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

Quote from Adrian Ashfield

The statement no control = no result is false period. That some skeptic requires it is another subject.

It is a meaningful statement, and it is not hard at all to understand the point LENR Calendar is making. Without a control in many benchtop experiments, you cannot draw meaningful conclusions. Your instruments may be miscalibrated, for example. And just as important: if the experimenter is seeking to argue that the results are meaningful, as he does in an experimental writeup, he must make a case for this, outlining how he showed that his instruments are not miscalibrated, and not expect his audience to simply go along with him. This is not arguing about the number of angels that can stand on a pin. This is basic science.

Are you telling me that ITER researchers are not performing any sanity checks on their experimentation?

Every serious scientist, programmer, accountant etc, use whatever ways they have at their disposal to double-check their work.

So I say no control= no result because any scientist with 2 ounces of curiosity would have run a control in this specific experiment.

If you are getting an unexpected result, such as COP=20000, you find ways to double check it before writing a whole theoretical paper and having pointless debates on a forum.

I know that Rossi probably hasn't done so because 1) it's not in the report ; 2) Rossi has shown contempt on the use of dummies in the past; 3) he spammed questions on the JONP about whether he used a control.

In case it isn't clear, "no control= no result" is a hyperbole; obviously Rossi did get a result from his experiment. If you manage not to take those words literally you will understand what I mean. Words from a professional to a student

Quote from maryyugo

Stop quoting Rossi. He's a liar and a criminal. Rossi and his cronies were personally and entirely responsible for an environmental disaster that cost dozens of millions of dollars to clean up. Where in the world is it documented that Rossi was not responsible for Petroldragon other than in Rossifiction? No wonder you swallow the QuarkX nonsense hook line and sinker! If you believer Rossi innocent in Petroldragon (and the Thermoelectric ripoff of the government) you'd believe anything.

BTW. the reason I "take out old topics" is that people like you persist in forgetting or misstating them. Again and again and again.

Hi Mary! Have yourself a big thumbs up! Not quite as big as the calorimeters you're used to but big nonetheless. I would have given you two thumbs up if you'd managed to squeeze Defkalion into that post.

Now, I have to take issue with you on not quoting Rossi. As you know Rossi says is what's the driving force behind the eCat. We would not be here if it was not for Rossi says. I say keep the quotes coming. You feel me?

Quote from IH Fanboy

Malcolm,

Rossi is right. Consider this handy-dandy diagram:

That is how I now understand it to be configured. So if you assign a resistance to R (REACTOR) other than zero, you are decreasing I. So by assuming that R (REACTOR) is zero, you are being conservative in the power calculatio

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@IHFanboy

Could you point out where you read that Rossi measures voltage across the reactor (or both the resistance and the reactor)?

By ignoring the resistance to the reactor, I agree that you are being conservative; however, if you ignore the voltage across the reactor, you are underestimating the input power.

Rossi owns two voltmeters that are giving close enough results. Why doesn't he use one of them to measure voltage across the reactor? It would have been a very simple sanity check.

Quote from Malcolm Lear

"If the E-Cat has a resistance, that makes our calculations more conservative

Wrong wrong wrong!!!

Its just the opposite. Here he is admitting they don't know the resistance, worst apparently it doesn't matter.

Rossi calculates input power using his 1R resistor and 0.1V a current of 0.1A. This gives 0.01W

However suppose the reactor had an impedance of 1K. Here we have given 0.1A a voltage of 100V and reactor power of 10W

Wow a COP of 1000 !!!!

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Correct Malcolm. Rossi's explanation (via Acland) just confirms what was 100% clear from the paper - he is completely wrong. IHFB I think does not understand electricity, or else is blinded by bias.

One common meme is the "Rossi could not be as obviously wrong as that" one. He has often so been.

Even Mats Lewan agrees with the skeptics here: (ECW)

"You simply don't need to know the resistance of the reactor to calculate the correct current. But you need to know the voltage over the reactor, or the voltage over the total circuit and over the resistance, to calculate the input power in the reactor."

To be fair, ignoring the reactor's resistance does make the calculations more conservative. As long as you are measuring the right voltage.

Anyhow, the spectrometer used reports the radiant power in W/m^2. So why not report that value?

Quote from IH Fanboy

Ohms law takes various forms. You can approach it from different angles, and at different times, using different measurement points, and still arrive at the same correct answer. The current will be the same through both the resistor and the reactor because they are in series.

My diagram above fits with Rossi's clarification of how they are determining the power.

To measure power you need current through the reactor (correctly measured from voltage across the resistor) multiplied by voltage across the reactor. There no other way (it gets more complex for AC but it still the same thing). The reactor voltage can be anything - 1V - 10V - 100V - 1000V and is no way predicated by the resistor voltage. Rossi gets it badly wrong.

Quote from LENR Calender

Are you telling me that ITER researchers are not performing any sanity checks on their experimentation?

Are you telling me that Rossi and his team of researchers are not performing any sanity checks on their experimentation?

Quote from LENR Calender

@IHFanboy

Could you point out where you read that Rossi measures voltage across the reactor (or both the resistance and the reactor)?

