Bellows Length Relative to the Adjustment of f-stops

I wonder if someone could explain something that, thus far, is eluding me. I've recently acquired my first view camera (Cambo 4x5) the complement of lenses consists of 135mm, 150mm and a 210mm.

I've been reading about anything I can get my hands on to learn the basics and I "think" I understand most of it. One thing that I'm still having trouble getting my arms around is the fact that I may meter a scene, but under certain circumstances (bellows length/extension) have to alter the indicated f-stop. I understand the concept of extended bellows resulting in less light to the film plane, but can't quite grasp under what circumstances this would be necessary.

I suspect most shooting will be various scenes at some distance, but I may try a portrait, say at 10 ft. or so. If someone could explain, in relation to the lens focal lengths I have, I would be very appreciative. Thanks.

You don't have to worry about bellows factor until the length of the bellows exceeds the focal length of the lens (ie with a 150mm lens, the distance between the lens and the film plane exceeds 150mm), which isn't going happen when focusing at ten feet with any of your lenses.

It will happen most often when shooting still life objects. Set up something small and fill the frame with it, and see how far your bellows has to extend to get focus, and you'll see first hand.

This may also help:

http://www.youtube.com/watch?v=vwfRA615Mx8

don't you mean 150mm from infinity focus position?

JBrunner said:

You don't have to worry about bellows factor until the length of the bellows exceeds the focal length of the lens (ie with a 150mm lens, the distance between the lens and the film plane exceeds 150mm), which isn't going happen when focusing at ten feet with any of your lenses.

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The "f/stop" is defined as the aperture diameter (actually it has to do with aperture AREA) divided by the distance to the film plane. The optics - the lens itself, and its focal length, really has nothing to do with the f/stop. "T/stops are different; the effects of the optical system, the glasses, etc., are considered.

Most of us use "f/stops". If a 150mm lens is focused at infinity ... Whoops ... if a "simple" 150mm lens is focused at infinity (and the diaphragm is placed at the optical center of the lens), the f/stop will be as indicated. As soon as we extend the lens (diaphragm) to a position farther than 150mm, the effect is to reduce the amount of light reaching the film plane - the equation:

Distance of aperture to film plane / aperture diameter = f/stop.

Holds, in any event.

I've just checked out:

http://www.youtube.com/watch?v=vwfRA615Mx8

... And he explains it very well, actually. I haven't downloaded his "scale and target", but I've seen similar, and they work very well - accurately enough for any sane ... (???). or at least, practical purposes.

Simply: (bellows draw / lens focal length) squared = exposure factor.

Bellows draw is measured from the film plane to the nodal point of the lens. The nodal point is usually around where the aperture is. This is not rocket science.

So (150mm bellows draw divided by a lens of 150mm) squared equals 1. An exposure factor of 1 means no adjustment of the aperture is needed.

So an exposure factor of 8 would be 3 f-stops?

rob champagne said:

don't you mean 150mm from infinity focus position?

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Yes, of course, which would be the focal length of the lens, 150mm. The long hand version was completed by Pinholemaster, and as noted it's not rocket science, unless you want to make it so.

When out in the field or studio/kitchen/lounge and you wish to know your effective f/stop number I use a simple equation to do so. I require only one piece of extra equipment, but in fact, use two to do so.

The first piece of equipment is a cloth tape measure available at any sewing or haberdashery shop. The other is a small credit card sized calculator powered by a solar cell inbuilt, that I acquired at a trade show as a freebie.

Assuming here, you are using your 150mm lens, and your light meter is telling you that at your chosen shutter speed, you require f11.

Focus first, and then measure the distance from the film plane to the centre of the lens.

In this instance we assume you have decided to get a same size image on your negative of the object being focused, therefore your extension from the film plane is 300mm.

Your light meter measured f/stop was f11, therefore your effective f-number = f/11 x 300/150 = f/22

In this instance you doubled the distance, so you lost exactly two stops of light.

Doing it again at ½ life size on the negative it goes like this:- your measured f/stop number was f11, your effective f-number = f11 x 225/150 = f/16.5

In this instance you only added 50% distance, so you lost exactly one stop of light.

This method works well enough for normal lenses, like those that you have. I also find it extremely easy to use as I use my light meter to give me shutter speed - f/stop combinations, which I suspect is what most of us do.

Mick.

Bellows extension

Another method to determine the amount of increased exposure you'll need.

