


Originally Posted by pbromaghin
The Cambo Legend conveniently has a scale printed on the rail that indicates the distance between front and rear standards.
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I got a perhaps dumb question...
I shoot landscapes with my Pacemaker, I often shoot with a just the one, on front, of the double Carl Zeiss, Protarlise Vll f 35 cm lens.. the focus at infinity is about 12" to the ground glass.. that's a long way... so one does not have to compensate, but when I focus closer of coarse, then the bellows is even further out.. then one may have to compensate a little.. but generally, close is 40 ~ 50 feet away with the landscapes I do, with just the one lens on, and have never really bothered to compensate.
Is the reason because I'm not doing a portrait, of a wide single subject, close up, but instead, many smaller items, that compose the image, and as a whole make the focal point?

hi peter k.
never any dumb questions mostly there are dumb answers !
i might be giving you a dumb answer, mainly because i don't do much if any close up work .
one of the best posts in this thread which simplifies bellows extension pretty well is this one
Originally Posted by Mateo
this can be easy if you want it to be
if you got a 5.6 inch lens and you extend the bellows to 8 inches you need to add a stop
if you got a 5.6 inch lens and you extend the bellows to 11 inches you need to add 2 stops
if you got a 210mm lens and you extend 320mm you ain't gonna be far off if you add 1 stop
if you got a 210mm lens and you extend 450mm you ain't gonna be far off if you add 2 stops
see the pattern, forget about inverse square law this is quick and dirty and if you can be more accurate you have some super calibrated lenses and need to calm down
so with your situation
you have a 12" lens ( lets say it is 11 inches for laughs and to simplify things even more )
if you extend the bellows of your 11" lens to 16" you open 1 stop ( think f 11>> f 16 > 1 stop difference )
if you extend the bellows of your 11" lens to 22" you open 2 stops ( think f 11 >> f 22 > 2 stops difference )
there are all sorts of other ways to calculate bellows extension
some rely on philosophy, and higher math others on disks and measuring magnification
but converting the lens focal length to inches making believe it is an Fstop and measuring
the bellows distance and making believe it is an Fstop seems the simplest way to do it, at least to me.
have fun!
( nice lens ! )
john

John, you really should pay attention to the post you're responding to.
I shoot landscapes with my Pacemaker, I often shoot with a just the one, on front, of the double Carl Zeiss, Protarlise Vll f 35 cm lens.. the focus at infinity is about 12" to the ground glass..
He's about at his camera's maximum extension. Per the bible, 10th edition, the 4x5 Pacemaker Speed Graphic's maximum extension is 12 3/4", the Crown's is 12 1/2". His lenscamera combination can't focus close enough for bellows factors to come into play. Your examples of extension are irrelevant, he can't get them.
After that the original post goes incomprehensible. This is an observation, not an insult.


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Well.. I'm the one who needs to pay attention, as didn't phrase the question correctly.. and got us all sidetracked into the camera, lens and the bellow distance of that lens, on that particular camera.
Alright, let me try again...
At the focal length of any lens or camera.. compensation does not have to be given, but as the subject is nearer, focusing causes the bellows to extend, and therefore causing less light to fall on the film surface and which requires compensation.
Understood.. But why, what causes this?
Is it because, the image circle of light, at the film plane has enlarged, distributing the same amount of light over a larger surface, causing it to be less bright?

The Inverse Square Law. I think this is what you are looking for. Google offers much better and clearer explanatiom than I can.
Bruce Osgood
If you can't find the answer in APUG then it probably is a really dumb question.
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Originally Posted by peter k.
Is it because, the image circle of light, at the film plane has enlarged, distributing the same amount of light over a larger surface, causing it to be less bright?
Yes.
The decrease in light intensity follows the inverse square law. If you have a unit distance "d", a unit intensity of light "I", and this is projected over a unit area, the intensity of light over that unit area is inversely proportional to the square of the distance from the light to the surface (or with bellows extension, the lens to film plane distance with the film area acting as the unit area).
I = 1/d^{2}
Notice if we plug the f/stop numbers in for "d" we get the following:
If d = 1, then I = 1/1
If d = 1.4, then I = 1/2
If d = 2, then I = 1/4
If d = 2.8, then I = 1/8
If d = 4, then I = 1/16
If d = 5.6, then I = 1/32
If d = 8, then I = 1/64
If d = 11, then I = 1/128
If d = 16, then I = 1/256
If d = 22, then I = 1/512
If d = 32, then I = 1/1024
Those intensity numbers are all one stop apart in terms of units of exposure. Notice the f/stop sequence (d) is increasing by a factor of 1.4x (~ the square root of 2) between one number and the next. (1 x 1.4 = 1.4; 1.4 x 1.4 = 2; 2 x 1.4 = 2.8, 2.8 x 1.4 = 4, etc.)
If you think about the distance units a certain way, you can figure out bellows extension factors fairly easily in terms of when a full stop increase will be needed. Every time the unit distance increases by 1.4 x a full stop of exposure is lost at the film plane over the unit area. If you started with a 150mm lens when the extension reaches 150mm x 1.4 = 210mm, you will need to compensate an entire stop of exposure. If you had an 8" lens then 8" x 1.4 = 11.2" extension would require a full stop increase.
This relationship is also very useful when determining exposure using lights of a specific intensity. Say for example you had a constant light positioned at 8' and you wanted it to be 1/2 as bright. Just move it back to 11.2'. Want it twice as bright? Move it to 5.6'. Have two lights of equal intensity and want one twice as bright as the other? Put one at 4' and the other at 5.6'. Or 4m and 5.6m. Or 4" and 5.6". Placement of lights to get certain lighting ratios just became a breeze.
Have a look at the table or dial of an electronic flash unit. What does it read for a distance of 10'? Let's say f/8. If so, then you'll lose onestop at 14' (10' x 1.4) and the flash will indicate f/5.6 for the exposure. You'll lose two stops at 20' (10' x 1.4 x 1.4) and the flash will read f4, three stops at 28' (10' x 1.4 x 1.4 x 1.4) and the flash will indicate f/2.8 and so on. If you know the exposure for the flash at 10 units of distance, you can easily figure out the other distances where the flash becomes full stops brighter or dimmer. (2.5, 3.5, 5, 7, 10, 14, 20, 28, 40, 56, 80...).
Just remember to multiply (or divide) distance by 1.4 to figure out where full stop compensation is required.
Last edited by smieglitz; 03142015 at 05:34 PM. Click to view previous post history.

ah ha... got it.. CLARITY ..
had the happening, but not the why,
read up on inverse square law on google, and put together with what I know and with your explanation, its now all falling in place...
will do some more study on it.. but thank you all, .. for the heads up.

