Bellows Length Relative to the Adjustment of f-stops

bobwysiwyg · Jul 13, 2008

But note that I gave the formula for a lens facing normally. The formula for a reversed lens is different.

Not to stray off my own topic.. but what is this about reversing a lens?:confused: Why would one do that, what does it accomplish? Yes, I'm new.

David A. Goldfarb · Jul 13, 2008

Asymmetric lenses are often better reversed for macro photography (this goes for any format). If the lens is optimized for a magnification ratio of 1:10 to 1:50, for instance, and you want to photograph something using a magnification ratio of 10:1 (10x), then you might get better results by reversing the lens, presuming you have a way to mount it reversed, stop down the diaphragm, focus, and shade the rear element from extraneous light to prevent flare.

bobwysiwyg · Jul 13, 2008

Very interesting. Thank you.

RobC · Jul 13, 2008

And for the original poster the math is:

1/v = 1/F - 1/u

where:

v = rear nodal point to image
F = lens focal length
u = subject to rear nodal point

So if F = 150 and u = (8*150) then we have

1/150 - 1/1200 = 1/171.5

which means required bellows extension is 171.5mm when subject is 1200mm and using a 150mm lens.

And then using (171.5/150)² = 1.3 time factor which is pretty close to a 1/3 stop.

Using 10 times focal length we get:

1/150 - 1/1500 = 1/166.7
then (166.7/150)² = 1.24 time factor which is less than a 1/3 stop.

And that's exact

RobC · Jul 13, 2008

bobwysiwyg said:
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:

I think you'll find that for the vast majority of 35mm camera lenses, the available close focus distance is reached before bellows extension factor reaches 1/3 stop value. For lenses designed for macro work which have much closer focussing available, then it is a factor to be considered.

So taking the 10 times focal length rule of thumb, then a 28mm lens would need to be able to focus closer than 280mm before bellows extension becomes significant enough to worry about.

RobertP · Jul 13, 2008

Print out a quick disc, (its free) stick it in your wallet. By the time most get their tape measure and calculator out you'll have the compensation figured and the shot already done. The last thing I want to do when I'm trying to be creative is crunch a bunch of numbers. If you're more than ten times the focal length away then don't worry about bellows factor.

RobertP · Jul 13, 2008

www.salzgeber.at/disc/index.html

bobwysiwyg · Jul 13, 2008

RobertP said:
Print out a quick disc, (its free) stick it in your wallet. By the time most get their tape measure and calculator out you'll have the compensation figured and the shot already done. The last thing I want to do when I'm trying to be creative is crunch a bunch of numbers. If you're more than ten times the focal length away then don't worry about bellows factor.

Very interesting as well. Going to give this a try, so much to learn [sigh].

Dave Wooten · Jul 13, 2008

A 160 mm lens at f/16 would have an aperature that is 10 mm. i.e. 160/16 = 10.

A 320 mm lens at f/16 would have an aperature that is 20 mm. i.e. 320/16 = 20.

If you therefore "pulled" your 160 mm lens to a 320 mm length and your meter tells you you need f/16 and you put it to f/16 on your 160 mm shutter scale you would be shooting through a 10 mm f stop. For a 320 lens, the length at which you are shooting, 320/16 = 20 mm dia f stop. You need to let in more light the more distance you have pulled your lens from the film. So you can just compute the diameter of your f stop needed. Make a little chart of your lens f stops and the diameter of each. example, 160 at f/8 = 160mm/8 = 20 mm, 160mm/16 = 10 mm etc.

so you know if meter says you need f/16, and your lens is "streatched" to 320 you will need 320/16 = 20 mm f stop size....what is a 20 mm f stop size on your 160 lens shutter ring? 160/20 = f 8

So your 160 mm lens with the bellows pulled to 320 is now a 320 mm lens and f 16 on a 320 mm lens is 320/16 = 20 mm. To get to 20 mm on your 160 lens, simply divide 160/20 and the answer is f8. Set your lens at f/8.

Ed Sukach · Jul 13, 2008

Dan Fromm said:
Ed, your first post confused me and your correction confused me more.

Are you trying to explain how to calculate effective aperture?

No. I am relating the empiracal definition of "f/stop". "f" does not represent "focal length at infinity" ... it does indicate the distance from the aperture to the film plane in any given situation. "Effective" indicates a "T" stop, with other factors (air-glass reflection, dispersion, etc.) considered.

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.

Sounds good to me.

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

See preceeding. "Effective aperture/ T/stop is far more accurate, but I would say "most of us" in our fumbling, inaccurate way, rely on what is etched into the lens scale ... "f/stop @ infinity focus".