Take the length of the lens your using and turn it into F stops, i.e. 120mm = 4" 150mm = 6", 210mm = 8 1/4", 300mm = 12"

If you have a 4" lens and when closed focused on subject matter the lens nodal to the film plane measures 5.6" then you add one stop of exposure, if it measures 8" then you would allow 2 stops of exposure, this is also known as "life size". In other words the size of the actual subject is recreated on the film plane as the exact size that it is in real life.

A 300mm lens measures 16" when close focused you would correlate to F11 to F 16, therefore add one stop of exposure, 19" when focused add 1.5 stops of exposure.

To those who suggest that a 12" lens does not exactly correlate to F 11. For years I used this system in a studio setting shooting nearly life size most of the time using Ektachrome film which has little tolerance for exposure error.

The beauty is you will not have to search for anything, F stop scale should be in your head to draw reference from.

Cheers

Pinholemaster said:

Simply: (bellows draw / lens focal length) squared = exposure factor.

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That's the method I use. It's very easy if you carry a pocket calculator in your camera bag. Actually, it's even easier with a slide rule!

I figure bellows factor by magnification. Any time you have a magnification of 1:10 on any camera in any format, you need to compensate for exposure (if you use TTL metering, you're compensating automatically). With large format, you get into bellows factor territory much more easily than with small formats--pretty much any time you're indoors and not using a wide lens, for instance.

I've posted instructions and a magnification-->exposure factor table here--

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

Thanks all for the replies and feedback. I think I get the idea. I'll just have to try some of this for a better grasp and understanding.

Ed Sukach said:

The "f/stop" is defined as the aperture diameter (actually it has to do with aperture AREA) divided by the distance to the film plane.

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I re-read this XX times, trying to get it right --- and did not!
This is incorrect - backwards. The correct equation is:

The distance from the aperture to the film plane divided by the aperture diameter = f/stop.

Example: 150mm to film plane / 75mm diameter aperture = f/2

Amazing! .. no one called it to my attention - nor the attention of everyone else. Are we slipping??

well Ed, to nit pick, it's entrance pupil diameter, not aperture diameter....

Ed Sukach said:

I re-read this XX times, trying to get it right --- and did not!
This is incorrect - backwards. The correct equation is:

The distance from the aperture to the film plane divided by the aperture diameter = f/stop.

Example: 150mm to film plane / 75mm diameter aperture = f/2

Amazing! .. no one called it to my attention - nor the attention of everyone else. Are we slipping??

Click to expand...

Ed, your first post confused me and your correction confused me more.

Are you trying to explain how to calculate effective aperture?

Most of us calculate effective aperture as (aperture set) * (1 + 1/(magnification/pupillary magnification)) and most of the time we ignore pupillary magnification. (aperture set) is the f/number, for example 11. Magnification is the size of the image on film/size of object, for example 1. In this case, the effective aperture is f/22. Doing the calculation doesn't really need a calculator.

Also please explain how to make sense of your formula with a lens whose focal length is longer than 150 mm?

Please don't tell me again that you worked in QC for a optical goods manufacturer. QC is about statistics, not about optical design.

Cheers,

Dan

Bob, most (if not all) of the above is probably correct. But one thing which I didn't see while browsing through the answers is that this comes into play much sooner than you'd expect, especially if you are used to 35mm cameras. (You've become aware of the fact that everything is manual on the 4x5" while you don't have to think of it with the full auto film-plane metering 35mm/digital camera.)
On a manual 35mm camera with external metering (they actually made such cameras in the -60'ies. ) you'd had to manually compensate exposure with 2 stops (4 times) if magnification is 1:1.
Now, that's the size of a postage stamp.
At 1:2 you'd need to double the exposure, i.e. 1 extra stop. That's the size of a matchbox.

The very same basic rules apply for 4x5 cameras, but they start to come into play much sooner. 1:1 is about the size of a slighly cupped hand. 1:2 (8x10") would almost be a full-body shot of my sleeping cat. (Which by the way is the only way of getting any sharp picture of my cat. ) As someone mentioned the bellows factor starts to chime in at around 1:10 (gives a factor of 1.21 which is close to 1/3 stop). The size of the subject is then 40x50". For b/w I wouldn't care (yet), but e.g. Velvia 50 is very fuzzy and there are no margins to play with.

To make a long story short: Whenever the long side of the picture is shorter than a yard, be aware. You will need some extra exposure.

//Björn

edtbjon said:

On a manual 35mm camera with external metering (they actually made such cameras in the -60'ies. )//Björn

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I certainly do remember them. My first 35mm was an Ashi Pentax SV. Still have it and have shot with it recently, though I never did any macro shooting.

Sounds like this exposure adjustment comes into play only when doing extreme close-ups or macro work?:confused:

bobwysiwyg said:

Sounds like this exposure adjustment comes into play only when doing extreme close-ups or macro work?:confused:

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On 35mm yes. On larger formats no.

