COP 17 Heater | Rosemary Ainslie

[Aaron]

bcm

Rosemary, I've already done this by saying in a much earlier post that if a good quality BCM is used to measure battery capacity before and after the test, a system running considerably overunity as yours is claimed to do, will be obvious. There is no need for complicated and time consuming test procedures. If you really are running at COP17 (thats 16 pints back into the fridge), then you should easily self-run the system. Aaron should be able to make this happen and video it for everyone.

Hoppy

BCM? .

You really need to get in the real world and do EXPERIMENTS. If you think BCM gives you a real answer, you have completely and totally disqualified yourself from knowing anything about batteries. Battery capacitor analyzers never reflect what the battery will be able to deliver. If you think they do, go do it, film it and show everyone.

You want to bastardize a real test procedure by even hinting at using a BCM to tell the story? LOL, you have got to be kidding!

TIMEd draw down on the battery as I described from batteries charged by a charger that does know when the battery is really full is MANY times more accurate that your recommended BCM test!

Also, you clearly do NOT understand open systems. Make this particular circuit closed loop to run itself? Are you kidding me? It is not over 100% efficient, it is over 1.0 COP. There is dissipation but the dissipation of source potential simply dissipates over a much longer period of TIME! Hoppy, I think you're losing it here. Most of the free work is constant heat production...I'm supposed to magically make that heat make up for the loss in the battery so the battery never goes down. Please tell me your joking.

No need for time consuming procedures. Yeah, lets not do real tests and experiments and take bogus shortcuts that won't tell anything. If that is your idea of science and testing, then you should really stick to something you can demonstrate proficiency at. Test procedures are NOT it.

Even with a BCM, do you not still have to draw down the battery in order to charge it back up to put a BCM to see if it is at full charge? So we magically wave our hand and pretend we depeted the battery, bypassing all time requirement then apply a charger, wave our hand and pretend it just got charged and then put a BCM on it? lol You even mention "before and after" the test but "There is no need for complicated and time consuming test procedures." What is this Orwellian double speak?

Doing that will certain eliminate long drawn out test procedure but I think I'll do it my way. Thanks for the advice but I'll stick to doing real tests.

This takes the cake hoppy. I was open to hearing what you had to say - please no more distracting suggestions.

[Aaron]

assuming

But I can see now how you get high COP!

Hoppy, stop assuming things for others and putting words in people's mouths. You take things out of context constantly as all the skeptics do. No more or you're not welcome in this thread anymore.

[Aaron]

measurements

You take the voltage across the resistor / resistor Ohm to get the Amp and you multiply by the battery voltage seriously ? you kidding us right ?

you said 1.2V across the resistor , so 1.2/0.25 = 4.8A and you multiply the 4.8A by the batterie voltage is just a non sense, im sorry but i cant beleive you wrote that. Its Voltage across resistor MUTIPLIED BY the Amp (Vr/Ohm) = The number of watt.

Wait, are you serious EgmQC?

What in the world do you think that 1.2V across the shunt resistor represents? About 24-25 volts from the battery! The entire point is that with the battery voltage moving over that resistor, at a certain resistance, it will have a corresponding potential difference across the resistor. Based on THOSE calculations on the "shunt" it tells what current is associated with what is leaving the power source.

The non-sense is actually what you have posted.

And Hoppy
"Rosemary, With respect, you really do need to properly understand the basics of EE before you launch into preparing valid test routines for your circuit. Hoppy"

It appears she is more qualified than you or Egm as you apparently don't even know how to measure what is delivered by the battery as evidenced by your agreement with Egm's analysis. Stop your ridiculous insults NOW.

[Aaron]

subversive activities

This was the original statement that Rosemary made which started the ball rolling and from there things appear to have got confused: -

TK - I'm hoping I can bend your mind around this problem.

The battery recharges, power through, voltage first drops - then a spike to, what was it - say 50 volts or thereby? At that same moment the value across the shunt say 0.4volts positive drops to about 1.2volts negative, (aproximate because I couldn't see the actual value). Then how do you work out the product of the energy available at that moment? In my reckoning it is 1.2/0.25 = plus/minus 4.8 amps. So. I need to be reasonably certain that the actual energy calculated at that moment as v*i = 240 watts BACK TO THE SYSTEM. (Again not shouting. Just emphasis)

Hoppy

Nice try but no cigar. Please don't pretend you could have mistaken multiplying the current by the voltage over the shunt resistor! It was specifically stated by Rosemary to multiply by battery voltage, you agreed with EGM. There was no confusion about what was being discussed.