Please don't tell me again that you worked in QC for a optical goods manufacturer.

Tell you what - PLEASE don't read my posts where I try to explain where I am "coming from". Sorry to have offended you so deeply ... (Note 1)

QC is about statistics, not about optical design.

I could argue about Quality ASSURANCE as opposed to Quality CONTROL --

Believe me, my job required a clear uderstanding of basic optical principles - starting with the ability to communicate with the Optical Designers - and getting my "terms" right.

Statistics are tools - very useful tools - used in controlling conformance to design and intent. There is much more to that activity.

Note 1: NOT!

Ed Sukach · Jul 13, 2008

Dave Wooten said:
A 160 mm lens at f/16 would have an aperature that is 10 mm. i.e. 160/16 = 10.
A 320 mm lens at f/16 would have an aperature that is 20 mm. i.e. 320/16 = 20.
If you therefore "pulled" your 160 mm lens to a 320 mm length ...

Yes, - you are correct.

David A. Goldfarb · Jul 13, 2008

RobertP said:
www.salzgeber.at/disc/index.html

The QuickDisc works on the principle of using magnification factor to determine bellows factor, just like the method I use. The QuickDisc is even simpler for large format, since the ruler measures in bellows factor, but it doesn't work for smaller formats where you might not have easy access to the groundglass to measure, or where the groundglass may be smaller than the target.

Steve Sherman · Jul 13, 2008

Be honest, anyone have a headache

epatsellis · Jul 13, 2008

Yup, on LF I use a quick stick, for smaller formats, either an educated guesstimate or quick measurement. Either way, I don't shoot transparency film, so if any exposure errors result, I prefer them to the over exposure side.

erie

RobertP · Jul 14, 2008

David A. Goldfarb said:
The QuickDisc works on the principle of using magnification factor to determine bellows factor, just like the method I use. The QuickDisc is even simpler for large format, since the ruler measures in bellows factor, but it doesn't work for smaller formats where you might not have easy access to the groundglass to measure, or where the groundglass may be smaller than the target.

Isn't this the large format and accessories catagory? It works fine on a 4x5 and it can be printed out to what ever size you want.

RobertP · Jul 14, 2008

I would love to see how lost some people would be if they were working with wet plate collodion. Where a light meter, scale, calculator, slide rule or pocket protector would do them no good at all.....lol

smieglitz · Jul 14, 2008

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

Cheers

It can be even simpler. Just think of the focal length as an f/stop series but shift the decimal point to make it more apparent. For example, with a 160mm lens think of 160mm extension as f/16, 220mm extension as f/22 320mm extension as f/32, etc. If your 160mm lens is at 220mm extension you need to give +1 stops exposure to compensate. If at 320 mm you'll need +2 stops. If 200mm probably about 2/3 stops. Simple, and you can pretty much eyeball this.

I've used 160mm because it translates a bit easier than 150mm, but the same principle holds. To figure out any next f/stop of extension for any focal length, just multiply the focal length by the square root of 2 (~1.414, but 1.4 is close enough). So, if a 150mm lens is used, a full f/stop of compensation would be needed at 150mm x 1.4 = 210mm, 2 stops more would be needed at 150mm x 2.0 =300mm, 3 stops more at 150mm x2.8 = 420mm, 4 stops more at 150mm x 4.0 = 600mm, and so on. (If you really want to use a calculator for extra precision, note each stop increase happens at an extension of 1.414^n where n will be the number of stops of compensation needed. But this is really over-complicating a very simple procedure.)

Another example: with a 90mm lens a full stop compensation would be needed at 90mm x 1.4 = 126mm, 2 stops at 90mm x 2 = 180mm, 3 stops at 252mm, etc.

smieglitz · Jul 14, 2008

RobertP said:
I would love to see how lost some people would be if they were working with wet plate collodion. Where a light meter, scale, calculator, slide rule or pocket protector would do them no good at all.....lol

Hey! I resemble that remark!

JBrunner · Jul 14, 2008

Dead Link Removed

mark · Jul 14, 2008

Leave it to the Australians to have a left brain / right brain test with a naked chick.

Dan Fromm · Jul 14, 2008

Ed, "T" stops report a lens' measured transmission, are used when getting exposure absolutely right is essential. Cine lenses are often, but not always, marked in "T" stops. I've never seen or heard of an LF lens marked in T stops, but this doesn't mean there are none, rather that if they exist they're not common.

f/stops are geometric, have nothing to do with transmission, and are used when calculating, e.g., depth of field. Effective f/numbers are f/numbers adjusted for magnification.

