George Egely's Magic Wand

A nearby skeptical electrical engineer like THHuxleynew would be ideal as a test companion. But I doubt he will want to be put in charge of testing something as risky as his skeptical reputation.

Quote from axil

Since power is being formatted as electrical current, little to no heat will be produced by the Egely device. It looked to me from the oscilloscope traces shown in Egely's video of the input and output power production, the COP for electrical production of the device was 2.

I don't think that it is as simple as you say. Especially the output isn't easy. The fastest spikes in the output voltage correspond to frequencies about 2 MHz and above that.

For accurate sampling this would require sampling frequency above 4 MHz.

Another idea would be to use low-pass filtering and use lower sampling frequency to get the average voltage by that but that doesn't give a correct result.

Simply because P=U²/R. The you would filter first the voltage and use the power of two that and then integrate the result over time. The result is not the same as in the fast sampling case to first calculate power of two and integrating the result over time after that. As we have seen, the spikes can be extremely high.

Because of this reason i almost expect that the calorimetry gives more reliable results.

This should be easy in practice because the unit has already a possibility to feed the whole output power to a resistor located inside a calorimeter.

I think the simplest way ei measure the output is going to be with a matched pair of resistances in a couple of oil baths- one on the input and one on the output. Scope traces are possible, but I only have a 1MHz scope, so they will be 'illustrative' rather than informative.

Quote from Alan Smith

I think the simplest way ei measure the output is going to be with a matched pair of resistances in a couple of oil baths- one on the input and one on the output.

I fully agree. The only thing to take care of is choosing a good resistor value: higher resistance will bring higher sensitivity, but will also cause undesired voltage drops that may be very harmful for the normal operation of the device (and/or load). The ideal criterion would be to choose the lowest possible value that allow sufficient sensitivity. Thermal sensitivity can be increased by decreasing the specific heat of the fluid and having a very good insulation. Temperature measurement should be done with PT-100 (possibly class A), not thermocouples or even fancy techniques (e.g. optical/IR) .

Quote from Alan Smith

I think the simplest way ei measure the output is going to be with a matched pair of resistances in a couple of oil baths- one on the input and one on the output.

Dumping heat through a resistor in the "primary" stage of the cell tells you nothing about the cell's function (especially in that circuit). That is power that is not passing through the cell.

You need to have a way to measure the power going into the unit. DC voltage & current going into the "pumping" oscillator would be measurable. You can measure the power loss of the oscillator as a separate item.

Measuring heat emitted from an output loading resistor seems reasonable. This technique could be used for the initial oscillator efficiency test, and also (later) for the whole unit.

Quote from Frogfall

You need to have a way to measure the power going into the unit. DC voltage & current going into the "pumping" oscillator would be measurable. You can measure the power loss of the oscillator as a separate item.

I suppose it would be possible to use a 12V DC PSU to drive an inverted to create 230V 50Hz 'synthetic AC, then it would be possible to measure on the DC side. Inverter efficiency is generally quite high. But I'm not sure why you doubt the validity of using a resistor in the input power circuit and measuring the Joule heating - surely everything passing through such a resistor contributes to the heat evolved?

Quote from Alan Smith

But I'm not sure why you doubt the validity of using a resistor in the input power circuit and measuring the Joule heating - surely everything passing through such a resistor contributes to the heat evolved?

Yes, that is fundamental. A calorimeter never loses track of even one joule. That is why high end power meters often include a resistor and thermocouple to measure wattage. Methods of sampling amperage and voltage can fail to measure very brief power surges, but a calorimeter cannot. The problem is that a calorimeter is insensitive compared to a voltmeter. A voltmeter can measure far smaller levels of power.

As far as I know, a capacitor never loses any power either, no matter how brief a power surge is. A LEC is connected to a capacitor to measure total output.

Quote from Alan Smith

But I'm not sure why you doubt the validity of using a resistor in the input power circuit and measuring the Joule heating - surely everything passing through such a resistor contributes to the heat evolved?

I really cannot believe I'm having this conversation

You have an unknown amount of electrical power entering the stage after the oscillator. You then have a (capacitor bypassed, inductor-choked resistor) circuit that dumps an amount of power, as heat, into some water (which you can measure), before the second stage.

By definition, that power (in joules per second) is not going into the second stage. Therefore, the power that that is going into the second stage is also unknown.

The electrical output power of the second stage then generates heat in a load resistor - and the rate of heat production can be measured (in Joules per second).

However, this output power (heat rate) is not related to the power (heat rate) bled from the first resistor. So it is impossible to use these two values to infer any kind of cell efficiency.

I watched again the presentation a bit from the beginning of it. To me the input-side situation is not that bad as explained.

If we ignore the inductor and diode, whose function is to protect the kV-level voltage source, we have 2 components left; a resistor whose label is G

in the drawing and a capacitor, whose label is H. What happens now:

1. All the input current flows through the resistor into the capacitor. Note that at this time-point there is no current yet to the cell. The cell acts like an open connection.

2. When the voltage over the capacitor reaches the required level the spark fires. During that time the cell acts almost as a short circuit from the diagram point of view.

