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Matt Blaze
20-Jan-2008, 20:48
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).

http://www.crypto.com/photos/misc/lock-key-650.jpg

Thanks

-matt

Alan Davenport
20-Jan-2008, 21:46
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
21-Jan-2008, 02:02
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
21-Jan-2008, 07:49
With a still life like that: Polaroid.

steve simmons
21-Jan-2008, 08:26
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
21-Jan-2008, 08:39
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
21-Jan-2008, 11:15
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
21-Jan-2008, 11:23
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
21-Jan-2008, 11:36
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
21-Jan-2008, 12:00
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

Matt Blaze
21-Jan-2008, 17:04
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

Just to confirm then, do you agree that this implies that the reproduction ratio method won't work under extreme tilt/swing, and the only method that gives an accurate exposure compensation is to measure the distance between the film/sensor plane and the lens nodal point, correct?

I apologize if I seem impossibly difficult, stupid, or pedantic here, but I'm honestly just trying to understand what's going on. I find the reproduction ratio method generally more convenient, but on some compositions that gives a ambiguous answers (off by many stops from one another), depending on where in the plane of focus the ratio is measured.

Thanks

-matt

steve simmons
21-Jan-2008, 17:30
Just to confirm then, do you agree that this implies that the reproduction ratio method won't work under extreme tilt/swing,

Where did I say this?

What I have said is that the relative difference in the distance the light travels when doing back swing and tilt is so much less than the extra distance the light has to travel when extending the bellows to allow you to focus on a very close subject is such that it is not necessary to worry about it.

How extreme a movement are you talking about? Have you ever really been in this situation ? Most beginners use movements that are too extreme. If you do one inch of back tilt, or swing, then the difference in distance from the center of the lens to each end of the film area is only 2 inches. But if you have doubled the bellows to do a 1:1 and have an 8" lens then the bellows extension is now 16" In this situation you are fussing over a change in the projection distance that is 1/8th of the bellows extension! If you do 2" of swing/tilt on the back then the difference in the projection distance is now 4", still only 1/4th of the bellows extension.!!


You seem to have a problem that I can not understand.

sorry

steve

Alan Davenport
21-Jan-2008, 17:45
Matt, as Steve says it sounds like you are making more of the problem than need be.

Yes, you can get a difference in bellows factor across the frame, due to movements. I question whether (in the real world) you'll run into such a disparity that you can't get a workable exposure (which seems to be your real question.)

This image had a 1/2 stop difference from the foreground (1/4 lifesize) to the background at infinity. To get there, I used more tilts and swings than I thought possible; the camera looked almost like in an ebay seller's ad. In the real world, it seems unlikely that you'll run into a multi-stop difference across the frame.

http://farm1.static.flickr.com/98/252158852_8e22e7310b.jpg

Matt Blaze
21-Jan-2008, 19:19
Just to confirm then, do you agree that this implies that the reproduction ratio method won't work under extreme tilt/swing,

Where did I say this?

What I have said is that the relative difference in the distance the light travels when doing back swing and tilt is so much less than the extra distance the light has to travel when extending the bellows to allow you to focus on a very close subject is such that it is not necessary to worry about it.

How extreme a movement are you talking about? Have you ever really been in this situation ? Most beginners use movements that are too extreme. If you do one inch of back tilt, or swing, then the difference in distance from the center of the lens to each end of the film area is only 2 inches. But if you have doubled the bellows to do a 1:1 and have an 8" lens then the bellows extension is now 16" In this situation you are fussing over a change in the projection distance that is 1/8th of the bellows extension! If you do 2" of swing/tilt on the back then the difference in the projection distance is now 4", still only 1/4th of the bellows extension.!!


You seem to have a problem that I can not understand.

sorry

steve

Steve,

I didn't say you said that. I said that I understood what you wrote to imply that, but wasn't sure you agreed, which is why I asked. My confusion is because of this: Suppose we tilt the lens (in a closeup photo) such that several parts of the image are at the plane of focus but are at different (but relatively high) reproduction ratios. How do we calculate the correct exposure compensation?

Measuring the subject reproduction ratio on the ground glass gives us different answers depending on which part of the image we choose to measure. But this method would be the best we could do if the the amount of light falloff across the image is nonuniform and depends on the reproduction ration.

Alternatively, we could measure the lens nodal point to film (or sensor) distance. This gives us a single, unambiguous answer, which is presumably correct if the light falloff is uniform across the image regardless of tilt, but not if it isn't.

At high reproduction ratios and a lot of tilt, and with high coverage super-duper grand angulon lenses, it might be possible to focus such that parts of the image at are 1:2 (1 stop) and other parts are at 3:1 (4 stops). That's a big exposure difference. It certainly doesn't come up every day for a typical landscape photo, but I work a lot on a tabletop and in the macro world.