By ignoring the resistance to the reactor, I agree that you are being conservative; however, if you ignore the voltage across the reactor, you are underestimating the input power.

Rossi owns two voltmeters that are giving close enough results.

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It comes from my interpretation of Frank's notes from Rossi's update. So yeah, admittedly, fourth hand. But it is what makes the most sense.

Quote

Why doesn't he use one of them to measure voltage across the reactor? It would have been a very simple sanity check.

How do you know he hasn't? Any person with an elementary education in these matters would probe and sanity check in a number of different ways.

Quote from LENR Calender

Even Mats Lewan agrees with the skeptics here: (ECW)

"You simply don't need to know the resistance of the reactor to calculate the correct current. But you need to know the voltage over the reactor, or the voltage over the total circuit and over the resistance, to calculate the input power in the reactor."

That is actually in agreement with me. THH apparently professes to know everything about electricity, but could not see the obvious. He doesn't even critique. He just insults.

Quote from THHuxleynew

To measure power you need current through the reactor (correctly measured from voltage across the resistor) multiplied by voltage across the reactor. There no other way (it gets more complex for AC but it still the same thing). The reactor voltage can be anything - 1V - 10V - 100V - 1000V and is no way predicated by the resistor voltage. Rossi gets it badly wrong.

You still don't understand. The entire circuit (1 Ohm resistor + reactor) is being treated as a single unit in terms of power and COP calculations. You are correct in that the current through the reactor can be correctly measured from voltage across the resistor. You can also measure the current by taking the total voltage drop (across both the 1Ohm resistor + reactor) and dividing that value by the sum of the resistances. And if you do the latter, and assign a value of zero as a resistance to the reactor, then what you end up with is a higher (i.e. conservative) current value. Then, when you go to calculate the power of the single unit (1 Ohm resistor + reactor), you will square the higher (i.e., conservative) current value and multiply that product by the sum of the resistances. Now, if you are still assigning a zero value to the reactor in the final power calculation, there is a non-conservative effect. But the squaring of the higher (i.e., conservative) current value overcomes the non-conservative effect, and you are left with a conservative input power value for the unit.

Quote from IH Fanboy

Are you telling me that Rossi and his team of researchers are not performing any sanity checks on their experimentation?

No, he is not telling you that. He is suggesting that R-Boy is telling you that.

Read the words.

This is so simple, just like the multiplication question that I asked you a couple of days ago that you ignored.

come on ihfb - answer simple question

Provide evidence that he did a control, or retract.

Up to you, but I know you always ignore the simple ones.

Pete

Quote from IH Fanboy

You still don't understand. The entire circuit (1 Ohm resistor + reactor) is being treated as a single unit in terms of power and COP calculations. You are correct in that the current through the reactor can be correctly measured from voltage across the resistor. You can also measure the current by taking the total voltage drop (across both the 1Ohm resistor + reactor) and dividing that value by the sum of the resistances. And if you do the latter, and assign a value of zero as a resistance to the reactor, then what you end up with is a higher (i.e. conservative) current value. Then, when you go to calculate the power of the single unit (1 Ohm resistor + reactor), you will square the higher (i.e., conservative) current value and multiple that product by the sum of the resistances. Now, if you are still assigning a zero value to the reactor in the final power calculation, there is a non-conservative effect. But the squaring of the higher (i.e., conservative) current value overcomes the non-conservative effect, and you are left with a conservative input power value for the unit.

Right, but that is not what Rossi says he has done. He says the voltage is that across the resistor, not across the whole circuit. Neither in the paper, nor in explanation to Frank, does he say otherwise. Also it would be a stupid way to measure input power. You are grasping at straws here - it only makes sense if you think Rossi's measurements are good (and therefore have to make this measurement good). Even Mats knows that Rossi's attitude towards measurements has been like Humpty-Dumpty's towards words. As for "team" it is quite clear that no-one competent has checked that paragraph in the paper, it is even less likely anyone would have checked the experiment.

@IHFB. Are you saying that you know Rossi does sanity checks? How? And why then is he on record saying that Controls (the grandmother of all sanity checks) are not needed?

@THH,

It is how I interpret Rossi's additional clarification to Frank.

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

As with all things in life, and with language, multiple interpretations are possible. I know this frustrates many here, but such is life. Language is imperfect, and sometimes people write inelegantly, especially when writing in a language that isn't your native tongue.

I interpret both Rossi's clarification and the line in the paper to mean that the two voltmeters are measuring the mV across both the 1 Ohm resistor and the reactor. "That is absurd!" you cry. But it is not. It is simply my interpretation based on the information that we have been provided so far. The "current passing through the 1 Ohm brown resistance" is the same as the current passing through the reactor. So when the author wrote that sentence, the probable intentional meaning is that the voltmeters are measuring the mV across the unit.

Quote from THHuxleynew

@IHFB. Are you saying that you know Rossi does sanity checks? How? And why then is he on record saying that Controls (the grandmother of all sanity checks) are not needed?

He has said many times on his blog that he is using a control reactor. Go and read it for yourself. (Yes, he has also mentioned in the past that a control is not needed, but it seems he had a change of heart.)