Rossi-Blog Comment Discussion

Mary,

Have no idea if this person is who he says he is, sounds like a typical logical statement from any # of the “Skeptopaths” that troll this site, including me.

I have been spouting same for years only to get shouted down by the “Rossi Faithful”.

Based on his ideas of how quickly Nobel prizes are awarded, I assume he knows as much about "real science" as he apparently knows about bankruptcy law.

JedRothwell ,

Regarding technical errors in Indictation of..., I am considering the following potential source of error (as yet unquantified):

Temperature averaging rather than RMS temperature, where power is calculated from temperature (from equal size areas). Since P is proportional to T⁴, a simple average biases the calculated power to a higher value.

In Indication of, the temperatures of the various IR camera detection areas are simple averaged.

Consider the effect, for example, of the temperatures 100 C in one area, and 300 C in an adjacent area. The average is 200 C. The RMS T would be 158 C. The power calculated for the average T could be 2.57 times that of the RMS T.

Am I mistaken here?

Something long Rossi introduces the opening. Mother Earth perishes from oil pumping out, it should be stopped. So it is necessary urgently new energy, oil is blood of the planet Earth!

Quote from Paradigmnoia

Since P is proportional to T⁴, a simple average biases the calculated power to a higher value.

If you project your calculations for alumina, then for temperatures above 400 C, the decrease in emissivity compensates somewhat for the bias for high temperatures.

Thus also taking emissivity into account brings the temperatures much closer.

LDM ,

Let's not complicate this discussion excessively until the simple case is examined.

For sake of discussion, let's consider the case where the hypothetical emissivity = 1.This allows the cancelling of much of the equation.

Adrian's comment is deeper than it looks at first glance. The bar will have a temperature gradient from the 100 C end to the 300 C end.

The IR camera will do its own RMS-like power determination by directly measuring radiant power for each measurement area. (Let's assume the 100 to 300 C bar is divided into three equal areas).

Then the camera assigns a temperature appropriate for the total radiant power measured in each respective measurement area. For the 100 C end, the temperature reported will not be halfway between 100 and 200 C. (In fact, the 200 C middle will span a temperature range from < 200 C to > 200 C). The temperature reported by the IR camera is effectively a proxy for the radiant power, which is a logarithmic function of T.

The radiant power along the 100 to C C bar will have a logarithmic distribution, even if the temperature gradient is perfectly linear. (In real life, the temperature along the bar will almost certainly have a logarithmic gradient also, but let's keep this simple as possible for now.)

Where I am getting hung up on is, if the T is a proxy for a logarithmic value, is simple averaging T akin to simple averaging of logarithmic values? (Leading to skewed results)

Or, because the IR camera (in theory) effectively does an integration of the radiant power in each segment to determine a temperature value, can the respective reported temperatures be simple averaged and the power calculated from that average give the same answer as summing the calculated power for each individual segment?

Or, do we need to use the RMS temperature of the segments to achieve the correct answer? (Or is this method perhaps more appropriate for spot T measurements?)

I might have to do some tests with the Optris software to see what actually happens.

Consider for a moment also the position a thermocouple would need to be at, within each segment, to match the IR camera temperature for each segment in the above example 100 to 300 C equal gradient, (and for a logarithmic temperature gradient) bar.

(In our mental experiment, the bar has the temperature gradient along the X axis, the Y axis is isothermal at all respective X coordinates, and there is no Z axis.)

RMS = root mean square?

What's the benefit of using that instead of a simple average?

Isn't RMS averaging basically just a way to normalise and include negative values? Ie. The RMS of a curve equals the simple average...?

@Adrian Ashfield ,

The IR camera will report a different T for different sized areas, even if the measurement areas centred on the same location, if there is a temperature gradient across that area. This is because the camera measures radiant power from the measurement area, not the temperature directly. The radiant power is increasing proportional to T4 across the temperature gradient. The camera therefore calculates a proportional root 4 T based on the total radiant power determined in the area measured.

It seems likely to me, after some consideration, that the power calculated for the total area from simple temperature averaging from the different IR camera measurement areas will indeed correctly reflect the power calculated from the averaged temperatures that are originally based on actual radiant power measured for the respective areas, as long as the temperatures being averaged from the different measurement areas are areas are of equal size. The camera itself is integrating the power over the measurement areas to a single proxy temperature representative of the total radiant power in each area.

