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  1. #11
    Troy Hamon's Avatar
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    Another thread on the topic is here. The simple manner in which you are trying to calculate exposure changes is pretty much what I do, but the accuracy of this method depends on the degree of enlargement. If you are working from 35 mm negatives and using your calculation for the change from 5x7 to 8x10, it will be closer in accuracy than printing from a 4x5 inch negative and going from 5x7 to 8x10, as pointed out by Michael Briggs and others. Good luck.

  2. #12
    RH Designs's Avatar
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    Quote Originally Posted by metod
    Richard your formula for f-stop printing is interesting, but isnít kind of hard to adjust f-stops in decimals numbers?
    Yes, hence the f-stop enlarger timer which does it all for you (shameless plug ) - http://www.rhdesigns.co.uk/darkroom/...fessional.html. When I get a minute I might draw up a table of print sizes vs exposure in stops, so you can simply look up the old and new print sizes and read off the exposure correction in stops - which will be the same whatever the original exposure time.

    Regards
    Richard

  3. #13
    RH Designs's Avatar
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    OK, here's the table:

    Print Length - f-stop factor - Twelfth stops
    1 0.00 0
    2 2.01 24
    3 3.18 38
    4 4.01 48
    5 4.66 56
    6 5.19 62
    7 5.63 68
    8 6.02 72
    9 6.36 76
    10 6.67 80
    11 6.94 83
    12 7.19 86
    13 7.43 89
    14 7.64 92
    15 7.84 94
    16 8.03 96
    17 8.20 98
    18 8.37 100
    19 8.53 102
    20 8.67 104
    21 8.81 106
    22 8.95 107
    23 9.08 109
    24 9.20 110
    Sorry about the formatting (or lack of it), the first column is the length of the print (inches, cm, miles, it doesn't matter), the middle column is the correction in f-stops, and the third column is the same correction to the nearest 1/12th stop for users of our timers.

    To use the table, read off the factors for the old print length and the new, and subtract one from the other. For example, if your existing print is 10x8 and your new print is 16x12, the correction factor is 8.03 - 6.67 or 1.36 stops. In 1/12 stops, it's 96 - 80 or 16 twelfths.

    Another example: 7x5 to 10x8 is 1.04 stops or 12 twelfths.

    It works both ways so if you're downsizing just subtract the correction instead of adding it.

    As others have reminded me, reciprocity failure is an issue for larger corrections so add 1/12 per full stop as an approximation.

    If you don't have an f-stop timer, you can use a scientific calculator - enter 2, press x^y, enter the f-stop factor, press =, then multiply your original exposure time by that figure. Or of course you can use an f-stop look up table, there's probably one on this site somewhere.

    This only works if all other aspects remain unchanged i.e the lens, lens aperture, etc.

    Hope it makes some sense!

    Regards
    Richard

  4. #14
    Lee L's Avatar
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    Richard,

    I put your data into the attached .pdf file for folks to print out and use. (Thanks to openoffice it was cut and paste.)

    Lee
    Attached Files

  5. #15
    RH Designs's Avatar
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    Thanks Lee. I'll put an expanded version on our web site in due course. The great advantage of f-stop printing is that you can make up similar exposure correction tables for all sorts of things, such as filter factors, exposure changes with paper grade, etc, and they're all independent of the original exposure time. Saves a load of test strips.

    Cheers
    Richard

  6. #16
    RalphLambrecht's Avatar
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    I agree, but the distance-squared law just doesn't work with the neg-topaper distance.
    Regards

    Ralph W. Lambrecht
    www.darkroomagic.comrorrlambrec@ymail.com[/URL]
    www.waybeyondmonochrome.com

  7. #17
    RalphLambrecht's Avatar
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    Feel free to use this:
    Attached Files
    Regards

    Ralph W. Lambrecht
    www.darkroomagic.comrorrlambrec@ymail.com[/URL]
    www.waybeyondmonochrome.com

  8. #18
    RalphLambrecht's Avatar
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    Quote Originally Posted by metod
    Jim, your calculation taking into account the height of enlarger is more precise than mine with the paper. I think Iíll stick to that.

    Regards.
    Metod

    Careful! The enlarger height is irrelevant, the lens height is what matters. This and reciprocity failure might explain why some had problems with accuracy of the proposed math.
    Regards

    Ralph W. Lambrecht
    www.darkroomagic.comrorrlambrec@ymail.com[/URL]
    www.waybeyondmonochrome.com

  9. #19

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    Quote Originally Posted by Steve Smith
    The time is proportional to the area of the image
    image ....
    ...it is proportional to a linear dimension squared.
    Directly proportional at that. Areas are the product
    of TWO dimensions. Using "A" dimension, as you suggest,
    is nothing more than a quick and dirty application of the
    inverse square law.

    You've made no allowance for the optics involved.
    There is or are more complex formulas which include
    the optics. As I've already mentioned the lens speed
    increases as it draws near the film plane. Just as with
    a camera lens. Lens to negative distance is a factor in
    the more complex but more accurate formulas. Dan

  10. #20
    Steve Smith's Avatar
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    Dan is quite correct.

    I was not suggesting that doing a square law conversion would give a highly accurate new exposure time for each size of paper but it will give a very good indication of a starting point without having to do a full range test strip each time.

    And using the distance from the paper to the lens is probably the most accurate linear distance to use as it is not affected by size of negative, size of crop on the paper etc. So long as everything else stays the same, lens, stopping down, etc, then it can be used as a fairly good guide.


    Steve.

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