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Thread: 90 inch bellows - what is exposure?

  1. #11

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    Re: 90 inch bellows - what is exposure?

    Funny thing, in open shade, I'd use 1 sec at f/22, by the time you stop 4 stops for the bellows and 2 stops to get you to optimum aperture!

  2. #12
    Drew Bedo's Avatar
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    Re: 90 inch bellows - what is exposure?

    So, lets see . . . .

    The focal length of the lens is 24 inches.
    The bellows extension is 90 inches.
    Each focal length equivalent of bellows extension will require an additional two stops of exposure, either by opening up the lens or lengthening the exposure time.



    So: 90 inches divided by 24 inches gives 3.75 focal lengths of extension.

    With two stops of additional exposure required for each additional focal length of extension that is 7.5 stops.

    (90/24)2=7.5

    Do I have that right?



    I don't know much about reciprocity issues, but this will be a long exposure.
    Drew Bedo
    http://www.artsyhome.com/author/drew-bedo




    There are only three types of mounting flanges; too big, too small and wrong thread!

  3. #13

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    Re: 90 inch bellows - what is exposure?

    A few useful magic formulas for closeup/macro:

    These apply to symmetrical lenses. The OP’s Apo Artar is symmetrical.

    f = lens’ focal length
    m = magnification
    Practically speaking the OP’s lens’ nodes are at its diaphragm.

    Extension (distance from the lens’ rear node to film) = f*(m + 1)

    Whence m = (extension/f) - 1

    Distance from the lens’ front node to the subject = f*(m + 1)/m

    Exposure adjustments:

    Effective f/ number = f/ number set * (m + 1)
    Exposure time increase factor = (m + 1)^2

    If the OP tells his meter he is shooting at his lens’ effective aperture the meter will suggest the right exposure time.

  4. #14

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    Re: 90 inch bellows - what is exposure?

    All good info, Dan. If someone's kit contains a lens that is known to be un-symmetrical it would be useful to know how the exposure correction is determined. For that I refer to an explanation published in my Leica Manual, excerpted here:

    EF = (1 + m/mp)^2 where mp is the pupil magnification; mp is the ratio of the diameter of the exit pupil to the diameter of the entrance pupil (seen from in front of the lens). It can be determined with sufficient precision by looking first into the rear of the lens and then into the front. Obviously, mp=1 for a symmetrical lens but will be less than 1 for an un-symmetrical lens, thus causing a higher EF due to reduced illumination at the film plane. Many lenses for MF or 35mm are likely un-symmetrical designs, and if the camera has through-the-lens metering EF would automatically be accounted for; a rangefinder user would not have that benefit. So the EF for an LF lens that is not known to be symmetrical or un-symmetrical can be determined after measuring its mp.

  5. #15

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    Re: 90 inch bellows - what is exposure?

    Quote Originally Posted by Drew Bedo View Post
    So, lets see . . . .

    The focal length of the lens is 24 inches.
    The bellows extension is 90 inches.
    Each focal length equivalent of bellows extension will require an additional two stops of exposure, either by opening up the lens or lengthening the exposure time.



    So: 90 inches divided by 24 inches gives 3.75 focal lengths of extension.

    With two stops of additional exposure required for each additional focal length of extension that is 7.5 stops.

    (90/24)2=7.5

    Do I have that right?

    I don't know much about reciprocity issues, but this will be a long exposure.

    No.

    (90/24)^2 Means to square the quantity (90/24)

    (90/24) = 3.75

    3.75 squared = 14.06

    That's the Bellows Factor. You have to increase the exposure 14 times. That's the equivalent of 3.8 stops exposure increase.

    To get the number of stops from the bellows factor, either estimate in your head:
    1 stop = 2x.
    2 stops = 4x.
    3 stops = 8x.
    4 stops = 16x.
    Since 14x is needed, it's a little less than 4 stops. You can safely round to 4 stops.

    Or, the number of stops = Ln Bellows Factor/Ln 2.
    You can also use Log base10. It doesn't matter.

    Ln 14 = 2.6.
    Ln 2 = 0.69
    2.6/0.69 = 3.78 stops

    Log base10 of 14 = 1.15
    Log base10 of 2 = 0.3
    1.15/0.3 = 3.8 stops

    Rich

  6. #16

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    Re: 90 inch bellows - what is exposure?

