed by academics and considered as sufficient - albeit not one academic was prepared to attend a demonstration to accredit those results.  But their objections in those days - were largely based on the 'understanding' or, dare I say it, 'belief' that the circuit - however it was configured - would not be able to 'recharge' itself.  Our measurements and battery performance put paid to that 'belief'.  Indeed - as shown by Bedini et al - the battery rather prefers to be recharged with that heavy duty spike.  It's just that we didn't even bother with the a diode to take that energy back.  It seems that the only path required was supplied by the internal body diode of the MOSFET itself.  In any event.  Our learned and revered were entirely satisfied that a battery could not recharge itself - however it was configured.  They were wrong.  LOL

I was told by sundry transistor manufacturers that the cost of developing a MOSFET at the tolerances we require - would cost in the region of 500 000 Euros.  That's a little outside our budget.

I'm going to see if I can improve that 'give and take' battery configuration to see if I can find some solution.  I still feel that it's a means to get that extra energy.  But it will, undoubtedly, need to be more closely defined and designed.  Perhaps one can use it in conjunction with MOSFET - somehow - to get the benefit of both.  Anyway.  I see more sleepless nights ahead.  LOL

Thanks for your input.  Much appreciated.
Kindest regards,
Rosemary
http://www.scribd.com/aetherevarising

Offline Rosemary Ainslie

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Guys - this is a big appeal to you all.  I've been approached by a broadcaster to do a documentary on this 'over unity' movement that's gaining momentum all over the place.  I only really know my own circuit. It's a little off topic but I'd be very grateful if you could give me some names of all those 'inventions' that have been proven to give over unity.  Especially as it relates to motors.  Steorn is the one that springs to mind.  Are there others?  I'd quite like to forward this so that he can do the required investigation.

Thanks,
Regards,
Rosemary

Offline fritz

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If we assume 3 identical 12V batteries  20Ah capacity, 25mOhm internal Resistance, identical charge condition (1+1 == 24V, 1 == 12V), I would expect the following:
Battery Voltage each 12.6V, 3times internal resistance = 75mOhm, Voltage loss diode+switch 2V-> voltage difference==10.6V, resistance 75mOhm
Well, that would give a theoretical value of 140Amps. The major impact will be the wiring and contacts involved (which is missing but would play the dominant role). I would expect something around 30 Amps.
The problem with 140Amps diodes is that they are not the fastest ones.

I would expect the transfered charge as u(diff)(10.6)*i(==u(diff)/(internal resistances+wiring)) *t(on)
->@ 320W (@30Amps) - adding some losses, heating up - we transfer maybe @250W

The diodes would heat up (@30Amps, 1V drop) with 30W each.

So I would assume that you blow any diodes and switches using 20Ah lead-acid with fast diodes at the first pulse.

To be correct - I would estimate these figures for current pulses >5ms.
In your circuit you have pulsed DC with  500us pulse duration.

I don´t know your batteries - nor do I have my own data or data from a battery manufactorer at hand so - nothing we can base on.

If we reach down to 1us - I would estimate the internal resistance in the area of few ohms. For 500us - a current from 2-5Amps would sound reasonable.

This is why I heavily suggest to determine the AC input/output resistance of batteries used with such setup.

How to determine AC impedance of a battery ?
Use 80´ties DC coupled HIFI-AMP+sinus generator+output capacitor.
(or industrial servo amp or similar)
Determine the internal resistance as function of frequency using low inductive precise test load (1Ohm).
Measure damping as function of frequency if feeding to battery terminal.
The battery impedance can then be calculated from that damping function corrected with the values from the 1Ohm test.

rgds.

fritz

 

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