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Drew Bedo
5-Feb-2016, 09:50
I have a 15" (280mm) Wolensak Tele Optar. Have never used it.

The infinity focus distance is ~8"-9" of bellows draw.

How should the bellows correction be calculated?

Dan Fromm
5-Feb-2016, 10:34
magic formulas of the day:

magnification = f*(1 + ((extension from the infinity position)/f))

effective f/ number = (f/ number set) * (1 + m)

Randy
5-Feb-2016, 11:08
I have the same lens...and had the same question a couple years ago...but knew better than to ask...

Peter De Smidt
5-Feb-2016, 11:33
I have a Calumet target and ruler. You put the target in the scene, measure it's size with the ruler, and read off the compensation. You can test your lenses and come up with a chart for extension, if desired. While Calumet is gone, a quick search brought up: http://www.brucebarrett.com/bellows.ps

Drew Bedo
5-Feb-2016, 14:02
Good stuff. Thanks to the responders. For critical precision work these methods are excellent.

Let me rephrase the question: The foxal length of the lens is 15"/380mm. the flange to film distance at infinity is 8". When focusing on something somewhat closer than the the horizon (but perhaps not a table-top still life) how is the exposure compensation figured?

Current every-day lenses are 90mm, 150mm and 210mm. With the 150mm, for every inch past infinity I normally give 1/3 stop more exposure. For the 210mm, for every inch past infinity I give 1/4 stop more exposure. This is a simple enough rule of thumb approach that works for me in the field. I can do this math in my head or on my fingers.

How may I use a similar methodology when using the Wollensak 15" telephoto?

cowanw
5-Feb-2016, 14:18
What a units salad mix of mm and inches you live with,
Well, 23 inches extension is 2 stops more.
15 inches is about 1 stop more. so about every 1 inch and 7/8 from infinity; add 1/4 stop.
same as for any 15 inch (380mm) lens

Drew Bedo
5-Feb-2016, 15:05
Right Bill. I don't understand 23 inches, or why 15" is 1 stop (if 23 is 2 stops) . Where do we start figuring the extension from? Not arguing, just not getting it. Feels like quick-sand.

Anyone out there using a Tel-Xenar ?

Dan Fromm
5-Feb-2016, 15:41
so about every 1 inch and 7/8 from infinity; add 1/4 stop.
Same as for any 15 inch (380mm) lens

pmfji

plaubel
5-Feb-2016, 16:03
I have no tele lens, but:
Without a lightmeter which is able to measure the light on the screen for creating a formula, I would measure the difference between focus /infinity, 8", and focus/ratio 1:1, probably 16" in your case, and would then create a simple formula.

Ritchie

cowanw
5-Feb-2016, 16:38
Infinity focus is 8 inches. Start from there and then figure as if the lens were a regular 15 inch lens.
The distance to 1:1 is 8 inches (infinity) and then one more focal length past infinity i.e 8 + 15 or 23. This is your 2 + stop mark.
Half of 15 is 7.5. and therefor 8+7.5 is 15.5; approximately half way between infinity and 1:1 (not precise). this is your 1 + stop mark
I figured the number of spaces to give a 1/4 stop increase as you were used to. i.e. 8 segments in two stops.
8 segments between 8 and 23 is about 1 7/8 segments. Which brings us close to 8 inches for regular exposure, 15 inches for 1 more stop and right on 23 inches for 2 stops.
I think Dan is peeved with me for pmfji (which I had to google)
sincere humble apologies, particularily if I am wrong.

cowanw
5-Feb-2016, 16:59
Here is Struan Gray on Jun 13, 2000
1. I focus to infintiy, mark the position of the front standard. 2. Focus closer, then note the additional bellows extension. 3. Add this additional extension to the lens fl, in this case 1200mm. 4. Use this value in the normal exp. compensation formula.

Note the I have converted this to your physical measurements as per your previous technique. The difference being the physical extension at infinity vs the optical focal length.
Dan's formula is, of course, correct

always however see my byline, just in case

Drew Bedo
6-Feb-2016, 07:52
Cowanw-Bill:

Thanks for the explicit description. I think I can work with that explanation. I'll get it out and do some careful measurements of the infinity distance. I haven't had this lens out in a while and never did get a careful measurement.

My thanks to all who contributed to this discussion. Regarding mixed measurement units: At one point in my life, I had to be able to convert between radiation levels expressed in Curies and Becquerels, which is much like the mm vs inch relationship. After all, " . . .a rose by any other name . . ."

Kevin Crisp
6-Feb-2016, 08:12
Does somebody have a link to that disc tool for this? Satz or Saltz something? I stopped worrying about these calculations when I found that on this forum, works every time I just keep it in my notebook.

Kevin Crisp
6-Feb-2016, 08:27
Found it: http://www.salzgeber.at/disc/

neil poulsen
6-Feb-2016, 11:44
Since it's all based on inverse square, the difference between the actual focal length and the flange focal length doesn't matter. Use the same approach as with non-tele lenses, assuming that the "focal length" is the flange focal length.

Drew Bedo
7-Feb-2016, 07:14
Since it's all based on inverse square, the difference between the actual focal length and the flange focal length doesn't matter. Use the same approach as with non-tele lenses, assuming that the "focal length" is the flange focal length.

So assuming that the flange-to-Film distance is the lens' focal length, an extension of half that distance would equal one stop additional exposure?

Example: Lets assume that the 15' telephoto lens does have an infinity flange to film distance of 8"(I have yet to actually determine this exactly). A four inch extension to 12" total would require a one stop correction?

That doesn't sound like the method described earlier by Bill and others. Did I misunderstand something Neil?

Dan Fromm
7-Feb-2016, 08:07
So assuming that the flange-to-Film distance is the lens' focal length, an extension of half that distance would equal one stop additional exposure?

Neil is mistaken. The film-to-flange distance isn't related to focal length.

Peter De Smidt
7-Feb-2016, 08:49
Seriously, use one of the free targets/rulers cited above. Set your camera up. Place the target 10 feet away from the camera. Measure on ground glass with the included ruler. Is the compensation to the 1/3 stop range? If not, move it appropriately. Measure. Once you get +1/3 stop, mark your camera, or a card, or....? Now move it a bit closer until you get +1/2 stop, then + 2/3 stop. This is a good empirical way of testing any formula work. It doesn't take very long, and then you'll know...

Drew Bedo
7-Feb-2016, 14:35
Ok, Think I've got it.

For every 1.8 inches (1-7/8" or 45mm) past infinity, the Exposure compensation will be 1/4 stop. A bit messy to do in my head .

For every 2-1/2 inches (63mm) past infinity, the exposure compensation will be 1/3 stop—probably what I'll use most for ease and simplicity in the field.

If this lens is used for anything like a close-up or table top still life, I will do the math.

My thanks to everyone for chiming in on this.

RSalles
7-Feb-2016, 17:54
I measure as this: each inch more bellows draw, 1/3 of f-stop to increase.
A 11 inch lens at infinity (11 inch bellows draw for instance) have a 1/3 of f-stop more exposure when the bellows is at 12" mark, 2/3 of f-stop more when the bellows is at 13" mark, a full f-stop when the bellows is at the 14" mark, and so on. After that reciprocity failure correction if required,

Cheers,

Renato

Goldman
8-Feb-2016, 08:51
Here how I calculate the effective F stop for bellow extension:

The New F stop= Indicated F stop (measured with light meter) x Focal length of the lens (mm) divided by the bellow extension (mm)

It is so simple. I used it for years and always have perfect negatives.

Thanks
Bill