Optical Calculations for Bellow Extension for Closer Focusing
As many of you know, I have recently built a 6x12 format camera with a 65mm SA lens. I originally intended to fit the lens to be focused at the hyperfocal distance for f16 but in the end, I fitted it with a helical focus mechanism.
I want to know if there is a formula for determining the amount of forward movement (or bellows extension if on a view camera) for the lens position to focus at a given distance.
The things I know already are:
Lens focal length: 65mm
Lens flange distance: 70.5mm
Suggested focus point for hyperfocal: 8.8 feet (5x4 film distance from dofmaster website).
So with these figures, is it possible to calculate how much further forward the lens needs to be from its 70.5mm flange distance in order to be focused at 8.8 feet or is it easier to put it on a view camera, focus and measure the distance?
Hyperfocal Focusing Table for 65mm f/4~f/45 lens for 6x12cm format (circle of confusion c = 0.083mm).
The entries are: aperture, hyperfocal distance (actual distance focused), the near limit of acceptable focus (1/2 the hyperfocal distance), lens displacement Δ forward of its infinity position.
Note: all focusing distances are from the lens center to the subject.
f/4, 12.79m, 6.40m, 0.33mm
f/5.6, 9.15m, 4.58m, 0.46mm
f/8, 6.43m, 3.21m, 0.66mm
f/11, 4.69m, 2.35m, 0.91mm
f/16, 3.25m, 1.62m, 1.33mm
f/22, 2.38m, 2.81.19m, 1.83mm
f/32, 1.66m, 0.83m, 2.66mm
f/45, 1.20m, 0.60m, 3.74mm
We must be careful of “focusing distance.” Some equations or the resulting tables take focusing distance = film-to-subject distance.
Others, such as hyperfocal distance equations and tables use subject distance = center of lens to subject.
For a subject distance taken as s = lens to subject, then the lens must be displaced
Δ = 1.61mm forward of it’s infinity position to focus on a subject 8.8 feet (2682.2mm) from the lens “center” to the subject.
For a 65mm lens of flange distance = 70.5mm, the flange needs to be 72.1mm forward of the film plane to focus on a subject 8.8 feet from the lens “center”.
The accuracy depends on the true focal length of the lens. Marked focal lengths can be +/- 2.4mm different than actual. An example is the 50/2.8N EL Nikkor. Its true focal length is 52.0mm (Nikon data).
When you use an online calculator or consult tables you can’t mix the data for different formats without changing the circle of confusion parameter to suit. The above table is strictly for 6 x 12cm format and the corresponding COC of 0.083mm.
The table for the same lens on the 4” x 5” format would look quite different.
Hopefully I computed the table correctly, but it’s not guaranteed.
Thanks for that. I wasn't going to confuse things by including the circle of confusion (!) but just wondered if there was a formula to give extension for focusing distance for a given focal length.
I was going to work out the actual hyperfocal distance separately.
"I want to know if there is a formula for determining the amount of forward movement (or bellows extension if on a view camera) for the lens position to focus at a given distance."
Before you do any 'cutting' it might behoove you to double check the 'REAL/TRUE/ACTUAL' focal length of the lens, since a majority of camera lenses I have used in the past 50-odd years have always been a few mm either above (or below the) marked focal length. If my 'greying' brain still recalls accurately, the 'standard' was a deviance of plus or minus 5% of the marked FL. I got "bitten" once... early in my career. about 17 yeras ago, I had suggest that an author (I was reviewing his paper for publication) double check the FL of the lens in use, since some of his calculations were based on microscopic examination the negative and measurement of 'blur' distances of travel. About six months later, I received notification that my suggestions had been valuable and had resulted in the modification of the results. Your situation may not 'need' such accuracy.... but it may be worth the effort to measure.
Quando omni flunkus moritati (R. Green)
Here’s how to calculate the extension, lens increment, and film-to-flange distance for a given subject distance s (reckoned from lens-to-subject).
If “subject distance” s = lens-to-subject distance, then
Total extension (film-to-lens) is
E = sf/[s – f]
Example: f = 65mm, s = 8.8 feet = 2682.24mm.
Total extension = 2682.24mm*65mm/[2684.2mm – 65mm] = 66.61mm.
Lens displacement (from infinity position) is
Δ = E – f
For the example, Δ = 66.61mm – 65mm = 1.61mm
Film-to-flange distance (at finite subject distance s) = flange distance + Δ
For a lens with flange distance 70.5mm, the film-to-flange distance at s = 8.8 feet is
Film-to-flange distance = 70.5mm + 1.61mm = 72.11mm.
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Thanks for the formulas (formulae?). As I have already made this with the focusing mount, I'm not going to modify it to be fixed but I was interested in learning if there was a mathematical way of working out the lens position.
This will however be of some use to me in re-calibrating the distance scale based on movement from the infinity position.