Calculating bellows extension/focal length at different focal distances?
I have a box camera. Right now, my 90mm lens is focused at about 12 feet. How much closer do I have to move it to the film plane to put it at infinity? I just realized I have no idea how to do this calculation.
I don't know about the actual calculations (i.e. the kind that people who know what they are talking about on a technical level do), however, if you want to focus a 90mm lens on infinity, I would start by placing the vertical plane in which the diaphragm resides roughly 90mm from the film plane. So, I'd say that your answer for how far back to move the lens is roughly 90mm subtracted from however far the lens is from the film now. Knowing 90mm lenses in practical use, I would say that you won't have to move it very far back; it'll be under 5mm, looking at the distance scale on my Technika.
Last edited by 2F/2F; 09-24-2010 at 10:24 PM. Click to view previous post history.
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How do you figure?
I just realized that I could put the lens on my speed graphic and measure. However, I would still like to know how to do the proper calculation of how bellows extension relates to focal distance. I know that the bellows extension is 2x the infinity distance when focused at 1:1 magnification, but I don't know any of the relationship in between.
pq = f^2
2.21mm = (90mm)^2/(3657.6mm)
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"The focal length of a lens is defined as the distance from the lens at which the image is focused when the object is at infinity".
So if you have a lens with a focal length of 90mm, infinity focus is when the film plane is at 90mm from the lens. That's how I understand it.
(Should be 2.3mm closer to the lens than where it is focused at 12ft, as the above posts agree)
Bellows Increment as a Function of Subject Distance and Focal Length
Here’s how to calculate the bellows increment from the data given, as I believe your concern was how to calculate it—not just the answer.
When you ask questions like, “my 90mm lens is focused at about 12 feet”, it’s important to carefully define the variables so that we all agree upon what we’re talking about.
To some that’s the distance from the subject to the front of the lens barrel. It’s easy to measure, but imprecise so far as calculations are concerned.
The film plane is the correct reference point, as it defines part of the optical system. With this definition of “subject distance” we have
f = 90mm.
k = 12 feet = 3657.6mm (film to subject distance).
We must first calculate the film to center of lens distance
i = [k - sqrt(k^2 – 4kf)]/2
The distance you want is
Δ = i – f (where i >= f in all cases)
For the situation you stated,
i = [3657.6mm - sqrt(3657.6mm^2 – 4(3657.6mm )90mm)]/2 = 92.3mm
Δ = 92.3mm – 90mm = 2.3mm
All this is given with the usual caveat about applying to lenses of approximately symmetrical design.
We decide whether to use the standard formulas or to consider pupillary magnification based on the lens angle.
Non-symmetric lenses are those with lens angles less than 20° or greater than 75° according to the Kodak Professional Photoguide, 1st Edition, 1st printing 1975, page 33.
Post #5 gives a simple formula that tacitly assumes that the stated “subject distance” is the center of lens to subject distance = q. If that’s the case then the formula given works perfectly. But note that it leads to a slightly different result than using “subject distance” = film to subject distance. That’s why we must carefully define each variable so that we’re all speaking of the same thing.
It was late, but I should have clarified that p is the distance to from the rear focal point to the film plane, and q is the distance from the front focal point to the subject plane, and is thus well suited to answer the question at hand. By itself, pq = f^2 is one of the handiest formulas around for those using view cameras, and is easily derived from similar triangles with the standard image/object Newtonian ray diagram for simple lenses.
Originally Posted by David William White
I agree. The solution in #5 is simple, elegant, and correct, where "subject distance" is understood as subject to front node (roughly the geometric center of the lens for a symmetric lens).
even if you do not know the exact calculations
Originally Posted by BetterSense
you could remove the back of your box camera and put a piece of
parchment paper, velum, wax paper inside and see where things focus.
then mark the inside of the camera / box and make a focus scale so you know
12 feet = "here"
infinity = "here"
4 feet = "here"