temperature measurement

Discussion in 'Digital Photography' started by Allan Adler, Feb 21, 2008.

Suppose I have a digital camera and an vacuum tube and I want to use the
camera to measure the filament temperature of the tube by studying a
digital photograph of the filament, saved as a computer file.

Is that possible and, if so, how?

I realize that the temperature varies along the length of the filament,
so hopefully this method would assign a temperature to each point of the
filament.

2. m IIGuest

I don't know if these are good enough, but they're a start.

http://www.r2d2u.com/htm pages/color firing chart.htm
http://www.lightbulbsdirect.com/Merchant2/graphics/00000001/colortemp_chart2.jpg
http://www.mediacollege.com/lighting/colour/colour-temperature.html

Getting the camera set properly may be a challenge. I don't know how
linear they are in their response.

mike

m II, Feb 21, 2008

3. Ilya ZakharevichGuest

[A complimentary Cc of this posting was sent to
[I never tried anything similar, but] I doubt very much that you will
be able to get any reasonable result without calibration.

On the other way, if you can measure similar temperatures by
independent means, it looks quite probable that you would be able to
calibrate your images, and get reasonable precision (I would say,
5-10%?).

Hope this helps,
Ilya

Ilya Zakharevich, Feb 21, 2008
4. Chris SavageGuest

Interesting question. What resources has your teacher suggested might
help with your assignment? rpd should not have been listed as a primary
resource.

Chris Savage, Feb 21, 2008
5. Roger N. Clark (change username to rnclark)Guest

Yes, not only is it possible, you could do it with great precision
with some calibration. The peak of the black body curve is in the
infrared and the red, green and blue filters they are on the blue
side of the filament black body peak. That means small changes in temperature
will result in a large change in intensity and a significant change
in color. You could compute the change in response given the
transmission of each filter and the spectral response of the detector
(I do it often with spectrometers, remotely determining temperatures).
Without knowing the spectral response of the filters, you would
need to calibrate the response using known temperatures (e.g. use
a blackbody; there are many around MIT; it is an instrument
that makes a precise temperature and lets the heat radiate out
a small hole). The digital camera must have raw output and you
use software that produces a linear response (e.g. dcraw).
With calibration, you could get temperature measurements
to a fraction of a degree.

Roger

Roger N. Clark (change username to rnclark), Feb 21, 2008
6. Don Stauffer in MinnesotaGuest

I also agree that the use of RAW is essential in this application.
JPEG is not known for its accuracy of color rendition, and this
application requires spot-on color accuracy. Incidently, filament
structures at operating temperatures ARE pretty close to black body.

Don Stauffer in Minnesota, Feb 21, 2008
7. Ilya ZakharevichGuest

[A complimentary Cc of this posting was sent to
Roger N. Clark (change username to rnclark)
I doubt this very much. Luminosity channel is useless, and the color
one is going to be quite noisy (and changing quite insignificantly
when the temperature changes in such a cold range as a filament). I
would be (pleasantly!) surprised if even 1% error is achievable with a
digicam...

Hope this helps,
Ilya

Ilya Zakharevich, Feb 21, 2008
8. GregSGuest

The biggest challange may be keeping the filiment from saturation. I suppose
room light should be fairly bright, else use a small F stop and or filter.

greg

GregS, Feb 21, 2008
9. GregSGuest

Assignment ??? Why does this have to be an assignment ?

greg

GregS, Feb 21, 2008

Thanks, that is very helpful. How cheaply can I get a digital camera that
will have raw output?

Since posting this, I obtained a copy of Michael Covington's book,
"Astrophotography for the amateur, 2d ed", and located the chapter
on using CCD cameras. Of course I'm not going to do astrophotography,
but maybe some of the information can be used to make sense of the
raw output. If there is a better book to look at for this application
(i.e. the filament temperature, with a digital camera), I'd be glad to

11. Roger N. Clark (change username to rnclark)Guest

There are many digital cameras, and I'm not sure which lower
cost ones have raw (that question has been asked in this newsgroup,
so you might look through the archives). Then the question is
how accurate do you want the temperatures? If you want
a degree or less precision, assuming you have the blackbody
references, then it is a matter of collecting enough photons
to reduce the noise. Here are a couple of quick calculations:

The RGB filters in a digital camera peak near:
blue: 0.47, green: 0.53, red: 0.60 microns
(that's 4700, 5300, and 6000 angstroms, respectively).

Filament temperature change in blue change in red blue/red
T=2001K / T=2000K 1.0076 1.006 1.016

T=3501K / T=3500K 1.0025 1.002 1.00054

(K = Kelvin, subtract 273 to get Celcius)

To see a change of 0.76% you would need to collect
(1/0.0076)*(1/0.0076) = 17,300 photons
To see a change of only 0.25% you need
(1/0.0025)*(1/0.0025) = 160,000 photons.

