​New E-Cat QX Picture and New Rossi-Gullstrom Paper (Very high COP reported with Calorimetry)​

Quote from DNI

That would be a possible way to do it. But that is not how it seems to be done in the report. The power in is calculated from "Energy input: V=0.1 R=1 Ohm → W=0.01" where V is the voltage over the 1 ohm resistor. Hence what seems to be calculated is the power in to the resistor. Not the power in to the QuarkX.

That is your misunderstanding, which is the same misunderstanding that THH has, in my opinion. There is a single resistive path. The two voltmeters are being used in a redundant fashion. They are measuring the voltage drop across a known resistance, then determining the power using one of the most basic and well-known laws. I don't even want to name it here, it is that well-known. THH is smug because he thinks I don't know what law it is, or how it is used. THH has no idea what my background is nor my education. He takes cheap shots and capitalizes on typos.

My success rate is quite high with my predictions. And I predict that the two voltmeters are measuring the same voltage drop across the total resistance of the single resistive path through the reactor. Let's see whether THH or I turn out to be right. My guess is--probably me again.

Quote from maryyugo

Rossi's style is to advance from one convoluted, hard to clarify, poorly set up experiment to a worse one. I think he has outdone himself this time. But the style, at least, is consistent. There are even more ways to mismeasure this mess than prior idiotic ecat versions.

Um, no. He has simplified it down to a DC system with a single path to measure input power. It doesn't get much simpler than this.

@maryyugo

You keep me healthy. I thank you.

Quote from Adrian Ashfield

First Jed maintains the E-Cat QX doesn't even exist, so the photo is apparently a figment of one's imagination.

Then THH can't resist speculating in his usual negative way, that one can't determine the power by measuring the voltage drop across a known resistor in series with the reactor, to get the current, and measure the output voltage pf the power supply. As it it supposed to be DC it should be fairly simple.

Also curious is where my last post on the settlement thread disappeared to

No Adrian, that is not my point. The issue is that Rossi gives his calculation (I've shown it above) and that is not what he says he does in words, nor what the equations he writes, and figures he give, say.

It is hardly negative speculation to read a research paper and go by the experimental details given? maybe you are used to substituting hope or prejudice for fact, but i find it simpler to reckon experimenters tell the truth (though often can make mistakes). Rossi is on public record as making very grievous mistakes in input power measurement - again this is not my negative prejudice, but fact. Therefore an equally grievous one here, as he claims he does, is quite reasonable.

Quote from IH Fanboy

That is your misunderstanding, which is the same misunderstanding that THH has, in my opinion. There is a single resistive path. The two voltmeters are being used in a redundant fashion. They are measuring the voltage drop across a known resistance, then determining the power using one of the most basic and well-known laws. I don't even want to name it here, it is that well-known. THH is smug because he thinks I don't know what law it is, or how it is used. THH has no idea what my background is nor my education. He takes cheap shots and capitalizes on typos.

My success rate is quite high with my predictions. And I predict that the two voltmeters are measuring the same voltage drop across the total resistance of the single resistive path through the reactor. Let's see whether THH or I turn out to be right. My guess is--probably me again.

I think maybe we misunderstand each other.

As I read the report there is a resistor (R = 1 ohm) connected in series with the reactor. Then the voltage across the resistor is measured (V = 100mV) (with two voltmeters). And the power is calculated as:

Pin= V^2/R = 0,01W

But this is the power to the resistor not the power to the reactor.

To calculate the power to the reactor we would need to know the voltage across the reactor as well.

What am I misunderstanding?

DNI,

In my mind, it is all very simple. I'm confused as to why there is so much confusion, although the description of the experiment in the paper could use some polishing.

If you look at the photo, you will see that the two voltmeters are measuring the same thing: the voltage drop across the reactor. The resistance path through the reactor is known. Therefore power can be easily calculated.

The brown resistor is not connected to either voltmeter in the photo. It is there and available as a sanity check. It is probably connected in series with the reactor. In other words, the voltage drop across the brown resistor can be measured to determine current through the reactor.

Quote from randombit0

Is a different type of experiment. Charge a capacitor with a known energy and discharge it in the device.

@RB: That is what R.Mills does. Do you know or believe that AR does the same?

IHFB: I share in the confusion. You need the total voltage into the system or better across the cell. An analogy is to measure the output of a iron by measuring the voltage drop across the cord. You can measure current but not power in that case.

For me, I did not understand how you can create a plasma with 100 mV. You could theoretically maintain a plasma with a low voltage if it was making energy. Therefore, just show that the voltage from the power supply is X and with X small and that would be impressive (at least to me).

All of this "experimentation" with plasmas is likely misguided. Has anyone seen excess power from a deuterium/hydrogen lamp in the past 100 years of making them? Consider how they work. They are certainly high-temperature systems. Have any extra atomic lines appeared in the spectrum from such a system? Helium or extra atoms would show this in spades. You could argue that a hydrogen lamp does not have LAH present and that is the key. Energetics and Sandia have seen strange things in plasmas but they are explainable by conventional physics and where not in the range of COPs > 100.

