Is there a formula or way to manually calculate the temperature of an object using an IR camera or Raytek gun if you take all temperatures using an emissivity of 1.00 on IR equipment?

 

Say using an e=1.00 on a P60, Infracam, Mikron, or other camera or even with a Raytek gun, etc… Say you record a temperature of 100 F and you know the background temperature is say 80 F. Is there a way to manually calculate the new temperature for an emissivity of .80?

 

I heard somewhere that there is no way for a person to use a formula to calculate this because all of this a proprietary to the IR camera maker and based on their own calibrated blackbody curves and that it is all different for each maker. And then I read a technical newsletter that some engineers in a refinery are using LAND pyrometers with an e=1.00 and manually using a formula to correct for the correct temperature.

 

What’s right and what’s not? Can you manually calculate the temperatures or can you only get corrected temperatures by using the IR equipment or software form the maker?

 

Steve

 

Here are 2 formulas I came across:

this one was posted sometime ago on a messageboard:

Correct temperature = (Instrument reading – Treflected) x (Instrument E/New E) + Treflected

 And this one was in a technical newsletter for furnace tubes: (I scanned the section of the newsletter with the formula. It should be attached as a jpg):

I would like to know if these and other formulas are legit? Or that there is no way to manually calculate without the IR equipment maker's calibration info. 

Steve

 

I dont think you will find a universally acceptable formula for this purpose suitable for using in the field, and if you do it will be more difficult than entering the correct values in the camera anyway. But I suggest you try these formulas for yourself, I just did, and for me the error is unacceptable, but if I had to use one, it would be the second one.

Why do you want to do this? Is there a specific reason?

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Bob Berry
BINDT, IRT Level 3 EL, Level 3 CV (ISO 18436)
Thermal Vision
8 Old Fair Green
Dunboyne
Co Meath
Ireland
bob@thermalvision.ie
www.thermalvision.ie

 

Bob, thanks for the reply. It seems as if no one knows the answer to this. It must be one of those misteries of Infrared. 

No immediate specific reason. But some reasons are:

1. Reviewing an IR report and need to find out the temp at a different emissivity setting.

2. Getting a temperature reading lets say at an emissivity setting of 1.00 or .95 and cannot get a temp. value on the same image at a different emissivity setting of lets say .50 because at the setting the temperature is found to be out of range for that image.

 Dear Mr. Snell or any other experienced instructor or even IR equipment engineers, is there a way to do this generically, or is it all based on the equipment's very specific calibration setup? How did people acheive temp readings before computer software? Was there a supplied conversion table?

 Steve

 

Steve, I think to Bob's point, the key is 'in the field'.   There are indeed ways to correct for emissivity in post processing.  I regularly use an E of 1.00 due to the varied surfaces I see and correct later.  It takes some horsepower to do it and the results are only as good as the data.  For example, if you take the image with an emissivity of 1.00 and intend to correct later, what will you use for 'true' emissivity?  In my case, I rely on calibration points with logging thermistors that are time synched to my imager as well as known emissivity markers.  I don't rely on lookup tables that say this metal will have this emissivity, because it isn't always so.

 The other aspect of this is what data you are using from your imager.  If you have a JPEG image, you will be able to figure out what the error is and possibly adjust the legend, but not spot values on the image.  If, on the other hand, you export the image data and want to massage it in software and plot it as a corrected image, you can, but the programming is not for the faint of heart.  Manipulating data like this also creates an image that may look identical to what was taken on site, but may no longer be admissable in certain cases as a true image.

You might have trouble getting any takers here to go through your options in any detail. You're on the right track trying to get ahold of one of the Snell gurus.  This would strike me as a good consult item.  FWIW, the second formula has some familiar aspects, however it looks like it has part of a Kelvin/Fareheit conversion (the 460 deg 0 to 0 correction).  I think you might be better off just using C or K as their degree widths are the same.

 -John

Steve,

 

If you are trying to get accurate or reliable measurements from a surface with an E value as low as 0.5 you are not going to have much success anyway, it is simply too reflective. My opinion is that you should do as much analysis as possible in the field, if it does go out of range due to changing the emissivity value, then you can change the range and take a new image.

