Time calculations when switching print sizes

Discussion in 'Darkroom Equipment' started by metod, Jul 4, 2006.

  1. metod

    metod Member

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    I am very curious how you folks calculate the exposure time change when switching print size in the darkroom. In my case, I take the longer size of the prints, square them up and then divide. So let’s say I am making my test print in 5x7 and switching to 10x8. I square up 10 and 7 and divide each other 10² : 7² = 2.04
    So my new time would double, or I open one stop the lens.
    This pretty much worked well for me. But is there a simpler way? What about those height markings on the enlarger column? Are they helpfull when rising or lowering the enlarger? Any ideas would be appreciated.

    Metod
     
  2. Ole

    Ole Moderator Staff Member Moderator

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    I'm lazy, and use lots of different sizes including 24x30cm, which isn't "double" anything at all.

    So I cheat, and use a EM-10 meter: Highlights are just almost-white when exposed for 20 seconds at an aperture which corresponds to "80" on the meter.

    Varies with the paper, of course...
     
  3. Steve Smith

    Steve Smith Member

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    Assuming you are using the same negative, your maths seems o.k. The time is proportional to the area of the image therefore it is proportional to a linear dimension squared.

    As for the height markings, I tend to make notes so I can get a starting point next time. E.g. I know that with a 6x6 or 6x4.5 negative with the enlarger head 16" high I get an exposure time of about 20 seconds with the lens stopped down two stops. In fact this height is the key to exposure rather than the paper size (assuming a normal negative).

    In this case, if I drop the enlarger to 8" I would expect four times as much light (in a quarter of the area) so the exposure should drop to 5 seconds.

    Steve.
     
  4. RalphLambrecht

    RalphLambrecht Subscriber

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    Steve

    Some paper sizes are designed to have 1 stop difference to each other (5x7, 8x10, 11x14, 16x20). You can also use the column height to apply the distance-squared law to calculate an exposure factor, but you need to take the lens-to-paper distances (not the negative-to-paper distances). In either case, don't forget to add some compensation for reciprocity failure. Rule-of-thumb is 1/12 stop for every doubling of exposure.
     
  5. PhotoJim

    PhotoJim Member

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    I use the height markings, and I use them the same way you use the paper dimensions. Suppose my enlarger is at a height of 60 cm. I move it to 90 cm. Suppose I have the first print at f/8 and 22 seconds. My new exposure time is 22 seconds * (90^2/60^2) which is 22*(8100/3600) which is 49.5 seconds.

    You can use inches if you like. I am more used to metric measurements. You can use mm if you prefer, too, as long as your units are consistent.

    This ignores reciprocity failure, which can be an issue with very long or very short exposures. It will result in underexposure, so if you notice that density is lower after making an adjustment, that might be why.
     
  6. RH Designs

    RH Designs Advertiser Advertiser

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    Hi Metod

    Your maths is correct. If you double the area of a print the exposure must increase by one stop and vice versa, because one stop equals a factor of two. If you double one dimension (as per 5x7 to 10x8) then the area increases four times, so you need an additional two stops.

    That formula is probably as easy as it's going to get.

    For those who use f-stop printing (and if not why not? :wink: ) use

    T = 2log(L2/L1) / 0.3

    where L2 is the new length of the print, L1 is the original length, and T is the adjustment in f-stops. E.g., if the new size is 14x11 and the old was 10x8, the adjustment required is 1 stop. If the new size is 16x12 and the old was 10x8, the adjustment is 1.4 stops, and so on.

    The height scale on an enlarger column isn't a lot of help unless it's marked in magnification rather than inches or mm. However, using the f-stop formula above you could (at least in theory, I haven't tried it) derive a scale for your enlarger column marked directly in f-stops, and then when you changed the height you could read off the difference directly in stops and use that as the correction factor.

