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  1. #11
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    Tom, they (EK) are still here and still making film to the same high standards, just at a reduced rate of output.

    PE

  2. #12
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    Well the issue I am having with the Kodachrome 25 sheet is this:

    If you look at the H-D curve on the sheet where the density=1, you see that the required exposure to get density=1 for both blue and green layers is pretty much exactly the same. Both blue and green layers as well as red all must have equal densities in order for the developed film to look neutral in color if a neutral object was shot under the correct lighting. So from this we know at the very least the blue and green layers must absorb very closely equal amounts of exposure to produce a neutral color in the developed film.

    But then when you look at the sensitivity curves, you see the blue layer has a big peak at about 420-430 nm, that peak will cause it to absorb a lot more exposure than the green layer, a lot more. So really it doesnt look like there is anyway blue and green will absorb the same amount of exposure if we are shooting a neutral object. but that is what must happen if the two layers are to develop equal densities. This is the problem.

    But maybe the color temperature of the illuminant compensates for this? No, the correct color temperature for this film is 5500K-5600K, which while slightly lopsided giving more weight to the green and red side of the spectrum, it is still approximately balanced. The green and red range of the spectrum hardly are getting much more exposure than the blue range. The big peak in the blue sensitivity curve at 420-430nm will cause it overcome the subtle imbalance of the spectral power distribution favoring the green and red range, as a result the blue layer still absorbs more exposure and so the two layers blue and green are not equal.

    So if Kodak is releasing accurate data sheets, then where is the flaw in my logic? What am I failing to account for?

  3. #13

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    With a Kodachrome 25 sheet, what difference does it make now?

  4. #14
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    not much, it just bugs me. And makes me wonder if this sheet is wrong, how many other sheets are also wrong including the ones for films Kodak still sells.

    Or it could mean that there is something wrong with my understanding of the how to interpret data sheets. I thought I knew how to read them, now Im questioning myself.

  5. #15

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    Typos are always possible, but speaking for myself I always figured the technicians at Kodak knew something I just wasn't considering. I mean, obviously Kodachrome was the best color film ever made, so that speaks for itself. Just my opinion, mind you. What you have read in the tech data in itself explains a lot when the results are self-evident. It may have been a simple reciprocity-effect in that layer or dye, or another variable the technicians compensated for. In my 40 some years in photography, I don't recall ever seeing any typo errors in a Kodak publication, though, that I could ever tell.

  6. #16
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    Ok, lets try this approach for you guys.

    The human eye integrates the area under the curve of any dye. The broader it is, the lower the peak must be and the narrower it is, the higher the peak must be. The same thing holds for the spectral sensitivity. So, basically, you have to make a "cutout" of the total dye curve all by itself and "weigh" it to see how much it is "worth" in the imaging process.

    If you look at the dye curves (spectral dye density curves), you will see that the cyan is very narrow, and the magenta and yellow are very broad. Thus, to make a neutral and the rest of the colors as well, you need more cyan. See the sensitometric curves just to the left of the dye density curves. The cyan is higher in contrast to the densitometer.

    So, when you view it, the Kodachrome appears neutral even though to the instruments there is apparently more cyan (or less magenta and yellow).

    This is both good and bad.

    But, it works!

    PE

  7. #17
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    I'm so truly sad we don't have Kodachrome anymore.
    Stop worrying about grain, resolution, sharpness, and everything else that doesn't have a damn thing to do with substance.

    http://www.flickr.com/kediwah

  8. #18
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    Alright, so Ive concluded that the specification sheets ARE accurate as Ive been seeing the same issue in specification sheets for other films and from other brands. Theres no way they can all be wrong.

    So the problem here is that I am missing something when interpreting the spectral sensitivity curves. I thought I know how to read them, I guess not. Can anyone help me out by explaining how I would use the log sensitivity value at a wavelength to determine how much light a layer is absorbing from light at that wavelength? Theres not a lot of information out there explaining how to read these data sheets, in fact theres really none.

  9. #19
    AgX
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    That is why I question the value of all those sheets. The majority of readers does not understand them, or at least not correctly.

    They are written by engineers for engineers.

    Concerning your curves-issue I posted on that very issue in detail some time ago.
    Last edited by AgX; 08-18-2013 at 01:27 PM. Click to view previous post history.

  10. #20
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    Retro...

    The sheets can be used that way. The vertical height of the curve is the response in LogE. The "Y" axis is thus the speed at that wavelength. However, the speed is the integral of the entire curve. And so the speed that you measure at one wavelength is infinitely small regardless of the "Y" value unless combined with all of the other values.

    This all is related to calculus, I'm afraid. And, the curves have to be smoothed out because they are bumpy due to the dyes themselves.

    It is easier for a B&W film than for a color film, due to overlap. Thus, some green light produces red exposure. How do you estimate that? You need to do spectral curves of each dye formed at each wavelength and then begin to calculate what must be done to improve color reproduction.

    To do that, you must also include a factor for the human eye vs the measuring system, and you must therefore consider the half band width of the imaging dyes.

    If I can find one, I might post a typical picture used for this type of measurement.

    PE

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