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Leonard Evens
30-May-2006, 09:25
Before going on, let me say that this is of little importance from a practical point of view, so if it offends you, don't proceed further.

It seems well known to those who engage in panoramic photographic that, in order to avoid parallax errors where adjacent shots are joined, you should roate about the center of the entrance pupil. For most lenses, with both sides of the lens in air and unit pupil magnification (or close to that), the center of the entrance pupil coincides with the front nodal point (or is close to it). But for some lenses, e.g., those of telephoto design, the pupil magnification may not be close to one, and so the entrance pupil will be some distance from the front nodal point.

This seems to be a contradiction. The nodal points are defined by the condition that for every off axis point in the subject, the ray from that point to the front nodal point makes the same angle with the lens axis as the corresponding exit ray from the rear nodal point to the image point. That means that subject points along the entrance ray will yield image points along the exit ray. That seems by definition to say the nodal point is the center of perspective and the proper point to rotate about.

I've looked for an explanation of this seeming paradox without much success. The only explanation I found which began to make sense suggested that the entrance pupil can in effect be treated as a peephole for a pinhole camera, but I didn't find the rest of the explanation convincing. After some thought, I've come up with the following explanation, but I'm not sure it is right. Anyway, here it is.

Consider the pencil of rays starting at the subject point and passing through the entrance pupil. They form a solid cone. with base the entrance pupil. The key point is that the front nodal ray may not be the central ray of this cone. In that case, the rear nodal ray will not be the central ray of the corresponding solid cone from the exit pupil to the image point. Now take into account that the film plane is never going to be set precisely at the image point. It certainly won't be if the subject point is not in the plane of exact focus, and if we are interested in parallax errors, at least one of two points which should line up will be at some distance from the plane of exact focus, albeit in the DOF region. That means we have to consider not points in the negative plane but image discs. One can make the argument that for two image points to appear to be lined up, the central rays of the corresponding cones to the exit pupil should be lined up, which means that the two image discs in the film plane will be concentric. If you rotate about the nodal point, you will mess up that arrangement.

So the conclusion would seem to be that it makes more sense to rotate about the center of the entrance pupil, although in point of fact it is not the actual center of perspective in the strict geometric sense usually studied in art theory. I haven't worked it out quantitatively, but it may be that you mess up the concentric relation of the image discs less by so doing.

I'm not sure I find my own explanation convincing. Perhaps it is obvious why the nodal point despite the equal angle property is not the center of perspective and I'm just missing something. If so, I would like an explanation. Of course, in most large format photography, the nodal point, the principal point (intersection of the principal plane with the lens axis) and the center of the entrance pupil are the same or at least extremely close to one another, so it doesn't make any difference.

A somewhat different question is the following. It is often asserted that the entrance and exit pupils can depend on the distance to the plane of exact focus (and hence to the corresponding image plane). This is equivalent to saying that the pupil magnification can vary when focusing. The entrance pupil is defined to be the image of the physical aperture as seen through the front elements of the lens and the exit pupil similarly with respect to the rear elements. Unless the distance between lens elements or distances to the physical aperture change while focusing, I don't see how the positions of the entrance and exit pupil can change. That may happen with zoom lenses or possibly some other small format lenses, but it doesn't happen with large format lenses. Am I missing something?

Michael S. Briggs
30-May-2006, 10:31
There are at several cases of panaromaric cameras. I think Leonard is addressing the case where a conventional camera is rotated to take a series of photos. It seems that the answer for this case is the entrance pupil. For a swing-lens cameras with stationary film, the lens is swung about its rear nodal point. This case is discussed in Applied Photographic Optics by Sidney Ray. He also mentions rotary cameras in which the camera rotates and the film is moved past a slit -- in this case he asserts that the correct rotation point is the front nodal point. Most of this is a diversion -- just we should be clear about which case is being discussed -- no doubt the case of swinging an otherwise conventional camera.

Leonard Evens
30-May-2006, 11:14
Michale Briggs said:
There are at several cases of panaromaric cameras. I think Leonard is addressing the case where a conventional camera is rotated to take a series of photos.

Yes that is what I am talking about.

It seems that the answer for this case is the entrance pupil.

Do you have a reference for that which explains why it should be the case?

For a swing-lens cameras with stationary film, the lens is swung about its rear nodal point. This case is discussed in Applied Photographic Optics by Sidney Ray. He also mentions rotary cameras in which the camera rotates and the film is moved past a slit -- in this case he asserts that the correct rotation point is the front nodal point.

I don't see how the latter case, a rotary camera, is different from rotating a fixed camera. In that case, he seems to be agreeing with me that the center of perspective is the front nodal point. That is just the paradox that is concerning me. The front nodal point is not the center of the entrance pupil if the pupil magnification is not one.

Ray's book seems to be the authoritative source, but I don't know where to get a copy of it, short of buying it---at least \$149 used. Northwestern doesn't have a copy.

Helen Bach
31-May-2006, 10:56
Rotary cameras are different from rotating a normal camera and taking successive frames because the displacement of the image on the film isn't an issue in the latter case.

If the nodal points and entrance pupil are displaced, rotating the lens about the rear nodal point (ie for rotary cameras) only works perfectly for distant objects.
From the second edition of Ray:

Struan Gray
31-May-2006, 12:58
Leonard, I went though a similar epiphany a few years ago, after prompting by Yves Coulombe, who sadly doesn't seem to be posting these days. The dawning of the light has been preserved for posterity here:

http://www.photo.net/bboard/q-and-a-fetch-msg?msg_id=003bju

[Sadly, I think the discussion in French that Yves refers to did not survive the change of forum at galerie-photo. If it did, I failed to find it.]

