View Full Version : Bellows correction with tele-photo?

Drew Bedo

22-Apr-2014, 06:16

I have a Wollensak 15' telephoto that I have not really used much. Infinity focus is at ~12". How is the correction for bellows extension made for this type of lens?

In the field, I have been measuring lens to film plane distances from the diaphragm for my other lenses . . .works pretty well. Where should measurements be taken from when using this telephoto?

ic-racer

22-Apr-2014, 06:23

I have a Wollensak 15' telephoto that I have not really used much. Infinity focus is at ~12". How is the correction for bellows extension made for this type of lens?

In the field, I have been measuring lens to film plane distances from the diaphragm for my other lenses . . .works pretty well. Where should measurements be taken from when using this telephoto?

http://www.brucebarrett.com/bellows.ps

Drew Bedo

22-Apr-2014, 11:24

Link won't work.

djdister

22-Apr-2014, 11:32

Link won't work.

It was a postscript file - I converted it with Adobe Distiller, so here it is in PDF.

114072

Randy Moe

22-Apr-2014, 11:34

I have one also and have never used it. Mine is mounted for a Speed Graphic. I don't think my clumsy exposures would notice the tiny correction factor involved. I am interested in any comments on using this lens.

ymmv

I have a Wollensak 15' telephoto that I have not really used much. Infinity focus is at ~12". How is the correction for bellows extension made for this type of lens?

In the field, I have been measuring lens to film plane distances from the diaphragm for my other lenses . . .works pretty well. Where should measurements be taken from when using this telephoto?

15' telephoto, sounds a bit strange for me. probably it is 15" telephoto. If you infinity focus at 12" bellows extension, that mean you have 3” at the front of the lens. These 3" must be taken in account when you calculating bellow extension factor. Say you setup the camera, focused and measured length of the bellows draw =16”. Than you total bellows draw for calculation will be 16+3=18”.

The 1:1 bellows extension always equals one focal length forward from the infinity focus position.

The infinity focus position is always the extension at which the rear node is exactly one focal length from the film.

That's the definition of the focal length of the lens in the first place. If that dimension does not correspond to the

location of the diaphragm, then the rear nodal point has moved, as with a telephoto lens.

While we commonly encounter a rear-node shift for telephotos, a similar situation may occur with very short FL lenses, where the rear node is actually behind the expected location, to move the lens farther forward at infinity focus.

The extension from infinity focus to 1:1 is always equal to the lens focal length.

- Leigh

Drew Bedo

22-Apr-2014, 13:39

Everyone: Thanks for the info.

So now, As I understand it, I will carefully focus at infinity, then measure the bellows extension and add the difference between this infinity focus distance and the nominal focal length of the lens.

If the infinity focused distance is 12" (it may be something else), 3" brings that back to 15". Now to focus closer; measuring the bellows extension at, say 14", I would add 3" and use 17 inches to calculate the correction factor . . .giving about 1/2 stop extra exposure.

Do I have this right?

blueribbontea

22-Apr-2014, 15:50

This sounds a bit strange. A telephoto lens requires less bellows extension than a long lens. The bellows extension factor is concerned with the distance from infinity focus to a point of closer focus, and it is the greater distances that kick in the rule of fours, isn't it? The fact that this lens is a 15 inch, unless there is something I do not understand, is less significant than the distance from the infinity bellows position to the bellows position at a focus of 10 feet. I use this lens on my 4X5 RB Graflex and can focus that close, without extreme bellows draw precisely because it is a telephoto design. So why wouldn't I not simply measure the difference between the lens focused at infinity and the lens focused at 10 feet, r.e. bellows draw? What am I not understanding here?

Bill

So why wouldn't I not simply measure the difference between the lens focused at infinity and the lens focused at 10 feet, r.e. bellows draw?

That's the simplest and most accurate method of determining the bellows draw in a real situation.

Calculating bellows draw involves several other factors, principally the position of the first and second principle planes.

When focused at any distance, the lens-to-film distance is from the second principle point (H'), while the

lens-to-subject distance is measured from the first principle point (H).

On a simple lens of relatively "standard" focal length, those two points may be close to each other, with the second being "behind" (i.e. closer to the film than) the first.

However, that's not an optical requirement. In telephoto lenses the two planes can be anywhere.

As an extreme example, the Zeiss 500mm/f8 Tele-Apo-Tessar:

The distance from the front of the lens to the film is 402mm, but the entrance pupil is located 486mm behind the front lens vertex. That puts the entrance pupil about 80mm BEHIND the film.

- Leigh

Drew Bedo

22-Apr-2014, 16:39

Ok—that exchange muddied the water a little bit for me.

