Posts by randombit0

Quote from Abd Ul-Rahman Lomax

Being "well seen" by the camera is not particularly relevant.

Quote from Abd Ul-Rahman Lomax

Yes. We know that from black body characteristics, and this was known before QM, so we do not know it "based on Quantum Mechanics," it would be more true that QM was based on the black body problem.

Dear Abdul,
this two phrases demonstrate that your field is probably Grammar or Literature, but not Science or Engineering.
Not a problem Writers and Journalist are nice people. I will try to explain some concepts with simple words.
First :
How much signal you receive on a detector IS relevant ! If the received signal is low because the detector is not "tuned" to receive it the experimenter risk to collect more noise ( random signals ) then significant data. This make the measure impossible. In technical wording is called Signal to Noise ratio and for some details look at https://en.wikipedia.org/wiki/Signal-to-noise_ratio.

Also. Before QM BB radiation was known only as an experimental fact with ( almost ) no explanation. Is just after QM that we really KNOW a complete physical explanation of that phenomenon and so we also know that there will never be a material with better emissivity !

If you need a more detailed explanation please ask.

Quote from Paradigmnoia

Lets see if this makes any sense and saves a thousand words:

In fact the picture is nice but NOT the interpretation you give that is totally wrong.
In the picture we ca see ( we know that already ) that Alumina is very well seen by the camera. That is an important fact because if you do not detect signal or detect a very small signal a good measurement could be impossible because of errors.
From our previous reasoning based only on Quantum Mechanics we know that for a given input power at steady conditions a non ideal emitter should be much hotter then a Black Body.
For any material so we have to inform the camera about the total emissivity so that the camera software can compute the temperature correctly. This is valid also for materials that have a good emissivity in the camera window. If we don't do that then the camera would treat the data from the non ideal emitter in the same way as a Black Body computing a much lower temperature.
It doesn't important that "the body emits less in a part of the spectrum the camera don't see" because this kind of considerations ( normalization to camera ( and optics) sensitivity ) are treated by the factory calibration process. A camera wouldn't work without the specific calibration files.
What we do using total emissivity is to inform the camera software of how much power the body emits compared to a Black Body.

Quote from Abd Ul-Rahman Lomax

The RB0 reaction is amusing, showing the major brain fault involved,

I can ensure you that my brain is ok. Also because I'm not alone.

Quote from Wyttenbach

I hope that nobody is satisfied with this findings except the people who adhere to scientific standards...

Interesting. I'm quite happy. May I ( we ) have the paper ?

Quote from Paradigmnoia

And an object hotter than a black body still has a mathematical equivalent to a black body at the same radiant power level which will have a temperature specific to that amount of power.

Paradigmoia try to write in plain english . What you are writing is pure non sense. I will try to rephrase it.
"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.

Quote from Paradigmnoia

The Optris simply measures the power in the spectral segmen

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.

Quote from Wyttenbach

2) Then at the point where you calculate the Energy output of the reactor at the measured T you have to use the low total emissivity!!!!!!

Oh Dear are you crazy ? When calculating temperature (from energy) you use one vale of emissivity and when calculating the energy back you use another ?
You would have no energy conservation in this way !
Of course we know very well the manual of Optris and even page 9 on which there are many formulas, not only one, all of them only with one value of emissivity !

Quote from THHuxley

I'd be interested for you to provide your reasoning.

Oh I see you have lost the short term memory...... I will repeat it for you even if I posted just few days ago.

Consider a normal resistor with at a fixed power in equilibrium condition in vacuum.
Then the input power and radiated power must be the same.
Consider first a Black Body.
If you integrate the Plank curve then you have the power density radiated from the surface of the resistor IBB , and multiplying than for the body area A you obtain the input power IBB*A .
Because Quantum Mechanichs impose that no material can have an emissivity higher then that of a Black Body if you have a gray or selective emitter body with identical geometry then to maintain Ib*A=IBB*A constant the body must be hotter then the Black Body in order to have Ib=IBB.
This simple considerations are independent from any detector. The ratio Ib/IBB is called total emissivity.

Hello again everybody,

Quote from Abd Ul-Rahman Lomax

Important for what?

Dear fellow, what game you are playing ? I will repeat some important points we agreed on.
Starting from first principles without considering any camera or detector we concluded that a Black Body Dog Bone would be much cooler then an Alumina Dog Bone, because the second has a lower total emissivity.

Paradigmoia has also demonstrated that Black Bodies and grey bodies could have a very similar emission in a selective spectral range even if the gray body is much more hotter than the Black Body. So to calculate the temperature from the integrated energy the total emissivity of the material is necessary if that spectral range the same of the detector.

Quote from Paradigmnoia

within the camera spectral sensitivity range only.