The point - ahem - is that although the nodal points are essential in locating the in-Gaussian-focus image of any point in subject space, the selection of which rays from the subject actually combine to form the image is done by the aperture. The privileged ray passing through the nodal points need not contribute at all.

If the front nodal point were the center of perspective you would expect that bodies in subject space lying on lines radiating out from the nodal point would be superimposed on the negative. However, for small stops a napkin sketch will quickly convince you that if the center of the entrance pupil is displaced from the nodal point, those bodies will put light onto the negative in different positions. Their perfect, in-focus representations are collinear with the exit pupil, but that doesn't mean that the ray bundles permitted to pass through the lens will necessarily pass through the film plane at the same place.

This is easy to see with the simplest possible asymmetric lens: a small hole in front of a singlet. The difficulty comes in accepting the results as general for complex, thick lenses, for wide apertures, or for LF-irrelevant exotics such as underwater objectives or oil-immersion microscope lenses.

"The flow of light" does also affect DOF and exposure compensation. Searching here and photo.net using "pupillary magnification" will turn up a host of discussions of the topic. Jeff Conrad's detailed DOF notes here at LF.info are clearer than anything I have read on or offline on that topic, and Emmanuel Bigler has posted some detailed descriptions of exposure compensation with asymmetric lenses at photo.net.

This is not just a theoretical exercise (although mostly so :-). Some of the LF aerial telephotos can have significant pupillary magnifications, as well as the early 200-400 mm telephotos from Meyer and - I think - Dallmeyer. Tessars are the classic example of a 'normal' lens that is asymmetric, but I doubt that they are asymmetric enough to cause problems in anything but oddball use. How many photographers make stitched macro panoramics on sheet film?

Leonard Evens
31-May-2006, 17:26
Struan,

Thanks for the explanation. I think I was trying to say something along the same lines.

The point is that any object points which are collinear with the center of the entrance pupil will produce image points which are collinear with the center of the exit pupil. So rotating about the center of the entrance pupil will not disturb parallax relations if we look at the cone with apex the object (image) point and base the entrance (exit pupil). Rotating about the front nodal point instead will disturb the collinearity of the rotated points with the center of the entrance pupil (and hence of the image points with the center of the exit pupil). What wasn't clear to me was that this actually made any difference for real lenses. As long as the blur circles of the image points in the film plane are within the coc, we shouldn't be able to tell the difference.

I've done some calculations, which I don't really trust, but they suggest that with a 500 mm lens and two points at roughly 10 and 20 meters from the lens, the parallax error from a 45 degree rotation might be of order of magnitude 1 degree. At a distance of 15 meters that would suggest a spread of about 260 mm. If the camera was focused at that distance, the magnification would be 29, so the two exit pupil rays would be off by about 260/29 ~ 9 mm where they intersect the film plane. That is a signficant shift. But, as I said, I don't really trust the calculation.

C. D. Keth
31-May-2006, 17:42
What sort of panoramas are you making that demand this sort of accuracy? I know from firsthand experience that for motion control in motion picture effects work (where the perspective must match to a T) it is accurate enough 99% of the time to use the center of the lens barrel at the front of the iris ring as the point of rotation. It's not millimeter accurate but it's plenty accurate to fool audiences and satisfy multi-million-dollar investors. It could also be (this is likely, actually) that I'm just not familiar enough optically with telephoto design lenses for LF work.

Helen Bach
31-May-2006, 18:41
What sort of panoramas are you making that demand this sort of accuracy?...

Well, Leonard's first sentence reads; "Before going on, let me say that this is of little importance from a practical point of view, so if it offends you, don't proceed further."

For me, this discussion is about understanding the principles behind what we do. This may or may not have practical benefit, but it certainly doesn't detract from anything. However, knowing that the entrance pupil is the centre of perspective of a lens, and given that the entrance pupil is extremely easy to locate in practice* makes me wonder why you don't do motion control using the entrance pupil instead of the front of the iris ring. I mean, it's just such a trivial matter in comparison to everything else to do with motion control, or to almost any aspect of cinematography...

*You look into the front of the lens and put your finger on the barrel where the aperture appears to be. How hard is that?

*******************

I find it a little bit puzzling that Leonard has re-written what I wrote in another recent discussion without mentioning the discussion, or my explanation of why it is the entrance pupil and not the front nodal point. No matter.

Though this is a re-write of that which has already been written, here is my version with less DIY required than my previous version. No disrespect to Struan for his excellent explanation. It's good to get things from two or more independent sources.

Perspective is a three-dimensional issue: it is not bound by the rules of in-focus images.

A ray passing through the front nodal point is sufficient to define the location of an object point in the plane of focus, but not of an object point that is not in focus, ie that forms a disc in the film plane.

A ray passing through the centre of the aperture defines the apparent centre of the disc, on the film, of an object point that is not in perfect focus. A ray that passes through the centre of the aperture is a ray that, before entering the lens, is directed towards the centre of the entrance pupil.

Therefore the rays that are directed towards the centre of the entrance pupil define the way in which the three-dimensional object space is translated into a two-dimensional image.

*******************

Swing lens cameras: The two types of swing-lens camera (lens swings, film stays still; lens and camera body swing, film moves in the camera) are analogous. In both cases the rotation should be about the rear nodal point, because swinging about the rear nodal point is the only way in which a distant in-focus point will stay in the same place on the film as the rotation occurs. The error introduced for near points, and for the change in the location of the entrance pupil (assuming that the entrance pupil is not coincident with the rear nodal plane), is limited by the width of the slit. The slit minimises the effective amount of rotation for any point on the film.

Best,
Helen

Struan Gray
1-Jun-2006, 01:14
...Their perfect, in-focus representations are collinear with the exit pupil, ....

I meant, of course, collinear with the rear nodal point.