One might reasonably conclude that this is just academic or too theoretical to be of any interest, or that I'm too much of a neophyte to be fooling with this, and I apologize for wasting your time if that's the case. But I'd still like to know.

Here's a quick and dirty example that I took just now for the purposes of example. Sorry about the low quality. It was taken with a 120mm/8 Super Angulon, with a BetterLight scanning back (which is a bit smaller than 4x5) on a Sinar P. The rear standard was vertical and the front standard tilted about 35-40 degrees down to keep the tops of the sample RFID casino checks in focus (or approximately so) across the image. It was shot wide open at f/8. (Some chromatic aberration is visible on the full size scan; this is not a macro lens, but it was handy). The lighting was from a soft source (a Kino-flo box) above the subject; I aimed for even illumination, although I may not have achieved that because of interference from the camera. The reproduction ratio at the bottom of the image is about 2.5:1 and at the top is about 1.4:1, give or take. That's a significant difference in exposure by the reproduction ratio method.

I don't regard this as a valid test; I didn't perfectly control the light, the lens is pushed beyond its design, and I didn't calibrate or double check all my measurements. And the results are unclear; here's obviously some light falloff at the bottom, but perhaps that's because I did a bad job lighting it evenly. I could probably spend some time and effort trying to conduct more rigorous experiments, but I'd like to know what the theory says I should expect.

If you don't know, there's no shame in that; you're certainly not obligated to respond or research the answer! I realize I'm asking about an esoteric subject.

Thanks

-matt

http://www.crypto.com/photos/misc/chips103-600.jpg

Emmanuel BIGLER
22-Jan-2008, 03:58
I realize I'm asking about an esoteric subject.

Not at all, but it is difficult to explain all the effects without a diagram.

So I have prepared two diagrams as attached pdf files that hopefully will clarify the situation even if they do not give to our faithful readers any detailed and quantitative answer in all cases... because it becomes complex in general.

The first diagram reminds us the usual correction formulae for wide-angle (WA) lenses (natural fall-off or natural vignetting effects, I prefer the terme of fall-off to vignetting, to me vignetting is something unwanted, a nuisance generated by an bad lens hood or something than intrudes in the light path)
Same is plotted for the close-up range (bellows factor) and shows where the image plane is located fo both cases. The general case of a wide angle used in close-up is not covered by the usual formulae.

The correction factor for a WA angle in infinity-focus is cos^4(theta) for most lense except... modern WA lenses for which a superb trick named pupillar distorsion will "block" the effect of one of the cosines, hence is will be something like cos^3 (theta) (the model is too simple to explain everything, see you manufacture's data sheets !)
For classical close-up images with no WA lenses the usual effect is (1+M)^2 or (d/f)^2

Now in general.... we have to combine d^2 and cosines together. And imagine that the film plane extends off the diagram plane, so we are in fact oblique in all directions !
The second diagram explain how to estimate in general the correction factor with respect to a given reference position. Usually, the reference position is on-axis in the focal plane, d=f for quasi-symmetric lens designs. This is what hand-held meters give you.
So in order to make things more manageable, we'll consider here only quasi symmetric lens designs for which the exit pupil is located at the rear nodal or principal point N'=H', one focal length ahead of the focal plane.
The exit pupil acts like a small light source disc that sends rays in all directions, this is the somple idea to keep in mind in order to imagine what happens without any complex computations.

If the film plane is really slanted with respect to the mean ray exiting from the centre of the exit pupil, there is an effect of oblique rays that can reach an attenuation factor of 70% when the film is at 45° from the mean ray, Simply because cos(45°)=0.707, the same amount of light is distributed aver a bigger surface.
Most ouf our readers living at 45° of latitude know that their climate is milder than at the equator, the reason, optically speaking, is the same. However the cosine does not explain everything, 45° is the latitude for Bordeaux in France as well as for Québec, Canada ; the climate is definitely not the same ;)

This effect occurs in the corners of a wide angle lens and is combined with the effect for increased distance d^2 between the exit pupil and film corners.
So if you compare the amount of light per square area received by a film in a certain position and if you apply a rear tilt, you get an additional attenuation factor due to a second cosine, cos( theta_2) on the diagram.

Now where are we in your practical situation.
First, if the tilt angle for the film plane with respect to the mean exit pupil ray is, say, 20°, which is already quite large as a rear tilt, the attenuation factor, cos(20°) is only .93, this usually negligible.

So my guess is that in your case the usual close-up formulae (1+M)^2 apply, there are no cosine effects at all, simply uses Steve Simmons's good rules, again non computer no chart is necessary !

