Re: Which scanners REALLY provide 36 bit output? HP?

From: Marko Cebokli (
Date: Wed Dec 13 2000 - 09:32:29 PST

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    Stephen Williams wrote:
    > said:
    > > (Yes, it's 3 - the luminance sensors (the 'rods') have the same
    > > sensitivity as the 'green' channel 'cones')
    > Not quite, they respond to a wider spectrum then the green sensors.
    > Though they are highly sensitive to green, they respond to most all
    > of the visible color range. I think.

    They do have the same curve (they also use the same sensitive compuond).
    None of the eye (or film/CCD) sensors is really a 'rectangular' band
    pass. They all have relatively wide responses, with the skirts extending
    all over the visual range.

    The eye curves are well measured and documented, you can find them
    tabulated and graphed in any serious book about imaging.

    The film curves can be found in manufacturer's brochures. (the same
    goes for CCD's)

    The linear approximation for the output of such a channel sensor is
    (let the S represent the integral sign and l lambda)

    R = S r(l)i(l) dl

    one such equation for each channel,
    where r(l) is the sensitivity curve of the sensor, and i(l) is the
    spectral intensity of the scene. Integration limits are over the
    full range where r(l) is nonzero.

    The eye and film can be highly non-linear, but CCD's are for practical
    purposes perfectly linear (= ideal image sensors, they caused a
     revolution in astronomical and other scientific imaging)

    > But do the spectra of those three channels cause the RGB sensors in
    > your scanner to respond similarly? That's a question for a photography

    The consequence of the above equation is that you have crosstalk
    between the channels, which depends on the shapes of all the curves
    involved, and these can differ between various films, photo-papers,
    printer inks, CRT phosphors and whatewer. Trouble guaranted! (and this
    was only the linear model ;-) There is a 3x3 matrix linking two such
    linear tricolor systems. Ideally, it would be a identity matrix, but as
    Murphy designed it, it never is.

    Even if the artificial sensor's curves would match the eye curves
    perfectly, you wuold still have crosstalk. That's why you can't really
    reproduce ALL the colors using a tricolor system - you're limited to
    the inside area of the triangle defined by the primarie's places in
    the chromacity diagram. Most notably, pure spectral colors are always
    outside. But most (virtually all) colors in nature DO fall within
    this triangle, and with some (a lot of--) tweaking, one can get very
    good results.

    Colorimetrists represent the 'outside' colors by assigning negative
    values to the channels. But that's only a mathematical trik, useful
    for computation, because in practice, you can't squeeze 'minus red'
    out of your monitor...

    When you try color metrics (measuring differences between colors),
    things get even more scary. Colorimetrisis found out that the human
    chromacity space IS 3D, but its metrics are non-euclidean....

    Do you still dare to scan? :-)

    Marko Cebokli

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