Helen, Leonard, please restate at will. I am a great fan of learning by discussion, and wish there was more time for symposium-style dialogue in today's formal education systems. It's one reason I like online forums so much, and I wish there had been discussions like this in the archives when I first started to confuse myself with this stuff. It is easy to find statements of fact like those in Ray or the Lens Tutorial; much harder to find someone who can put things into context or explain with words and concepts instead of squiggles and equations.

The classic application for this degree of nitpicking is photogrammetry of, architectural interiors or historical artefacts. If you wish to have an accurate photographic record of your collection of Etruscan terracottas or, say, the House of the Satyr at Pompeii, you need to care about where your camera is seeing from.

Struan Gray
1-Jun-2006, 01:14
...Their perfect, in-focus representations are collinear with the exit pupil, ....

I meant, of course, collinear with the rear nodal point.

Helen, Leonard, please restate at will. I am a great fan of learning by discussion, and wish there was more time for symposium-style dialogue in today's formal education systems. It's one reason I like online forums so much, and I wish there had been discussions like this in the archives when I first started to confuse myself with this stuff. It is easy to find statements of fact like those in Ray or the Lens Tutorial; much harder to find someone who can put things into context or explain with words and concepts instead of squiggles and equations.

The classic application for this degree of nitpicking is photogrammetry of, architectural interiors or historical artefacts. If you wish to have an accurate photographic record of your collection of Etruscan terracottas or, say, the House of the Satyr at Pompeii, you need to care about where your camera is seeing from.

Emmanuel BIGLER
1-Jun-2006, 02:12
Hi Gang

Welcome to the nodal-pupil-panoramic club

First we'll use principal or nodal points as being equivalent when the entrance and exit media are air. The question of panoramic stiching underwater will be left for another time ;-)

Struan and Helen have explained almost everything with pure ASCII text. A true challenge indeed.

Now it is time for diagrams and images.
Let us start with a geometricall ray tracing (see attached pdf file).

The problem is to define what we want in panoramic stiching with a regular camera.

We want that two points B1 and B2 "aligned" on view #1 (two points are always aligned !! so we should add "aligned with something in the camera") stay aligned on view #2 after rotating the camera. But what is the meaning of 'aligned' on film ?

When "B1 and B2 show as aligned on view #1", it means that, focusing on B1, the centre C'2 of the blurred image of B2 coincides with B'1, sharp image of B1 on film. The centre of the blurred image C'2 is located on the ray that connects the centre of the exit pupil P' with the sharp image B'2.

So the only way that the alignement on film is not destroyed is that in object space, B1, B2 and the centre of the entrance pupil P stay aligned while rotating the camera.
This condition can be fulfilled only if we rotate the camera around P.

P is located close to a H when the pupillar magnification ratio is close to unity.
The pupillar magnification ratio is fixed when the position of the iris relative to the position of the lenses does not change when focusing the camera.

This applies to all lenses without moving groups like view camera lenses and most traditional fixed focal lengths lenses. However in zoom lenses or macro lenses with floating elements, the pupillar magnification ratio can, of course, change, but not if we simply move the whole lens for focusing. The trouble is that is most modern zoom lenses, groups move also when focusing, not only when zooming... My understanding is that this change of pupillar magnification ratio, and hence the position of the entrance pupil, can be dramatic in a trans-standard zoom lens, changing from >1 in wide-angle-retrofocus position to <1 in long-focal-tele posistion.

Fortunately, we LF users ignore those trans-standard kind of zoom lenses as being unacceptable for our purpose ;-) So we escape from the dramatic situation of a diabolic entrance pupil that moves according to its fantasy !!

The change in pupilar magnification ratio is probably very minor in a floating-element macro lens.

I have plotted an arbitrary case with a pupilar magnification ratio equal to 1.5, this is the case for modest retrofocus lens designs. A kind of lens unknown to large-format users so I hope that the moderators will forgive me ;-)
In this case I have chosen to place the exit pupil half a focal length (f.l.) in front of H', therefore everything else is fixed, the position of the entrance pupil has to be 1.33 times the f.l. in front of H and the entrance pupil diameter 1.5 times smaller that the exit pupil.

Please note the weird ray tracing connecting the edges of the pupil together through the principal planes.
The green dashed ray is there only to tell us where the sharp image B'2 is located. It crosses the principal points H=N and H=N' and exits at the same angle as the entrance angle.
If there is a key-statement in this delicate question, this is the one: H=N and H'=N' have little to do with parallax issues. "little' since the position of the sharp image B'2 defines the conical projection for the blurred image C'2 ; so actually the position of H and H' plays, of course, a certain role in the question.

What about panoramic cameras like the Noblex&#174;
They simply do not fulfill the parallax-free condition.
They are designed for far-distant objects and the axis of ratation should be located close to the rear nodal point to keep the image sationnary w/respect to film.

If a similar camera was designed for macro work, the rotation point shoud be different.
For example at 1:1 ratio with a perfectly symmetric lens, the proper rotation axis for a stationnary image on film should be the centre of symmetry of the lens, located mid-way between H=N and H'=N', not H'=N' like for far-distant objects. And this still-to-be-designed camera would not satisfy the parallax-free condition either, since in this case the entrance pupil is located in H=N (symmetric lens). Except when H' and H are close together, a situation that actually occurs not only for the single thin lens element, but also in some "thick" coupound lenses like the 150 mm Zeiss Sonnar (again off-topic, sorry).

Now that we are convinced that the centre of the entrance pupil is the right place to rotate for parallax-free panoramic stitching, it is time for an example "how precise should we be"

Sorry again if this is not large format, but Mr Carlos Manuel Freaza has done the test for a twin lens rolleiflex. He manages to show the parallax effects with objects located close to the camera, a few metres, and he is able to show that the Rolleiflex should not be rotated around the regular tripod socket, but a few centimetres in front, where the entrance pupil is located and where Rollei engineers in the old times had purposedly (God bless Them) placed the rotation point of the (long discontinued) Rollei panoramic attachment.

