Film RMS noise in LF vs. digital full well

Discussion in 'Digital Photography' started by Ilya Zakharevich, Oct 22, 2006.

  1. This post is a coproduct of a discussion in a different thread
    ("Again: film vs digital"). I think it deserves a separate thread.

    However, the results below are so unintuitive, that it is probable
    that some goof have been made; please be watchful when you read this.
    In particular, the digital noise-equivalent sensitivity of print film
    turns out to be 2 orders of magnitude higher than of the slide film;
    do not think this matches reality...

    ============================================

    Conclusions: if one believes the RMS noise numbers provided by
    manufacturers, MF digital cameras of today should beat slide film in
    formfactor 4x5in as far as noise is concerned. However, they are very
    far from matching noise of print film in formfactor 4x5in.

    ============================================

    RMS noise of film measures the variation of film density when averaged
    over a HUGE circle, 48um in diameter. Well, this circle is huge when
    one thinks in terms of 35mm shots; but film is not a useful media in
    the 35mm formfactor. And starting from f-stops close to f/45, the
    diffraction pattern becomes close in size to this 48um circle, so this
    number is very applicable to LF photography.

    How this number, RMS noise, translates to metrics useful in digital
    world? The principal measure of digital noise is the full well. So
    how to translate RMS to an equivalent full well?

    As Roger reads the Kodak definition, it is 1000 * (RMS of variation of
    log_base10(density of film)). [This is not how I initially read it,
    but it is a reasonable interpretation, and now I believe it is
    preferable to my initial reading of "relative variation of the
    (linear) density of the film".] See for yourselves in

    http://www.kodak.com/US/en/motion/students/handbook/sensitometric6.jhtml

    Translating to linear-density space, "RMS number" R corresponds to RMS
    variation of linear density of R*log(10). Next, we want to translate
    it to linear luminance. So one needs to divide it by the contrast

    CONTRAST = | d log Density / d log Exposure |

    Contrast of the film depends a lot on the exposure; by Kodak
    definition, RMS is measured at log_base10(density) = 1. For slide
    film, contrast at this density is about 1.6; for print film, about
    1/1.6.

    Combining all this together, one gets the luminance S/N ratio
    corresponding to RMS number R as

    S/N = 1000/log(10)/(R/1.6) ~ 700/R for slide film

    S/N = 1000/log(10)/(R*1.6) ~ 270/R for print film

    The next question to analyse is to find in which exposition zone is
    the density with the log_base10(density) = 1. The curves
    density/vs/exposure for 3 channels (R,G,B) are very different. But
    usually G is between R and B, and, moreover, green is the visually
    most important. So we assume that exposure corresponds to this
    density of the green channel.

    For typical print film (I looked up Kodak Gold 100), this density
    corresponds to the exposure of approx. 0.3% of the max. For typical
    slide film (I looked up KODAK PROFESSIONAL ELITE Chrome Extra Color
    100 Film), this density corresponds to the exposure of approx. 9% of
    the max. Although these numbers will vary with the film, we will use
    these numbers in what follows.

    Assuming full well of N electrons, the S/N ratio at 9% of the max is
    0.3 sqrt(N), at 0.3% of the max it is 0.055 sqrt(N). Comparing with
    numbers above, one gets

    Equivalent full well = 5.4e6/R^2 for slide film
    Equivalent full well = 25e6 /R^2 for print film

    Typical values of R for slide film are about 11, for best print films
    about 3..5.

    Example: Comparing 40MP 36x48mm sensor to an 4x5in Velvia 50 shot
    (RMS=9). The scale factor is about 3; so the 48um circle on film
    corresponds to diameter=12um circle on digital sensor. This is
    approximately 2.6 pixels; applying the formula above, one needs about
    26K full well of one pixel to get similar noise at midrange tones.
    Such a full well usually corresponds to ISO=200 sensitivity; thus one
    needs about 3^2 * 200/50 = 36x smaller exposure time for the shot when
    one shots digital vs film.

    Example: Do the same with print film (RMS=3, as for Fuji Pro 160S, see
    http://www.cacreeks.com/films.htm
    ). Likewise, one needs full well about 1M to get similar noise in
    zone I. Assuming maximum full well of one shot about 60K, one needs
    to stack together about 16 shots to get an equivalent noise. On the
    other hand, with the scale factor 3, one needs the f-stop 3 times
    larger to get an equivalent image on the focal plane. So these 16
    shots would have about 1.8 times longer total exposure (assuming
    instantaneous sensor readout). If this full well is achieved at
    ISO100 (vs ISO160 of the film we compare with), the total exposure
    time is going to be about 3x longer for digital.

    Enjoy,
    Ilya
    Ilya Zakharevich, Oct 22, 2006
    #1
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