No more subversive garbage.

[Aaron]

test procedure

HH is right but now I've had time to think in more depth about the proposed test procedure, lets look more closely at this. Rosemary says: -

OK this is how I would suggest the test be conducted if there are no storage scope meters available.

Set the duty cycle and check the temperature of the resistor when it's
stable.

I assume this is done using the test battery?


Then apply the same resistor to a variable power supply and adjust the voltage until the same temperature is found and stable over the same resistor

Then do a v^2/r analysis to determine the wattage dissipated at the start of the experiment.

So correct me if I'm wrong, but the 'control' is actually based on the stable temperature of a resistor originally heated by a battery that will be used in future stages of the test and therefore very likely to have a different capacity and condition as the test progresses. If this is the case, then how can this be considered a control?

Then record the start time to run the experiment until the battery is depleted to say, 11 volts from a 12volt supply or 22 volts from a 24 volt supply.

Don't you mean total run time?

I assume to do this stage of the test, the battery used to establish the control dissipation wattage must be re-charged. How does the test procedure ensure that the energy placed into the battery during recharge does not alter the total energy contained in the battery prior to the control wattage being established? I mean, how do you know when to stop charging given that a battery requires considerably more charging energy to replace that taken on discharge?

Then recharge the batteries and apply a resistor in series with them to draw down the same amperage as recorded at the start of the experimental test.

The same amperage? Don't you mean battery terminal voltage as this is all I read above as being recorded as a reference point?

As above, how does the test procedure ensure that the energy placed into the battery during recharge does not alter the total energy contained in the battery, this time prior to the first experiment being run?

Then recharge the batteries and apply a resistor in series with them to draw down the same amperage as recorded at the start of the experimental test.

As above, how does the test procedure ensure that the energy placed into the battery during recharge does not alter the total energy contained in the battery, this time prior to when the series resistor was applied?


Then rerun both tests.

Does this include doing a new v^2/r calculation?

If you've got two sets of batteries - run them concurrently until one or other hits that critical voltage level. Then recharge both and swap them, control to experiment and vice versa.

How does the test procedure ensure that a second set of batteries would have produced the same v^2/r control result as the first set of batteries?

Please note that of the above questions don't even address the battery conditioning issue which I've now dropped.

Hoppy

Temp is based on what the Ainslie circuit runs at on the battery.

If you actually did the tests yourself, you would see that over many hours, the Ainslie circuit keeps the resistor at a very stable temperature. Over that many hours, any deviation from the control is insignificant.

You make many assumptions on battery charging. I'll make it easy for you. I'm using a 1AU charger from Energenx. All the Energenx charges takes live readings of the battery on EVERY pulse and based on its programming, it knows exactly when the battery is back to its full capacity and then stops charging. You have absolutely no idea how smart these chargers are as they are the smartest chargers on the planet hands down no contest. For one, you assume it must be total run time but let me tell you what works. Please do not debate this unless you're willing to buy one Tesla Chargers | World's Most Efficient, Effective & Advanced Battery Chargers and do the tests yourself to see.

You clearly don't understand how to do an honest battery run test. I'll tell you how. If the battery is full charged to 25.50v with the 1AU for example.... You can take total time into account but what is more accurate is to run the Ainslie circuit from full charge and then start recording the time from the time it drops to say 25.01V or whatever until it reaches 24.80 or whatever. Then, run the control from full charge and then start tracking time when it hits 25.01V until it reaches 24.80. That way you are ensured your are not dealing with just some static charge on the top and you have let it run long enough to really get deep into the battery. All your points about high cop at the front and the going down are completely moot by doing this.

Plus you keep repeating experiment, control, experiment, control on the same batteries over and over and you see the results are consistent.

For the control test, you put enough resistance on the battery after they're charged to match the WATTAGE that the DC supply showed was necessary to get the resistor to the same temp the Ainslie circuit runs at. It is no more complicated than this.

Again, the Ainslie circuit running for many hours will be extremely close to this. It will drop only within 3F or so over say 10 hours. There is something called a margin of error that accounts for imperfections in just about any kind of test. So apply one if you want. And you need to consider the spread between resistor and ambient is also very consistent.

You should drop your battery conditioning options as they are completely and totally irrelevant when you know how to drain the batteries and then charge them with the right charger.

[Peter Lindemann]

DC-to-DC Converters

Indeed Hoppy.

.99 over and out.

To Hoppy and Poynt99,

Apparently you guys have finally played your last cards, and shown the readers here the astonishing limits to your knowledge. The simplest DC-to-DC converters are an "inductive fly-back charging system" and almost all of them operate at electrical COP>0.9. This is true of the smallest units made that deliver less than ONE WATT. The idea that you guys openly admit to not being able to build one that works better than COP=0.6 is remarkable.

All I can say is.... can't you two find something more interesting to do than bothering us?

Peter

[Hoppy]

To Hoppy and Poynt99,

Apparently you guys have finally played your last cards, and shown the readers here the astonishing limits to your knowledge. The simplest DC-to-DC converters are an "inductive fly-back charging system" and almost all of them operate at electrical COP>0.9. This is true of the smallest units made that deliver less than ONE WATT. The idea that you guys openly admit to not being able to build one that works better than COP=0.6 is remarkable.

All I can say is.... can't you two find something more interesting to do than bothering us?

Peter

I meant to say this type of flyback system, meaning using a circuit and an inductor of this open design. Of course there are more efficient types of flyback designs and of course I can build them!

I'm looking forward to see how you and Aaron handle the testing for Rosemary.

Hoppy

[Hoppy]

BCM? .

You really need to get in the real world and do EXPERIMENTS. If you think BCM gives you a real answer, you have completely and totally disqualified yourself from knowing anything about batteries. Battery capacitor analyzers never reflect what the battery will be able to deliver. If you think they do, go do it, film it and show everyone.

You want to bastardize a real test procedure by even hinting at using a BCM to tell the story? LOL, you have got to be kidding!

TIMEd draw down on the battery as I described from batteries charged by a charger that does know when the battery is really full is MANY times more accurate that your recommended BCM test!

Also, you clearly do NOT understand open systems. Make this particular circuit closed loop to run itself? Are you kidding me? It is not over 100% efficient, it is over 1.0 COP. There is dissipation but the dissipation of source potential simply dissipates over a much longer period of TIME! Hoppy, I think you're losing it here. Most of the free work is constant heat production...I'm supposed to magically make that heat make up for the loss in the battery so the battery never goes down. Please tell me your joking.

No need for time consuming procedures. Yeah, lets not do real tests and experiments and take bogus shortcuts that won't tell anything. If that is your idea of science and testing, then you should really stick to something you can demonstrate proficiency at. Test procedures are NOT it.

Even with a BCM, do you not still have to draw down the battery in order to charge it back up to put a BCM to see if it is at full charge? So we magically wave our hand and pretend we depeted the battery, bypassing all time requirement then apply a charger, wave our hand and pretend it just got charged and then put a BCM on it? lol You even mention "before and after" the test but "There is no need for complicated and time consuming test procedures." What is this Orwellian double speak?

Doing that will certain eliminate long drawn out test procedure but I think I'll do it my way. Thanks for the advice but I'll stick to doing real tests.

This takes the cake hoppy. I was open to hearing what you had to say - please no more distracting suggestions.

Aaron

Its not a battery capacitor analyser, its a battery capacity analyser. Read up on how they work. John Bedini uses one to measure his batteries before and after discharge

I know that its under 100% efficient, well under. but its not me saying that 16pints can be put back into the fridge. Perhaps you would like to explain to Rosemary how this is possible.

Hoppy

[Hoppy]

Hoppy, stop assuming things for others and putting words in people's mouths. You take things out of context constantly as all the skeptics do. No more or you're not welcome in this thread anymore.

Skeptic or realist Aaron; do you understand the difference.

Hoppy

[EgmQC]

Wait, are you serious EgmQC?

What in the world do you think that 1.2V across the shunt resistor represents? About 24-25 volts from the battery! The entire point is that with the battery voltage moving over that resistor, at a certain resistance, it will have a corresponding potential difference across the resistor. Based on THOSE calculations on the "shunt" it tells what current is associated with what is leaving the power source.

The non-sense is actually what you have posted.

And Hoppy
"Rosemary, With respect, you really do need to properly understand the basics of EE before you launch into preparing valid test routines for your circuit. Hoppy"

It appears she is more qualified than you or Egm as you apparently don't even know how to measure what is delivered by the battery as evidenced by your agreement with Egm's analysis. Stop your ridiculous insults NOW.

Aaron, if you dont read the following post whats the point you try to make in this post ? maybe if you had read this one http://www.energeticforum.com/63397-post1616.html ill be able to understand what you mean.

EgmQC

[Harvey]

Inductive Anomolies

The classical approach to the magnetic field produced by current flowing through and inductor, treats the field as conservative. In short, conservative field do not care what the path is or how long it takes because when you get back to start everything nets to zero. Reality has supported this view with a very strong and solid foundation. It has been proven many times over.