See images here, the title could be : "The Ultimate Quest for the Panoramic Parallax-Free Guitar"

"Rolleiflex 2.8C rotation axis: Technical folder, test"
http://www.rollei-gallery.net/itar/folder-5285.html

As a conclusion.
A kind of Secret of the Old Masters, lost for decades, unveiled and re-discovered thanks to the Internet. Am I kidding ? Certainly yes, however it should be mentioned (sotto voce) that Hasselblad brochures had to be updated to correctly mention the entrance pupil and its proper position for panoramic stitching, after initially mentioning incorrectly the entrance nodal point.
Do not ask me why this was corrected some day ;-);-)

To actually see the diagram, you should click on the thumbnail below, the pdf diagram will open readable in the pdf viewer attached to your browser.

Emmanuel BIGLER
1-Jun-2006, 06:34
A similar diagram but with a pupillar magnification of 0.5 like in a telephoto is attached below. (the vintage Telomar actually has such a ratio)

And to continue Helen's remarks about swinging lens cameras, the two examples I know are the Noblex&#174; Pro 6/150 U
http://www.galerie-photo.com/noblex-guigue.html
which has a 60 mm lens (tessar-like ??) and the Alpa Rotocamera which has a 6.8-75mm Grandagon-N lens. The rotocamera is irrelevant since it can record more that 360 degrees ;-) however stiching three Noblex shots to make a full 360&#176; panorama could be possible !!

Now I can tell to Leonard that I followed the same route as him and it took me some time to see what happens.
I know that Struan will not sleep quietly if he does not read in the original French text the exchanges of arguments we had with Yves Colombe.
So here are the references.
http://www.galerie-photo.org/a-f-498.html Yves Colombe is right from the beginning
http://www.galerie-photo.org/a-f-4cb.html I disagree and I try to argue about the properties of nodal points
http://www.galerie-photo.org/a-f-4d8.html Yves tries another explanation
http://www.galerie-photo.org/a-f-4dd.html I am not convinced, and I argue (but I am wrong, I still do not understand that we deal with blurred, defocused images and not sharp images)
http://www.galerie-photo.org/a-f-4ec.html Yves tries again to convince me
http://www.galerie-photo.org/a-f-4f1.html Eventually I am converted to the True Faith.

If somebody finds in a reference textbook diagrams similar to the one I have re-drawn, I'll be happy to know since up to now I've never seen the explanation with a ray tracing.

Leonard Evens
1-Jun-2006, 08:10
Helen,

I was a bit confused about the whole thing, but I did understand that the crucial issue was the one you described. I should have noted that you already said it. But there was something else bothering me. The argument shows that rotating about the center of the entrance pupil preserves an important property and hence the entrance pupil can be regarded as the center of perspective if what concerns you is introducing parallax in the negative by rotation. Your argument shows why rotating about the center of the entrance pupil doesn't affect that.

But it doesn't tell you why the front nodal point is not also the center of geomtretic perspective. Indeed, in the strict geomtric sense, it is. In art theory, you regularly use arguments relating similar triangles in object and image spaces. The corresponding triangles using the centers of the entrance and exit pupils would not be similar.

So one part of the resolution of my "paradox" is that the term center of perspective is being used with two different meanings.

The second part of the resolution of the "paradox" for me was that I needed to be convinced that rotation about the front nodal point actually makes a real difference, if the pupil magnification is significantly different from 1. The corresponding image discs do get shifted, but it wasn't immediately clear to me that they were shifted enough to get outside a plausible coc. I statisfied myself that such would happen by doing some calculations, so I'm now happy. I haven't studied Emmmanuel's latest contribution in detail, but I think he makes that clear also by diagrammatic means.

Emmanuel BIGLER
1-Jun-2006, 08:45
Leonard
In the two diagrams I have plotted and they are as accurate as possible, except that I did not take into account a give f-number, in the case of pupillar magnification ratios of 1.5 and 0.5, the difference between the ray crossing the nodal points and rays crossing the centres of the pupils is not that big.

However as far as parallax effects are concerned, this can be visible, see the visual example of the Rolleiflex and the Guitar (the parallax effects are visible, but small)

But now that everything is clear, I'd like to submit the community something more relevant to architecture LF photographers who are considering to abandon the large format camera and switch to a small format digital reflex camera body fitted with a tilt and shift lens.

Another surprise. Something difficult to explain but again related to non-unit pupillar magnification ratio.

In all classical textbooks it is stated that if the building (= a vertical plane) is vertical and if the film plane is placed parallel to the object plane, those parallels on the object will stay parallel on film. Scheimpflug's rule requires then that the lens plane should be parallel to the two others, however if you tilt the lens plane, blur occurs but no distorsion as long as the object plane is kept parallel to the film plane.

When using a stong retrofocus lens like a tilt+shift 24mm for the small format, with a pupillar magnification betwwen 1.5 and 2, then the above mentioned rule is no longer valid.
Of course if the three planes are parallel, no distorsion occurs. But if the object and film plane are parallel, a tilt of the lens plane generates a distorsion in the image due to the fact that the "centre of perspective" = entrance pupil according to definition #1 is no longer located at the nodal points which are the other centre of perspective according to definition #2.

This effect can be simulated with Oslo and is quite strange.
If means that if you point your tiltable lens upward with the camera body horizontal and the film vertical, the blurred image you get of a vertical plane will be distorted with a retrofocus, whereas it will not be distorted with a lens with unit pupillar magnification like most view camera lenses.

I thing this effect is a good illustration of the two different meanings of centre of perspective/centre of projection.

Michael S. Briggs
1-Jun-2006, 20:21
Hi Gang

..............

