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01-12-2018, 03:43 PM   #1
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Calculating effect of flange focal distance on infinity and minimum focus distance?

OK... apologies in advance for what's probably a dumb question, but here goes...

If I use a particular lens with a camera that has a slightly different (larger or smaller) flange focal distance, and I don't compensate for that difference with the adapter, is there a way of calculating the maximum and minimum focus distances of that lens when used in this configuration?

An example (though by no means the only one I'm interested in) would be using something like a 50mm M39-mount Zenit SLR lens on a Pentax M42-mount film camera... The flange focal distance for the lens is 45.2mm, while for the camera it's 45.46mm. If I simply fit an M39-to-M42 step-up ring to the lens and mount it direct to the camera, I know that, since the lens is sitting too far away from the film plane, it won't be able to focus to infinity, and will focus closer than it would on the "correct" camera. But is there any way I can calculate (presumably based on focal length?) what the actual maximum and minimum focus distances should be?

This is for a project I'm working on regarding adapted lens use on various manufacturer's cameras. Any assistance - even if it's to tell me how badly I misunderstand this (!) - would be greatly appreciated.

Thanks in advance


Last edited by BigMackCam; 01-12-2018 at 04:03 PM.
01-12-2018, 05:39 PM - 1 Like   #2
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IIRC, the full answer is complicated.

A less-than-full answer may be adequate, however, at least for block-focus primes.

The Wikipedia article on "Focal Length" may be of help, particularly the equation in the section on "Thin Lens Equation". In all likelihood, your lens will require the more involved the discussions that follow that section, but the general form of the equation should allow an understanding of how a few tenths of a millimeter difference in flange focal length might affect maximum distance. Minimum distance is, of course, determined by the same considerations.

Focal length - Wikipedia

Hint: A change in flange focal distance directly changes the lens-to-object distance the same amount at both the nominal infinity and MFD points on the focus ring.


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6 Days Ago - 1 Like   #3
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I suspect that many if not the same formulas used for macro calculations would give you the information you need. The problem being most of the formulas I've seen are based on a "thin" or simple lens model for simplicity's sake placing the optical plane or center of the lens near the front element. Changing the extension of the lens is essentially the same as changing the flange distance.
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The relevant equations are on Page 4 of Doug Kerr's excellent paper on extension tubes (which is basically what you're doing by using a camera with a longer flange length than the lens was meant for). Even a mathematical doofus like me could just about understand it:

http://dougkerr.net/Pumpkin/articles/Extension_tubes.pdf

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QuoteOriginally posted by Not a Number Quote
Changing the extension of the lens is essentially the same as changing the flange distance.
+1.

Off the cuff, from my own experience, your zenith M39 is unlikely to focus into double digit metres distance. I recall trying a konica 28mm f3.5 on my pentax, it was focus to a few m only. Telephotos however are more amenable, I have had no problem getting test pics of the castle with various 300mm/400mm, the moon however is another matter.
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If D is the added distance to the flange and F is the focal length of the lens, the maximum focus distance will be:

MaxFD = F^2/D + 2*F + D

Note that the F^2/D term dominates the equation and implies that even tiny offsets to the flange severely limit the Max FD for a wide angle lens. Even a tiny 0.26 mm addition to the flange with a 28 mm lens will limit the focus to about 3 meters. The effect is a very strong function of focal length. That same 0.26 mm addition to the flange with a 400 mm lens will limit the max focus to about 0.6 kilometers (which would be indistinguishable from infinity). Of course if the lens can focus "past infinity" then it might be able to handle a small offset in the flange.

The Minimum FD with a flange offset is more complicated but I can give that to you if you want some nastier math! Overall, the MinFD is not affected very much by flange offsets. As a rough example: a 28 mm lens that focuses from 0.5 m to ∞ might become a lens that focuses from 0.44 m to 3 m with a 0.26 mm flange offset.

(P.S. These are the formulas using the "thin lens" approximation. For more complex multi-element lenses the overall effect remains the same but the exact numbers on MaxFD might be different by a focal length or two. The MinFD numbers might be very different with a "thick lens" because the focal length of a more complex lens may shift at the lenses' MinFD.)
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Many thanks, all, for the assistance. It's very much appreciated as always

QuoteOriginally posted by photoptimist Quote
If D is the added distance to the flange and F is the focal length of the lens, the maximum focus distance will be:

MaxFD = F^2/D + 2*F + D
Splendid. This is exactly what I need

I note the limitations of using "thin lens" approximation. For my application, absolute accuracy really doesn't matter.

