Bellows extension

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pbromaghin

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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.
 

RalphLambrecht

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

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:smile:
 

peter k.

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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?
 

removed account4

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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

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
 

Dan Fromm

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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 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.
 

removed account4

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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 :smile:
.. 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 ! :smile: )

:smile: john
 

peter k.

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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?
 

smieglitz

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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/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.
 
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peter k.

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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.
 

Alan Ross

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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.
 

RalphLambrecht

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Some Mamiya Cs also have those bellows factor markings that are indicated by the parallax correction bar. Even the early ones that don't technically have them marked (like my C33) will give you an idea, as they still have the parallax correction bar. Yes; a very simple and good feature. I love my C33 for close-up pix (although I probably would not if I did not have the Paramender).

For the OP, these are the basic considerations that create the "problem":

1. Your light readings give you an exposure that is accurate at infinity, to which your lens' f stops are calibrated.
2. Amount of light falling diminishes (exponentially) as distance from a light source increases.
3. When a lens focuses on something closer than infinity, the distance between the lens and the film is greater than when it is focused at infinity. (Turn the focusing collar on any external focus fixed-length lens to see this.)
4. From your film's point of view, the lens is considered the light source, not the light source itself.

Now, apply this stuff to an actual situation:

A. Due to 3 above, when you focus on anything closer than infinity, you are moving you lens farther from the film than it is at infinity.
sounds complicated
B. Due to number 4 and number 2 above, this makes less light fall on the film versus that which falls when the lens is at infinity.
C. Due to number 1 above, your calculated exposure will actually provide less exposure any time you are focused closer than infinity.

If you add more specifically to number 2 above, you learn how to calculate how just much light falls off using the inverse square law. This law means that if distance from the source doubles, the amount of light falling is quartered. Vise versa: If distance from the source halves, amount of light falling quadruples.

You need to state this as an equation so you can plug your measurements into it. You compare the square of the extension at infinity, where f stops are accurate, to the square of the extension at wherever your shot is focused. To state this mathematically, to get correct results with your calculations, the measurements all have to be squared.

"Compare" means state as a ratio, which can also be stated as an act of division. When you divide, you end up with a correction factor that tells you how many more times the amount of light that you needed at infinity you will need to make up. It will always be greater than 1x, so, that helps you remember that you always need to be dividing the longer measurement squared (actual extension) by the shorter measurement squared (extension at infinity AKA the focal length of the lens).

Algebra states that when you have something raised to a certain power divided by something raised to the same power, you get the same solution if you remove the exponents and perform the act of division, and then raise the result to the power of whatever that removed exponent was. This just means that you can remove a step by squaring the factor at the end rather than by squaring each measurement individually before dividing.

The minimum amount of compensation you can provide is, of course, limited by the precision of your exposure controls. If you end up with a factor of 1.10 in the end, you need 110% of the amount of light that you decided would give the correct exposure. That is not even 1/3 stop more. I guess you could estimate that 10% more with "in between" an f stop adjustment or by adding a little time if you have exposure times that are long enough to time, but for the most part, 1/3 stop, and often 1/2 stop is as precise as you can easily get. So, basically, you just decide what your cutoff is for when you add that first 33% more light (or 50% more light if you only have 1/2 stop precision). I simply round to the next highest or lowest 1/3 stop. This means if my factor hits the 1.17 mark, I round up to 1.33 and add the 1/3 stop, but below that, I go with the exposure that would be correct for infinity.

To get your measurements, it first helps to determine what your measuring spot on the lens will be. Go somewhere outside and focus on something that is at infinity focus. Something way in the distance, miles away would be good. Now, say you have a 150mm lens. Measure 150mm from the film plane and remember (or mark) where exactly 150mm from the film plane is on the lens case. In the future, that is the spot at which you take all your measurements. You know what the point of reference is at infinity, and to be accurate, you need to always measure to that point.

This is picking nits. You could just measure to the diaphragm and you would be fine.

So, you know the bottom of the division problem without even having to measure it. It is the focal length of your lens. You just need to measure the actual extension, divide it by the focal length, and square the result to get your factor.

For instance, say you are using a 360mm lens to shoot a portrait. You are focused on the subject. Measure from the film plane to your lens. Say it is 550mm. So, you divide 550mm by 360mm and get 1.53. Multiply that by itself and you get 2.34. You need to give the film 234% of the exposure you have decided is appropriate. Multiply the shutter speed by 2.34, or open up to let in 2.35 times more light. I always just call this 1 1/3 stops and am happy, but it is not technically accurate, I believe. Someone else should chime in with exactly why it isn't.
 

iamthejeff

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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?
 

markbarendt

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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.
 

markbarendt

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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.
 
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RalphLambrecht

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simply put, the exposure reading from your light meter will only be accurate when the camera is focused at infinity. To focus on objects closer to the camera the lens must be moved farther from the film plane, and less light falls on the film. So additional exposure needs to be provided to make up for this light loss. The farther the bellows is extended, the more compensation is required.
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.
 
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a simpler way

The system I learned was very simpole and straight forward. For evey 25% beyond the infinity focus extension you simply add one-half stop to your exposuree. I carry around a cloth measuring tape in my camera bag. Works everytime.

eric
This approach is the only one that makes sense to me, how can you beat it in the field? Oh how I am going to “stir it...” This subject makes my head hurt.
 

RalphLambrecht

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This is a pretty common question, but I can't seem to find a good thread on it on APUG, so let's make this a sticky and hope everyone chips in to give all the standard answers.

There are several methods for calculating bellows exposure factor.

I determine bellows factor in relation to magnification of the subject. You can do this by putting a ruler in the scene and measuring the size of the ruler on the groundglass to determine the magnification ratio, which I do for still life setups, but for most situations I can usually estimate the magnification of the subject in relation to the size of the format. For instance, on 8x10", a tight headshot--head and shoulders--is about 24" across, so 8/24 is a magnification factor of 1:3. Then I have a table that I keep taped to my light meter, my notebook, and the back of each camera, to convert (and to remind myself to convert) magnification factor to bellows factor. I've posted that table in DOC format here--

(there was a url link here which no longer exists)

There are several copies of the table on the same page, so you can cut them out and attach them to everything.

I like this method because it works well with any format, even if you can't put a ruler up to the groundglass, and I can have a single method for every camera I use.
the simplest way to determine the exp comp factor with view cameras is placing a little target into the scene and measure it with a template on the view screen.Send me an email to rwlambrec@gmail.com if you like a detailed write up on the subject; happy to send it to you for free.
 

jeffreyg

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When I started using 4x5 a number of years ago I took the formula for figuring exposure compensation and did it for each of my lenses according to stops and marked a small retractable ruler with a different colored marker for each lens ( ie 1/2, 1, 1 1/2 stops and so on) and have attached it to the lightmeter strap with one of those light pull chains. I take my reading and then measure from the lensboard to the film plane. The length doesn't matter because I can read the Fstop change on the tape. If using a filter I add that in. I only had to do the math once and forgot about the formula.

http://www.jeffreyglasser.com/
 

DWThomas

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If you open the calculator on your iPhone and turn it sideways it turns into a scientific calculator.
Ha! I've had an iPhone for several years and never knew that. Thanks! One might think I would have found it by accident, but since I retired in 2002 I tend to avoid activities involving calculators. :unsure:
 

Bruce Osgood

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Ha! I've had an iPhone for several years and never knew that. Thanks! One might think I would have found it by accident, but since I retired in 2002 I tend to avoid activities involving calculators. :unsure:
Me too, I only wish I'd known this 15 tears ago.
 
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