The charge loaded in the capacitor discharges and creates a current through the cell. The value of the input resistor can be neglected during this phase

3. When the voltage over the capacitor drops enough the short circuit state ends and the resistance of the cell reverts back to high again.

4. The input current flows again into the capacitor.

According to the simplification and functional description it is valid to assume that all of the energy that has been used to fire the cell (i.e. input energy)

has to flow through the resistor and can be measured with input-side calorimeter.

Quote from Rjzk

According to the simplification and functional description it is valid to assume that all of the energy that has been used to fire the cell (i.e. input energy)

has to flow through the resistor and can be measured with input-side calorimeter.

I am with you there. There is no free lunch, every mW the device uses comes out of the wall socket and enters the machine via the input side resistor.

I had missed this controversy. I see that the concerns raised by Frogfall are technically valid but, from the point of view of the bottomline of what we want to prove, somewhat irrelevant.

We want to know if there's excess energy. If we start with a discharged device, and can measure accurately how much energy is being fed to it, and how much is dissipated as heat, we will know if there's excess energy or not.

We will not get, at all, a precise image of when and where the excess energy is being produced, as this is pretty much a black box test, but we will know if we can get more heat from the device than the electric energy being fed to it, and that's all what the aim of this test is, and what has been asked for.

Quote from Curbina

We want to know if there's excess energy. If we start with a discharged device, and can measure accurately how much energy is being fed to it, and how much is dissipated as heat, we will know if there's excess energy or not.

I think Frogfall was thinking about an investigation of the device. A misunderstanding -The plan is a simple in-out test.

Sorry but i need to correct my final conclusion. I was too fast and didn't think the situation correctly.

(As a thought example think about a very low resistance and a high capacitance. The capacitance can store a lot of energy, but there is very little power loss in the resistor)

The functional description should be OK, but the energy consumption should be calculated according to the charge difference in the capacitor at each loading cycle.

The input voltage over the capacitor varies between a certain minimum voltage U_min and maximum voltage U_max.

The energy of the charge at the minimum is E_min = U_min²*C, and at the maximum E_max = U_max²*C. The difference in these energies tell how much energy is stored and

released to the cell during each cycle.

We know now all required information:

- voltages from the oscilloscope display

- capacitor value

- cycle time also from the oscilloscope to calculate the average power

I suppose the oscilloscope will tell the voltages and cycle-time with the required accuracy to get at least the rough estimate of the input power.

Quote from Rjzk

I suppose the oscilloscope will tell the voltages and cycle-time with the required accuracy to get at least the rough estimate of the input power.

I think that would require a 4MHz scope- the output has a lot of high-frequency jitter. Measuring such a dirty signal is fraught with problems. With a claimed COP if 2 to 10 depending on input settings it is possible to keep things simple.

Quote from Rjzk

Sorry but i need to correct my final conclusion. I was too fast and didn't think the situation correctly.

(As a thought example think about a very low resistance and a high capacitance. The capacitance can store a lot of energy, but there is very little power loss in the resistor)

The functional description should be OK, but the energy consumption should be calculated according to the charge difference in the capacitor at each loading cycle.

The input voltage over the capacitor varies between a certain minimum voltage U_min and maximum voltage U_max.

The energy of the charge at the minimum is E_min = U_min²*C, and at the maximum E_max = U_max²*C. The difference in these energies tell how much energy is stored and

released to the cell during each cycle.

We know now all required information:

- voltages from the oscilloscope display

- capacitor value

- cycle time also from the oscilloscope to calculate the average power

I suppose the oscilloscope will tell the voltages and cycle-time with the required accuracy to get at least the rough estimate of the input power.

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The enery stored in a capacitor is E = (1/2)*C*U²

Quote from LDM

The enery stored in a capacitor is E = (1/2)*C*U²

Thanks for the correction. Division by 2 was missing.

Next step would be to formulate the problem in such a way that the input calorimeter could be used here also.

Can anyone please remind me about the Magic Wand : What the input and output currently and projected?

I guess it's at

DC in.
Some heat out.
Direct electricity output ... varies per resistor/matched inductor/matched inductor + tube

And here.

Direct conversion. I like it.

Quote from Curbina

I had missed this controversy. I see that the concerns raised by Frogfall are technically valid but, from the point of view of the bottomline of what we want to prove, somewhat irrelevant.

We want to know if there's excess energy. If we start with a discharged device, and can measure accurately how much energy is being fed to it, and how much is dissipated as heat, we will know if there's excess energy or not.

We will not get, at all, a precise image of when and where the excess energy is being produced, as this is pretty much a black box test, but we will know if we can get more heat from the device than the electric energy being fed to it, and that's all what the aim of this test is, and what has been asked for.

You are right. I was also concentrating too much on the internal behaviour of the input oscillator and cell interaction.

To be specific, if we measure the total the electric energy feed from the battery to the unit and subtract from that (the loss in the resistor) measured energy in the first calorimeter, we get the actual energy feed in the cell.

This should be quite near the E_in to the cell.

The output calorimeter gives directly E_out. So then the COP should be simply E_out/E_in