QuoteOriginally posted by photoptimist Quote
The Minimum FD with a flange offset is more complicated but I can give that to you if you want some nastier math! Overall, the MinFD is not affected very much by flange offsets. As a rough example: a 28 mm lens that focuses from 0.5 m to ∞ might become a lens that focuses from 0.44 m to 3 m with a 0.26 mm flange offset.
I'd love to see the math, if you can spare the time and it won't run into pages. A formula or three I might be able to deal with... (famous last words ).

Thanks again!
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QuoteOriginally posted by Dartmoor Dave Quote
The relevant equations are on Page 4 of Doug Kerr's excellent paper on extension tubes (which is basically what you're doing by using a camera with a longer flange length than the lens was meant for). Even a mathematical doofus like me could just about understand it:

http://dougkerr.net/Pumpkin/articles/Extension_tubes.pdf
Thanks for contributing this to the discussion. Doug Kerr has an amazing amount of good stuff. It is too bad that much of it was formerly hard to access directly on his Web site due to bad links (now corrected). His paper on split prism rangerfinders as applied to PDAF is one of the best explanations on the Web of how it works and how lens aperture affects operation.

The full contents of the Doug Kerr "Pumpkin" may be found at:
The Pumpkin
The page organization is a little confusing, but goes like this (top to bottom)...
  1. List of links to detail article descriptions/links(pdf) for each of the major headings
  2. Outline of the major headings with article list for each (probably the most useful way to avoid endless scrolling)
  3. Detailed article descriptions/links(pdf) organized by major heading


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QuoteOriginally posted by photoptimist Quote
The Minimum FD with a flange offset is more complicated but I can give that to you if you want some nastier math! Overall, the MinFD is not affected very much by flange offsets. As a rough example: a 28 mm lens that focuses from 0.5 m to ∞ might become a lens that focuses from 0.44 m to 3 m with a 0.26 mm flange offset.
That is my experience and understanding as well. Doug Kerr has an article that provides equations as well as the supporting calculus and math stuff for those who actually took those classes and remember how to apply what was taught.
http://dougkerr.net/Pumpkin/articles/FilmShift.pdf


Apologies to Gary Larson



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6 Days Ago   #10
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"... for those who actually took those classes and remember how to apply what was taught were listening"
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QuoteOriginally posted by BigMackCam Quote
Many thanks, all, for the assistance. It's very much appreciated as always



Splendid. This is exactly what I need

I note the limitations of using "thin lens" approximation. For my application, absolute accuracy really doesn't matter.



I'd love to see the math, if you can spare the time and it won't run into pages. A formula or three I might be able to deal with... (famous last words ).

Thanks again!
LOL! OK! You asked for it! ...... Sticking to only three equations is tough, though......

For various reasons, the distances indicated on lens barrels are the total distances from the camera's focal plane to the subject. This probably makes sense for doing flash-distance calculations and for replicating tripod-to-subject setups. But it makes the math for changes in flange distance really messy -- the dreaded quadratic formula is involved!

The basic thin-lens formula for the camera-to-subject distance is:

C = F*(2 + m + 1/m)

where C is the camera-to-subject distance, F is the focal length of the lens, and m is the optical image magnification. If you know F and have C set to the MFD, then you can calculate the corresponding m. But it involves the quadratic formula. The value of m at the MFD is the one in the following quadratic equation:

0 = m^2 - (MFD/F - 2)*m + 1

I'm not going to turn the crank on that formula here because: 1) it's a mess and 2) it's often not right for non-thin lenses. For example, the quadratic formula blows up if you plug in the numbers for the Pentax 100mm f/2.8 Macro lens. For that lens, the quadratic equation solution involves imaginary numbers which must have something to do with those Pentax pixies!

Instead, it is better to estimate the maximum subject magnification at minimum focus distance by another means. Maximum subject magnification can be found either in the specs of the lens, on markings on the lens barrel, or by taking a picture of a ruler at the minimum focus distance (just remember to use the correct FF or APS-C frame size in estimating the magnification). Then you can use the fact that a flange displacement of d will change the magnification by exactly d/F. A wee bit 'o algebra with the thin lens equation can compute the new minimum focus distance, MFD', from the original MFD with an adjustment for flange offset, d, with focal length, f, at the max magnification, m:

MFD' = MFD - F*(1/m - 1/(m+d/F) - d/F)

Note that this calculation still has the thin-lens approximation in it but it should be fairly close if d is not too large.
5 Days Ago   #12
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QuoteOriginally posted by photoptimist Quote
LOL! OK! You asked for it! ...... Sticking to only three equations is tough, though......
Thank you very much! I appreciate the info (and it wasn't quite as long as I'd feared!)

Having read your post once, I'm now going to have a glass of wine, then come back to it again. I'm sure it'll make perfect sense then
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