There are a few adjustments to what you'd expect from this, but it is a good 1st pass approx.
@THH: Then you should ask Optris to correct the manual... What they certainly not will do!
They clearly state:
Since per the Stefan Boltzmann law, the electric signal of the detector is as follows:
U ~ εTobj4
I propose before You, next time, write up something, read first the manual of the tools, that were used...
Paradigmoia try to write in plain english . What you are writing is pure non sense. I will try to rephrase it.
Let me rephrase it: For every object that is hotter than a black body, (which must therefore be a grey body or a selective emitter), there is an equivalent Brightness Temperature.
The Brightness Temperature is the temperature that a black body would have to be in order to duplicate the same amount of power output as a non-black body at some other temperature.
The Brightness Temperature can be calculated for individual wavelengths, groups of wavelengths, or as much of the EM spectrum as can be measured.
Display More@THH: Then you should ask Optris to correct the manual... What they certainly not will do!
They clearly state:
Since per the Stefan Boltzmann law, the electric signal of the detector is as follows:
U ~ εTobj4
I propose before You, next time, write up something, read first the manual of the tools, that were used...
Wonderful. You now have the electric signal of the Optris. I'm sure that will be a lot of help to you, without the Optris signal to temperature map.
Paradigmnoia wrote:
Paradigmoia try to write in plain english . What you are writing is pure non sense. I will try to rephrase it.
Paradigmnnoia has been attempting to express what is correct about your claims, which you made quite complex, and quite unnecessarily so, unless your necessity is creating confusion. Pure nonsense, rephrased, would still be pure nonsense. However, what he wrote is equivalent to this rephrasing, which does happen to be more clear, not only than Paradigmnoia's expression, but also than your own prior expression. Congratulations.
Quote"A gray body or a selective emitter must be hotter then a Black Body if it radiates the same total power." This is due Quantum Mechanics, a law of Nature.
Does not regard any "mathematical description" but merely physics.
Sure. The mention of Quantum Mechanics, though true, is irrelevant and unnecessarily confusing here. We could tell the story of how QM was invented, and it was from the black body problem. It is radically off-topic. It is not at all necessary to consider quantum mechanics in understanding the issue of gray bodies and temperature and total power.
QuoteParadigmnoia wrote:
Now to be clear. All you have written, the plots and the rest just demonstrate that without the correct epsilon, the total emissivity, the same factor that you are using to produce your plots.
Without it, what? Do try to be clear, RB0. It does take reading over what you write to notice typographical errors and emissions and brain faults.
Assuming you meant " ... the plots will be incorrect," total emissivity, the value of the emissivity for a material, cannot be used to produce spectral plots, unless the body is a gray body. If it is a gray body, the spectral plot will be purely a function of the black body radiation for a body, reduced by the fraction that is the total emissivity value, because a gray body has equal emissivity in all bands, that is the definition of a gray body.
Here, you simply repeat your error, ignoring that, for his plots, he used spectral emissivity, not total emissivity. My opinion is that Paradigmnoa used more complex language and analysis than necessary, in describing what he did. But he is basically correct, and what you have been claiming, as far as I recall it, is way out on a limb. Not even close. You were going to explain "mathematically, " and you have not explained at all.
The camera measures band emissivity. In order to produce an accurate temperature reading, it must have the band emissivity for the material, and for the temperature. That need for temperature to be known is what led to the recursive analysis from the Lugano researchers. They picked an emissivity, read the temperature, then changed the emissivity to match the temperature, and continued this until the value settled. However, that process would not work if begun with an incorrect band emissivity, it would not settle on the correct value. They would be using the wrong emissivity curve, it would need to be for the camera band.
No expert has been willing to stake his or her reputation on validating the Lugano thermometry and calorimetry. Anonymous internet pretenders do not count for this. I am not claiming that critique is valid because of the reputation of the critic, and critics can be anonymous and still point out fact. However, the claim has been that Lugano was the work of "expert professors," and they were not experts in calorimetry, at all. It was out of their expertise, and as a result, they made some newbie errors. That's all. It's totally understandable in a way, except not in one way: did they consult with experts? Did they ask for expert review of their paper before publishing it?
Because materials may vary in emissivity, because of details of manufacture, it has been highly recommended to calibrate with the actual material, and that user guide gives methods.
One more point: RB0 did not completely quote the statement of Paradigmnoia, contributing to the point being lost. Here it is:
The Optris simply measures the power in the spectral segment, which is all of the IR range it can detect, and then the camera determines what the black body (emissivity 1.0) equivalent is.