When you use 1 as an emissivity, you are telling the camera that it is looking at a blackbody, and that all the radiated energy comes from the object itself. It therefore does not compensate for reflected radiated energy at all, as you lower your emissivity setting, you simultanously increase the reflectivity within the camera and the camera increases the compensation for reflected energy. TRY both formulas yourself, with an image using an emissivity of 1 and then change your E to an assumed correct new value, do your calculation with both formulas, then change your values on the image, either in the software or on the camera, if your formulas are correct they will calculate the same new value that the camera displays, I did this already and got a significant error with both formulas compared to the camera software.

John,

We should not rely on table values at all especially for metals.

The second formula works in Rankin, the width degree is the same as Farenheit.

---

Bob Berry
BINDT, IRT Level 3 EL, Level 3 CV (ISO 18436)
Thermal Vision
8 Old Fair Green
Dunboyne
Co Meath
Ireland
bob@thermalvision.ie
www.thermalvision.ie

Thanks for the useful information. Bob, you are right that the above formulas gives an answer different than the IR camera or software.

Maybe I tried to explain myself the wrong way, It seems as if my question is being changed to emissivity errors and shiney materials. The examples above were just examples. My main point is what is we do not have access to the IR equipment or software. What if we cannot change range because all we have is this data on a report and no equipment or software. What if we do not have any digital data or image but simply just numbers on a sheet of paper.

example: Is there a correct calculation to find the new temp.:

Recorded Temp= 125C emissivity= 1.00  Background= 30C  Distance= 1 meter  Transmission=100% Rel Humidity= 50%

What is the new corrected temperature if you change the emissivity to 0.74? All other variables remain the same.

My question again, is there such a formula? Or is it all based on the equipment's calibration that is different for each model and maker equipment?

Steve

 

Hi Bob.  I think we are agreed on the use of tables. 

Rankin makes sense of the second formula, it's a scale I haven't used in a very long time.  Thanks.

-John

Steve, using your example, it looks like you have a table of apparent temp readings.  I would hesitate to try any back-driven solution based solely on this data.  Even if you find a formula that works to your satisfaction, you would need some way to cross check your results for a test point or two, based on a contact temp. 

Applying an emissivity correction is a non-linear adjustment to the data, as you can tell by the exponents.  This means that using a rough guess for emissivity, such as a table lookup post event, is going to produce unacceptable errors that themselves vary in magnitude across your data.  In this case, it's probably best just to stick with apparent temps, or if you need better data, revisit the test with calibration points, preferably with contact temps to cross-check.

FWIW

-John

Steve,

Measuring temperatures radiometrically requires you to have control of several variables if you want to get an accurate temperature. First let's agree on what is an accurate temperature. Most all radiometric imagers available today specify an accuracy of +or- 2C or 2% whichever is greater. The specification also states that this assumes you are looking at a blackbody source at 30C. As you move away from 30C and decrease emissivity the accuracy also decreases.

Remember that cameras do not measure temperature, they quantify the energy hitting each detector element. The Stephan Boltmann equation describes this relationship:

The emitted energy from a surface is equal to q/A= es  T⁴    If you wish to calculate the temperature at a different temperature you can write the equation    º

 

 

Let’s assume you used an emissivity of .95 and recorded a temperature of 250 ºC , but afterward you realize the emissivity is really .89. We can get rid of s because it is on both sides of the equation (edited 11:10 a.m. EDT, 7/30)

 

 

 

This methodology allows you to calculate the adjusted temperature. As stated by others you can achieve the same result in-camera or in software if you saved your image as a 12 or 14 bit file.

 

We always hesitate suggesting the use of formulas by people who may not understand the relationships between all the variables. As has been discussed by others above, the effect of background increases as the emissivity decreases, meaning that the lower the emissivity the more certain you must be of the accuracy of the background signal.