It should be noted that in the model, the shape and tilt angle of the object plane with respect to the lens bord plane has, surprisingly no direct effect.
The only indirect effect you'll get, again if the rear tilt angle is moderate, is a possible change of magnification within the image.
In the image just posted above, I cannot tell you whether the darkening effect observed at the bottom of the image is due to a cosine effect or is due to a d^2 effect. Probably both.
Objects in the foreground are projectd on film with a magnification which is visibly bigger than those in the background. Hence we could explain the effect by the (1+M)^2 formula.
But depends on how big you rear tilt is set in this example.
Of course we assume that the objects are evenly illuminated and that the gradual darkening of the paper is due to the camera and not to the incident light distribution.

Matt Blaze
22-Jan-2008, 07:12
Emmanuel,

Thanks so much for the lucid and patient explanation. The various effects are now much clearer in my mind. Ignoring the second cosine effect, (M+1)^2 is clearly equivalent to (d/f)^2 at all parts of the image. I was neglecting to measure d to the part of the film plane on which the image is actually projected (and was incorrectly just measuring to the center of the groundglass, as I had understood from previous references). So by either method, under tilt we expect the light loss to be different in different parts of the image (depending on magnification), since the distance between the lens and the film will vary widely across the image.

Fortunately, my camera was still set up from last night. and so I could re-measure distance from the lens to different parts of the film plane. And sure enough, d/f was (approximately) equal to M+1 at each part of the image. The light falloff can thus be explained by either method (although (M+1)^2 is probably easier to work with in practice in such cases).

So for such setups, there will be different compensation required in different parts of the image, and one could use either nd-grads or arrange the lighting accordingly.

Thanks again. It's always nice when theory and practice manage to agree on something!

-matt

Emmanuel BIGLER
22-Jan-2008, 07:35
Matt, I am really happy if my explanations are meaningful to your experiments.

Additional notes:
- I should not have used "rear tilt" ; in fact the relevant tilt angles for light fall-off effects is the total tilt angle between the lens board and the film plane, equal to the sum of the front tilt + rear tilt on the camera, in the usual sense for view camera movements.

- on the previous slide #2 the formula for close-up should read : d^2 / f^2 = (1+M)^2

- For those inclined to really esoteric things ;) I have made an additional slide (see below) explaining how the luminance of the aerial image, equal to the luminance of the source, explains the above mentioned formulae. The notion of luminance is something totally non intuitive and perfectly lambertian sources hardly found in our photographic subjects (an approximation of such an ideal source could be a perfect matte screen covered with fine powder of magnesium oxide, illuminated by a perfect large-size diffusor)

Leonard Evens
22-Jan-2008, 09:54
Emanuel,

I am confused about your formulas. What are S_{EP} and S_{IM}?

Emmanuel BIGLER
22-Jan-2008, 10:12
Oups ! "A" like Area in the figure became "S" like Surface (the French version) in the text by mistake.
I have attached the corrected version.

CG
22-Jan-2008, 11:14
The example with three "coins", up close, on a white background, in studio, is a worst case scenario, and demonstrates that compensation could be needed under some circumstances. One's eye expects even lighting to really look even under this condition.

While most folks just don't tilt so much, or under such a demanding setting, it can be an issue. It would seem that for such major tilts, if you want near perfect evenness of exposure, there are three practical ways to approach the conundrum.

Most simply, pump more light into the nearest areas where the repro ratio is highest and exposure is lessened most. IE, fake the lighting to create even exposure at the film or sensor. Either use a behind the lens exposure probe to get a spot reading of the light actually hitting the film, in essence a incedent reading using a white card or a gray card at various places within the setup, or, second, use a digital back and a couple of exposures to work it out empirically while setting the lighting to be brightest where the falloff is worst.

Third, use a set of grad filters to even the exposure out to a workable degree. It would take only three or four various soft grads to get you a reasonable compensation; 1/2 stop, 1 stop, 1 1/2 stop etc.

While you have formulae that explain the losses up close, a fast and acceptably accurate working plan is what one needs to get work done. Most folks won't want to consult a spreadsheet to shoot a closeup.

Maybe others can chime in with other better plans.

C

Matt Blaze
22-Jan-2008, 14:36
Matt, I am really happy if my explanations are meaningful to your experiments.


Yes -- extremely helpful and exactly what I was looking for. Thanks again.



Additional notes:
- I should not have used "rear tilt" ; in fact the relevant tilt angles for light fall-off effects is the total tilt angle between the lens board and the film plane, equal to the sum of the front tilt + rear tilt on the camera, in the usual sense for view camera movements.


Right, I understood that (which is the only thing that makes sense, given that the camera doesn't know which standards are tilted, only their relationship to each other and the subject).



- on the previous slide #2 the formula for close-up should read : d^2 / f^2 = (1+M)^2


And so log_2 {(M+1)^2} (or log_2 {(d/f)^2}) gives the correction in f stops.