The images show the spectral power between 7.5 and 13 μm for three different temperatures being all the same, but the spectral segment power at any level can be converted to a black body equivalent, from which the temperature of the black body equivalent can be determined, and then from that the temperature change due to emissivity
I can ensure you that my brain is ok. Also because I'm not alone.
We are always the last to know, unless we have cultivated friends who will confront us when we lose it. It also can be useful to have people who don't like us and will confront us for that reason. They will try hard to find anything wrong, and we can then consider that carefully. My life really started moving when I learned to do that.
... the power of T is much smaller than you would get from the Planck integral. Wyttenbach should have seen that you cannot generalise from the behavior wrt total power to the behavior wrt power in a narrow band far from the Planck curve peak.
What I wrote, though stated with more detail. The error had an obvious source, an easy misunderstanding. I would have caught me, easily, because I "know" that thermal radiation varies with the fourth power of the temperature. That's something I remember from basic physics. Right? That, I might easily overlook, is total radiation, not radiation in a band.
(And the user guide is not wrong. It states a basic relation then proceeds to modify it according to various factors. The 4th power formula given is not given as the formula for determining power from the camera reading. The manual then goes on to discuss the complications, in following pages, leading up to page 12, where calibration is described.)
@Wyttenbach,
I found another calculator that is a bit simpler. At least it calculates the band radiance instead of me having to manually sum the wavelengths up. (I'm glad this agrees with my earlier manual power matching within a watt).
Wyttenbach wrote:
Wonderful. You now have the electric signal of the Optris. I'm sure that will be a lot of help to you, without the Optris signal to temperature map.
No, he doesn't have that signal. Wyttenbach has misread the manual, and quotes that formula out of context. They are developing the formula for the signal. What he quoted is not the formula, that was the first step in an explanation. The actual formula is much more complex. For those who want to follow this, here is the user guide: http://www.optris.com/applicat…ds/Zubehoer/IR-Basics.pdf The quoted material is from page 9.
Because I previously described this, and Wyttenbach apparently didn't see it, he might have me blocked, so it could be useful if someone else quotes this. The final formula for the signal does not use 4 as the exponent, but n, and then explains that n can be a number between 2 and 17. Then the guide says that the complex math is stored in the EEPROM in the device, basically a lookup table. To function properly, the band emissivity must be entered, and since band emissivity can vary with temperature, differently for different materials, if the material is not a well-documented standard material, calibration is necessary. That is covered later in the guide.
Damn! We can be soooo sure we are right! Me too. I do this from time to time. Like yesterday. Fast track: Make a mistake, DGAF but correct it ASAP. Trying to avoid all mistakes is essentially dumb, because mistakes are the fastest way to learn. That is, if we can admit them!
This particular thread alone proves LENR is an energy transform process,
the keystrokes alone must involve several hundred watts,
what at a circle jerk,
Quote from Wyttenbach
@THH: Then you should ask Optris to correct the manual... What they certainly not will do!They clearly state:Since per the Stefan Boltzmann law, the electric signal of the detector is as follows: U ~ εTobj4 I propose before You, next time, write up something, read first the manual of the tools, that were used...
You have lost my respect.
Do fixed views triumph over easily verified experiment? Surely not, for somone with an enquiring mind.
The Optris document you refer to is not a manual, but a backgrounder - explaining how the Optris camera calculates temperature. It makes no recommendation about operator practice. But in any case it agrees with me, you seem to have omitted to read the sentences directly after the passage you quote:
QuoteSince infrared thermometers do not cover the total wavelength range, the exponent n depends on the wavelength λ. At wavelengths ranging from 1 to 14 µm. n is between 17 and 2 (at long wavelengths between 2 and 3 and at short wavelengths between 15 and 17).
Making mistakes is only human - refusing to consider clearly stated contrary evidence (the band calculator) is also human - but in my book despicable.
EDIT - I see Abd has also posted this!
Because I previously described this, and Wyttenbach apparently didn't see it, he might have me blocked, so it could be useful if someone else quotes this. The final formula for the signal does not use 4 as the exponent, but n, and then explains that n can be a number between 2 and 17.
Before we all loose ground I propose you ask Optris!
The exponent (2-3) (n is about 4 for 7.5 um) you mention is for metal measurement and given independent of the camera type and based on low temperatures measurements. This value might not be valid for the IR only cameras with high temperatures. Further on the exponent (n) values are stored together with the temperature, thus they have a multi dimensional backgroud.
(The one big reason for lower (exponent) n's lays in the ambient temperature correction, which is very large at low temperatures, we don't have...)
I'm pretty sure that the IR only cameras at high temperatures follow the Plank law very closely, as the newest sensors are optimized for 7.5 to 13um.
