temperature measurement

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

  1. Allan Adler

    Allan Adler Guest

    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
    Allan Adler, Feb 21, 2008
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  2. Allan Adler

    m II Guest

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

    http://www.r2d2u.com/htm pages/color firing chart.htm

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

    m II, Feb 21, 2008
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  3. [A complimentary Cc of this posting was sent to
    Allan Adler
    [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,

    Hope this helps,
    Ilya Zakharevich, Feb 21, 2008
  4. Allan Adler

    Chris Savage Guest

    Interesting question. What resources has your teacher suggested might
    help with your assignment? rpd should not have been listed as a primary
    Chris Savage, Feb 21, 2008
  5. 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 N. Clark (change username to rnclark), Feb 21, 2008
  6. 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. [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

    Hope this helps,
    Ilya Zakharevich, Feb 21, 2008
  8. Allan Adler

    GregS Guest

    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.

    GregS, Feb 21, 2008
  9. Allan Adler

    GregS Guest

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

    GregS, Feb 21, 2008
  10. Allan Adler

    Allan Adler Guest

    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
    know about it.
    Allan Adler, Feb 22, 2008
  11. 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

    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:
    The graph is in electrons which are produced by the incident

    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:

    Roger N. Clark (change username to rnclark), Feb 22, 2008
  12. [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).
    Wrong. blue/read = 1.0016.
    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.]
    Instead-of/in-addition-to this, one can do spacial averaging instead
    of multishot averaging.
    Or harder, depending on the precise form of the spectral curves. My
    guess would be "harder".

    Hope this helps,
    Ilya Zakharevich, Feb 22, 2008
  13. Allan Adler

    user Guest

    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. [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

    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?

    Ilya Zakharevich, Feb 22, 2008
  15. 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 N. Clark (change username to rnclark), Feb 23, 2008
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