No Paradigmoia, we can discuss this privately, but because of the reasoning we have done is interdependent from any detector (so valid for all detectors) you can treat gray and selective emitters in the same way.

Quote from Paradigmnoia

What is being matched is the effective emissivity of the microbolmeter to the emissivity of the object

This is a good point. You here demonstrated that Alumina is in fact one of the best materials to be used with the camera because have a good emissivity in rhe camera sensitivity range.
Because of the first point you stressed (with gray and BB) total emissivity is necessary to calculate the temperature.

Quote from Paradigmnoia

compare that segment of radiant power to that which a black body produces in the same small segment of the entire IR band, in order to calculate a temperature.

No Paradigmoia. the comparison is made on the total integrated spectrum. Otherwise even a simple measure for a gray body would impossible as you have shown.

Quote from THHuxley

So how could total power (summed over all frequencies) be relevant? Similarly how could total emissivity be relevant?

Total power is relevant as well as total emissivity for the reasoning explained.
Remenber that we started supposing a normal resistor in vacuum in equilibrium condition so that the radiated power must be equal to the input power-

Quote from Paradigmnoia

will all generate the same blackbody temperature of 1273 K, or 1000°C according to the Optris.

That's why the total emissivity parameter is so important !

Quote from Paradigmnoia

However, selective emitters can be approximated by grey bodies.

Thank you Paradigmoia. So if we need to measure Alumina we can treat it as a gray body with the correct emissivity.

Quote from Paradigmnoia

That should say "Objects that do have a better emissivity are cooler than an object with low or selective emissivity at the same level of total power"

Very good so you agree that a Black Body "Dog Bone" would be much cooler then an Alumina "Dog Bone" at the same level of total power.
That's why Alumina emissivity must be used. To measure the correct temperature.
I'm very glad that you finally agree on this point.

Quote from zorud

I am sure quite a few folks here are still waiting for your derivation...

Ok boys we can start !

Quote from Grafiker

Does this help?

Thank you for your contribution but NO, It does NOT help. The Optris window is not part of the reasoning. What we are writing about is physical considerations that motivate what Paradigmoia has found experimentally.
Objects that do have a better emissivity are cooler then object with low or selective emissivity.

Quote from THHuxley

In fact, since the emissivity within the IR band is identical for A and B, for the same IR band output we must have the same temperature.

No. You and Lomax are missing the basic physics.
Again ( and again)
We have two system a) ideal BB b) Alumina with constant (known) input power 100 at equilibrium in vacuum ( no convection)
The two system are geometrically identical.
1) because of equilibrium condition input power must be equal to the output power.
2) because of QM emissivity of b) can't (no typo is NOT POSSIBLE) higher then that of a) (ideal BB) at any lambda ( or frequency )
3) because of 1) the integral over all lambda of the emitted power density must be equal so that integrating again over the body surface (identical for the two bodies ) we obtain the input power 100.
4) because of 2) and 3) then we obtain that b) must have a much higher temperature then a.

If you need a simplified explanation:
5) because the actual shape of the Plank curve is not relevant to our reasoning we can suppose it constant up to a frequency Lth, Integrating the curve for a BB a) is in that case equal to calculate the area of a rectangle with base from 0 to Lth. The height of the rectangle is function of T and the total area give us the total emitted power density.
6) suppose that you have a selective emitter b) that can emit only in the interval [La,Lb]
7) so we compare an area of a rectangle with a long base [0, Lth] wit one with a much shorter base [0<La, Lb<Lth]. Because of 1) the two area must be equal.
This implies that the height of b) rectangle must be higher and so its temperature.

Next time we will make a complete derivation of Plank law with all the math.

Quote from Paradigmnoia

You are on a reckless path to certain self destruction.

Oh Oh ! We (about 1000 employees )are all scared by you.(a student!) ( Buh ! Treat or Trick ?)

Quote from THHuxley

If you think differently, pray

Even if we are not related with "Bert", I will try to answer you.......
You have missed totally the point ! I will try to illustrate it (again and again) for you ( and all others boys )

Try to think that IF you are testing a normal resistor in vacuum (so no convection and conduction) in equilibrium condition the same power ( let say 100 per unit area in arbitrary units ) must flow in and be radiated. In that condition temperature is constant. Suppose that at first you have a black body resistor. Then it will reach a temperature T so that the integral on all spectrum of the emitted radiation curve ( a perfect Plank law curve ) must give 100.
A gray body would of course have to reach a temperature so that the integral of the emitted radiation must reach 100. That is because we are in equilibrium and because the gray body radiates in a less efficient way his temperature will be higher then the temperature of a black body.
Now consider a non gray body.Remember that for any frequency any body can't radiate more power then a black body. That is Quantum Mechanics.
This means that if you have a body that radiates only in a window of the spectrum at a temperature T is not possible that at any frequency the radiated energy per unit area can't be higher then Plank curve.
But if the integral (limited to the emission window) of the curve must be 100 then we have that the temperature T of the body must be much higher then the temperature T of an ideal Black Body emitting the same total power per unit area.
All that you have from:
1) Equilibrium conditions
2) Quantum Mechanics.
Total emissivity, that is the ratio of two integrals, express exactly the equilibrium condition.
If you need we can go on. I can make a complete derivation of Plank law for you and also make a complete calculus of the integrals.