Sorry again if this is not large format, but Mr Carlos Manuel Freaza has done the test for a twin lens rolleiflex. He manages to show the parallax effects with objects located close to the camera, a few metres, and he is able to show that the Rolleiflex should not be rotated around the regular tripod socket, but a few centimetres in front, where the entrance pupil is located and where Rollei engineers in the old times had purposedly (God bless Them) placed the rotation point of the (long discontinued) Rollei panoramic attachment.

See images here, the title could be : "The Ultimate Quest for the Panoramic Parallax-Free Guitar"

"Rolleiflex 2.8C rotation axis: Technical folder, test"
http://www.rollei-gallery.net/itar/folder-5285.html

......

Kudos to Mr. Freaza for conducting this experiment, but let's be clear about what he has shown. He tried rotation axes 23, 64 and 90 mm from the film plane, obtaining parallax free results with the 64 mm displaced axis. Apparently 64 mm is the location of the entrance pupil. (Is 23 mm the position of the tripod socket?) But unless this lens has a huge displacement between the entrance pupil and front nodal point, his test doesn't experimentally show that the entrance pupil is better than the nodal point.

Leonard started his question by suggesting that it might not be important from a practical point of view. To setup an experiment which could distinguish between the entrance pupil and front nodal point by showing that rotating about one causes parallax effects and rotating about the other doesn't, one would need a lens in which these points were significantly separated.

Emmanuel BIGLER
2-Jun-2006, 01:41
But unless this lens has a huge displacement between the entrance pupil and front nodal point, his test doesn't experimentally show that the entrance pupil is better than the nodal point.

Michael has a good point and I agree with him.
The demonstration however shows that moving the rotation axis by a few centimetres with a 80mm lens can yield visible parallax effects with objects located several decimetres, a few feet, in front of the camera. More difficult is to estimate the sensitivity of the method, moving by 20% of the focal length yields visible effets for sufficiently closely-located objects, can we easily scale this to a LF lens ?
My understanding is that Mr. Freaza has a planar lens for which I do not have the data ; if it were a 2,8-80 xenotar, we could'nt conclude either since the position of the entrance pupil is not mentioned in Schneider's vintage brochures, but for sure, yes, yes Michael, you're right, the separation between the entrance pupil and H is definitely, and without doubt, indetectable by this simple parallax experiment.

Leonard started his question by suggesting that it might not be important from a practical point of view. To setup an experiment which could distinguish between the entrance pupil and front nodal point by showing that rotating about one causes parallax effects and rotating about the other doesn't, one would need a lens in which these points were significantly separated.

Agreed 100%. So let us continue our Gedankenexperiments.

I see good potential candidates in the LF telephoto family, the vintage Voigtländer Telomar which has a pupillar magnification factor of .5, or a Schneider 360 Tele-Arton lens, or a recent apo-tele-xenar for which the data are available (.75).
In the 360 tele-arton, the principal planes are located in air in front of the lens. The entrance pupil, as Helen says, can be seen with the naked eye (I own this lens, so I've an idea ;-) and is located quite deeply inside the lens, somewhere around the shuter or even deeper.

From the vintage Schneider brochure we find that H' is located in air 353 mm in front of the focal plane (the actual f.l. is 353, not 360) and the point H is located 74 mm in front of it i.e. about 430mm in front of the focal plane whereas the entrance pupil is located somewhere near the shutter, about 210-250 mm in front of the focal plane. So in this lens the separation between H and the entrance pupil is something around 200 mm, a really significant fraction of the focal length.

As far as the apo-tele-xenar 400mm is concerned, things are simple, HH' is negligible (5mm), the pupillar magnification ratio denoted by beta'_p in Schneider's charts is .75, hence the entrance pupil is located about 1/3 of the f.l. = 130 mm behind H ; H being located close to H' i.e. about 400 mm in front of the focal plane, again quite far in air in front of the first lens vertex.

But who will attempt panoramic stitching with the 400 mm LF telephoto !! ;)

Emmanuel BIGLER
2-Jun-2006, 03:14
I have uploaded somewhere a diagram obtained with oslo-edu, the original oslo file was designed by Fabrice, a professional lens designer and a regular contributor of our French LF forum.

For those who can read French, the story starts here by the discovery of the strange behavior of the Canon TSE-24mm tilt+shift lens by Mr; J.P. Planchon.

http://www.galerie-photo.org/n-f-69259.html

The file with the results of the oslo simulation is too big to be stored here so feel free to ask a copy when this temporary link expires.
http://cjoint.com/?gcmiNXAOUm

Oslo® has built-in a lens element which is an ideally thin aplanat lens.
Fabrice, the designer, simply asked this aplanat to make the image of a slanted plane and we record the image in another slanted plane parallel to the object plane. The optical axis is horizontal in the diagram. An aperture is set somewhere and closed down quite heaviliy and the simulation does not take diffraction into account.

In the first diagram, the aperture is located close to the lens plane, so the pupillar magnification ratio is 1, the image of the grid is blurred but not distorted, we get what we could expect from a pinhole photography.
In the second diagram we move the aperture at some distance in front of the lens, the pupillar magnification ratio significantly departs from unity, the image is still blurred, of course no hope to have it sharp, Scheimpflug "Rules", but is now affected by keystone distorsion, although the object and film plane are kept parallel.

And this effect shows up with the 24mm Canon TS-E lens.

Note in the second case the strange ray tracing, the exit ray does not cross the input ray on the lens plane, this is because in a theoretical aplanat the principal planes are spheres centered to the focal point, the radius being equal to one focal length.

Leonard Evens
2-Jun-2006, 08:34
Emmanuel,

I don't read French well enough to want to wade through a long technical formum discussion. I did look at the diagram and your explanation above in English. Is the point that for keystoning and similar matters you do need to use the nodal point?