Remember that because of light lost to reflections, real lenses usually transmit less light than their geometric f/stop numbers indicate. Since the really lossy reflections occur at air-glass interfaces, this is one of the reasons that lens designers used to try to minimize the number of air-glass interfaces. Hence, e.g., the Aldis Uno, Dagor types, Protars, ...

If you're going to lay down the law, please follow long-established conventions.

Your empirical definition of f/stop is (a) unconventional and (b) nonsense. If you meant to say that the f/number is (diameter of the entrance pupil)/(focal length), you should have said so. You missed the point of my question about focal length. A lens can't form a sharp image if the lens rear nodal point is less than the lens' focal length from the film plane. Please revisit your definition with this inconvenient but true fact in mind.

Cheers,

Dan

Ed Sukach · Jul 14, 2008

Dan Fromm said:
Ed, "T" stops report a lens' measured transmission, are used when getting exposure absolutely right is essential. Cine lenses are often, but not always, marked in "T" stops. I've never seen or heard of an LF lens marked in T stops, but this doesn't mean there are none, rather that if they exist they're not common...

Imteresting continuation of the discussion. Now you are repeating what I had said all along... as if you had agreed with it from the beginning.

f/stops are geometric, have nothing to do with transmission, and are used when calculating, e.g., depth of field. Effective f/numbers are f/numbers adjusted for magnification

Again, I have described "f/stop" according to what I have been taught as a "Standard" definition - one that has been accepted in the Optical Industry.
There are other means of describing other attributes ... "T/stops" - indeed more accurate - are one of them. If you choose to modify definitions the end result will be a decrease in the understanding of "what is gong on."

Reflect for a moment on the original question: "What effect does focusing have on f/stop"? As the lens is focused the diaphragm is (usually) moved AWAY from the film plane ... therefore increasing the ratio of "distance to film plane" to "diaphragm" and increasing the numerical "f/stop".

Remember that because of light lost to reflections, real lenses usually transmit less light than their geometric f/stop numbers indicate. Since the really lossy reflections occur at air-glass interfaces, this is one of the reasons that lens designers used to try to minimize the number of air-glass interfaces. Hence, e.g., the Aldis Uno, Dagor types, Protars, ...

... ??? Quite an echo in here. Why do you repeat?

Trivia question for the moment ... One camera manufacturer *DID* produce a 35mm camera with the lens marked in "T/stops". Unfortunately, as you have repeated, "T/stops" are always numerically greater than "f/stops" and the great unwashed masses out there perceived their lenses as "slower" than those found on other cameras .. and "slower" translated to "inferior."
They had quite an advertising campaign, with Joan Crawford as their primary spokesperson. Care to venture a guess as to their identity?

If you're going to lay down the law, please follow long-established conventions.

Let me see if I can force this discussion up a notch - something other than a constant repetition of what was said before...

I AM following ... etc.

..Your empirical definition of f/stop is (a) unconventional and (b) nonsense. If you meant to say that the f/number is (diameter of the entrance pupil)/(focal length), you should have said so. You missed the point of my question about focal length. A lens can't form a sharp image if the lens rear nodal point is less than the lens' focal length from the film plane. Please revisit your definition with this inconvenient but true fact in mind.

... Inconvenient? How - why - did the discussion suddenly expand to include "nodal points" and whether or not a lens COULD be focused?

"It is TRUE .. because *I* say it is true..."? Nice try ..

I did NOT "mean to say that f/diameter had ANYTHING to do with entrance/ exit pupil." Why did you assume that I did? I say that "f/stop" is one thing and that "T/stops" and "Effective f/stops" are another... and it is dangerous to confuse them.

- I missed the point" about "focal length.."? Why don't you try restating it ..? It seems to me hat I understand what you are saying - and I am disagreeing with what you are saying.

Cheers, to you, too,

Dan Fromm · Jul 14, 2008

Ed, were you serious when you wrote the following?

"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"

This statement, copied direct from an earlier post of yours in this thread, is what I commented on. It is flatly wrong. Now, what did you mean?

Cheers,

Dan

JBrunner · Jul 14, 2008

Dan Fromm said:
Ed, were you serious when you wrote the following?

"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"

This statement, copied direct from an earlier post of yours in this thread, is what I commented on. It is flatly wrong. Now, what did you mean?

Cheers,

Dan

Actually, with a simple lens, where the aperture denotes the nodal point and the distance to the film plane is focused at infinity, it works.

It's to early for a guy with a clockwise spinning naked lady to extrapolate the math for close focus.

Dan Fromm · Jul 14, 2008

Jason, it works if you mean that the lens' focal length is 150 mm and the aperture (or entrance pupil) is 75 mm. But that's not what Ed wrote.

Sorry,

Dan

Bellows Length Relative to the Adjustment of f-stops

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