I am looking for a practical, in-the-field understanding of how to use this lens. R.K. has given a clear and concise answer to my follow-up question, but now I feel as though I am standing at the edge of a pond of quicksand looking at a bunch of floating hats.

Leigh: do you concur with R.K.?

blueribbontea

22-Apr-2014, 21:49

Didn't mean to muddie up the water for you, Drew. This Wollensak 15 inch was made to be used on Speed Graphics, unless it's a different animal than I think it is. I've been using it for years and exposing on close in shots as if what mattered was the multiplication factor based on the actual infinity focus, which is not 15 inches. That made sense to me but it is quite possible there is something here I did not understand. There are lots of such things, in fact. The explanation of adding 3 inches up front and so on though just confused the heck out of me. So my own little pond was plenty muddy. I'll be glad to get some authoritative FINAL ANSWER on this. It has been, by the way, a good lens for me.

Bill

A brief explanation of "bellows extension" and exposure correction...

The lens throws a cone of light. Its apex is at the second lens node (designated H'). It's base is at the film.

By definition, the height of the cone at infinity focus = the true focal length of the lens.

The illuminated area on the film is called the "image circle". Its diameter is one of the published lens specs.

The area of the image circle = pi times the radius of the circle squared = pi * r * r.

Every time you double the IC area you must increase the exposure by one stop to compensate.

If you double the height of the cone, you double the diameter of the image circle, so you quadruple its area.

A simple rule to remember is that for every extension = one fourth of the lens FL, increase by 1/2 stop.

Example: for a 200mm lens, extension from infinity focus and compensation are:

50mm = 1/2 stop

100mm = 1 stop

150mm = 1 1/2 stops

200mm = 2 stops

This is why f/numbers (1, 1.4, 2, 2.8, 4, 5.6, etc) increase by a factor of 1.414, which is the square root of 2.

Each stop changes the area of the image circle by a factor of 2.

- Leigh

Doremus Scudder

23-Apr-2014, 02:42

... A simple rule to remember is that for every extension = one fourth of the lens FL, increase by 1/2 stop.

- Leigh

Applying that to Drew's situation:

For every extension that is one fourth of 15 inches, add 1/2 stop exposure.

Drew, just measure from infinity focus, which you have cleverly marked on your camera bed somewhere.

3 3/4" more: add 1/2 stop

7 1/2 inches more: add 1 stop

11 1/4 inches more: add 1/2 stop

15" more: add 2 stops

Heck, you could even estimate with your fingers and come within 1/3 stop of correct.

In Jim Galli's words, "Easy peasy!"

Best,

Doremus

Emmanuel BIGLER

23-Apr-2014, 04:59

Hello from France

You are right, the bellows factor with an asymmetrical lens like a telephoto is slightly different from a regular, quasi-symmetrical lens design.

It is quite simple to compute, the only additional parameter that you need to know is the pupillar magnification of your lens Mp = (diameter of exit pupil)/(diameter of entrance pupil).

Some manufacturers like Schneider Kreuznach, do provide the value of this parameter in their detailed technical data-sheet, but unfortunately, old archives on the German Schneider-Kreuznach web site are no longer directly available, however, they are stored in the Web archive "wayback machine".

The general formulae for a telephoto of pupillar magnification factor Mp, of a given (image)/(object) magnification M, is very simple:

M = ext / f

ext = additional bellows extension beyond the focal point; f = focal length. This formula M=ext/f is universal, valid for all lenses even very asymmetric.

Bellows factor X times = (1 + M / Mp )2

The origin of this somewhat cryptic formula is explained in detail here in this article in French,

http://www.galerie-photo.com/telechargement/pupilles.pdf

but you can just have a look at this graph in pdf (http://cjoint.com/?DDxn2qdu8Yl)

(this is fig. 30 of the article (http://www.galerie-photo.com/pupilles-objectif-photographie.html#htoc11), I have translated the legends into English).

This graph is a good summary of the differences between a retrofocus (Mp > 1) a quasi-symmetrical lens (Mp ~= 1, all standard LF lenses) and a telephoto (Mp < 1). For the Schneider 360 mm Tele-Arton, Mp ~= 0.57. This yields a difference of approx one f-stop at 1/1 ratio. But in principle an assymetric lens should never be used at 1:1 ratio!!

Hence in most usual cases like e.g. M = 1:5 = 0.2 or smaller (i.e. object is located further away than 6 times the focal length) you can safely ignore the additional correction with respect to a quasi-symmetrical lens.