On another but related matter:

I did a google search on "Zeiss entrance pupil" and came up with several examples of published data, including positions and sizes of the entrance and exit pupils. Lens theory says that the displacements from the corresponding principal planes should be f*(1-p) and f*(1 - 1/p) for the exit and entrance pupil respectively. I checked the consistency of the published data with theory and found some significant discrepencies. Jeff Conrad did a more extended search and found that sometimes the lens data fits theory quite well and sometimes it doesn't.

Neither of us understands why there should be such differences. One conjecture is that all the gaussian optics used in the theory depends on the paraxial approximation, and for real lenses it is not valid. Do you have any thoughts on the matter? More generally, it seems clear for many large format lenses that we use rays which are far from the lens axis, but eveything still seems to work properly. Do you know anything about this, and do you have a reference which someone like me could understand without in essence becoming a lens designer myself?

I hope Jeff doesn't mind, but I will attach a zip file containing his spreadsheet of results. I doubt a problem will arise, but just in case remember it is his intellectural property.

Jerry Fusselman
5-Jun-2006, 00:34
Terrific discussion, but suppose I am not at the moment interested in panoramic stitching. Instead, I have a perspective I like in 4x5 with m = 1/3 that implies a focal length of 300 mm. If I want to duplicate that perspective in 8x10, should I choose a lens in the larger format to duplicate u or u + Xep, where Xep = f(1/P - 1)? I.e., which center of perspective do I use? Yes, in my example, every lens involved is a telephoto, and Xep > 0.

It seems to me if the answer is u + Xep, then the sizes of distant objects will change, meaning the perspective has changed---against my wishes.

And if the answer is u, then out-of-focus points that were aligned in the 4x5 image will no longer be aligned in 8x10---meaning that here too, the perspective changed against my wishes.

It seems that, unless with my new lens both u and u + Xep are by happenstance unchanged in the 8x10 setup, I cannot really duplicate the perspective. So it appears that I cannot perfectly reproduce the perspective in general, but which variable held constant seems to do the best job of maintaining perspective...and why?

I wonder if the answer is that I should duplicate u and that I might be happy that at the same time I will get a little more visibility around the sides of defocused subject in the larger format? (With my lenses, the Xep will be larger in the larger format.)

Leonard Evens
5-Jun-2006, 08:04
Jerry,

You have posed the problem quite well. I want to think about it some more, but here are my initial thoughts. First assume that what you are photographing is in a single plane on which you focus. Then it seem to me you should use the distance to the front principal plane, which in air will be centered on the front nodal point. It is the distance to the principal plane which determines the location of the exact image plane and the magnfication. I couldn't tell if, for 4 x 5, m = 1/3 referred to the film magnification or the magnification in the print. In the former case, with f = 300, that translates into u = 1200. For 8 x 10, if we ignore margins for the moment, the magnfication in the film, for a comparable 8 x 10 print, should be 2/3, which would result from 1200/f = 3/2 + 1 = 5/2 or f = 480 mm. If you use modiified values for u, either for 4 x 5 or for 8 x 10, you are going to change the relative magnifications in the plane of exact focus, but nothing else. In some of the examples in Jeff's spreadsheet, this could be signficant. My guess, however, is that for any large format telephoto lenses, the value of xep would be small compared to 1200 mm, so the effect should be more modest.

So the question is to decide what happens in the DOF region. My intuition, is that this shouldn't change things that much, but I have to think about it a lot more. The exact image plane is still in the same location, but in calculating DOF, you have to use distance to the exit pupil, rather than to the back principal plane. For relatively close distances, as in your case, the amount in focus on either side of the exact subject plane should be very close to Nc(1 + m/p)/m^2. with N = 16, c = 0.1, m = 1/3, and p = 1/2, this would yield (16 x 0.1 x 5/3)/ (1/3)^2 = 24 mm. Changing c to 0.2, and M to 2/3 yields (16 x 0.2 x 7/3)/(2/3)^2 = 16.8 mm. In either case there isn't much DOF. What is going to happen is that points in the DOF region that line up with the front nodal point will produce image discs in the film plane which are not concentric. I would have to analyze how noticeable this parallax effect would be, but my guess is that for points that close to the plane of exact focus, the effect would be negligible. In principle, if you stopped down far enough to yield signficant DOF, points in the distance would appear shifted in angular position relative points in the plane of exact focus. But you can't stop down that far in any case.

Since I haven't actually done the calculation of the parallax shift, it might be that I am entirely wrong about it in the DOF region. I will do the calculation, but perhaps someone else will find something wrong with my analysis first.

Helen Bach
5-Jun-2006, 10:51
...I did a google search on "Zeiss entrance pupil" and came up with several examples of published data, including positions and sizes of the entrance and exit pupils. Lens theory says that the displacements from the corresponding principal planes should be f*(1-p) and f*(1 - 1/p) for the exit and entrance pupil respectively. I checked the consistency of the published data with theory and found some significant discrepencies. Jeff Conrad did a more extended search and found that sometimes the lens data fits theory quite well and sometimes it doesn't.

Neither of us understands why there should be such differences. One conjecture is that all the gaussian optics used in the theory depends on the paraxial approximation, and for real lenses it is not valid. Do you have any thoughts on the matter? More generally, it seems clear for many large format lenses that we use rays which are far from the lens axis, but eveything still seems to work properly. Do you know anything about this, and do you have a reference which someone like me could understand without in essence becoming a lens designer myself?

I hope Jeff doesn't mind, but I will attach a zip file containing his spreadsheet of results. I doubt a problem will arise, but just in case remember it is his intellectural property.

Leonard,

Just a quick note.

I've had a look at your data, and some more Zeiss data. The apparent error appears to be caused by the lack of precision of the given data: it is only to three sig figs. A small change to the pupil diameters, for example, could remove the discrepancy.

