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  1. #11
    BetterSense's Avatar
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    I'm familiar with how to calculate the "bellows factor" for a camera. But I've never heard bellows factor corrections mentioned when it comes to enlargers. This is probably because print exposures are usually arrived at through trial-and-error, so any bellows factor calculations are just unnecessary because knowing the effective aperture of the system just isn't important.

    However, there are formulas to calculate a changed print size that are based on the inverse square law, or on the changes in magnification--these two methods are geometrical and effectively do the same thing. However, I feel those formulas fail to provide the proper warning that they are correct only if the bellows factor changes by only a 'small amount' between the two print sizes. In the vicinity of 1:1 magnification, there could be quite a large difference in "bellows factor" and thus effective aperture between the two enlarger head positions. So in addition to applying the inverse square law, or magnification-based formulas, the user should also calculate the changed in bellows factor between the two print sizes and adjust the aperture ring of the lens so that the enlarger is operating at the same effective aperture at both print sizes. Only then will the magnification-based, or inverse-square-law-based formulas be correct.
    f/22 and be there.

  2. #12
    David A. Goldfarb's Avatar
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    No. Here's another way of thinking about it. Say you change the focal length of the lens with another lens of the same design (to minimize the effect of transmissive light loss due to factors like the number of glass-air surfaces) and make two prints of the same size from the same negative. The distance from the negative to the paper will be different and the length of the bellows will be different, but the exposure will be exactly the same. "Bellows factor" and magnification factor are two ways of describing the same optical phenomenon.
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  3. #13
    BetterSense's Avatar
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    The distance from the negative to the paper will be different and the length of the bellows will be different, but the exposure will be exactly the same.
    If so, surely, that is the case only if the lenses are set to the same F/number. We agree on this I hope.

    What's throwing me off, is that of course the two different lenses will be focused at different bellows draw, and different distance from the baseboard. You are effectively telling me that even though that's true, that both lenses, when set to the same F/number and focused, will suffer then same 'bellows factor' as each other. This is entirely plausible, but until I see the math I don't believe it.
    f/22 and be there.

  4. #14
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    Quote Originally Posted by David A. Goldfarb View Post
    It's purely a function of magnification and how wide an area you're spreading the projected image over compared to the negative size. Now think about the standard paper sizes--5x7, 8x10, 11x14, 16x20, etc. Notice anything familiar about those numbers? 5, 8, 11, 16? They're approximately the standard f:stop series, so when you go from one standard size to the next, the difference in exposure is going to be about one stop.
    Just like in camera exposure, given the 1:2 relationship that exists between f/stops (and shutter speeds), if the OPs exposure is 15 sec at f/8 for an 8x10 enlargement, then increasing magnifation to 11x14 (an effective one stop reduction in light intensity) would require a doubling of time to 30 sec. While I always understood that increased magnification affects exposure time, it never really clicked in that way with the standard print sizes, cool.

    Or, I guess he could keep the same 15 sec enlarging time and double the light intensity by opening up to f/5.6.

  5. #15

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    I am not sure but I suspect that this has something to do with it. and that is that the camera is a reverse enlarger.
    the subject in a camera is to the print in an enlarger.
    But the light is coming in from the opposite side.
    i.e. light from small negative to bellows to lens to large image for the enlarger. The camera
    is light from large image to lens to bellows to small negative. Having the light come from the opposite side of the lens/bellows combination must make a difference? Maybe?
    "There are a great many things I am in doubt about at the moment, and I should consider myself favoured if you would kindly enlighten me. Signed, Doubtful, off to Canada." (BJP 1914).

    Regards
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  6. #16
    David A. Goldfarb's Avatar
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    Quote Originally Posted by BetterSense View Post
    If so, surely, that is the case only if the lenses are set to the same F/number. We agree on this I hope.

    What's throwing me off, is that of course the two different lenses will be focused at different bellows draw, and different distance from the baseboard. You are effectively telling me that even though that's true, that both lenses, when set to the same F/number and focused, will suffer then same 'bellows factor' as each other. This is entirely plausible, but until I see the math I don't believe it.
    Yes, this is what I'm saying, but I'll let someone who can explain the math better than I can do that.
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  7. #17

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    I am not prepared to do a long post but keep in mind 1 thing. When you shoot with the camera you take the bellow extension (or distance from lens to film) into account but not the subject to lens distance. When you enlarge, the negative is now the subject so you don't care so much for the bellow extension there. What you need to know is the distance between the lens and the paper. So to calculate the effective aperture you use the distance betwen the lens and paper and not the distance between the negative and the lens. Because the effective aperture get smaller with bellow extension the effective aperture of the enlarging lens also gets smaller as you move the lens away from the paper. This is the reason why when you make large print you need to open the lens up or increase the exposure time. If the the effective aperture doesn't change then when you make a 4x6 or when you make an 8x10 you would use the same aperture and exposure time because the negative brightness is the same isn't it? Yes when used in a camera at infinity, enlarging lens transmit the same amount of light as a typical camera lens at the same aperture.

  8. #18
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    Your diffraction and the amount of light falling on the baseboard are related to your Effective Aperture which is calculated as follows:

    Ne = Ni(1 + m)

    Where,
    Ne = Effective aperture number (f-stop)
    Ni = Indicated aperture number on lens barrel
    m = magnification

  9. #19

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    I will add one further thing, the inverse square law only applies after the light has passed through the lens
    "There are a great many things I am in doubt about at the moment, and I should consider myself favoured if you would kindly enlighten me. Signed, Doubtful, off to Canada." (BJP 1914).

    Regards
    Bill

  10. #20

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    Close Ups

    Quote Originally Posted by cowanw View Post
    Having the light come from the opposite side of the lens/bellows
    combination must make a difference? Maybe?
    See the enlarger as a close up camera; a close up camera in
    reverse. Camera or enlarger, the same rules apply.

    The lens sees as subject a flat lighted film and projects what
    it sees upon a sheet of light sensitive paper, or film. Dan

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