Quote from THHuxley

Randombit0 was tiresome

I'm not getting tired. I was abroad working so I found the time to answer just now. I'm working elsewhere not here.

Quote from Paradigmnoia

You continue to repeat a false statement as though it is fact.

Are you writing about yourself Paradigmoia?
I see that you must continue "to not understand" and just repeat what you must repeat.
You intentionally ignore experimental facts ( from the Lugano Report ) ant try to generate false evidence.
If you want to really measure the emissivity of a brick then you must follow a strict scientific procedure.
Is not just like playing with kids !
Seems that the motto here is that anyone can do any measure in better way then experienced scientists.
Nice dream. Reality is much harder.
Regarding the table you linked and the page of Optris manual.
We of course know those sources as many more others. I have tried to explain you when and why use narrow band emissivity and when total. I have explained to you that our company normally use narrow band IR measurements to obtain the emissivity of the ground from satellites. You can detect a lot of interesting things with that method.
But I see that you completely ignore what I'm writing. Apart for dislike it.
No problem. We have time.
BTW emissivity has nothing to do to the gas (air) or other materials that are between the camera and the measured object.
Emissivity is a property of the object surface. Then if you are observing it through others materials (e.g. Quartz glass or Alumina window) win then you must correct the measure for "transparency" of those materials adding the energy that was absorbed by them. Lugano people have done no correction in order to avoid temperature overestimation.
So their measure is probably slight lower then real.

Quote from Paradigmnoia

randombit0,
Emissivity charts are an assembly of information.

Yes Paradigmoia, as any book, any encyclopedia or the whole Internet.

Quote from Paradigmnoia

The Optris can no more see the entire IR band than your eyes can see the whole IR band.

I have tried to explain this point to you is a question of calibration. Regarding human eyes: all of you has attempted to evaluate temperature from color ( of a photograph ! ) so in fact you are using some sort of calibration against the black body and even note that Alumina is partially transparent at high temperature (read MFMP site).

Quote from Paradigmnoia

if you cannot differentiate between total and partial (spectral) then you might find difficulties in all sorts of things. Like totally full pipes and partially filled pipes, or totally dry steam and partially dry steam, or totally fulfilling contracts and partially fulfilling contracts.

Is very difficult for me to understand your surprising and foolish statement.
Are you thinking that I'm Rossi ? ( from the fact yo used the word "contract". )
For obvious reasons I will neither confirm nor deny any information about my identity.
But I can tell you that our company knows very well what emissivity is and how to use it. Spectral and Total.
In the Lugano case Total emissivity was the one to use.

Quote from Paradigmnoia

I suspect that the specific heat capacity changes with the change from solid powder to a gas, liquid and solid fuel combination is enough that the core temperature increases with the same power input,

Paradigmoia. Solid fusion or liquid vaporization are endothermic process. You need extra heat to make the phase transition. Condensation and solidification are exothermic process. Heat energy recovered ( but not generated ) by those process.
Heat pipes https://en.wikipedia.org/wiki/Heat_pipe work exploiting this phase transitions.
If you provide the same amount of energy to two pipes, one empty and one containing e.g. some water, the filled one should be cooler because of specific heats.

Quote from Paradigmnoia

excess temperature is not excess heat

What a confusion ! Temperature is quite a different concept ( definable e.g via statistical mechanics ) then heat. Temperature is not "transported", while heat, that is a for of energy, is. Refer e.g. to Zemansky "Heat and thermodynamics : an intermediate textbook" or other textbooks.

Quote from Paradigmnoia

The total emissivity then is unimportant to the Optris, and wrong to use for its emissivity function.

WOW ! you are saying that ALL (not only Optris) IR camera manufactures are wrong and you are the only one right ! Come on In the table that YOU cited the vast majority of emissivities reported is Total. Even for Brick ( a non metal ) that is in the same table of Aluminium Oxide.
The fact that the sensitivity of a detector is limited in a window is considered in his calibration data. Band emissivity data wold be extremely difficult to use because one should consider also the exact band sensitivity and response of the detector, that is different for each pixel in a camera, for each optic mounted on the detector.
This would make extremely difficult, if not impossible, to compare measures from different devices.
Emissivity tables should also be different for each detector.
Using integrated functions instead, after a proper factory calibration ( that contain all the data regarding each part of the detector ), the measure is much more simpler and possible.