Were you looking for a derivation of the f*(1-p) and f*(1 - 1/p) displacements? I'll post them if you wish.

Best,
Helen

Jerry Fusselman
5-Jun-2006, 11:11
Were you looking for a derivation of the f*(1-p) and f*(1 - 1/p) displacements? I'll post them if you wish.

I would love to see them!

Helen Bach
5-Jun-2006, 14:17
Here are the two derivations. They are drawn roughly to scale for a Zeiss 60 mm f/3.5 CFi lens. They show the bundle of rays for an object at infinity (to show the relationship for the exit pupil) and for an image at infinity (for the entrance pupil). I hope that they are legible.

Best,
Helen

Leonard Evens
5-Jun-2006, 15:12
The arguments are also contained in Jeff Conrad's excellent Large Format Photography page article "DOF in Depth". It is basicaly the same as what Helen sketched in her response above, but he fills in all the details, and it is easy to read.

As to Helen's explanation of why the data is off for some of the lenses, it does appear that it has to do with the sizes of the entrance and exit pupils, as she says. The values given are not consistent in those cases with the rest of the data, which does appear to be consistent. I suspect that there was some error, either in measuring the sizes directly or in computing their sizes or even in listing the results. In the last example in the spreadsheeet, for example, the actual pupil magnification appears to be something like 0.346 rather than 0.369. This seems to me to be too large to be explained by the number of signficant digits in the values given for entrance and exit pupils.

Emmanuel BIGLER
6-Jun-2006, 04:54
Emmanuel...Is the point that for keystoning and similar matters you do need to use the nodal point?

Yes. Definitely.

We can describe the situation as follows.

Consider an object plane and an image plane, parallel and reasonably located so that it does make sense to try and find an image within the limits of a reasonable Depth of Field (DOF). For example, the object will be located quite far, the image close to the lens focal plane.

When the lens, even a very complex, thick, compound lens, has a unit pupillar magnification, in those conditions, the image is immune against residual keystone effects when you tilt the lens provided that the object and the film plane are parallel.

This is the basic rule of all photographic textbooks. In fact it appears that it is a particular property of lenses with unit pupillar magnification.

Of course ordinary, basic, keystone effects would occur as soon as the object and film plane are not parallel.

When the lens has a non-unit pupillar magnification, and with the object and film plane parallel, the image will exhibit residual, non-conventional, keystone effects when the lens is tilted.

This effect occurs with the Canon 24mm TS-E lens for which (I have double checked with what was measured by the photographer who discovered the effect) the pupillar magnification ratio is above 2, someting like 2.5, which is unusual for a non fisheye photographic lens.

On another but related matter:

I did a google search on "Zeiss entrance pupil" and came up with several examples of published data, including positions and sizes of the entrance and exit pupils. Lens theory says that the displacements from the corresponding principal planes should be f*(1-p) and f*(1 - 1/p) for the exit and entrance pupil respectively. I checked the consistency of the published data with theory and found some significant discrepencies. Jeff Conrad did a more extended search and found that sometimes the lens data fits theory quite well and sometimes it doesn't.

Neither of us understands why there should be such differences. One conjecture is that all the gaussian optics used in the theory depends on the paraxial approximation, and for real lenses it is not valid. Do you have any thoughts on the matter? More generally, it seems clear for many large format lenses that we use rays which are far from the lens axis, but eveything still seems to work properly. Do you know anything about this, and do you have a reference which someone like me could understand without in essence becoming a lens designer myself?

I hope Jeff doesn't mind, but I will attach a zip file containing his spreadsheet of results. I doubt a problem will arise, but just in case remember it is his intellectural property.

Leonard I have no clue for such discrepancies.

I have looked at what Prof. Kingslake says about pupils in 'Fundementals of Lens Design'. He does not says much details, however he clearly says that the actual position of pupils and therefore the pupillar magnification should be understood as paraxial-gaussian only.

So if we follow Kingslake, and why should'nt we, the only fuzziness I can imagine for the position of pupils in a lens with fixed groups are the dependance vs. wavelength and temperature dependance of gaussian parameters !!

As far as the formulae for displacements are concerned, may be we could write them algebraically : f(1-p) and f(1/p - 1). When p>1, both pupils move to the entrance of the lens (negative displacement), when p<1 both pupils move to the film direction (positive displacement). We could also note that there is no mystery, we would get the same formulae for any couple of conjugated images, it is a direct consequence of the regular object-image equations written algebraically : 1/x'= 1/x + 1/f ; with p =x'/x.

Since it is unlikely that the pupill cross the foci, in fact those displacements are of the same sign and there is no upside/down inversion between the entrance and the exit pupil. Hence p =x'/x algebraically as well and p is positive in any reasonable photographic lens ;-)

Another consequence of the displacement of the pupils is that the f-number of an asymmetric lens has to be defined w/respect to the entrance pupil. I let our readers do the derivation, simply start from the fundamental photometric equation, basically the squared sine of the angle under which the radius of the exit pupil is seen from film in the focal plane, and combine with the displacements of the pupils w/respect to the principal planes, displacements expressed as a sole function of f and the pupillar magnidication ratio p !! And a miracle happens, everything simplifies when expressed as a function of the entrance pupil diameter !

Helen Bach
6-Jun-2006, 05:18
Emmanuel,

I don't think that there are any significant discrepancies. If you do a quick analysis of the potential error produced by using imprecise data you will see that the numbers are consistent with the formulae. The pupil diameters are the most critical. It's that old problem of subtracting a number close to unity from unity.

Furthermore, we don't know which of the data are taken directly from Zeiss information and which are calculated (I assume that the pupil magnification has been calculated, by the way). If any numbers have been calculated, even by simple addition or subtraction, then the potential error from using rounded or truncated numbers is increased.