Think about the magnification ratio. A head and shoulders portrait on 35mm is a magnification of about 1:24--no significant exposure factor. The same portrait on 4x5" has a magnification factor of about 1:6--about 1/3 or 1/2 stop exposure factor. The same portrait on 8x10" has a magnification of about 1:3--about 2/3 or 1 stop exposure compensation (with slide film a little underexposure is usually better; with neg film a little overexposure is usually better).

Ah, thanks for the clarification. Not sure these tired, old eyes can discern a 1/3 stop difference, but I'll give it a try.

bobwysiwyg said:

Ah, thanks for the clarification. Not sure these tired, old eyes can discern a 1/3 stop difference, but I'll give it a try.

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Your eyes may not notice, but your film will...

I can't remember the exact number but I have seen posted, on several occasions, that when the subject is closer than x * lens focal length, then you need to adjust for it. I'm sure someone can work out the maths for it.

(p.s. I think it is around 8 times focal length )

We all instinctively know that the farther you get from a light source (any light source), the less light reaches you. This same thing happens inside of your camera with the light heading from the lens to the film.

We all know this happens, but how much? If you look at how light behaves with this cool thing called the inverse square law, you see that light fall off is not linear. The law tells us that increasing the distance 2x results in 1/4 of the light. This also means that halving the distance results in 4x more light. You apply this knowledge to your bellows, and you get your necessary amount of additional exposure over what you would need at infinity.

The way I learned it, the law states: The amount of light falling on a subject is inversely proportional to the square of the distance from the object.

With all this info, you can figure out a correction factor for what your exposure would be at infinity focus. So, you just state a ratio of your actual bellows draw to your infinity bellows draw. But remember that light falloff is not linear. The law states not that it is distance, but the *square* of distance to which amount of light is proportional. So, you square the result, and that is your factor. Say your 210 lens is drawn out to 300mm. 300mm:210mm = 1.43. Square that. 1.43x1.43=2 (roughly).

Since your exposure factor rounds to 2, you would need twice as much light as you do at infinity in this case. So, half rate your film, open up one stop, double your time; however you want to do it. You can call it whatever you want, but, basically, you just need double exposure.

You don't necessarily need to find an effective f stop, although that is technically the accurate way to look at the situation. However, it is also way I personally find the most confusing and error prone in "the field". You could call it an increase in time (which I usually do), you could use a lower EI (which I never do), you could just use the existing f numbers and open up (which I sometimes do).

At any rate, no matter how you achieve it, you simply need more exposure when you focus closer than infinity, which is what your f stops are calibrated to. An f stop is not physical. It is theoretical. It is not a physical aperture diameter that can be measured, but a ratio of bellows draw to aperture. Thus, any time you focus at anything other than infinity, your bellows draw changes from infinity. However, the actual physical apertures stay the same sizes. These two things are what determines f stop. Therefore, if one of them is changing and the other one is fixed, f stop must be changing as well.

In the end, just remember: (bigger number/smaller number) times itself. Then you have an exposure factor, and you can apply it however is most convenient for you.

rob champagne said:

I can't remember the exact number but I have seen posted, on several occasions, that when the subject is closer than x * lens focal length, then you need to adjust for it. I'm sure someone can work out the maths for it.

(p.s. I think it is around 8 times focal length )

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Yes I just confirmed this with the maths.

Subject at 8 times or more focal length distance will require LESS than 1/3 stop correction. Subject at 7 times (or less) focal length will require 1/3 stop or more correction.
So as a very quick mental guide as to whether you need to worry about it, then roughly speaking, anything less than 10 (KISS) times focal length from camera should have bellows correction applied.

Rob, if you look a little up the thread you'll find the magic formula. We ogres sometimes share them.

Not a rule of thumb. Exact. But note that I gave the formula for a lens facing normally. The formula for a reversed lens is different. Finding it, if you think you need it, will be a good exercise for you.

Dan Fromm said:

Rob, if you look a little up the thread you'll find the magic formula. We ogres sometimes share them.

Not a rule of thumb. Exact. But note that I gave the formula for a lens facing normally. The formula for a reversed lens is different. Finding it, if you think you need it, will be a good exercise for you.

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Yeah I know the formula but for the original poster it will be useful to know that if the subject is 10 or more times the focal length from the camera, then he doesn't need to worry about bellows factor. What could be simpler than that. 150mm lens, subject at least 1500mm from camera = no bellows factor to worry about. And that mostly can be done by eye without using a measuring stick.