I Don't Understand the Luminous Landscape "Expose Right" Article

Discussion in 'Digital Photography' started by jim evans, Feb 25, 2005.

  1. jim evans

    jim evans Guest

    I Don't Understand Luminous Landscape "Expose Right" Article

    Thomas Knoll and Michael Reichmann coauthored the "Expose Right"
    article on Michael's Luminous Landscape website.

    Last fall I emailed Thomas Knoll this question. After he did not
    reply, early this year I emailed the question to Michael Reichmann.
    He did not reply either. I have posted a copy of the question I sent
    Reichmann at http://factsfacts.com/LuminousLandscape.htm.

    Since I have not received a reply from either of the authors of the
    article in question, I am posting my question here.

    I will try to state my question simply here. This is intended only as
    a summary statement. Please read the complete question.

    The table in the article shows the tonal values of an image shot with
    a digital camera as a geometric distribution, and implies that the
    tonal values of film are distributed linearly. Are the tonal values
    of film really distributed linearly in this context?

    If you reply please do so as requested in my question to Reichmann.
    That is, please either agree that my "Film" column in the table is
    correct, or give the 5 values that should be in the Film column.


    jim evans, Feb 25, 2005
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  2. jim evans

    Bill Hilton Guest

    I Don't Understand Luminous Landscape "Expose Right" Article

    Anyone wishing to understand the article could run this simple test ...
    pick a fairly monotone subject (sky, gray card, etc) and shoot it in
    RAW mode at ISO 400 metered properly, then set your exposure
    compensation +2 and shoot it again, then -2 and shoot it again. When
    you convert the three RAW files don't make any exposure adjustments to
    the first image, adjust the +2 image by -2 stops, adjust the -2 image
    by +2 stops. You should have roughly equivalent exposures by now if
    you convert these ... check the noise levels of the three "equivalent"
    exposures" and you'll see what he means.

    You could have gotten the same improvement that the +2 image gives by
    shooting as metered at ISO 100.

    Try it and see.
    Bill Hilton, Feb 25, 2005
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  3. jim evans

    Scott W Guest

    The article is discussing the behavior of digital cameras not film. How
    many levels you have in film is not so clear cut, you can deal with the
    SN of film and get some sense of it, Roger Clark has done a lot of work
    on this, see this link


    Scott W, Feb 25, 2005
  4. jim evans

    JPS Guest

    In message <>,
    This should be true, but some RAW converters treat RAW data based upon
    their absolute values, and not the values after a purely linear
    "exposure" adjustment. ACR, for example, nails the highest RAW values
    to 255 in the output, no matter if you use -4 "exposure".
    Not exactly; ISO 400 at +2 EC is usually better quality than ISO 100 at
    0 EC, if the RAW data doesn't clip.
    JPS, Feb 25, 2005
  5. jim evans

    paul Guest

    And figure 5 on this page:
    which does discuss film/digital

    I still find that one shocking, particularly given the implication that
    digital has less info in the shadows than digital highlights. But still
    digital has more information in the shadows than film.

    But that is not really what the luminous landscape article is discussing
    I think. They are talking about how RAW digital files carry more info in
    the highlights due to their being linear data of simple intensity.
    That's the best I can understand anyways. Brighter means bigger number
    in that linear scheme so there are more significant digits to assign
    finer gradations of tonal range. Once it's converted from RAW, that
    extra info is dropped. During the RAW conversion process, there is an
    opportunity to sneak up & expand that highlight information before
    dropping it.

    One result of this effect (if I understand correctly -and I'm unsure) is
    that an unmanipulated RAW image looks too dark & muddy. Most of the
    information is crammed in the highlights. A curve needs to be applied to
    the image to increase contrast back to what our non-linear eyes expect
    to see.