QuoteI'm pretty sure that the IR only cameras at high temperatures follow the Plank law very closely, as the newest sensors are optimized for 7.5 to 13um.
So - are you saying the radiance from a black body at 800C -> 1400C in the 8-13um band scales as T^4?
I think you are just scared to use the web calculator to find the answer, like I did, and Paradigmnoia did (actually we used two different web calculators!
Radiance (can be) measured in W/(cm^2St). And you can see from my post above that for a 50% increase in T radiance increased by about 2X. The exact exponent is 1.6. Not 4.
Now, instead of avoiding this, tell me which link in this reasoning you challenge. The web radiance band calculators? The figures P and I get from them? You think Watts are the wrong unit here to be scaled by epsilon?
Your argument about optimised sensors misses the point. A PERFECT band sensor (7-13um) would vary as per the change in the Planck curve at those wavelengths. This is nowhere near to T^4. Why should it be? The optris manual which you love (though I think it is a bit sloppy on the theory, not needed for use of the equipment) points out that the exponent for its sensors is highly variable.
I want you to engage with the maths here and understand precisely why that surprising exponent of 1.6 appears. It is not so difficult - and you can verify it yourself using a web calculator.
Finally - here is the Wikipedia page on Planck Law approximations:
https://en.wikipedia.org/wiki/Planck%27s_law#Approximations
You want the Rayleigh-Jeans Law which shows that the exponent is actually 1 in the long wavelength limit. That makes 1.6 look about right, given that the Planck peak at these temperatures is around 1.5u - much shorter than 7-13u. Notice that the higher the temperature, the larger the difference between peak and 8-13u, and therefore the closer to Rayleigh-Jeans we get. (The opposite of what you speculate above).
Quote(The one big reason for lower (exponent) n's lays in the ambient temperature correction, which is very large at low temperatures, we don't have...)
So work it out. You are wrong. Turns out 800C is not a low temperature! For 800C vs 30C ambient we have a T ratio of 1070/300 = 3. T^2 gives us 9 which means the correction for ambient will be at most 10%, not enough to change power law much. Go check the TC paper where he does all this stuff. Ironically, if you were right and T^4 was the dependence the correction would be even smaller!
Radiance (can be) measured in W/(cm^2St). And you can see from my post above that for a 50% increase in T radiance increased by about 2X. The exact exponent is 1.6. Not 4.
@THH: Sorry you mix up Wattage/ with temperature measurement... Of course is the wattage lower at longer (shorter) wavelengths, but this has nothing to do with the measurement process, which simply counts IR-photons.
You should carefully read the Optris formula and try to understand why e.g. at lower T "n" goes down, its not that difficult.
(1 – ε) · Tamb4 is the correction of the sensor source current for the ambient, which is very high if ε is low and T measured is close to t ambient. But in the Ecat case with ε high and T high this correction is negligible.
Wyttenbach, your attachment to being right, here, is leading you into error after error. The faster you recognize this, and acknowledge the errors, and there are blatant ones, the faster you can move ahead with real discussion.
THHuxley wrote:
@THH: Sorry you mix up Wattage/ with temperature measurement... Of course is the wattage lower at longer (shorter) wavelengths, but this has nothing to do with the measurement process, which simply counts IR-photons.
So another error. The measurement process does not count photons, though some sort of count could be inferred from it. It measures absorbed power that raises the temperature of the detector element, which has a crafted thermal response in the detection band. That is, "wattage" is used to measure radiance, it's a bolometer, that is what a bolometer does.
QuoteYou should carefully read the Optris formula and try to understand why e.g. at lower T "n" goes down, its not that difficult.
Now, without ever admitting prior error, Wyttenbach effectively acknowledges that there is variation in "n," that it might not be "4" as he claimed was established on that user guide page. He had pointed to that page as proof that he was right, but was misquoting the formula he gave, taking it out of context. I am not going into "why" it goes down, nor the conditions. I am simply pointing out that a massively defective argument was presented, about something relatively simple to verify, what the user guide actually says.
Standing firmly on that false argument, probably because it is what he expected from a general understanding of radiated power, he insisted that others had not read the manual. Look, this is embarrassing, I get it. However, if we want to progress, admitting errors is essential, and the faster we do it, the better. Most of us were raised under conditions that led to the suppression of the admission of error, thinking it looks bad. It looks much worse to display massive attachment to being right. There are trolls here who will jump on an admission of error, and, for years, will pull out the original error as some sort of proof of bad thinking or whatever.
Trolls. Never allow life to depend on what trolls will do.
(1 – ε) · Tamb4 is the correction of the sensor source current for the ambient, which is very high if ε is low and T measured is close to t ambient. But in the Ecat case with ε high and T high this correction is negligible.