.

DAK2,

Welcome to the forum. Lots of new participants as of late.

Please carefully read my comment above and let me know specifically what you don't understand.

Quote from IH Fanboy

In my mind, it is all very simple. I'm confused as to why there is so much confusion, although the description of the experiment in the paper could use some polishing.

If you look at the photo, you will see that the two voltmeters are measuring the same thing: the voltage drop across the reactor. The resistance path through the reactor is known. Therefore power can be easily calculated.

The brown resistor is not connected to either voltmeter in the photo. It is there and available as a sanity check. It is probably connected in series with the reactor. In other words, the voltage drop across the brown resistor can be measured to determine current through the reactor.

I think it's quite clear in the report that it is the voltage across the resistor that is measured.

Page 18 in the report:

"In the left in the figure there is two voltmeters that measure the mV of the current passing through the 1 Ohm brown resistance."

But lets say the report is wrong and you are right. That the voltage measured is the voltage across the reactor. Then it's is correct to calculated the power as they have done in the report only if the resistance in the reactor is always 1 ohm. But how do we know this? Because Rossi say so? This would be as as stupid as when Rossi on previous occasions claimed that all water was vaporized without measuring it. It would have been so simple to measure both the voltage across the resistor in series and the voltage across the reactor. And with those two values it would have been simple to calculate the power in to the reactor without the need to trust any "Rossi say" about the resistance in the reactor.

It seems like we have two different possibilities. Either it is the voltage across the resistor that is measured (as it is described in the report) and then the calculation of power is wrong. Or it is the voltage across the reactor that is measured as you think. And then Rossi once again has made a stupid test set-up that has to trust "Rossi say". When it would have been very simple to also measure the voltage across the resistor and eliminate "Rossi say" from the calculations.

I said some years ago that there is only three possible alternatives:

1. The e-cat works and Rossi wants to prove that it works. But Rossi is extremely incompetent and refuse to listen to advice.

2. The e-cat works but Rossi don't want to prove this without doubt.

3. The e-cat doesn't work as Rossi claims.

I think this is still valid and my guess is still number 3.

Looking carefully at the photo, its clear both meters are reading the 100mV across the resistor and not the reactor. This agrees with the paper so cannot see any ambiguity. I cannot see any way the reactor power input can be determined by voltage across a series resistor, looks like more Rossiscience. To maintain a fairly steady 100mV whilst feeding a plasma suggests to me that the power supply is running in constant current mode.

The red handles are probably also for adjusting electrode gap.

Yes, my thoughts as well. I wonder if Rossi used 2 pencils in the prototype

IH Fanboy:

"If you look at the photo, you will see that the two voltmeters are measuring the same thing: the voltage drop across the reactor."

Malcom Lear:

"Looking carefully at the photo, its clear both meters are reading the 100mV across the resistor and not the reactor."

Seems like you both see what you want to see. A bit like when the disussion was about frames or no frames in the window. In my opinon it is not possible to see in the picture if the voltage is measured across the resistor or reactor. The green connector seems to be connected to both the resistor and the reactor. And the red connector disapear behind the box. Can any of you explain how you come to your conclusion?

To me it really doesn't matter what can be seen in the picture or not. It is clearly described in the report Page 18:

"In the left in the figure there is two voltmeters that measure the mV of the current passing through the 1 Ohm brown resistance."

can

"The red handles are probably also for adjusting electrode gap."

Prosit! from an Italian Genius dell'arte di arrangiarsi

Quote from DNI

I think it's quite clear in the report that it is the voltage across the resistor that is measured.

Page 18 in the report:

"In the left in the figure there is two voltmeters that measure the mV of the current passing through the 1 Ohm brown resistance."

That is consistent with what I wrote above. Please read carefully.

Quote

It seems like we have two different possibilities. Either it is the voltage across the resistor that is measured (as it is described in the report) and then the calculation of power is wrong. Or it is the voltage across the reactor that is measured as you think. And then Rossi once again has made a stupid test set-up that has to trust "Rossi say". When it would have been very simple to also measure the voltage across the resistor and eliminate "Rossi say" from the calculations.

It is both. The photo shows both voltmeters measuring the voltage across the resistor. At some point in time (not shown in photo), at least one of the voltmeters can be used to measure the voltage across the brown resistor to determine the current through the reactor, in order to sanity check the power calculation of the reactor. It is that simple.

Quote from Malcolm Lear

I cannot see any way the reactor power input can be determined by voltage across a series resistor, looks like more Rossiscience.

The brown resistor is in series with the reactor. To determine the current through the reactor, you measure the voltage drop across the brown resistor and apply Ohms law. Then once you have the current value through the reactor, you can check your power calculation for the reactor, again using Ohms law.