Bellows Compensation for Landscapes

[BradS]

I think you are correct. The math for factoring the exposure is:

Lens Focal Length^2 / Bellows Length^2

Where: LFL = 5" (25) / BL = 10" (100) Compensation would be 4 stops.

Rounding is acceptable as it is not a precise issue -- good enough for government work.

In you r example, the result is the bellows factor. A bellows factor of four is equivalent to two stops. (or, you could multiple the shutter speed by four).

[BradS]

Bellows compensation only comes into play when the bellows draw is longer than the focal length of the lens.

I teach large format photography and this is the most mis-understood and mis-used bit of information in photography. To simplify it, I tell my students that if they are shooting anything outside other than a flower or similar close up, they will not need any compensation. I also tell them that EVERY TIME they shoot their large format camera indoors, they will need to compensate for bellows extension.

Best regards,
Randy

ah....but...if you're focused on anything closer than infinity, the bellows is longer than the focal length of the lens.

[Leonard Evens]

Brad,

If you focus on anything sufficiently far away, although in principle there is a bellows correction factor, it is so close to one that you can ignore it. The usual definition of close up is within 10 times the focal length. But even at 10 times the focal length, the correction factor is 1 + 1/19 = 1.111. You square that to get the change in time, which means you would change the time by about 23 percent. Or you would open up by slightly less than a third of a stop. While it is important to get exposure right, it is not rocket science. The usual margin of error in measuring exposure, setting it, development, etc. is large enough that even a shift of 1/3 a stop probalby won't matter. While at 10 times the focal length, a particularly finicky person might want to make a correction, if the distance is significantly greater than 10 times the focal length, the bellows factor will be so close to one that the correction can safely be ignored.

[BradS]

If you focus on anything sufficiently far away, although in principle there is a bellows correction factor, it is so close to one that you can ignore it. The usual definition of close up is within 10 times the focal length. But even at 10 times the focal length, the correction factor is 1 + 1/19 = 1.111.
***

Yes, I completely agree with you up to here.

***
You square that to get the change in time, which means you would change the time by about 23 percent. Or you would open up by slightly less than a third of a stop.

Ooops! Although, the factor you calculated is numerically correct for the "object distance is 10 time focal length" example, this is a multiplicative factor. To properly use this (admittedly insignificant) factor, you have to multiply the indicated exposure time by the factor. So if, for example, the meter says that you should use f/11 and 1/125 of a second, then to apply this bellows factor, you would multiply the 1/125 second by 1.111 and quickly discover that it is pretty safe to ignore the bellows factor (in this case). Alternatively, to convert this (or any multiplicative) factor to stops you take the log of the multiplicative factor and divide that number by log 2. When you do that, you find that the bellows factor is about 1/6 stop (I get 0.152 stops in this example). Again, not significant enough to be concerned about.

I completely agree, that if you've focussed on something farther away than, say 8~10 times the focal length, then the bellows factor may safely be neglegted.

The equation that I always use is:

bellows factor = (image distance)^2 / (focal length)^2

which is of course, the same as:
bellows factor = { (image distance) / (focal length) }^2

[Bruce Osgood]

In you r example, the result is the bellows factor. A bellows factor of four is equivalent to two stops. (or, you could multiple the shutter speed by four).

HMMMM, perhaps that will explain some things.

Thanks,

[Leonard Evens]

Yes, I completely agree with you up to here.



Ooops! Although, the factor you calculated is numerically correct for the "object distance is 10 time focal length" example, this is a multiplicative factor. To properly use this (admittedly insignificant) factor, you have to multiply the indicated exposure time by the factor. So if, for example, the meter says that you should use f/11 and 1/125 of a second, then to apply this bellows factor, you would multiply the 1/125 second by 1.111 and quickly discover that it is pretty safe to ignore the bellows factor (in this case). Alternatively, to convert this (or any multiplicative) factor to stops you take the log of the multiplicative factor and divide that number by log 2. When you do that, you find that the bellows factor is about 1/6 stop (I get 0.152 stops in this example). Again, not significant enough to be concerned about.

I completely agree, that if you've focussed on something farther away than, say 8~10 times the focal length, then the bellows factor may safely be neglegted.

The equation that I always use is:

bellows factor = (image distance)^2 / (focal length)^2

which is of course, the same as:
bellows factor = { (image distance) / (focal length) }^2

There seems to be some confusion here. In my example, if the subject distance is 10 times the focal length, then the ratio of the image distance to the focal length is 10/9 ~ 1.11. Its square, which you call the bellows factor and agree should be used to correct the exposure time, is approximately 1.23. So I don't see what the 'Oops!' is about. When I say you should change something by 23 percent, I mean you should multiply it by 1.23 or equivalently find 23 percent of the value (which is the change) and add it to the original value. I believe this is the ordinary meaning in English of the statement.

Alternately, you open up by the log to base two of what you called the bellows factor. Using orindary lograithms, this is the log of that bellows factor divided by the log of 2, which is the same as the log of the ratio of image distance to focal length divided by log 2, all multiplied by two. In this case it is

2 x log(10/9)/log(2) ~ 0.3 stops.

P.S. To get the ratio of the image distance to the focal length from the ratio of the subject distance to the focal length, take the latter value and divide it by that value minus one. If the ratio of subject distance to focal length is 10, the ratio of image distance to focal length is 10/(10-1) = 10/9.

[BradS]

There seems to be some confusion here. In my example, if the subject distance is 10 times the focal length, then the ratio of the image distance to the focal length is 10/9 ~ 1.11. Its square, which you call the bellows factor and agree should be used to correct the exposure time, is approximately 1.23. So I don't see what the 'Oops!' is about. When I say you should change something by 23 percent, I mean you should multiply it by 1.23 or equivalently find 23 percent of the value (which is the change) and add it to the original value. I believe this is the ordinary meaning in English of the statement.

Alternately, you open up by the log to base two of what you called the bellows factor. Using orindary lograithms, this is the log of that bellows factor divided by the log of 2, which is the same as the log of the ratio of image distance to focal length divided by log 2, all multiplied by two. In this case it is

2 x log(10/9)/log(2) ~ 0.3 stops.

P.S. To get the ratio of the image distance to the focal length from the ratio of the subject distance to the focal length, take the latter value and divide it by that value minus one. If the ratio of subject distance to focal length is 10, the ratio of image distance to focal length is 10/(10-1) = 10/9.

Ah, yes. You are exactly correct. The "Ooops!" is entirely mine. Sorry for screwing this up and thanks for the follow-up.

Brad.

[neil poulsen]

Versus having to worry about calculations at the time, I built a table parsed by focal length that I carry in a small notebook that gives me f-stop corrections for the distance between two predetermined points on my front and back standards. I measure the distance and look up the correction. It's very quick in the field.