    Maybe someone can comment on whether the diagrams below correctly
    explain the situation. Normal scenes appear to our eyes with most of the
    complexity in the middle ranges and not a lot going on in the highlights
    or shadows. Digital RAW format contains extra highlight data which is
    generally useless unless you stretch it to the left before clipping to a
    normally curved format. So yes RAW has less info in the shadows but it
    has all it needs as long as you don't need to adjust the contrast later.
    I think film is in the same boat on the shadow end of the chart, in fact
    according to the figure 5 mentioned above, film has less shadow detail
    and less room for adjusting highlight detail.

    | | |
    | | | |
    l | | | | |
    e | | | | | |
    v | | | | | | |
    e | | | | | | | |
    l | | | | | | | | |
    s | | | | | | | | | |
    shadow highlight


    | |
    l | |
    e | |
    v | |
    e | |
    l | | | | |
    s | | | | | | |
    | | | | | | | | |
    shadow highlight

    2. NORMAL (FILM, EYE, TIF, JPEG) -Bell Curve

    Or I could be misunderstanding. I'm just trying to learn as I go.
    paul, Feb 26, 2005
  6. jim evans

    Scott W Guest

    The raw file holds the light values for the pixel in a linear format,
    the normal jpg format is going to hold the data in a non-linear format
    to fit more dynamic range in the 8 bits. This is done with the gamma of
    the image, the light level is proportional to the pixel level raised to
    the gamma power. A common gamma that is used is 2.2, this is what
    photos that are in the sRGB color space use. This also turns out to be
    the gamma of most PC monitors, MACs use a lower number of 1.8.

    So it has nothing to do with our non-linear eye and everything to do
    with our non-linear monitors.

    Scott W, Feb 26, 2005
  7. jim evans

    Bill Hilton Guest

    John Sheehy writes ...
    We must be talking about something totally different, because this is
    total nonsense ... in the test shots I mentioned (monotone subject --
    in my case a heavily overcast sky -- at 0, +2 and -2) a straight
    conversion using ACR of each image leaves me with these RGB values for
    the brightest part of the sky ... 60/67/79 for the -2 shot, 129/140/165
    for 0, 223/229/246 for the +2. Think about it ... if what you say is
    correct (it isn't) then every monotone image would get mapped to white,
    which obviously doesn't happen.
    Run the test and compare the noise levels. That's what I did. ISO 100
    looks the same as ISO 400 at +2 corrected for exposure with Capture One

    Bill Hilton, Feb 26, 2005
  8. First, the digital camera response isn't a "geometric distribution".
    The response is linear. But the terms in the left-hand column of your
    chart are logarithmic. The top row is labelled "within the first
    f/stop", which by definition contains all tones between 1/2 of maximum
    and maximum. Naturally, because the sensor and A/D converter are
    linear, half of the available codes are assigned to this half of the
    brightness range.

    The problem with this approach is that the *eye* isn't linear. It sees
    the step from full brightness to 1/2 brightness as being about the same
    size as the step from 1/2 to 1/4, which is the same as from 1/4 to
    1/8. So if you have a limited number of bits available (e.g. 24 bit
    colour, with only 256 codes per colour), you're better off to use a
    nonlinear transformation to spread the available codes so that each stop
    of range gets something like the same number of codes. This is in fact

    Second, what do you mean by "the tonal values of film are distributed
    linearly"? Film itself is analog; it has no levels or steps. Digitized
    film *does* have levels, but the distribution of them depends on the
    software running the film scanner. Some film scanners output data that
    is proportional to film density, which is the logarithm of
    transmittance. In this case, there will be about the same number of
    code values per each one-stop range of input brightness. Some scanners
    output linear data - but it's linear in film transmittance, not original
    scene brightness. And there are other possibilities.

    Basically, your question makes no sense unless the right column is about
    *some specific film, scanner, and software* combination. Film itself
    has no levels.
    Your question is unanswerable as posed. There is no answer that would
    apply to all film scanners.

    Dave Martindale, Feb 26, 2005
  9. jim evans

    Drifter Guest


    I don't understand why you are trying to make a comparison to film.
    That article is strictly about digital photography, one of the quirks
    of using an electronic sensor, and how to compensate for it.

    It has nothing to do with using film, it implies nothing about the
    behavior of film. There is no connection to using film.