I'm not following this. However, I think that is not the final correction, but was stated using total emissivity, not band emissivity.
THC and Paradigmnoia have used similators -- calculators -- to show that the conclusion of an exponent of 4 is completely inconsistent with what is actually calculated. An exponent of 1.6 is suggested, which seems quite low, below the level given as a minimum in the user guide page (but not much different, the minimum given is 2 and that might simply be approximate. After all, the range given is 2 - 17, and 1.6 would round off to 2.)
The real point of all this is that the emissivity must be based on calibration, or taken from sources where it was calibrated for the exact material, at the camera band. Because the band sensitivity could vary from detector array and camera conditions, general material emissivity tables can't be used with accuracy. A real calibration is needed specific to the camera, at least. That's the fundamental take-home point.
Wyttenbach is making what he considers to be valid points. To accuse him of being a Troll merely because he is arguing rightly or wrongly (in this specific instance I don't much care which) against the arguments that THH and others have proposed is very unfair IMHO.
Wyttenbach is making what he considers to be valid points. To accuse him of being a Troll merely because he is arguing rightly or wrongly (in this specific instance I don't much care which) against the arguments that THH and others have proposed is very unfair IMHO.
Who accused Wyttenbach of being a troll? Alan, I think you misread my comment about trolls, which was clearly not at all about Wyttenbach, himself, but rather was about how trolls may treat an admission of error.
Wyttenbach is not a troll. Period. He is also not anonymous, and actually has credentials.
Who accused Wyttenbach of being a troll? Alan, I think you misread my comment about trolls, which was clearly not at all about Wyttenbach, himself, but rather was about how trolls may treat an admission of error.
Obviously not as 'clearly not about Wyttenbach' as you imagine it to be. But I am glad you corrected the misunderstanding.
Abd Ul-Rahman Lomax wrote:
Obviously not as 'clearly not about Wyttenbach' as you imagine it to be. But I am glad you corrected the misunderstanding.
Only a very casual reading that assumed any mention at all of "troll" would be about Wyttenbach would lead to that imagination. And Alan frequently reads very casually and jumps to conclusions, this is not the only incident. When I wrote "clearly," I do mean that if one goes back and reads the full commentary, "trolls" wasn't about Wyttenbach, but about what trolls do when one admits error (or makes mistakes, whether corrected or not). There is nothing about that language indicating trollhood for Wyttenbach, almost the opposite, and Alan did not quote anything. The only evidence is his own knee-jerk reaction, which he acted on with his post. At least he didn't green-ink the thing.
No apology at all. Typical, again.
Wyttenbach is acting here like a pseudo-skeptic - holding onto a fixed view in the face of overwhelming contrary evidence, which he does not botehr to investigate (e.g. the band power calculators).
Abd has answered his latest struggles: I don't disagree but will give a simpler comment:
Quote@THH: Sorry you mix up Wattage/ with temperature measurement... Of course is the wattage lower at longer (shorter) wavelengths, but this has nothing to do with the measurement process, which simply counts IR-photons.
The IR bolometer does (more or less) count photons, and hence measures radiant power. There is a weighting got from the fact that shorter wavelength photons carry more energy. But that does not matter: there are many other factors altering the bolometer spectral response curve which Paradigmnoia will post again for you if you ask. It is vaguely flat.
It does make a difference to the magical 'n' value how the response is weighted (TC took this into account using a "typical" bolometer response from Bob H). Let us imagine this might be completely wrong, and investigate the two extreme cases of a detector over the range that massively over-weights low and high wavelengths respectively.
We can do this with the spot radiance from a BB calculator at 7 and 13u.
1073K (=800C)
7um: 0.122
13um: 0.0177
1673K (=1400C)
0.293
0.0342
ratio of radiance 13um : 1.93
ratio of radiance at 7u: 2.4
ratio of T: 1.55
n (at 13um) 1.5
n (at 7um) 2
My previous value of 1.6 looks a bit low to me 
Maybe I forgot the C->K conversion using the web thingy? Also maybe I was using 8u as limit not 7u. Anyway I stick with my original statement of "roughly 2".
You can see that the bolometer spectral response over this band does alter the results, but they are all in the ballpark I was claiming, and far outside the ballpark Wyttenbach needs to claim to preserve his COP=2 fiction.
We are still making a few assumptions (ambient radiance correction - small, change in e with T - small but still exists at 7-13u band since the high (7u) end begins to drop and this gets more significant at higher temps). And maybe a few more little ones I can't remember, check the TC paper for a decent initial summary of them.
Lets see if this makes any sense and saves a thousand words:
