COP 17 Heater | Rosemary Ainslie

[Aaron]

ac and dc

Aaron states twice in his post that it is DC voltage across the shunt.

It looks to me that the scopemeter is in AC coupling mode as indicated by the little sine wave under the "rms", i.e. the "V~".

.99

The scope shows the AC and DC at the exact same time, period. I see that you seem to be asking questions that will confuse people more than help them.

Put the scope across the shunt and you see BOTH the AC and DC. But you use the DC number to calculate what leaves the battery. If you want to refute this, do it elsewhere.

[Aaron]

in resonance

Hi everyone,

Here is a picture of a scope shot from a Fluke 123 Scopemeter showing the DC voltage across the shunt.

This is the picture TK does not want you to see because it shows over 1.0 COP immediately - when you compare to AC RMS across load.

The ac rms across load squared divided by load resistance (10 ohm) is the wattage dissipated there.

The dc across shunt divided by shunt resistance is amps X battery voltage is watts leaving battery.

If he shows that, there is more watts being dissipated at load than is leaving battery. It is clearly over 1.0 COP.

Ask TK to show this as well as the wattage at load!

I needed to add that the over 1.0 COP is there when the circuit is in resonance. You can tune for resonance with any scope and when in resonance, your will get the gains in the battery with extended running time.

It absolutely can be done with the schematic I posted.

Make sure to use the updated one where I removed the snubber cap - what I called the snubber cap just means to me is snubs the spikes on the 555 circuit. I understand it is normally called a filter cap - I used to call it a buffer cap on the Bedini circuits or smoother cap. Anyway, everyone should get the idea that without the cap on the 555, the "transient noise" from that circuit is more likely to cause oscillation.

I removed the fixed resistors - just making a note again so it doesn't get drowned out in the transient noise that has been posted here.

If you build the circuit using the schematic I posted, you can get over 1.0 COP when it is tuned to resonance and self-oscillation isn't even necessary to get the gain. Oscillation just boosts the gain you do get.

PLEASE UNDERSTAND GETTING THE CIRCUIT IN RESONANCE SHOULD BE YOUR PRIMARY GOAL.

Even moreso than oscillation because you can still get the gain.

[poynt99]

The scope shows the AC and DC at the exact same time, period. I see that you seem to be asking questions that will confuse people more than help them.

Put the scope across the shunt and you see BOTH the AC and DC. But you use the DC number to calculate what leaves the battery. If you want to refute this, do it elsewhere.

It is a straight forward question Aaron.

Is the current the same magnitude and direction in the battery and shunt at all times or do the two currents vary from each other?

.99

[MileHigh]

Aaron:

Your statement, "This is the picture TK does not want you to see because it shows over 1.0 COP immediately - when you compare to AC RMS across load."

Honestly I don't see it and I don't get it, so I will ask you a second and final time: Can you crunch the numbers for all to see in a posting to show over 1.0 COP "immediately" with your data?

More importantly, if you are going to have the DSO for only one week time will be precious, so what is your action plan for using it to make measurements? It would be most interesting to know what your plans are. It may create a debate, so much better to have it now as compared to when you are actually using the DSO since your time with it is so short.

Finally, can you explain what you mean by resonance in this circuit? Many versions of resonance have been discussed on this thread, can you be more precise? What is the source of the resonance, what will the approximate frequency be, and what resonance effects will manifest themselves to create an OU situation? I will point out to you that Rosemary's paper does not mention resonance and just makes power-in and power-out measurements. In TKs clip where the current drops to near-zero when he gets the MOSFET to go into defibrillation you accused him of faking that. In contrast, a few days before that clip that's what I predicted would happen. If resonance is the key then please enlighten us.

I am pushing the resonance issue because there is a kind of cult around resonance, just like there is a kind of cult around inductors and the voltage spikes they can create. Lots of people seem to believe that resonance unlocks a process that creates OU, as you are stating about this circuit. I am asking you to flesh out that concept with a few paragraphs.

MileHigh

[Aaron]

measuring INPUT wattage

What is at the "shunt" resistor is a very accurate representation of the actual current leaving the battery and is widely accepted as such. Any margin of error is not significant enough to make a difference.

Take the dc reading across the shunt and divide by the shunt resistance.
That is the amperage the battery is giving up at the voltage indicated on the batteries.

Multiply the amperage by battery voltage and that is the wattage.

Wattage x Time is joules of energy and the above test shows how many joules of energy are leaving the battery to run the circuit at X temperature.

If you have an alternative to Ohm's Law, feel free to spell it out in another thread.

----------

For OUTPUT wattage at the resistor being "dissipated", put probe across load and take rms ac voltage squared and divide by 10 ohm inductive resistor and you have the wattage there.

That is a valid way to compare input watts and output watts.

The battery draw down method is another.

However, the battery draw down method should be done at the same time as the power readings on the scope so that the gains are corroborated even more so. The wattage gain will be apparent and the TIME gain is apparent by the battery lasting longer with the Ainslie circuit.

It is not just more watts at the resistor at any given time compared to input watts, it is MORE running TIME with the SAME HEAT PRODUCTION....at a LOWER wattage.

Gains in both will be apparent and running time COP comparison may be different than wattage COP comparison. If wattage cop is 2.11 for rough example battery draw down could be more.

[Aaron]

resonance

What is the source of the resonance, what will the approximate frequency be

It is irrelevant because each circuit will have it's own resonant frequency. More than one resonant frequency, you have to sweep thru either on time or off time of timer pots to synchronize. Meaning, adjust roughly the duty cycle you want then move the off time pop and watch the scope zoomed out. when you see a repeating pattern, you're in resonance.

Vice versa, set any off time and move the on time until you see the pattern.

All patterns will be different for everyone.

Because you ask what the frequency would be, it indicates you don't understand resonant circuits.

[MileHigh]

Aaron:

About your statement about the output power measurement:

> For OUTPUT wattage at the resistor being "dissipated", put probe across load and take rms ac voltage squared and divide by 10 ohm inductive resistor and you have the wattage there.

This is not correct. When I read between the lines the really tough issue that remains unresolved is that you still don't seem to fully understand how an inductor works.

What I hear in your statement is that you are considering the inductive resistor to be the same as a normal resistor for your power calculation. So we all know that the RMS voltage squared divided by a pure resistive load gives you the power being dissipated in the resistor. Of course the instantaneous current through a pure resistor is simply the instantaneous voltage divided by the resistance. That is the implicit train of thought in your statement.

In this case that is NOT truly what is happening. The current is not necessarily proportional to the voltage so VRMS-squared/R does not hold in a generalized case.

During the MOSFET on time, there will be less total (current x time) than the VRMS-squared/R formula is based on. Therefore this part of the measurement will exaggerate this part of the power measurement.

During the MOSFET off time, the inductor part of the inductive resistor is now discharging through a diode, which has a negative differential resistance after you overcome it's (simplified) threshold voltage. Therefore the VRMS-squared/R formula doesn't even apply here.

There are two straightforward ways to measure the power being dissipated through the inductive resistor + diode setup. You can do it the thermal way, or you can add a second shunt resistor in series with the diode and the record the inductive resistor voltage and the diode voltage separately. That way you have the second current waveform, the inductive resistor voltage waveform, and the diode voltage waveform, and go from there.

I think that you would agree that the thermal route would be a lot easier.

I hope that you don't freak out and instead argue the merits of what I said. I am willing to listen.

MileHigh

P.S.:

> Because you ask what the frequency would be, it indicates you don't understand resonant circuits.

There is no need to be like that, it's like Peter trying to claim that .99 and Cloxxki were useless because of a statement they made about pulse charger effeciency. You know that I could say a lot of things about resonance if I wanted to.

[witsend]

Using the metric of your choice (DC, rms, ave) for both, is the current the same magnitude in the shunt as it is in the battery at any given instant?

.99

Poynt - how would you determine current flow back through a battery? This is precisely the poynt where new age and classical part company. But if you're asking does the voltage across the battery correspond to voltage across the shunt - of course. How could they be different?

But how then does one compute the 'spike' over the battery which doubles the instantaneous voltage of the battery? Or for that matter the small negative spike that dips below battery voltage? This speaks to the need for averages to resolve this question. However, with the use of the Tektronix this question will be categorically answered and - from experience of this machine - the advantage is somewhat understated by averaging.

[Hoppy]

Aaron

Could you please show a scope shot taken between pin 3 and ground of the 555.

Hoppy

[Hoppy]

The issue here is the claim of COP17. This according to my understanding of what Aaron and Peter are claiming, means that energy not input by the operator / user (from somewhere) is doing work and this energy is seperate from that supplying the circuit from its battery power supply. This therefore means that the heat radiated from the inductive resistor - the 'free' energy' as Aaron describes it, needs to be quantified in terms of energy dissipation over the length of the test. This value can then be compared to the energy being taken by the circuit as measured using the shunt resistor. I see no attempt at present to calculate this 'free' thermal energy. Rosemary acknowledges that the circuit is less than 100% efficient in electrical terms, so it now needs to be proved that the energy radiating from the resistor in the form of heat is many times greater than the energy being taken by the circuit.

Hoppy

[Harvey]

At the risk of posting something erroneous with one eye open after a long day...here goes.

On the Scope pic there by Aaron, we see a DC value and can actually calculate that as an average DC value across the 0.25 ohm sensing resistor. But that DC reading is really the flat line in between the spikes. So the time period that we can apply the DC value is not 100%, this cascades back to the amperage and the total battery wattage as all being reduced by that percentage that is less than 100%. The AC waveform (the spikes) take up the remainder of the time. Each spike nets to zero so during that period there is zero DC voltage across the 0.25 ohm sensing resistor. For example, from zero we see a rise in voltage but it comes back down to zero so for that period the voltage nets to zero. The same is true when we see the negative going spike, from zero it goes down and then comes back up and for that period we have a net to zero volts. So we must exclude those from the DC average used to calculate the battery draw wattage. We must then calculate each spike to determine the current that flows during each spike period. We can see that there is a smaller wider positive spike and a longer narrower negative spike. These each, independently, reflect on the Battery drain wattage.

Ok...I need sleep

Cheers,

[poynt99]

Current not voltage

Poynt - how would you determine current flow back through a battery? This is precisely the poynt where new age and classical part company. But if you're asking does the voltage across the battery correspond to voltage across the shunt - of course. How could they be different?

The voltage across the battery does not change substantially over a number of operating cycles, yet the shunt voltage will and does. They are different.

However, that is not what I am asking. Once again:

Is the current the same magnitude and direction in the battery and shunt at all times or do the two currents vary from each other?

.99

[poynt99]

At the risk of posting something erroneous with one eye open after a long day...here goes.

Ok...I need sleep

Cheers,

Indeed. Try again later.

.99

[witsend]

poynt - could you explain the 'completely fair' comparative test proposed on OU.Com?

[witsend]

Aaron:

About your statement about the output power measurement:...I think that you would agree that the thermal route would be a lot easier...I hope that you don't freak out and instead argue the merits of what I said. I am willing to listen...

...There is no need to be like that, it's like Peter trying to claim that .99 and Cloxxki were useless because of a statement they made about pulse charger effeciency. You know that I could say a lot of things about resonance if I wanted to.

MileHigh - The experimental requirements are well in hand and under expert advisement. Relax. And no-one objects to valid comments. But look back. You'll maybe acknowledge that some of the requirements by our contributors were not exactly valid and certainly not classical. And many questions remain unanswered.

Until you find, in due course that our numbers are being misrepresented - then please refrain from advising. Until the data is actually put on the table I'd prefer it that you keep your concerns to yourself as they are both premature and inappropriate. You take a generalised statement out of context and use it to parade an ignorance on the part of the experimentalists here that is positively insulting - whether intended or not. I've said it before. We've enjoyed the general sense of being underrated in terms of the team's abilities which has - for some reason - helped rather than hindered us. But I would ask that you now look around you. I think there is sufficient evidence that there are talents this side of this argument that are also equal to that argument.