Anyone already figured it out?
Depending on the camera you can use a piece of ground glass to focus and mark the distance.
No need to involve the pitch; see below. Ditto for focal length.However, in order to make the scale, you would also need to know the diameter and thread pitch of the helicoid. The latter would require a disassembly, but maybe you can transfer the current scale onto paper to infer the pitch.
Beg to disagree, but in order to not leave false information on record:The spacing between the number is logarithmic -- that means 1 meter and 2 meter are further apart on the scale than 10 meters and 20 meters
+1You can do it graphically; if you measure the distance along the lens scale, around the barrel, (actual millimetres along the scale) from the infinity mark to each of the marked distances, you can plot a graph (on squared paper or in Excel) of the marked distance on the scale in feet or metres (say d) against the physical distance along the scale in millimetres (say x). It should be a reciprocal curve (d proportional to 1/x); and you could work out its exact formula; but you might as well just draw the graph, and read off from it the scale distance for the extra marks you want to make. I think it'll be accurate and precise enough for distances you're going to estimate.
These are based on Newton's equations, with distances measured in object and image spaces from heir respective foci. A typical MF folder with front element focusing has the distance measured from the lens; a typical "modern" 35mm camera has the distance measured from the film plane, with significant differences between the two cases at short distances; for instance, for 1:1 reproduction, the subject distance is 2f from the lens, and 4f from the film plane. And then there are real-world thick optical systems.Assuming that you can accurately measure or estimate the subject distance s (subject plane to first nodal point of the lens), let
f = focal length of lens
s = subject distance
Δ = the distance the lens is displaced forward of its infinity position.
Δ = (f^2)/(s – f)
At infinity, Δ = 0
Beg to disagree, but in order to not leave false information on record:
a) The scale is not logarithmic.
b) If the scale were logarithmic, 10m-20m would be just as far apart as 1m-2m;, they are much closer.
c) The scale is such that (to a good enough approximation) the inverses of the marked distances are in a linear progression; i.e. for example, 1m (1), 1.5m (2/3), 3m (1/3) and infinity (0) are equally spaced.
Now I have a piece of 8x10 ground glass, how do I cut it so I can use it on the 35mm?
I read somewhere about sharpie fix for the faded RF patch since my thread about fix lens RF. I'm not sure if it works.
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