- For those inclined to really esoteric things ;) I have made an additional slide (see below) explaining how the luminance of the aerial image, equal to the luminance of the source, explains the above mentioned formulae. The notion of luminance is something totally non intuitive and perfectly lambertian sources hardly found in our photographic subjects (an approximation of such an ideal source could be a perfect matte screen covered with fine powder of magnesium oxide, illuminated by a perfect large-size diffusor)

Terrific; thanks. These slides deserve to be archived somewhere; perhaps someone can link to them in the tutorial section on largformatphotography.info.

Best

-matt

Matt Blaze
22-Jan-2008, 14:41
While you have formulae that explain the losses up close, a fast and acceptably accurate working plan is what one needs to get work done. Most folks won't want to consult a spreadsheet to shoot a closeup.



Well, yes and no. While few photographs demand or benefit from an exact lighting or exposure calculation, these equations are not particularly onerous, and understanding them (even without making accurate measurements of the lighting, bellows length or reproduction ratios for any given exposure) provides the basis for developing an intuition about how a setup can be expected to behave and what kinds of corrections will be required. At least that's how I work.

Best

-matt

David A. Goldfarb
22-Jan-2008, 15:05
I don't see myself making these calculations in the settings were they might be relevant (pretty much exclusively tabletop work), but I think it is an interesting explanation of a phenomenon that is worth being aware of. In practice, I could see myself looking at a Polaroid and adjusting the lighting to compensate for falloff of this type, and knowing in principle why this type of falloff would be happening would provide some insight into what might happen if, say, I were to recompose and use less extreme tilt/shift. Knowing about this possibility, I might be more likely to make meter readings at the groundglass using a booster probe, and that could save a Polaroid ($15 for 8x10") or two.

Emmanuel BIGLER
23-Jan-2008, 08:50
I agree with David G. and C.G. ; and as an amateur using the view camera with film and mechanical shutters, I accept of course the electronic hand-held meter ;) but I would definitely be reluctant to carry anything else but a paper chart or the Rodenstock specialized slide rule in the field ! ( I belong to one of the last generations where we had to learn how to use the slide rule at school !!)

This yields a question about the use of boosters in the image plane or similar devices ; I have in mind a device marketed in the seventies-eighties by Sinar, based on a Gossen meter analyzing the ground glass. Arca Swiss had a device with silicon light-metering sensors embedded in the ground glass.
I have spoken of such devices with French professionals who had worked with a view camera on a daily basis, none of them were really enthusiastic about them and recommend the spotmeter aiming at the subject to analyze the scene + corrections, the classical close-up corrections for example.

Well, one could argue that often professionals in the past were reluctant to new tools when they had elaborated good lighting & metering methods that worked for them, but my feeling was that metering on the ground glass brought also some specific problems and never convinced professionals...
And there was the instant polaroid image for preview ! This seemed, on the conrtrary, to be a favourite method in studios in the past... at a nominal cost as explained by David.

Drew Bedo
10-Feb-2008, 10:55
Hello All;

I know that there is a practical need to "get it right " in one shot when working in LF, yet with a complex set-up i.e. extream macro and large tilt/swing, has any one tried bracketing to see what actually happens and by how much?

CG
11-Feb-2008, 00:35
Well, yes and no. While few photographs demand or benefit from an exact lighting or exposure calculation, these equations are not particularly onerous, and understanding them (even without making accurate measurements of the lighting, bellows length or reproduction ratios for any given exposure) provides the basis for developing an intuition about how a setup can be expected to behave and what kinds of corrections will be required. At least that's how I work.

Best

-matt

I don't see what gets accomplished beyond a theoretic knowledge by going to some set of equations that give you a perfect explanation almost no one will actually use.

Isn't it enough to know that the extension and repro ratio at each area of the gg determines the correct light loss? There are several methods for working it out, that are, in the real world, exact. I sort out my losses from that. You, I'm sure do also.

It's implicit in your example of the "coins". Measure the coins. Measure the images. You now have the repro ratio. The calculations are the basic ones. No more specialized calculations are needed.

I get concerned that photography tends to devolve into an exercise of expertese-itis and esoteric technical legerdemain. I'm enough of a geek to get that and to appreciate it, but to do a photograph I'm a believer in finding the approach of greatest and simplest practical utility. Bellows extension and repro ratio calcs or behind the lens metering - take your pick - are necessary and sufficient.


...a device marketed in the seventies-eighties by Sinar, based on a Gossen meter analyzing the ground glass. Arca Swiss had a device with silicon light-metering sensors embedded in the ground glass.
I have spoken of such devices with French professionals who had worked with a view camera on a daily basis, none of them were really enthusiastic about them and recommend the spotmeter aiming at the subject to analyze the scene + corrections, the classical close-up corrections for example

The probe meter is not for everyone, but it was a very practical answer to these sorts of annoying exposure issues - and the probe takes care of bellows extension and tilt etc all at once.

Best,

C