Quote from Paradigmnoia

The total normal

.......

Quote from Paradigmnoia

The GIGO principle applies here.

My Goodness !
You are quite repetitive in your errors !
We have discussed that a long number of times ! and I presume we will discuss it for a long time more. All detectors are calibrated using Total Emissivity for the material. If that value is known that's the value to use.
I will send you a complete textbook on Thermography next time .
And also you are making a big confusion among a general math method and a procedure to measure emissivity if you are able to make two narrow band measures.
Lugano people used the general math method using the correct alumina curve.
Because the method is general if you use the wrong curve, i.e. the Ascona rainfall or any other curve, you will obtain a wrong number.
But at least I see that you agree for one important thing when you say:

Quote from Paradigmnoia

At about 200 C, the spectral emissivity near 7 to 14 um and total emissivity are very similar. The emissivity found at that time would have been close to the emissivity that should have been used for the rest of the test, for the camera.

then you say that you would use a value near 0.65 ( the value at 200 °C ). This is much less then the 0.95 claimed by others.
I really can't belive that you think that this value should not change with temperature when even MFMP has found that alumina became transparent (i.e. emitting less radiation !) at high temp.
Also is nice to find that you agree with me that the Inconel would not disturb the measure. So only the alumina curve should be used as the Lugano people has done.

Quote from Paradigmnoia

Again, the alumina on inconel emissivity was the total hemispherical emissivity in the Wade report.

And that is exactly the the value to use and that why is reported in emissity tables. Eventually, as the Lugano group has done, using the right emissivity vs temperature function.
I see that you select facts and report only what is not against your opinions. Should I remind you that the Lugano group had measured the emissivity of the Alumina pipes and found a value compatible with the literature ?

Quote from Paradigmnoia

The thickness of the coating of alumina and zirconia on the Inconel should also be taken into consideration

As you cite this is true just for very thin layers. Due that most likely (read the Lugano report) the thickness of the Alumina was more then 0.4 mm just the Alumina emissivity should be taken in account.

Quote from Paradigmnoia

The iteration method requires multiple IR intensities or multiple IR bands

No Paradigmoia this means that you are not understanding the math. Iterative methods and fixed point iteration methods are quite general and were successfully used by the Lugano group. Please read Wikipedia for more reference.
The methods you are citing are used for example for measure terrain emissivity from satellite data. We do that normally in climate and environmental research. Is not the Lugano case.

Quote from THHuxley

Your comments here are trolling, pure and simple.

To be qualified as a Troll by one of the most active trolls is a compliment ! But no, thanks, I'm already engaged!

And how can you be sure that none of the Lugano team was expert in Thermography ?
Did any of you know that one member of the team is expert in criminal scientific investigation and has quite a good knowledge of chemistry, physics and almost any kind of laboratory technique ?
A very nice guy to talk at indeed !
And also the paper was seen and reviewed by colleagues of the Authors before the publication.
And still you affirm that NONE of those Academics was sufficiently expert!
But I see also that you want to discard any opinion of experts and qualified people !
Are you a member of the Flat Earth Society ? http://www.theflatearthsociety.org/home/index.php</a> <br>

Quote from Paradigmnoia

See the Optris version of the ε below.

O Dear Paradigmoia,
Thank you very much for the link to the "Infinite Energy" page. I see from that page that MFMP used the value of a generic "CERAMIC" emissivity as a first guess, tacking it from the manual of the Optris PI camera. This is quite wrong. CERAMIC means a vast number of materials so that number could be referred to a generic ceramic not to pure Alumina.
In fact fro the table of the book IR-Basics of Optris (not the manual of the camera)we see that the tipical Total emissivity of alluminium oxide is extremely low, with reported values (0.2; 0.3; 0.46; 0.16) that are even lower then the actual values used by the Lugano group.
Note also that those single values do not take in account the variation of emissivity with temperature.
Because this dependancy is important the Lugano group correctly found a measurement in literature and referred to that.
Without referring to previous work any measure or scientific activity should start from scratch, and we would have no progress.
It is so quite natural and not, certainly a scandal, to refer to literature from NASA for the emissivity of Allumina on Inconel.
Measures have value independently from the year on which they have been made.
So what's the problem ? Are you in trouble because your ideas are falsified by NASA ?
And about the iterative method: This is quite a normal graphical method to solve an equation, is called Fixed Point Iteration Method.
Try to search it with Google.
The ref in the Lugano paper show that this method is applied also in the thermography field.
So my dear Paradigmoia what's the problem ? The more we study this topic the more we see that Lugano procedure was correct.
Sorry if this hurts you.