    "I've been here, I've been there..."
    Drifter, Feb 26, 2005
  10. jim evans

    paul Guest

    I didn't get all of what you are saying but I was reading some other
    articles about a way of converting RAW in a linear format to 16 bit TIF.
    Done that way, the image looks almost pure black but when curves are
    applied, it can be adjusted to get more info out of high contrast
    images, less blown highlights. The thing is it usually messes up the
    colors so the technique is to make a B&W image with the linear
    conversion then do another normal one for color & merge the two.

    I think Dave is on to something about the logarithmic scale of stops
    versus linear digital capture though that seems such an extreme
    difference. Probably it has more to do with that than my bell curve
    histogram idea.
    paul, Feb 26, 2005
  11. jim evans

    jim evans Guest

    Versus what? If it's a quirk, what doesn't have this "quirk?" Film

    jim evans, Feb 27, 2005
  12. jim evans

    jim evans Guest

    Okay that's fine, have it your way. My question still remains. In
    the full luminosity range of *film*, is half of the tonal range in the
    first F/stop or is the range more geometrically distributed such that
    1/5 of the tonal range is in the first F/stop and 1/5 in each of the
    other five F/stops. Said another way, if the luminosity range were
    divided and in the table would an equal amount go in each row. And,
    if it is not linearly distributed (like the digital image) and it is
    not geometrically distributed, what is the distribution?
    I'm talking about film. I'm not talking about film scanners or the
    results of scanning film.

    jim evans, Feb 27, 2005
  13. jim evans

    paul Guest

    It's a tip on how to take advantage of 'bonus' data hidden in RAW
    digital files.
    paul, Feb 27, 2005
  14. jim evans

    paul Guest

    The same technique could be used when scanning film theoretically, if
    you wanted to make adjustments.
    paul, Feb 27, 2005
  15. jim evans

    Drifter Guest

    Okay, obviously you want to have a discussion about film. In that
    case maybe Rec.Photo.Digital isn't the place for you.

    "I've been here, I've been there..."
    Drifter, Feb 27, 2005
  16. The problem with this question is that you are trying to measure
    something that doesn't exist. You want to know how many gradations
    there are allocated to each stop of exposure range, but there aren't any
    gradations in film in the first place.

    Suppose you have a scene where the intensity of objects ranges from 0 to
    1 in some particular units. So the light intensity is 0.5 for one stop
    down from maximum, 0.25 for two stops down, 0.125 for 3 stops down, and
    so on. It does not make sense to ask whether half the tonal range is
    between 0.5 and 1.0 (the brightest f/stop). There are infinitely many
    values between 0.5 and 1, just as there are infinitely many values
    between 0.25 and 0.5, and betweeen 0.125 and 0.25. There is no
    meaningful way of "counting" or "measuring" the number of values
    assigned to each stop of brightness range.

    It's only when you quantize the range to a number that you can ask this
    question, and the answer depends entirely on the way you do the
    quantization. If you map intensity linearly to an integer in (say) 0 to
    4096, *then* there are 2048 values used for the first stop, 1024 for the
    next stop, and so on. But if you quantize the logarithm of intensity,
    then every stop gets exactly the same number of code values. Or if you
    represent intensity using a computer floating-point number, you get
    exactly the same number of code values for each stop.

    Now, film's tonal range is compressed from the original scene. Every
    time the original scene brightness increases by a factor of 2, the light
    transmittance of the negative is reduced to about 0.66. It isn't 0.5
    because of the lower contrast (gamma) of the negative. But film
    transmittance is still a continuous thing. It doesn't make sense to ask
    whether the first stop uses "the same amount" of the range as the second
    stop of exposure range, because there's no meaning to "amount".

    If you want, you can graph the response of film: the amount of light
    that comes through the film vs. the amount of light the film was exposed
    to. If you plot this graph on log-log scales (the usual way
    characteristic curves are shown), you'll see that each stop of exposure
    uses about the same fraction of the curve length. But that's precisely
    because you chose a log-log scale in the first place, which shows each
    stop of exposure as the same size on the X axis. If, instead, you drew
    exactly the same data on a linear graph, the brightest one stop of the
    exposure range would take up half the width of the graph. The
    difference in apparent importance is purely due to the graph scale you
    chose, not anything to do with the function you're graphing.

