Calculating bellows factor with large tilts and/or swings?

[Matt Blaze]

What is the correct way to calculate the exposure compensation for a closeup photo where the focal plane consists of more than one reproduction ratio? For example, in the image below (a quick photo of a old padlock and key from my collection), the lens was swung such that the reproduction ratio in focus on the left is about half of what it is in focus on the right. If calculating by the ratio of the subject to the image on the groundglass, which part of the image should be measured? If calculating based on the actual bellows extension, which part of the bellows? (Is it accurate to simply measure from the center of the film/sensor plane to the center of the lens nodal point?)

I can't find the correct way to calculate bellows factor when large amounts of swing or tilt are used in any of my usual references (e.g., this site, Stroebel's book, etc).



Thanks

-matt

[Alan Davenport]

I've had a half-stop difference between the foreground and background so I just exposed for the foreground. I guess if the bellows factor was more extreme than that, you might need to add some supplementary light to the foreground, or maybe use a graduated ND filter to even things out.

[Emmanuel BIGLER]

I agree with what Allan says.

There is not much to do in terms of complex computations. There could actually be some really weird effects with extreme tilts in macro, say, when you tilt by more than 20°, a situation that never occurs for landscape.
If we do not take into account slanted ray effects that would occur with a wide-angle lens and large tilt angles (Jack Dykinga has a few words about that, sometimes, he says, there is a compensation effect of the natural ight fall in wide-anle LF lenses due to a large tilt), I would simply try to estimate the "local" bellows factor for different points of the image by placing a target like the Phillip Salzgeber's Quickdisc ....

http://www.salzgeber.at/disc/index.html

.... at different places in the field and measure its image on the ground glass.
This might not be very easy however but would definitely suggest whether the range of bellows factor is manageable or not.
I like very much the idea of the Quickdisc, no computer, no electricity, no maths, no fee, no software licence, no subscriptiopn to mandatory & costly monthly updates !
Definitely the best (very specialized, though) computing tool for the XXI-st LF photographer

(Is it accurate to simply measure from the center of the film/sensor plane to the center of the lens nodal point?)
Definitely yes for quasi-symmetric lenses i.e. most LF lenses be they "general purpose" or "macro / apo-repro" all except telephotos have their nodal points quite close to each other and located at a distance of a few mm close to the lens board ; and formulae for bellows factors in quasi-symmetric lenses are the same as for a single thin lens element.
The degree of precision required in those photometric corrections does not justify to precisely measure distances with respect to the actual position of the rear nodal point.

The Quickdisc will save you the burden of actually measuring distances on the bellows itself ; however you have to measure the diameter of the quickdisc image on the ground glass.

When the lens is quasi-symmetric, the Quickdisc gives the right answer by connecting the magnification ratio to the bellows factor.
There would be a small error with a telephoto, the Quickdisc is not designed for strongly asymmetric lens designs. You can safely ignore this subtle effect with respect to the main effects which are the range of magnification ratios for various elements of the images and of course, and even more important, the range of incident light values falling on the subject. This can vary very quickly in the image, for example if you use a single amateur flash unit located near the camera, instead of a good and large softbox.

[David A. Goldfarb]

With a still life like that: Polaroid.

[steve simmons]

Reproduction ratio is the size of the object in reality vs the size of the image on the gg. If they are the same it is 1:1 and the bellows extension will be twice the focal length of the lens.

Some people do the math, some use a device, me, I just measure the bellows. If I have extended it 25% longer then the focal length of the lens I add 1/2 stop, 50% more then the focal length of the lens I add one stop, 75% more I add 1.5 stops, and twice as long I add two stops.

Works every time.

I would not worry about any differences in repro ratio between the front and back of the subject.


steve simmons

[darr]

Reproduction ratio is the size of the object in reality vs the size of the image on the gg. If they are the same it is 1:1 and the bellows extension will be twice the focal length of the lens.

Some people do the math, some use a device, me, I just measure the bellows. If I have extended it 25% longer then the focal length of the lens I add 1/2 stop, 50% more then the focal length of the lens I add one stop, 75% more I add 1.5 stops, and twice as long I add two stops.

Works every time.

I would not worry about any differences in repro ratio between the front and back of the subject.


steve simmons

I have done the math running an HP 48GX and Bob Wheeler's Vade Mecum (and tape measure), and also use the method Steve describes above. Steve's method is just as good. Just my 2 cents.

[Matt Blaze]

I appreciate all the responses, but I fear my question may have been misunderstood. I understand how to calculate exposure compensation based on bellows length vs. focal length and based on subject size vs. reproduction size. These techniques are well known and widely documented (e.g., in Stroebel and on this site). Either method is simple and easy to apply when the front and rear standards are approximately parallel and when the camera focus is for a single reproduction ratio.

But these methods give indeterminate answers when extreme swings or tilts are used to focus on subjects over a wide range of reproduction ratios. The image above was a (not terribly extreme) example of this; I could imagine arranging subjects and swinging the lens so as to focus on reproduction ratios from 1:2 to 2:1 in the frame. Would the bellows factor in that case be 1/2 stop, 3 stops, somewhere in between, or something else?

Is it the case, as Alan's answer implies, that the bellows light falloff is uneven across the image in such a case, and the only answer is to use ND-grads or to arrange the lighting to compensate? If so, does the usual reproduction-ratio-based formula accurately give the degree of lighting compensation required?

I'm mainly interested in the theory here. In practice, I can get by trial and error (which is what most of the responses here seem to be advocating). But I find I do better if I understand the theory behind what I'm doing, and my references fail me here.

Thanks

-matt

[steve simmons]

You are making this more complicated than it really is.

None of the responses is advocating trail and error. They are giving you ways to determine the necessary exposure. All of them will work.

The answer is in the bellows extension. Light does not travel well. It diminishes as it goes farther. F22, or 23 or 11, whatever, is only the indicated f-stop at infinity. As you extend the bellows less of the light will hit the film plane. You have to compensate for this. Even with extreme swings and tilts the center of the lens is in approxiamtely the same spot. I said approximately so lets not quibble. This is what will determine the necessary correction.

Again, all of the methods will work. Just shoot, don't worry.

steve simmons

[Matt Blaze]

Steve,

I'm happy to just shoot and deal with approximations, but I also like to understand the theory behind what I'm doing. It's just the way I think, and understanding the theory helps me predict how different setups will turn out.

Anyway, I appreciate your response, but I'm still confused; your answer seems to imply that the reproduction ratio method will not work under extreme tilts and swings, but that the bellows length method will because illumination will be near uniform across the image. Is that correct? In that case, the theoretical measurement should be from the lens nodal point to the center of the film (or sensor) plane?

Thanks

-matt

[steve simmons]

Your movements will not be nearly as much as a little more or a little less bellows extension. Again, you are making this too complicated. If you swing or tilt the lens you are not really moving the optical center of the lens so you are not asking the light to travel any greater or lessor distance. If you swing or tilt the back you may be making a 1 or 2 inch change, again very little compared to extending the bellows another 25-50-100%.

This is the best explanation I can offer. If you need more then someone else will have to chime in.

steve