Center of perspective and nodal point

[Emmanuel BIGLER]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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.