Experimental Evidence on Rossi Devices

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I pointed my laser pointer from my IR gun at a 2.5 mm thick pure alumina slab. The red laser spot is visible on the opposite side, although weakly. That means that with a power of < 5 mW, 630 to 670 nm light can easily pass through alumina.

Not precisely. It means that most of the radiation at this wavelength is scattered by 2.5mm alumina. Thus, at this wavelength, the alumina is mostly opaque, with most of the power scattered and therefore subject to the alumina emissivity not that of any interior object.

Note also that the total fraction of radiant power at even 1um or longer wavelength, for 800C, is > 99%.

https://www.sensiac.org/extern…d_radiance_calculator.jsf

And longer wavelengths mean more absorption. Your experiment has shown that in fact the alumina total emissivity will be to a good approximation that which determines reactor radiated power.

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This is close to a strong lithium spectral line. Which could mean that if the lithium inside were to produce strong coherent light at this wavelength, it would go mostly unnoticed by the Optris camera using a 750 to 1300 nm band, and therefore it would go unmeasured.

As you've just shown this frequency is mostly scattered, weakly passing straight through. In any case the optical area of the fuel is much smaller than that of the reactor, radiation correspondingly less. And Li does not emit coherent light when you heat it up to 800C. This argument is wrong 3 ways over.

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The other lithium spectral lines are much shorter in wavelength, and would definitely not be seen by that Optris. This just by way of example.The temperature of the alumina is related to how poor the alumina transmits heat, not how much energy is contained in the device at any given moment.Therefor I posit that the temperature that Thomas has calculated is the lower bound.

You are arguing that the fuel emits energy otehr than as a black body. That is highly unlikely. But, in any case, te reactor tune is there to absorb most of such energy and re-emit it according to alumina's total emissivity.

Whether the optis "sees" higher frequencies is irrelevant. What is sees allows accurate determination of temperature. The alumina (from your experiment) mostly scatters frequencies much higher than the Planck peak at 800C. What is not scattered will be emission from the fuel and that is mostly nickel. There is no conceivable reason why there should be strong spectral lines. Ask MFMP to measure this if you don't believe me.

So: your experiment has shown that the alumina emissivity book value is a good approximation to the power out. The temperature is known whatever. The Lugano clauclations hold good which is sort of expected since they get (without fudging) within 10% of the expected (COP=1) answer.

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Which assertions are not true, in your opinion?

Your assertions here:

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If the reactor was outputting the bulk of its heat in the wide range of IR outside of that detected by the Optris camera, then it could have been easily been outputting 3kW, or even much more. The temperature measured by the camera is not necessarily effective at measuring the total output, although it could be. But this requires that the alumina cylinder visible to the IR camera acts like a glowing pure alumina cylinder, not a device of unknown composition and complex construction, with several metals and gasses inside doing who-knows-what.

And look back on the thread to my detailed reply explaining why these assertions are false. Short version: "alumina at IR cam wavelengths is highly opaque and has understood emisivity".

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Urban Eriksson, I do not dispute that the Lugano test is seriously flawed. I do dispute the assumptions used in hypotheses about what the device was actually doing, since the real data is so poor. Simply modelling a complex device as if it were merely a solid stick of alumina is a gross oversimplification, and can only lead to grossly oversimplified conclusions. Conclusions that are only applicable to the limited case regarding only the assumptions used. That does not mean that any particular conclusion, including the one derived by Thomas, is wrong per se. There just is not enough information to prove or disprove that any conclusion is, ahem, conclusive."With three parameters I can fit an elephant"

My analysis does not have any free parameters. And it fits the "acceleration" to within 0.5%, when none of the material characteristics can alter much between the two temperatures over which acceleration was claimed.

There is uncertainty from the alumina total emissivity, under the assumption (which you have just supported with your IR pointer experiment) that it is mostly opaque that uncertainty becomes small. I reckon my +/- 30% error estimate is about right - perhaps a bit conservative.

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Simply modelling a complex device as if it were merely a solid stick of alumina is a gross oversimplification, and can only lead to grossly oversimplified conclusions.

At IR cam wavelengths we know anything inside the alumina will be scattered by alumina and so the temperature can be obtained accurately subject only to the band emissivity. Because this is close to 1 variation in samples etc will not chnage it much at all, and the temperature measurement is safe.

The power out depends on the effective emissivity of the reactor at 800C. It is true that thin alumina can have enmissivity affected by interior details. But the alumina scatters power and the scattered power gets absorbed as heat and re-radiated at a known emissivity from the alumina. This is around 0.5. +/- 30% gives us 0.65 to 0.35 which is pretty safe given that alumina is more opaque at longer wavelengths, so sensitivity to internal emissivity is very partial, and that must worst case vary between 0 and 1.

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The temperature talked about here is measured by the camera and is that of the alumina surface, since the camera operates at wavelengths that do not "see through" alumina. If high temperature black body radiation is created within the alumina body it can partly be transmitted through the alumina. If it is passing straight through or if it is scattered on its way through does not make a difference if absorption is not involved. An analogy here would be a matte light bulb which has a scattering layer on the inside of the glass. Measuring with a heat camera on a matte light bulb would give a correct value for the temerature of the glass bulb. Still more power is radiated from within the lamp to the surroundings than would be anticipated if one would calculate the radiated power based on the temperature of the glass bulb.

It is not credible for there to be a high temperature gradient between the alumina and internal (higher temperature) objects since alumina conducts heat well (independently of radiation).

My point was that at plausible internal temperatures most of the radiant power is absorbed by the alumina. I agree there is uncertainty about how much. Antoine was claiming his infra-red pointer experiment proved 630nm was not absorbed by alumina - but it did not because he observed a weak spot through the alumina.

However I'll give you this. Scattering is not the same as absorption and if the total (from all angles) 630nm power emerging from the alumina is comparable to that entering - which might be true even with a "weak" spot visible - this experiment does NOT show that the alumina is mostly opaque. But, equally, it does not show it is not mostly opaque.

Actually I think this is something we should be able to get definite data from unless absorption (as opposed to scattering) is highly dependent on microstructure. What matters is 2um-6um where most of the 800C black body radiant power is found.

From this: http://ntrs.nasa.gov/archive/n….nasa.gov/19670003469.pdf (page 61)

It looks like alumina (tested at 0.78mm thick) has low absorption over this range which means that the absorption could be low for the reactor, and therefore the power out more uncertain.

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No personal insults. You are making progress toward civilized social behavior. Keep up the good work. Your FUD is only helping Rossi keep a low profile. Please increase your efforts to protest Rossi in this way.

(1) As somone clearly wanting to proote LENR research i'm very surprised you want to promote Rossi
(2) Rossi lives on internet comment. As with politicians, there is no such thing as bad publicity. after all, he has nothing substantial with which to promote himself.

Axil -

Let us suppose I am an LENR fan. I interpret all the marginal data as indicating nuclear reactions without noticeable nasties.

Rossi claims to have scaled this up. Fine: I am disposed to be sympathetic.

But then I look at Rossi's 15 or so demos, all of which have glaring holes. I note that Rossi has been asked (by friends) to close these holes and has not done so. I look at the fact that Rossi's patents depend on him having got stuff to work - and the Lugano test used for this purpose showed an electric heater.

I have to reckon he is 100% incompetent or deliberately pretending to have nothing - in some cases both.

You have to have a special sort of mind to turn that into a judgement that he has any technology.