For Mp ~= 0.57 like in the 360 Tele-Arton, at M = 1:5 = 0.2, the Tele-Arton would in principle require only 1/2 f-stop of additional exposure with respect to a quasi-symmetrical lens formula.

The basic formula for Mp = 1 is simply :

M = ext / f

Quasi-Symmetrical Lenses Bellows factor X times = (1 + M )2

And you can have a good estimation of it by using Herr Salzgeber's "analog calculator" ! 0% Electrizität ! 100% Recycliebar, Umweltfreundlich!

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

http://www.salzgeber.at/disc/disc.pdf

Good morning Dr. Bigler,

I don't understand the magnification factor in the equation.

It's my understanding that the lens focal length = distance from the rear node (H') to the image (at infinity focus) for all lenses regardless of the design, from a single "thin lens" to the most complex multi-element optic.

Is this not correct?

- Leigh

Emmanuel BIGLER

23-Apr-2014, 08:17

Good morning Dr. Bigler,

Good afternoon, Mr. Leigh (I'm at GMT+2 Western European Continental Daylight Saving Time)

It's my understanding that the lens focal length = distance from the rear node (H') to the image (at infinity focus) for all lenses regardless of the design, from a single "thin lens" to the most complex multi-element optic.

This is perfectly correct.

But before we continue, let's have a look at this diagram that I also have uploaded to the forum's database,

tirage-tele-arton-360-EN.pdf (http://cjoint.com/?DDxqR1n3Q4P)

mixing real photos of my 360 Tele Arton with some distances (in mm) added on the picture.

The image magnification factor M is defined as: M = (image size) / (object size).

This definition is of course valid for all lenses and is even valid for slightly de-focused images.

It happens that there is a magic formula linking the image magnification factor M for a perfectly focused image, to the additional bellows extension "ext" beyond the focal point:

ext = M . f

where 'f' is the focal lenght.

And this formula is valid for all lenses, provided that the image is sharp.

For example with the 360 Tele Arton (effective focal length = 353 mm according to Schneider's specs), if you have to add 90 mm (actually : 353/4 = 88-1/4 mm) beyond the focal point (the infinity-focus reference position), then you are working at M = 1/4 = 0.25.

The image is 4 times smaller than the object.

In this situation where M=0.25 the approximate bellows factor formula (1+M)2 predicts that you have to multiply your exposure time by (1 + 0.25)2 = 1.56; this corresponds to plus 2/3 of a f-stop.

The more precise formula taking into account the pupillar magnification factor Mp ~=0.57 reads as

(1 + (0.25 / 0.57))2 = (1.44)2 = 2.07; this corresponds to about one full f-stop. Hence the error between the classical formula valid for symmetical lenses and the "true" formula for Mp~= 0.57 is about 0.4 f-stop.

Unfortunately, I do not have the specs for the 15" Wollensak telephoto we are discussing here. But if the flange focal distance is 12", there should not be a big difference w / respect to a perfectly symmetrical lens.

For the 360 Tele Arton the flange focal distance is only 210 mm for a focal length of 353: this makes a big difference.

So for this Wollensak lens I would simply use Herr Salzgeber's test target or I would use the approximate formula

(1+ (ext/f))2 where "ext" is the additional bellows draw w/respect to the infinity-focus position.

The advantage of working with "ext" instead ot any other distance is that you do not have to care for the actual position of the principal (or nodal) planes. And if the pupillar magnification factor, instead of 0.57, is something in between 0.57 and 1, e.g. 0.70, then you can safely ignore the general formula and use the simplified formula (1+ (ext/f))2 which is used in Herr Salzgeber's quick disc target & direct-reading scale for the bellows factor.

With M = 0.25 and Mp = 0.70, the general formula predicts X 1.84 whereas the simplified formula predicts X 1.56, the difference is only 1/4 of a f-stop.

Drew Bedo

23-Apr-2014, 09:11

Thank you everyone,

The talent pool here is truly deep and wide. I have learned a good deal from the responses in this thread. The relationship of the factors affecting exposure within the camera have never been more clear to me.

My take-away for photographing in the field: My anticipated method will be to focus at infinity and note the distance from the ground glass to the lens board (I know that this is not H'). when focusing at something closer, I will make the same measurement and note what multiple or fraction or the marked focal length (15 inches) that measurement is . . .and use that to calculate the additional exposure. A little pre-calculating gives 1/3 stop additional exposure for each 2 1/2 inches of extension beyond infinity.

In a studio setting (our living room when my wife is not home) I will use this procedure supplemented with focal plane metering with a Sinar Booster-1 probe.

Is this a workable strategy for landscape and still life imaging?