 

For further reading see section 8.6 in Experimental Methods for Engineers, McGraw Hill copyright 2001

---

Rob Spring, P.E.
ASNT NDT LIII #65375
The Snell Group
rspring@thesnellgroup.com
800-636-9820

Rob -

You forgot to convert to absolute temperature!! 

The forumla used, as I know you are aware, is based on absolute temperature, either Kelvin or Rankine, and does not work with Celsius or Fahrenhiet. 

The correct calculation, converting to Celsius, would be:

T' = (250+273)(0.95/0.89)^1/4 - 273

or 258.6C.

(Using 273 as the approximate conversion to K from C)

Jack

---

Jack M. Kleinfeld, P.E.
Kleinfeld Technical Services, Inc.
Bronx, NY
718-884-6644
JKEngineer@KleinfeldTechnical.com
come see what we can do for you: http://www.KleinfeldTechnical.com

Jack,

You are absolutely correct that absolute temperatures must be used.

Thanks for catching my error.  I've updated my post above.

Rob

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Rob Spring, P.E.
ASNT NDT LIII #65375
The Snell Group
rspring@thesnellgroup.com
800-636-9820

thanks everyone for your input. But unfortunately when I plug in the numbers none really match up with the temps the equipment or software gives me. They are close but not a match. Which leads me to the conclusion that you cannot manually calculate the new temp. without the maker's calibration information or input as to how they do it for your particular model of equipment. You can come close if you can live with a larger degree of unknown error.

I have also found that although the formula that jack and rob mentioned on ly seems to work (close but not same) if the background temp. is lower than the target temp. The formulas do not take background temperature into concideration whatsoever. So if the background is higher than the target the temp should decrease when you lower the emissivity value, but the formula does the oposite.

 I think rob has got it right when he states that IR equipment do not measure temps but radiant energy. And I think that each equipment will measure different amounts of raidant energy from the same object and it all comes down to how the equipment maker adjusts each equipment in order to convert the radiant energy it sees to an actual temperature.

So would that mean for every ir camera embedded into each radiometric image is the actual claibration formula or data of that detector so that the software knows how to apply temperature changes for camera/image whenever you change a variable?

Example: you take an image with a P65 serial number 01 and the same image with a P65 serial number 02. The formula/correction the software uses will be different for both imagers even though they are the same model but because they each have a unique calibration programing because they each measure the amount of radiant energy of an object differently and uniquely?

 what do you think?

Steve

Steve,

 Here's the story. The Stefan Boltzmann equation represents the radiant energy leaving the surface of the object you are viewing. It is not the energy hitting the detector. Between the surface and the detector is air and the lens of the camera both of which are not 100%  transmissive. Anything that is not 100% transmissive must have a degree of absorption and there for emission. Because of this fact the camera manufacturers must use sophisticated calibration curves and corrections to calculate temperature. Each IR lens is in reality one of a kind requiring the camera manufacturer to calibrate each camera to each lens.

 You also bring up another very important fact of radiometry. When the background is warmer than the target, regardless of the emissivity of the the target (unless it is a blackbody) there will be a much greater impact on temperture than if the background was cooler. The reason for this is that reflected background temperature is a 4th power function as is the emitted radiation. This is easy to demonstrate by reflecting something very warm off of high emissive surfaces, even skin.

I know of no manufacturere that uses software calibrated to each camera. The corrections are made in the camera software. The software works on the corrected data.

 Rob

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Rob Spring, P.E.
ASNT NDT LIII #65375
The Snell Group
rspring@thesnellgroup.com
800-636-9820

Rob, your comment on energy leaving the surface, vice what reaches the camera, is bang on.  If I may, I'd like to divert the discussion slightly into the area of air transmissivity.  You mention complex calibration curves for this.  I often work in conditions where air transmissivity is compromised by high humidity and condensation nucleii. I'm using compensation algorithms that work OK, but would really like to have a gold standard for doing this.

Are there public domain algorithms for mapping the energy drop through less than perfect air with distance and rel humidity (and maybe aerosol content) as variables?

 Thanks for your help.  Sorry for the sidetrack.

-John