


Bellows compensation question
I have been reading Stroebel's View Camera Technique, a tough but rewarding slog. There's a lot of math (mostly geometry) but it explains in great depth just how cameras and lenses work. On page 184 there are a couple of equations for calculating bellows extension exposure compensation. I think I know what is going on in them, but am not 100% sure.
1) Effective fN = Marked fN X (ImageDistance/Focal Length) = fN X v/f
Question 1: Does the Focal Length (f) in the divisor equal the infinity focal length of the lens (which is always the same for a particular lens), or is it the measured length when focused on the image? Infinity focal length seems to make the most sense.
Question 2: Is the "Marked fN" the current aperture setting of the lens? Again, this seems to make the most sense.
2) Exposure Factor = (Image Distance)^{2} / (Focal Length) ^{2}
Question 3: Just what does one multiply against the resulting "Exposure Factor"?
In my fiddling with the numbers, if I have been doing it right, it does appear that multiplying the Effective fstop by the Exposure Factor does yield the Marked fstop.
A politician is a man who will double cross that bridge when he comes to it.
Oscar Levant

Wow, 138 views and no comments.
A politician is a man who will double cross that bridge when he comes to it.
Oscar Levant

Basically, bellows factor is (bellows extension/lens focal length) squared. The resulting number is your factor. You then multiply your metered time by that factor to arrive at your "corrected" exposure.
For an 8 inch lens, focused to 12 inches you arrive at 12/8=1.5, 1.5 squared=about 2, so you'd double the time. You could also keep the time the same and open up 1 stop.
Going back to an 8 inch lens, 16 inches becomes 4x, 24 inches= 9x, and 32=16x. You can see the exponential increase in factor is related to the linear increase in length. 2x the length is 4x the time, 3x the length is 9x the time, and so on.
I hope this helps and isn't too far off. It's always worked for me, although I don't often do too much close up focusing.

Just get an App on your phone that works it out. Life's too short...

Q1: I believe it is the atinfinity focal length. But it's not usually the easiest way to get the answer you want; just use the extension ratio as a multiplier. Say you're at 1:1 macro, the extension ratio is 2 (lens twice as far from film as when at infinity), therefore the effective fnumber is twice what is marked on the barrel. If you compute exposure using the effective fnumber instead of the marked fnumber, you've performed the bellowsfactor compensation.
Q2: yes.
Q3: time. Again, this is often not the easiest way to arrive at the answer you want because you probably cannot measure the image distance. For me at least, I generally start the bellows compensation calculation from magnification (M = image size / object size). And if you've computed exposure using effective fnumber instead of marked fnumber, you've already performed this correction.
bellows factor = (1+M)^2
Originally Posted by pbromaghin
it does appear that multiplying the Effective fstop by the Exposure Factor does yield the Marked fstop.
Nearly. The extension factor is E=M+1 and the effective focal length is f*E. The physical aperture does not change, therefore the effective fnumber increases by a factor of E. Exposure goes as the square of fnumber, hence the exposure correction factor of (M+1)^2

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I started to reply, but as my response was more of a 'this is what I do in the real world situation' and less addressing your questions directly, I ended up deleting it incase I came across arrogant or dismissive.
...however, since the 'get an app' post has been made, I think I'm safe
Mathematics, geometry and physics are all great fun, for a while I used to calculate my bellows extension by way of formula, because it made me feel like a big man. It didn't take long for that to get tired and now I just keep 30cm (because that's all the bellows I have) of a sewing tape measure in my kit. I have marked on it where all the fstops fall (in inches, 5.6", 8" etc), I also have marked where my four lenses focal lengths are. I set up, I focus, I measure film plane to lens board, if it's past the focal length marking for the lens I'm using, it's quick and easy to make a visual check of how far over and adjust exposure accordingly.
...that's made me a bit lazy, I couldn't even tell you the formula now!
____________________________________________
My goal in life, is to be as good a person as my dog already thinks I am.

My take on it:
The physics of it is simple, most of the time, when we are focussed on something far away, the distance between the lens and the film is approximately the focal length of the lens, and the image circle is about the same size. When you start to focus on something that is very close, and now the lens is further away, and the light that it is gathering is spread over a bigger image circle, so we have to increase exposure to compensate. Most of the time, you don't have to worry and you can ignore the bellows factor.
When do we need to start worrying?
1) with B&W film, I usually overexpose some, so I feel that I can tolerate 1 stop of underexposure before I start running into problems  I get one stop of underexposure when the image circle is twice the area as it is at infinity  which means that the diameter is a factor of SQRT(2)  or about 1.4 larger. The image circle diameter is proportional to the lens to film distance, so if I set up my camera for a close up and the lens  film distance is more than about 1.4 X the focal length (eg with a 150mm lens, 210mm) then I go through the calculations and adjust my exposure  or more likely, I toss in a seat of the pants adjustment. If the distance is 1.4X the focal length, I add one stop of exposure, at 2X the focal length, 2 stops, etc.
2) With color film, (transparency) I am more sensitive to exposure, and I get out the calculator to work out the exact right exposure with the equations.....
The other way to do it is to move to a camera with a built in meter, where you can meter off of the ground glass, and then it is all taken care of automagically for you.

Originally Posted by polyglot
Q3: time. Again, this is often not the easiest way to arrive at the answer you want because you probably cannot measure the image distance. For me at least, I generally start the bellows compensation calculation from magnification (M = image size / object size). And if you've computed exposure using effective fnumber instead of marked fnumber, you've already performed this correction.
Sometimes even guessing magnification ratio is not that easy. Therefore I made an excel sheet for all lenses I had.
All I need is to focus, measure distance between standards and find a nearest compensation factor. E.g. if the distance between standards after focusing Symmar 150/5.6 is 200mm, I have to expose more by 1 1/3 EV. Row '3' values (distance when focused on infinity) was measures, the rest are calculated.

Wow, thanks for all the great replies. There are so many large brains here, and seeing the same goal approached from so many different directions really helps the understanding. That, after all, is entirely the point of my pursuit of film photography. It is an intellectual challenge to learn and understand an activity that pretty much appears to be pure magic from the outside and to hopefully become skilled at it. Think Zen Archery (Kyudo). As far as getting an app, if I wanted to follow that approach to photography, I would be shooting digital.
A politician is a man who will double cross that bridge when he comes to it.
Oscar Levant

