Bellows correction with tele-photo?

[Leigh]

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]

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]

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

[Leigh]

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]

... 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]

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 M_p = (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 M_p, 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 / M_p )²

The origin of this somewhat cryptic formula is explained in detail here in this article in French,
http://www.galerie-photo.com/telecha...t/pupilles.pdf

but you can just have a look at this graph in pdf
(this is fig. 30 of the article, I have translated the legends into English).

This graph is a good summary of the differences between a retrofocus (M_p > 1) a quasi-symmetrical lens (M_p ~= 1, all standard LF lenses) and a telephoto (M_p < 1). For the Schneider 360 mm Tele-Arton, M_p ~= 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 M_p ~= 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 M_p = 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

[Leigh]

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]

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
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 M_p ~=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 M_p~= 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 M_p = 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]

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?

[Tin Can]

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