    Both graphs will tell you exactly how much film density (or
    transmittance) you can expect for a given amount of exposure. But
    neither give you an objective answer to "how much of the range is used".
    Each graph *suggests* an answer to that question, but both are lies.
    Your question has no true answer at all when applied to film. There's
    no unbiased way to ask "how much of the film range is between these two
    points". If you use a linear definition of "how much", you'll find
    that the brightest stop gets about half the range. If you use a
    logarithmic measure of "how much" (and density is a logarithmic
    measure), you'll conclude that each stop of exposure uses the same
    fraction of film's range. But either way you'd be misleading yourself;
    the answer is completely determined by the measuring scale you

    It's only when you digitize film that you can meaningfully talk about
    "how much" with respect to the pixel code range.

    Dave Martindale, Feb 27, 2005
  17. It is a "quirk" of any system that linearly digitizes a continuous quantity
    (light) into a discrete representation. It applies to digitized audio
    too. Since film doesn't quantize brightness, it does not apply to film.
    That doesn't mean that film is better because of this.

    The article compares two different approaches to working with a digital
    camera, and shows that one produces better results than another. That's

    If you want to compare film to digital images, you need to take a
    broader look at how various "quirks" of BOTH imaging methods influence
    the quality of the final image. The quantization of brightness in a
    digital camera has one sort of effect on the image, and it may or may
    not be visible depending on the circumstances. The grain of film adds
    another sort of error to film images, one that's not present in digital
    images, and again it may or may not be visible.

    But this article really doesn't address film at all. If you wanted, you
    could publish the same sort of advice for film, but it would be for
    different reasons. (It's generally better to overexpose negative film
    slightly, but underexpose positive film. This could be called "expose
    right" and "expose left" respectively).

    Dave Martindale, Feb 28, 2005
  18. jim evans

    jim evans Guest

    I'm comparing film camera performance to digital camera performance.
    Using your logic, if I were to post in rec.photo.film they would say,
    "Obviously you want to have a discussion about digital cameras. In
    that case maybe Rec.Photo.Film isn't the place for you. "

    jim evans, Feb 28, 2005
  19. jim evans

    Drifter Guest

    Maybe they would. The whole "film vs digital" thing was thrashed to
    death a long time ago.
    If you like film, use film.
    If you like digital, use digital.
    If you like both, use both.
    But for pete's sake stop trying to make a digital-to-film comparison
    of an article that was written about how to best use a characteristic
    of digital cameras. As several of us have tried to tell you, there is
    no useful equal to this behavior in film just like there is no useful
    equal to chemical mixes in digital. (could you imagine trying to set
    up a written table equating developer mixes/times to Photoshop
    settings? Jeeze, you'd go totally bonkers first).

    Unfortunately when any of the responses here try to point out the
    "apples/oranges" element out to you, your response has essentially
    been a repeat of "but what about film" so I don't really know what
    else to say in order to help you understand.

    "I've been here, I've been there..."
    Drifter, Mar 1, 2005
  20. Actually, that would make somewhat more sense. All you'd need is data
    showing how developer concentration and development time affects the
    developed film (primarily the change in gamma) and then map it to
    equivalent Photoshop settings. In fact, if you know that a certain
    development change boosts gamma by a factor of 1.1, you can get almost
    the same effect in Photoshop by setting the middle box in Levels to 0.9
    (the reciprocal of 1.1).

    Or if you have the complete characteristic curve of the film and
    developer together, you could convert that into a Photoshop Curves
    operation. This is a bit messier because the axes of a film response
    graph are base-10 logarithms of light intensity and transmittance, while
    in Photoshop the axes are pixel values which are roughly intensity to
    the 0.45 power. But a spreadsheet will take care of the conversions

    Dave Martindale, Mar 1, 2005
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