However, it has also been proven that when a magnetic field is changing, it is no longer to be evaluated as a conservative field. This means that Kirchoff's current law cannot be applied to fields that are subjected to change unless it can be clearly defined that the change will be fully reversable within the evaluation. Watch Professor Lewin's Proof. If you, fully review both parts of this demonstration, you will see that the evaluation is path dependent and must be done differently than the conventional "the current through the circuit must be the same everywhere" approach.

It is these differences that often result in confusion and misunderstandings between 'trained' engineers, and 'real life experimenters' that have results that seem to defy what the engineers have been taught.

I see banter here regarding these very issues and felt compelled to share this information. I hope it helps to unify both sides of the prism so that the viewers from both sides can agree as to what is moving through the prism. One side says it is white light and the other side says it is colored light, but both are right.

So, while it may be tempting to say that the current through the 0.25Ω sensing resistor is the same as that flowing through the HEXFET and the InductoRistor ( ), the reality is that they are not the same and they are very path dependent. Some of that current is inductively traded for voltage in the self inductance of the load and can never make it into the sensing resistor because of inherent delay of current flow after voltage application in the inductor. That energy, is instead trapped behind the HEXFET and left to ring back and forth between the battery and the magnetic field until it has been fully dissipated as heat. So the sensing resistor, does provide a good idea of the current in the HEXFET, but a poor idea of the current in the inductor - especially at any given instant. The strength of the magnetic field is synchronized in time with the current flowing through the inductor, and a field strength sensor would probably be a better indicator of the current through the inductor for a given instant.

[witsend]

another note for TK

Hi TK - thanks for the 'off thread' science lesson. Re post 837. Can you please advise the relevance of the point in your second para 'hard copied' hereunder.

...The instantaneous power dissipated in the shunt is then (4.8 x 4.8)(0.25) or about 6 Watts.

I fail to see the relevance.

Then - who programmed the 4th Trace on your LeCroy?

Re the next post 838. Are you seriously proposing that the 'spike' that first reaches -1.2 volts and then returns to zero - somehow 'cancels out'? I would have thought that both moments have energy associated with it. Not so much a 'cancelling out' but an addition to the energy delivered to the battery.

Kindly explain your comment that the waveform over the battery looks like it is recharging but 'may not be recharging' - 'mind' - or words to that effect. And, more to the point - why do you think the battery is showing the same waveform pattern associated with recharging batteries yet it is not recharging?

[Harvey]

Food for thought:

If I shine 1 Joule of light through a glass window, will the glass change temperature? If so why? Is any of the light lost in the process?

If I allow that same light to reflect off of highly polished mirrors, capable of reflecting 100% of the light, on each side of the glass, so that the same light penetrates the glass repeatedly, will the light eventually cease to exist?

What if I chose a frequency of light with a wavelength of say 750nm. Would that change anything?

If UV light becomes visible when reflecting off of Fluorescent Paint, is it possible to convert other photons into Infra Red by passing them through a ceramic material?

At what point do you consider a magnetic field as a form of light? Or do you? Why, or why not?

[Hoppy]

Aaron, you say in post 1629 that: -

"Most of the free work is constant heat production...".

If its free (energy not input by the user), how are you going to measure this heat to compare its energy value with the electrical energy input to the system to prove the high COP claimed?

Hoppy

[Harvey]

another note for TK

Hi TK - thanks for the 'off thread' science lesson. Re post 837. Can you please advise the relevance of the point in your second para 'hard copied' hereunder.

...The instantaneous power dissipated in the shunt is then (4.8 x 4.8)(0.25) or about 6 Watts.

I fail to see the relevance.

Then - who programmed the 4th Trace on your LeCroy?

Re the next post 838. Are you seriously proposing that the 'spike' that first reaches -1.2 volts and then returns to zero - somehow 'cancels out'? I would have thought that both moments have energy associated with it. Not so much a 'cancelling out' but an addition to the energy delivered to the battery.

Kindly explain your comment that the waveform over the battery looks like it is recharing but 'may not be recharging' - 'mind' - or words to that effect. And, more to the point - why do you think the battery is showing the same waveform pattern associated with recharging batteries yet it is not recharging?

I would also like to add the following question for TK regarding the Battery waveform:

If the flatline represents the battery voltage e.g. ~24V then the spikes above and below indicate a change in battery voltage at the point the probe is connected, right? So, in your opinion, does the spike above the line represent charging, discharging or neither? Same questions for the spike below the line.

EDIT: And isn't that E squared over R?