    Regards
    Richard
     
  7. RalphLambrecht

    RalphLambrecht Subscriber

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    This math is correct, but only if you use the lens-to-paper distance (after focusing) and not the negative-to-paper distance. You can use the enlarger column if it is marked in magnification (mine is for several focal lengths), but you can't use the height of the negative above the paper. The distance-squared-law only works for the lens-to-paper distance. Anything else is likely to produce significant errors in exposure, and reciprocity is only of secondary influence.
     
  8. dancqu

    dancqu Member

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  9. lee

    lee Member

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    I use a Metrolux II about half the time and I have a gismo that plugs into the timer and reads densitys at the orignial size and then you move the enlarger head and re read the neg and it makes the correction for you.

    lee\c
     
  10. metod

    metod Member

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    Lee, it must be a joy to use that gizmo of yours….maybe one day I should get it.

    Anyway, thanks to all for the great suggestions. Richard your formula for f-stop printing is interesting, but isn’t kind of hard to adjust f-stops in decimals numbers?
    Jim, your calculation taking into account the height of enlarger is more precise than mine with the paper. I think I’ll stick to that.
    Reading up some stuff about the light I discovered another interesting formula. I found it actually in the book about plants and calculating the intensity of light for the plants to grow.
    It is I = L⁄D² where I = intensity, L=light output and D is distance.
    So I would calculate the Intensity 1 for the first print, then taking into account the distance change for the second print, calculate the Intensity 2. Then divide each other to get the needed ratio. A few different ways to get to the same point.

    Regards.
    Metod
     
  11. Troy Hamon

    Troy Hamon Member

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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.
     
  12. RH Designs

    RH Designs Advertiser Advertiser

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    Yes, hence the f-stop enlarger timer which does it all for you (shameless plug :smile: ) - www.rhdesigns.co.uk/darkroom/html/stopclock_professional.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
     
  13. RH Designs

    RH Designs Advertiser Advertiser

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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
     
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  15. Lee L

    Lee L Member

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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
     

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  16. RH Designs

    RH Designs Advertiser Advertiser

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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
     
  17. RalphLambrecht

    RalphLambrecht Subscriber

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    I agree, but the distance-squared law just doesn't work with the neg-topaper distance.
     
  18. RalphLambrecht

    RalphLambrecht Subscriber

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    Feel free to use this:
     

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  19. RalphLambrecht

    RalphLambrecht Subscriber

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    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.
     
  20. dancqu

    dancqu Member

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    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
     
  21. Steve Smith

    Steve Smith Member

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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.
     
  22. RH Designs

    RH Designs Advertiser Advertiser

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    Indeed. The OP was asking if there's a simple way, and the inverse square law is a simple way if the size change is not too large. By way of a check, the difference between what the formula predicted and what I actually measured with an Analyser when changing from a 10" width to a 16" width was less than a tenth of a stop - barely noticeable except for the finest quality work, and if you're working to that quality level you'll want to do another fine test strip anyway :smile:

    Regards
    Richard
     
  23. Gerald Koch

    Gerald Koch Member

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    See if you can get a copy of the Kodak Darkroom Handbook. There is a calculator for this purpose in it, sort of a circular slide rule, which is very handy. Lots of other stuff too.
     
  24. RalphLambrecht

    RalphLambrecht Subscriber

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    Richard

    Just use the inverse square law with the lens to paper distance, and there will be no difference to your analyzer measurement. You can check in 'Way Beyond Monochrome' on page 291 (same graph as I posted here before). There are no short cuts required for this. It's pretty simple as it is.
     
  25. dancqu

    dancqu Member

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    I'll point that out also. The reason the calculation for
    35 mm is more accurate is due to less of a change in the
    speed of the lens when going to 8x10. To 5x7 a 35mm is
    5 times enlarged while the 4x5 is only 1.4 times enlarged.
    For the 4x5 little more than a same size print. Think of
    the bellows extension. At a 1:1 ratio a lens, camera
    or enlarger, is down 2 stops. Dan
     
  26. RH Designs

    RH Designs Advertiser Advertiser

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    Just done a rather more scientific check and you're absolutely right - if you substitute lens-paper distance for print length in my previous post then it will work accurately.