Posts by Paradigmnoia

I just bolted down the two horizontal clamps, identical to those used for the QX.

It might be possible to get an idea of the length of the reactor body by comparing the handle angles, distance apart, etc., from the demo photos.

It is clear that the clamp rods are very nearly fully extended. And rusty. The bolt head, just beyond the more clean cinch nut at the end of the rod looks really rusty.

Edit: It is also clear that the clamps are not precision made... one has a 5 mm longer rod than the other. Also I had to shim up the RH clamp so that the rods would meet at the same height.

Quote from Adrian Ashfield

Sure, Rossi gives demos every other month. when was the last one again?

Was it the first 1 MW plant in Sep 2011? If he demonstrated that again I must have missed it.

I must have missed the massive production that will stop the useless flapping of gums in the blogs, since the last demo....

Since I have those exact clamps, perhaps I should test their resistance to the ends of the horizontal bar to the base plate? The clamps swivel over a set of rivits. Probably when squeezing something the resistance should be very low. But just a bit loose, where they might arc against the QX ends, the resistance could be something...

Edit: A quick test shows the expected quite low to negligable resistance. The highest resistance is when the clamps are backed off all the way and it becomes loose.

Also where the bolt threads into the end of the actuator is a potential problem spot. The threads are crappy inside the clamp actuator rod, and quite loose on the threads of standard boots. The bolt would have to be cinched in there tight, and not rusty to make a good connection.

Quote from Adrian Ashfield

There will be no further demo. Our next public action will be the launch of the product.

Warm Regards,

A.R.

So you can bet on most of the future speculation on this forum will be cod's wallop.

Rossi has said that before, in regards to previous E-Cat versions.

Quote from fusio

Why arent physicist adressing the elephant in the room??

Who on earth can state that there is no voltage drop in the reactor????

True, plasma can have low conductivity BUT to generate plasma you have to have VOLTAGE!!! and thereby resistance. His avoiding this in his calculations and thats the only reason he can state energy production with COP.

Why arent Mats Lewan adressing this?

Im really frustrated, Rossi is still trying to kill "lenr" with his stupid demonstrations and you are all fuelling this muppet show by showing up and giving him spotlight.

LENR is real but you are all mislead by the greatest puppest master...

So I suggest you all get your head out of the sand, go back and focus on real data, real experiments and real theory.

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my Durapot slab has a COP of 63600 using the 1 ohm method.

26 mV DC across 1 ohm resistor in series with slab.

43 watts out of slab.

In the simplest comparison, there was only one QX device, measured two different ways.

The two measurement methods and their respective output calculations should agree with each other, subject to the relative uncertainty of each measure. Therefore the magnitude of the uncertainty of the respective output calculations can be qualitatively assessed by comparison to the magnitude of the mean of the two output results calculated by different methods. If the magnitude of the either output calculation exceeds the magnitude of the mean of the output calculations, then neither output calculation can be considered to be representative without further data from this device. With only one example of each method, using this demonstrated device, there is no way to determine which is more representative than the other, although both might be considered representative if they closely agree.

Continued....

The sympathetic members of the audience then rationalize the discrepancy somehow, which then Rossi chooses the most flattering (to himself) version of, and later supplies as the correction, validating the sympathetic audience members, endearing him to the sympathetic audience some more.

The mezzanine heat exchanger is a possible example.

Lewan still presumably has his vision after looking directly into the reactor...

We can dream up many scenarios, some more plausible than others.

The spectrometer is supposed to be calibrated against a light of similar radiant power (check the manual, available online), etc.

The most plausible scenario is that Rossi makes up the number he wants to use for output, then calculates backward to obtain the parameters that would support this value, then designs the demonstration so that this can be reproduced. And occasionally he screws that up, leaving the audience guessing.

My trips to the arctic can now be explained. We build hydroelectric facilities that use the flow of water through the thin crust of the edge of flat Earth. This upsets the Asgardians a bit, since it interferes with water entering their flat world on the other side of our own.

Quote from can

I eventually gave a look at the indicated time. I think Rossi jumbled because the spectrometer didn't work as he wanted and he was getting irritated:

Compare with this from earlier on (which was consistent with what he said in Italian with Lewan in the lost audio section):

But maybe he's not measuring just the area of the plasma?

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Well then, if we go back to mm from cm the output is miniscule.

The majority of the math he used is straight from the Gullstrom-Rossi report, including the conversion of the Stefan-Boltzmann equation to suit cm² instead of m².

And then we have a few orders of magnitude problem with the fluid heating demonstration. Perhaps his reactors simply prefer to heat liquids uber alles.

Here is a start :

http://tinyurl.com/y8dkwc7b

Putting the 800 ohm resistor in series seriously restricted the current flow, and the voltage was reported to 11.52 V across the entire system. Strangely, the 1 ohm resistor was reported to have about the same voltage drop as before. (This is what I heard in the video, but I see differently posted above). If we try and use the current from before, 0.09 A, the 800 ohm resistor would have to dissipate ~6.5 W, which is not possible for the physical size used. It would burn, and rapidly.

If there is an AC component to the voltage, or rapidly fluctuating DC voltage, (or both) then the DC readings from a cheap voltmeter will be next to useless. The real voltage will be all over the place, and a cheap meter will display what it can, and there is zero guarantee that it will be representative. The 800 ohm resistor probably would introduce a lot of noise into the AM signal.

Anyways, just to summarize the reported radiant power (since this was getting mixed up here and elsewhere earlier), Rossi says at 2:27:00 or so, that the area of the reactor is 0.6 cm x 0.08 cm * 2 * pi and the temperature of 2036 K (not so well measured perhaps).

That makes 48.7 W, and the radius of the plasma/reactor to be 0.08 cm, or a diameter of 0.16 cm (1.6 mm)

(48.971 W if I don't skimp on all the decimal places in the area)

Quote from Alan Smith

Have a word with Mats Lewan - he has this in his possession according to his blog.

I asked him on ECW. My posts are moderated there, so it takes a while to hold a conversation.

(and I am being fairly well-behaved there)

So, no word on the watt rating of the 1 ohm resistor? No good photos of it so far.

I was just fiddling around with this, so now trying the link...

http://tinyurl.com/ydh866vm

@Mary Yugo ,

I have already demonstrated that 25 to 30 mV DC across a 1 ohm resistor is possible while passing 1.75 A through it. The AC voltage across the 1 ohm resistor at that time was about 1.3 V true RMS. And there was 25 to 30 V true RMS across the whole circuit at that time.

The handheld voltmeter, when set to DV volts, won't give any indication of AC volts.

The Klein (orange) meter imaged in the Gullstrom-Rossi report reports DV volts, but not by default. It defaults to AC volts, and so must be changed to DC manually.

The 1 ohm resistor rated wattage sets the current limit to some degree. They can pass a fair bit of current, but they will get hot. How hot can be used to estimate the current. A 10 W ceramic resistor will reach about 70 C at a 3 W load.

Quote from can

Paradigmnoia

Even if the electrodes themselves were not made of Ni, the gap is still supposed to contain Ni, Li and Al (and also Mn according to Gullström) in some form, which would eventually coat anyway the inner surfaces as they get vaporized by the plasma formed. I suspect this is the intended mode of operation and that it wouldn't really work like a regular HID lamp, where this would be normally considered a deleterious effect.

If for example the coating formed a conductive pathway between both electrodes, wouldn't it cause a short circuit that would vaporize it back into a plasma? Wouldn't the same happen for any other pathway formed between both electrodes with the evaporated metals?

For all intents and purposes this would work by repeatedly short-circuiting itself, minus the initial HV pulse.

Or at least, that's how I imagine it would "work"

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The HV pulse might be short-circuited, and simply not fire. Just like the 800 ohm resistor did not explode into a plasma.

An analogy might be a spark plug with carbon across the gap. Carbon coatings down the insulator and back up the threaded end can also short circuit a spark plug. That route can be 2 cm long on some plugs.

The gap is required in order to get the voltage high enough, and so it might be with the QX system. But who knows...

Hmmm..... I do have an old analog MSD unit in a box somewhere in my collection of stuff.

EDIT: The vaporized elements added to HID bulbs assist the bulb operation, as long as they vaporize and then turn to gas each time it is fired up. Some increase the internal pressure, some condition the output color, and some, like Xenon, condition the plasma voltage by changing the resistance of the gas with temperature. Xenon gas has decreasing resistance with temperature, the opposite of most gasses.

can , Eric Walker ,

Nickel melts at a lot lower temperature than the plasma is reported to be at.

Eventually the nickel will melt, boil or sublimate, and end up coating the "bulb" area. Even a fine dusting will be conductive at some point, creating starting problems. The coating will also reflect heat inwards, leading to heat failure. Nickel is a great IR reflector.

Tungsten is used in many systems for a good reason.

The reflected radiation is equal to being directly radiated, I would think. I don't see how it is additive.

On a flat surface, the radiation is hemispherical from a diffuse emitter. When ribs are added, the total area of hemispherical emittance does not increase per steradian when the view distance is larger in magnitude than the rib dimensions. The degree or roughness/diffusiveness is slightly increased, which improves emissivity slightly. If the surface was perfectly diffuse, the ribs would make no difference to the roughness, and could make it less.

The tips of the ribs are heated more, due to absorbing some of the radiated heat (when there is a gradient from rib tip to valley) whereas in a flat emitting surface, self absorption of radiation is not possible. This does not improve emissivity much (if at all) because it interferes with the gradient that is required to remove heat from the rib by conduction, increasing the valley temperature, but does increase the radiant power of the tips and valleys slightly, because it is hotter.

The benefit of the ribs is primarily given to improvement of convection rates, where the extra surface area will have a relatively large effect. (But is still greatly inferior to radiation above around 650 C).