Bellows extension

The Cambo Legend conveniently has a scale printed on the rail that indicates the distance between front and rear standards.

 

To keep it really simple: for B&W film especially, the exposure compensation that works well in the field is: add 1/3 of a stop for every inch of bellows extension over the focal length of the lens that you are using. EG.: with a 6" lens (150 mm) on the camera and with a bellows extension of 9" (film plane to nodal point of the lens) the exposure compensation would be plus 1 stop. I have used this "rule of thumb" for 49 years and it has served me well.
(PS: I don't like thin negs. in BW. If in doubt give a little extra exposure: develop according to the scene brightness ratio.

I use a Zone VI, 4x5 with 75, 150 and 240 Nikkor lenses. I also use a 6x7 roll film back which is especially good with the 75 mm Nikkor.

Enjoy, don't over think it, keep it simple. Shoot film, print lots!

Dan Gordon

 

the easiest way to deal with this is to place a target discinto the subject and measure it with a special disc ruler on the view screen. the ruler will tell you how the exposure has tobe xhanged to compensate for the bellows extenson. send a private email to rlambrec@ymail.com and I'll send you a free pdf of the whole thing

 

I forgot that the Mamiya tlr's have a bar that comes down into the focus screen that tells you the exposure correction as you focus on close objects.

 

The Quick Disc takes all the pain out of calculating for large format macros - it also works with Mamiya TLRs. http://www.salzgeber.at/disc/

 

worked well for me toofor LFand even on the small Hasselblad screen

 

to whom it may concern,I have a CMBO LEGEND 4X5 IN PERFECT CONDITION AND WITH IT'S ALUMINUM TRAVELLING BOX FOR SALE;It's a monster butvery easy to use and control;any reasonable offer will be accepted1BEST FOR PICK-UP IN CENTRAL FLORIDA

 

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

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.

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 lens-camera 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.

 

thanks dan

yeah i saw that
.. but figured just in case he had some sort of
you know ... modification to his pacemaker .. to make it a 24" bellows extension LOL
you never know ( trying to cover for my cluelessness and poor reading abilities ! )

john

 

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.

 

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/d2

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 one-stop 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.

 

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.

 

When a lens is focused at infinity, the distance from ground glass to lensboard is just about the same as the lens focal length. When the lens is focused on a closer subject, the distance from the lens (board) is increased, and the amount of light stiking the film is diminished because the lens is farther away. Calculating the correct exposure compensation is beyond easy! All you need is a small tape measure and a meter showing f-stop numbers! Let's say you have a 150mm lens, and after you have focused on your subjet, the lensboard measures 180mm from the ground glass. Just think: what is the difference between f15 and f18? It's about ⅔ of a stop! You need to give ⅔ stop more exposure to compensate for the bellows extension! Same with any other focal length/extension. A 210mm lens with bellows extension of 320mm would be f21 to f32 = just over 1 stop more exposure needed! If you want the actual formula, it is the Extension Squared divided by the Focal-length Squared.

 

What I ultimately did was work backwards and create a cheat sheet. I listed all of my lens focal lengths, for each I did the math of (fl squared x extension factor= new extension factor squared) and listed the new extensions (rounded to nearest whole number). When I take a table top shot I measure the extension and look for the nearest one on the list for that lens and apply the factor to my iso. Film has enough latitude to allow for a bit of slack in the math.

E.g., for my 210mm lens I made the list:
240mm 1.3
270mm 1.6
300mm 2.0
340mm 2.5
375mm 3.0
420mm 4.0

I arrived at these by multiplying (210x210=44,100) multiplied by the desired factor, (44,100x1.6=70,560), square root of 70,560 is 265mm and just rounding up a bit to make the chart easier to use. If I measure close to 270mm with my measuring tape I apply a factor of 1.6. The cheat sheet is laminated and in my camera bag.

 

Another newbie question. I am about to go try out my first 4x5 camera and I just wanted to make sure I got the bellows extension stuff figured out. I will be using a 135mm lens shooting landscapes, so I will more than likely be focused at infinity. Even with my bellows fully compressed, there's still a few inches between the lens and film planes. Do I need to factor this distance in when calculating my exposure?

 

No.

 

Thank you for the thorough explanation

 

Giggle. It was concise.

The focal length in your case is 135mm or roughly 5.3 inches. That is the distance from the measured point in the lens (often, but not always, at the lens board) to the film when focused at infinity.

The lens when focussed at infinity is as close to the film as it can be and still be in focus. This is true for normal lenses, telephoto and retro focus lens designs are different. The terms tele and retro are used differently in the vernacular of large format cameras than in miniature format cameras, that's a different lesson. Most Large format lenses are normal designs.

When focused at the "designed" length the "f" number Marked on the lens is correct. 135 at f/5.6 means the aperture is open about 24mm in diameter (just shy of an inch). Look at the lens and you'll see that's true. At any other focussed length the "f" number marking on the lens is wrong.

To focus on subjects closer to the camera the lens needs to move away from the film, so for example lets say you focus on something very close and from the measured point on the lens to the film is 270mm, then the math needs to be adjusted because the aperture is still 24mm. 270/24=11.25. So when focussed at 270mm the working "f" number (at the 5.6 marking point) would be 11 not 5.6 as the lens is marked.

 

correct.if yoursubject magnification is less than 1/10 you can safely ignore it;at 1:1 you alreadyhave to add 2 stops of exposure;you can google for exposure scales on the web or email me at rwlambrec@gmail.com and I'll send you one;can't post it here due to size of file.