Best,
Helen

Leonard Evens
6-Jun-2006, 08:19
Helen,

Look again at Jeff Conrad's spreadsheet, which I attached previously as a zip file. Look at the last column, the Tele-Apotessar 2.8 300 mm lens. It is true that H16 and H17 are calculated by "subtracting" other values which were taken directly from the Zeiss data. But in some cases that "subtraction" actually amounted to addition because of the signs, and in any case, the quantities being subtracted were not particularly close together. So I don't see how your explanation works in that case, unless you assume very large errors in the measurements.

For the next to last column, the 50 mm f/1.4 Planar, the two values being subtracted are closer together, but the larger is still roughly 2 to 3 times the smaller. Some quick error analysis suggest that the relative error in the difference could be 4 times the relative error in the larger for G17 and 6 times it for G16. Errors in the last significant digit wouldn't be large enough to account for the actual discrepency, since it could produce in each case at worst a 2 percent error in the final difference. On the other hand, given the relatively small values, it is possible that the measurements were off by enough to account for the discrepancy, but the measurement errors would have to be quite large, more than a mm, to account for the differences.

I still think the errors for those two lenses are probably in the sizes of the entrance pupils. Either they were measured or calculated incorrectly, or else they were listed incorrectly. But there may also be errors in some of the other listed values.

In passing let me note that for the Tele-Apotessar, the values in rows 16 and 17 for entrance and exit pupil positions relative to the principal planes are very close to satisfying the lens equation, i.e they are conjugate positions, which suggests they are in fact accurate. For the Planar, the sum of the reciprocals gives a focal length closer to 49 mm, so it is possible something else is wrong.

Helen Bach
6-Jun-2006, 09:32
Leonard,

I looked at the spreadsheet you attached, and also at all the current data on the Zeiss website. Only the Tele-Apo-Tessar and Planar on Jeff's spreadsheet show a difference in the calculated values that cannot be explained by simple rounding error. What is interesting is that, for both lenses, there is a value for the pupil magnification that does satisfy both the two displacement formulae, and it requires less than 2% change in the pupil diameters to acheive.

Best,
Helen

Leonard Evens
6-Jun-2006, 09:59
Helen,

I think we are now on the same page. It was only the last two columns that Jeff and I considered anomalous. The others were so close that we considered them to be in agreement with theory. As you say, in those cases, the small differences are explainable through observational erros or errors in the computation such as rounding error.

Emmanuel BIGLER
10-Jul-2006, 04:59
From Michael Briggs:
But unless this lens has a huge displacement between the entrance pupil and front nodal point, his test doesn't experimentally show that the entrance pupil is better than the nodal point.

News from the front after staying silent for a while.

I did some tests on a 360 Schneider Tele-Arton telephoto lens.
Anybody who owns a telephoto any format can do the same, it is extremely easy to show.

Combining my own measurements with the published Schneider data,

http://www.schneiderkreuznach.com/archiv/pdf/tele_obj_67.pdf

this is what I find

- from my estimation the entrance pupil (EP) is located approx 130 mm behind the front filter mount i.e. well inside the bellows at about 6 mm behind the last lens vertex

I determined (roughly) the position of the PE simply by plotting on a piece of paper a couple of rays aiming at the iris stopped down to the maximum allowed = f/45

- from Schneider's specs the principal/nodal points are located in air in front of the front filter mount
H=N = at 137 mm in front ; H'=N' at 63 mm in front ; effective focal length = 353

I have also checked for the position of H and H' by finding the position of the front focal point and what I found, no surprise, matched Schneider's specs.
Reversing the lens, I found about 500 mm between the (now reversed) front filter ring and the focal plane. The image should have been terrible, but you can actually focus and see the image without problem except that my bellows was not long enough to shield from stray light.

So anybody can redo those experiments even without any original manufacturer's specs. Only the actual value for the focal length is difficult to determine within a few millimetres of precision.
For example I tried to use the x = f tan(theta) approach, by measuring the displacement "x" of the image of a distant object for a given angle of rotation "theta", estimated on a pan/tilt head with graduations every 5 degrees. I found a value of the f.l. between 330 and 380 mm which is a bit rough for an interval of confidence ;-) the nice thing with the f-theta approach aiming a distant object is that you are not annoyed by the position of the principal planes. They can be anywhere.

So with this good ol' telephoto lens we have everything in hand to get a robust separation between the nodal points and the entrance pupil.

Combining both data, I find that the distance between H and the PE is approx 267 mm ; according to basic formulae this yields a pupillar magnification factor PM equal to approx 0.57

Well to date I have no images to show you but I rotated the camera around H, H', the PE and the focal plane (this is easy with a monorail camera). The object was a foreground at 3 metres and the background, a building at 200 metres.

So it worked exactly as expected.
Rotating around H (137 mm ahead of the filter ring) or H' (63 mm ahead of the filtre ring) yielded visible parallax effects when rotating the full image width of 120 mm in 4"x5",
Rotating around the focal plane (290mm behind the front filter ring) yielded visible parallax effects in the opposite direction,

Rotating around the PE (130 mm behind the front filter ring, approx 6mm behind the last lens vertex), minimized the parallax effects.

As expected, no mystery and no surprise.

I must confess that I was a bit dissapointed ;-). I would have preferred to find something exotic or unexpected ;-)

A perfectly useless experiment ;-) just for fun.

I was doing the experiment yesterday evening while listening to the Soccer World Cup Final in Berlin. I found the result before our national team lost the final against Italy. ;-)

In order to see the foreground and the background together I had of course to stop down to f/22 which is the normal working aperture for this kind of lens. At full aperture 5.5 DOF is not enough to see the foreground @3 metres. But no need to stop @f/45 to see the effect.