Large sensor DSLRs collect up to about 80,000 photons/pixel,
while small sensor P&S cameras collect less than 10,000 photons/pixel.
You can do the job with essentially any digital camera with raw
by simply taking multiple exposures and adding the images together
(I think the book you reference above explains stacking).
Free programs like registax can do it. For example, if you have
a small sensor camera that collects only 5,000 photons/pixel,
simply 30 to 50 frames and add them together to boost your
signal-to-noise ratio. If you have a DSLR, only a few frames
would need to be averaged. You could also average multiple
pixels over an area of the filament to derive the average
temperature of that area.

If you remove the IR filter, which is done in DSLRs for astrophotography,
the wavelength lever-arm increases and the temperature problem becomes
easier.

If you want only a few degrees precision, then single frames
can do the job. To see how many photons a typical camera records,
see Figure 1 at:
http://www.clarkvision.com/imagedetail/digital.sensor.performance.summary
The graph is in electrons which are produced by the incident
photons.

You also want enough magnification so that the filament is
more than one pixel across, otherwise you will also have
1/r squared effects.

If you really want to read up on the topic, with all the gory
equations that include the 1/r effects and when they cancel, I suggest
Clark, R. N., 1979, Planetary reflectance measurements in the
region of planetary thermal emission: Icarus, vol 40, p. 94-103.

If you have enough wavelength range and the subject is sub-pixel,
you can simultaneously solve for the temperature and the area
of the hot spot. Example:
http://pubs.usgs.gov/of/2001/ofr-01-0429/thermal.r09.html

Roger

Roger N. Clark (change username to rnclark), Feb 22, 2008
12. Ilya ZakharevichGuest

[A complimentary Cc of this posting was sent to
Roger N. Clark (change username to rnclark)
It is quite obvious that 1K precision is a blue moon dream. About 5K
precision may be achievable (with some tricks).
This checks with

h = 6.62607e-34; c = 299792458; k = 1.38065027360697e-23;
f(l,T) = l *= 1e-6; l^-5/(exp(h*c/k/T/l)-1)
df(l1,l2,T) = f(l1,T)/f(l2,T)
ddf(l1,l2,T) = df(l1,l2,T+1)/ddf(l1,l2,T)
There is no question of seeing a change of 0.76%, since absolute
numbers are not going to be measurable. Basing on your numbers, the
relative change is at best 0.15% per 1K change. So with about 60K
photons collected per site (possible with an add-on filter which would
equalize the electron count at R and B sites, and bracketing), one has
an error of about 0.57% in measuring blue/red ratio, would would give
the error 4K at best conditions. [This is indeed better than waht I
initially expected.] If one does not use filters, or with hotter
filament, the precision drops 3x-10x.

[This is applicable if pixel-precise results are needed. With coarser
temperature map, one can get better precision.]
of multishot averaging.
Or harder, depending on the precise form of the spectral curves. My
guess would be "harder".

Hope this helps,
Ilya

Ilya Zakharevich, Feb 22, 2008
13. userGuest

You would use RAW files and the three color filters. Of course,
the main problem is that you would need a calibration lamp,
and the temperature would be wrong if the objects being measured
were not "gray" bodies. They don't have to be black, but if they
are not gray, you'd need to know or calibrate the color.

With calibration and RAW, over a fair number of pixels,
I suspect that 1 to 5 degrees C could be measured. Seen by
eye through a calibrated pyrometer (that's what these things are called)
1 degree C difference is easy to see.

With calibration the total intensity is a more sensitive measure.

Doug McDonald

user, Feb 22, 2008
14. Ilya ZakharevichGuest

[A complimentary Cc of this posting was sent to

How would you explain it? 1C difference corresponds to (at best)
0.15% color shift (as measured by R/B ratio). AFAIK, eye is not good
in noticing better than 1% intensity shift; should be yet worse with
colors.

Are you sure that what you see is not a change in intensity (which is
useless in this context due to impossibility of calibration)?
??? How would you calibrate intensity with a digicam?

Puzzled,
Ilya

Ilya Zakharevich, Feb 22, 2008
15. Roger N. Clark (change username to rnclark)Guest

I left out a 0 when I typed in the 1.016. It should be 1.0016
blue/red = change in blue / change in red.
I gave the highest temperatures and hardest examples. If your
filament is lower in temperature, the problem becomes easier.
A simple way to look at this is consider 3 heat sources:
1) a high intensity tungsten light,
2) a 60-watt light bulb,
3) an electric burner on a stove.

The color difference between the tungsten lamp and a 60 watt
light bulb is small but noticeable to your eye. The stove element
glows a nice red and has huge color difference from the 60-watt
light. So as temperature drops, the color difference per
degree drop increases up until you can no longer detect the
thermal emission.

Roger

Roger N. Clark (change username to rnclark), Feb 23, 2008