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  1. #31

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    Apr 2003
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    Conservation of light, i.e., what area is the light spread over, is certainly pertinent. The other physical factor is that as the lens is refocused the distance of the entrance pupil from the film changes. For the small print with the small enlargement factor, the light is spread out over a small area, but the lens is focused farther from the negative and so the entrance pupil is farther from the film and therefore captures a smaller fraction of the light that is emitted by the film. This effect is usually termed "effective aperture". This is still conservation of light -- as the entrance pupil is farther from the film, some of the light rays no longer enter the pupil.

    The effective aperture change when the lens is far from the film is also why one must correct exposure when taking a closeup photo. This is most commonly encountered by large format photographers ("bellows extension"). (Cameras with through-the-lens metering do the correction automatically.) In that application, the equation is highly accurate.

    For the enlarging application, the equation is implicitly making some assumptions about how the film emits light, which of course depends on the illumination system of the enlarger. So the equation might break down in extreme cases, such as point source enlargers or very wide apertures. It should be most accurate for diffusion enlargers with the lens at reasonable apertures for making a print.

    For someone who wants a calculator/computer implementation of this, there is a paper dial "computer" in the Kodak Darkroom Dataguide that gives the change in exposure time going from one magnification to another. You can probably find this inexpensively on eBay or internet used book sites.
    Last edited by MichaelBriggs; 12-08-2005 at 10:55 AM. Click to view previous post history.

  2. #32
    Troy Hamon's Avatar
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    Hi Michael,

    My apologies. As you astutely pointed out, I do in general prefer intuitive mathematics. I am also aware that there are times when the simplified version of an equation loses its transparency. Perhaps that is what has happened in this instance. As you suggest, I certainly have not read the references you provide.

    I understand the change in effective aperture, and apologize for my denseness in not recognizing it as you stated it originally. I had never considered it as a factor with enlargements, but I can see how it would apply. I am interested a bit because I've used my intuitive approach for some time and have never had to recalculate exposure, perhaps I ought to go back and compare some of my prints of different sizes for tonal variation.

    On the other hand, perhaps I won't, since it seems to be working...

    Happy printing.

  3. #33

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    I find the following explanation a little more succinct and clear. I realize this is just a restate of what's already been posted, but perhaps in more clear, laymen's terms. I can't remember where I found this, but i've been using it for about 5 years and works for me, unless you make a dramatic change to the photo's aspect ratio

    ---

    The exposure needed to make a print is proportional to the picture’s surface area. Thus, if you change print size, you can eliminate the need for new test strips by calculating the change in area mathematically. A pocket calculator facilitates this.

    1. Divide the new print size by the old one to get the change in width.
    2. Multiply this answer by itself—square it—to get the change in area.
    3. Finally, multiply the original exposure time by the change in area to get the new exposure time.

    For example, if you were adjusting from a 4x5 to a 16x20 print size:

    Old print width = 5 inches
    Old exposure time = 4 seconds
    New print width = 20 inches

    New width / old width = 20 / 5 = 4.0 (change in width)
    4.0 * 4.0 = 16.0 (change in area)
    Change in area x original exposure = 16.0 x 4 = 64

    New exposure time = 64 seconds

  4. #34
    Max Power's Avatar
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    Quote Originally Posted by Matt5791
    One question I have is:

    Say I decide that 30 seconds exposure is about right for my 5X5 print I am about to make.

    Is there any calculation to enable me to decide on comparable exposure if I increase the size of the enlargement to say 8X8?
    Matt:
    I found this chart Enlarging Ratios to be extremely useful. I believe that it is what you are looking for. I have used it myself and it works very well.

    Hope that this helps,
    Kent
    Max Power, he's the man who's name you'd love to touch! But you mustn't touch! His name sounds good in your ear, but when you say it, you mustn't fear! 'Cause his name can be said by anyone!

  5. #35

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    Quote Originally Posted by johnnywalker
    Is there any reason why you can't simply calculate it
    using the square inches of the print? If a 4x5 (20 sq. in)
    print takes 20 seconds, then an 8x10 (80 sq in) will take
    80/20 = 4 times as long at the same f stop.
    Perhaps this matter has already been mentioned. As
    the negative's distance from the baseboard increases
    and the focus is adjusted by bringing the lens closer
    to the negative, the speed of the lens increases.
    Just as one would expect from a camera lens.
    So, with your example the time might be
    3.5 times as long.

    I also recommend the EM-10. If calibrated it can
    be used as a densitometer. Same exposures are
    obtained by adjusting the F stop. Dan

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