    My reasoning would be: f/11 with 24 inches focal length means an entrance pupil of 2.18 inches (24 / 11). Then I would measure the new lens/ground glass distance and calculate the actual diaphragm in this situation based on the entrance pupil that remained the same. If the 90 inches (bellows extension) is representative of that conjugated foci, then 90 / 2.18 = f/41.28, that I would round to 45 and get the same fours stops as Jerry and Johnny calculated above. I guess it is all the same. When distance doubles, area quadruples in the base of a cone light.

  7. #17
    Tim Meisburger's Avatar
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    Re: 90 inch bellows - what is exposure?

    There is a lot of unnecessary calculation going on in this thread. Just remember this formula: f number = focal length divided by internal lens diameter (more properly called entrance pupil). Lets call that F = fl/p. For most LF lenses you can simply measure the entrance pupil with a ruler and you will be pretty close, but it is also possible to calculate it if you have two of the three numbers, as lungovw has above. So

    fl = 24, F=11, so p = 24/11 or 2.18.

    Now the trick to all this is to remember that the fl stamped on the lens is its focal length at infinity. But in your case, you are not using the lens at infinity, you are using it at 90 inches, Consequently, your focal length for this image is 90 inches!

    So, pop that in our equation (F = fl/p) and it becomes F = 90/2.18, or an actual F number of 41 (or 45, as suggested above). Just determine the correct shutter speed for the subject a F45, and the film will be correctly exposed. Then, if its a long exposure, you can deal with reciprocity normally.

  8. #18

    Re: 90 inch bellows - what is exposure?

    Goodness. Forget the frigging math. Have the subject hold a Calumet bellows correction wafer on her cheek and measure it with the corresponding ruler on the GG. DONE. It does not matter what the lens is or the format or anything. The proof is in the pudding.

    For the life of me the willingness and ability of people here to make this excessively move complicated that needs to be is mind boggling. The objective here is to make photographs. This is not math class. Onward!

  9. #19

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    Re: 90 inch bellows - what is exposure?

    Quote Originally Posted by Tim Meisburger View Post
    There is a lot of unnecessary calculation going on in this thread. Just remember this formula: f number = focal length divided by internal lens diameter (more properly called entrance pupil). Lets call that F = fl/p. For most LF lenses you can simply measure the entrance pupil with a ruler and you will be pretty close, but it is also possible to calculate it if you have two of the three numbers, as lungovw has above. So

    fl = 24, F=11, so p = 24/11 or 2.18.

    Now the trick to all this is to remember that the fl stamped on the lens is its focal length at infinity. But in your case, you are not using the lens at infinity, you are using it at 90 inches, Consequently, your focal length for this image is 90 inches!

    So, pop that in our equation (F = fl/p) and it becomes F = 90/2.18, or an actual F number of 41 (or 45, as suggested above). Just determine the correct shutter speed for the subject a F45, and the film will be correctly exposed. Then, if its a long exposure, you can deal with reciprocity normally.
    Jim,

    Respectfully, I disagree. The gyrations that have been suggested over the years for understanding a very, very simple relationship are just amazing.

    All to avoid dreaded MATH. (Shudder!).

    What is so hard to comprehend? What is so hard to do?

    The length of the bellows when extended.
    Divided by the lens focal length.
    Squared.

    That's all there is to it. One silly measurement.
    That's the Bellows Factor. Thats the amount the exposure needs to be increased.

    1. Two.
    2. Four
    3. Eight
    4. Sixteen
    5. Thirty Two
    Those are f/stop relationships. Pick the value closest to the Bellows Factor. Guess an in-between value when necessary. How many steps were necessary to get to that value? That's the number of f/stops to open up.

    This is not higher math. It's not hard.

    If the Bellows Factor is eight, we need 3 stops more exposure.
    If the Bellows factor is 20, we need more than 4 stops, but less than 5 stops. Call it 4-1/3 stops. Take the shot.

    It's a direct, basic embracing of the actual situation.

    Just measure the darn bellows and everything else falls into place.

    Anyone who does it two or three times will never forget it. The only special piece of equipment needed is a little tape measure. I carry a spring-loaded cloth one made for sewing. It doesn't weigh a thing. Takes up no room in the camera case. Very low tech. Works great.

    Rich

  10. #20
    Tim Meisburger's Avatar
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    Re: 90 inch bellows - what is exposure?

    Interesting. So by direct calculation I end up needing two stops, and you three. I hope the OP reports the final result! (its such a large sheet of film, I think I would mount a 4x5 reduction back and then test. But I'm cheap!).
    Last edited by Tim Meisburger; 14-Aug-2017 at 04:50.

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