I was mostly wondering if this assumption is correct.

When I get in the right mood, I'll do up a spreadsheet and see how it hangs together.

Zeus46 ,

RMS is probably not what I actually mean.

Something more like the logarithmic mean or geometric mean is more like what I am thinking of.

It's been a very long time since I did any of that sort math

P..noia wrote

"The camera itself is integrating the power over the measurement areas to a single proxy temperature representative of the total radiant power in each area."

"

In-camera or manufacturer provided calibrations and temperature conversion algorithms may provide sufficiently low measurement uncertainty, however these results are often ‘hidden’ within the camera operating software and are not readily available for thorough uncertainty analysis, or, these algorithms are known but their form inhibits analytical means for uncertainty analyses. "

nvlpubs.nist.gov/nistpubs/ir/2016/NIST.IR.8098.pdf

Interesting...

I averaged the temperature from the 40 temperature cells from Indication of..., (page 9), and got the average T of 711.5425 K, same as the report (rounded to 711.5 there)...

I calculated the radiant power, and arrived at the answer I got before of 1505.727 W (rather than the reported 1609 W, which I have commented on previously).

Then I divided the area 0.1036 m² by 40, and calculated the radiant power for each of the 40 temperature cells, individually, and summed the power of all those cells.

I have calculated 1607.7 W that way...

So, FWIW....

the geometric mean of the all the cell temperatures results in 706.45 K, leading to 1463.13 W

and the RMS of all the cell temperatures is 715.997, leading to 1543.78 W

(All power calculations do not include the room temperature corrections, subsequently applied in the report, page 10)

RobertBryant ,

As long as the camera does whatever it does consistently, whatever internal algorithms it uses is of little consequence.

I am not discussing the uncertainties or errors caused by the IR camera itself, but those that might be caused by errant calculation methods using the data generated by the IR camera.

Average of 40 cells = 711.5425 K (711.5 K report).......... 1505.727 W over total area (1609 K report).......... 1607.711 W summing 40 cells individually calculated

Average of 20 cells = 710.695 K (710.7 K report)............ 1498.556 W over total area..................................... 1590.225 W summing 20 cells individually calculated

Average of 10 cells = 709.445 K (709 K report)............... 1488.093 W over total area..................................... 1552.993 W summing 10 cells individually calculated

Cell # and individual cell temperatures equals those as described in the Indication of ...report, page 9.

I just noticed that in the report, the average of the T^4 was used to calculate power (2.73694 * 10¹¹). This means an effective T of 723.2964 K

Therefore in that case, total P = 1607.711 W

Bahhhh....

If one uses the geometric mean of the T⁴ for each of the cells (instead of averaging them), in all three versions (10 cells, 20 cells, and 30 cells), and then calculate radiant power, all three agree very closely, with the following resulting temperatures (from least cells to most) : 706.4538 K, 706.3596 K, and 706.7204 K

The calculated power is then (same order) :1463.113 W, 1462.333 W and 1465.323 W

I don't know if this more or less accurate then what was presented in the report, but considering that we are discussing the same object sliced and diced different ways, the pieces should all add up to about the same. Obviously, this adds up to a ~143 W reduction in the reported power, but that is not my point at present.

(It certainly demonstrates how finicky these calculations can be).

The question remains: which method is most correct ?

Ok, then. Thanks for your help, everyone that contributed.

The upshot of my wandering thoughts (above) is that the arithmetic mean of the numeric values for T⁴ of all individual cells, entered into the Stefan-Boltzmann equation for the total area is equivalent to the summed power of the individual cells when those cells are calculated independently. Good to know.

Averaging raw temperatures would seem to be poor practice and seems to skew the power calculations to a lower value.

It is interesting that increasing the number of measurement areas seems to increase the overall power, but I accept that the metal frame blocking part of the measurement area is the main culprit in this case, as the area of unaffected cells is increased in size when the total number of cells is increased.

So I thereby retract my earlier statements about the total power being calculated mathematically incorrectly in the Indication of... report. That part seems fine. Now I fully understand what was done in that section.