Randy Moe

23-Apr-2014, 10:22

Vindicated.

I have one also and have never used it. Mine is mounted for a Speed Graphic. I don't think my clumsy exposures would notice the tiny correction factor involved. I am interested in any comments on using this lens.

ymmv

But before we continue, let's have a look at this diagram that I also have uploaded to the forum's database...

Thank you Dr. Bigler. I'm still digesting the text of your reply.

I have two questions regarding the first attachment.

The distances H -> F = H' -> F' = 353mm suggest a 1:1 reproduction ratio.

However, given the lens FL = 360mm, shouldn't those distances be about 720mm?

Also, in the telephoto designs with which I'm familiar, H' is closer to the subject than H.

For example, in the Zeiss 500mm Tele-ApoTessar, H is 80mm behind the film (away from the subject).

Does the Tele-Arton use a different design?

Thank you, Sir.

- Leigh

Emmanuel BIGLER

24-Apr-2014, 07:19

Thank you

You are welcome.

I have two questions regarding the first attachment.

I realize that I made a small mistake (a classical cut-paste error when you forget to modify the pasted text) the exit pupil for the 360 tele arton is located about 89 mm behind the front filter ring.

Here is the corrected file "tirage-tele-arton-360-EN.pdf"; the exit pupil (in magenta color) is located about 89 mm behind the front filter ring (http://cjoint.com/?DDypIDFgZ0N)

I have also attached the same diagram here in the forum's database as a JPEG image.

The distances H -> F = H' -> F' = 353mm suggest a 1:1 reproduction ratio.

However, given the lens FL = 360mm, shouldn't those distances be about 720mm?

In fact F is the objet focal point and F' is the image focal point. Those points are not conjugate of each other.

I have used the classical F and F' symbols as in all French textbooks on geometrical optics, but I realize that those good old textbooks suffer from an inconsitency. Usually A and A' denote an object and its conjugate image. F and F' are the exception to this rule. What I have drawn for F' comes from Schneider's datasheet; and for F, I have estimated "by hand" the position of this reversed focus F myself, by reversing the lens and aiming at a distant object. In principle the image quality of such an asymmetric lens once reversed should be terrible, but actually in the real world is not terrible enough to prevent the approximate pointing of the reversed focal point F. One thing which is granted of course for all lenses is that F is ahead of H by one focal length and that F' is behind H' by one focal length. here, 353 mm.

An of course if you reverse a lens, its focal length is the same. This is one of the reasons why I try to avoid using the sentence "a retrofocus is a reversed telephoto" because one could dream that by simply reversing a long telephoto lens, you could get a wide angle retrofocus lens with a much shorter focal length;-)

Also, in the telephoto designs with which I'm familiar, H' is closer to the subject than H.

For example, in the Zeiss 500mm Tele-ApoTessar, H is 80mm behind the film (away from the subject).

Does the Tele-Arton use a different design?

Certainly yes. To the best of my knowledge there is no general rule regarding how H and H' are located in a telephoto design. In the 360 Tele Arton, H and H' come in this order but those principal points can be "crossed" i.e. H' can be ahead of H, like in the telephoto lens you are referring to. In the 360 tele arton, HH' is relatively small and positive: (137 - 63) = +74 mm. However according to old Schneider's specs, for the 250 tele-Arton, HH' = -32.5 mm, i.e H' is ahead of H unlike the 360 tele-arton.

Why, I have absolutely no idea! The two lenses share the same name but are obviously a different design: the 360 tele arton is not the same as a 250, everything being scaled by a factor 360/250!

I have looked at the following Zeiss datasheet for Hasselblad V-system cameras and I have found, I m not sure that we are speaking about the same references:

500mm f/8 Zeiss tele-tessar C Nr. 1046093 - HH' = (413.5 - 125.5) = + 288 mm

500mm f/8 Zeiss tele apotessar CF Nr. 104615 - HH' = (373.7 - 370.1) = +3.6 mm

In both cases, HH' is positive but the actual value is very different between the older C model and the newer CF model.

Note that those lenses exhibit a pupillar magification factor of about 22 mm (exit ) / 62 mm (entrance) = 0.35 which is much smaller than for the 360 tele-arton and would yield noticeable additional bellows factor corrections w/respect to a symmetrical lens design, if ever used in the close-up range (something that probably nobody ever experienced ;) )

ic-racer

24-Apr-2014, 07:25

'Bellows factor' is a change in effective aperture related to magnification. Projecting an object of known dimensions onto the ground glass is an easy way to quantify the effect. Personally, however, I use a film-plane light meter in these situations.

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