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Photo Forum / Film Photography / Medium format / January 2007how to read the distance scale on the lens?
think of a lens, which gets distance scale on the focuing knob/ring, there is a distance mark on it, say m feet, next to the marke there is is another mark that is n feet. so, my question is where is the x feet, supposing m < x < n and ( x - m ) / ( n - x ) = k ? i believe there is some math right here, but i dont know. thanks. - woody > think of a lens, which gets distance scale on the focuing knob/ring, > there is a distance mark on it, say m feet, next to the marke there is [quoted text clipped - 6 lines] > - > woody It is not always a linear scale. You just have to guesstimate. My own solution for this is to stop down to about f/8, to give me extra depth-of-field, in case I was off on the distance measurement. I've never had a problem when I used that technique. You just want to stay away from shooting wide-open, as this giges you no wiggle room. > think of a lens, which gets distance scale on the focusing knob/ring, > there is a distance mark on it, say m feet, next to the mark there is > is another mark that is n feet. so, my question is where is the x feet, > supposing m < x < n and ( x - m ) / ( n - x ) = k ? i believe there is > some math right here, but i dont know. 2xy/x+y gives the mark half way between two marks x and y. So for the mark half way between 30 and 45 units (feet of metres) we get 2 x 30 x 45 / 30 + 45 = 36 Also a mark half way between infinity and mark x is 2x. So the position halfway between 50 feet and infinity is 100 feet.  Signaturepirwin@ktb.net Just to clarify .. (I'm not really good at math) >> think of a lens, which gets distance scale on the focusing knob/ring, >> there is a distance mark on it, say m feet, next to the mark there is [quoted text clipped - 3 lines] > > 2xy/x+y gives the mark half way between two marks x and y. 2xy/(x+y) > So for the mark half way between 30 and 45 units (feet of metres) > we get 2 x 30 x 45 / 30 + 45 = 36 2 x 20 x 45 / (30 + 45) = 2700/75 = 36 > Also a mark half way between infinity and mark x is 2x. > > So the position halfway between 50 feet and infinity is > 100 feet. > Just to clarify .. (I'm not really good at math) and lousy at typing accurately ... >> So for the mark half way between 30 and 45 units (feet of metres) >> we get 2 x 30 x 45 / 30 + 45 = 36 > > 2 x 20 x 45 / (30 + 45) ^^ 2 x 30 x 45 / (30 + 45) > = 2700/75 > = 36 [quoted text clipped - 3 lines] >> So the position halfway between 50 feet and infinity is >> 100 feet. > > think of a lens, which gets distance scale on the focusing knob/ring, > > there is a distance mark on it, say m feet, next to the mark there is [quoted text clipped - 14 lines] > -- > pirwin@ktb.net thank you give the math. now we can get position of 1/4, 1/8, ..., 1/(2^n) > > > think of a lens, which gets distance scale on the focusing knob/ring, > > > there is a distance mark on it, say m feet, next to the mark there is [quoted text clipped - 14 lines] > thank you give the math. now we can get position of 1/4, 1/8, ..., > 1/(2^n) Assuming I haven't made any errors, and assuming that rotation of the lens barrel is proportional to the movement of the lens, the mark for x is found as follows: Calculate k = [n(x - m)] / [x(n - m)] (should give 0 <= k <= 1 for m <= x <= n) Now if the distance from m to n on the barrel is 1, the distance from m to x is approximately k. For example, taking m = 30, n = 45 and x = 40, k = (45*10)/(40*15) = 0.75 of the way from the 30 mark to the 45 mark. When the focal length of the lens is not small with respect to the subject distances then use: k = [(n - f)(x - m)] / [(x - f)(n - m)] SignatureMatthew Winn [If replying by mail remove the "r" from "urk"] > > > > think of a lens, which gets distance scale on the focusing knob/ring, > > > > there is a distance mark on it, say m feet, next to the mark there is [quoted text clipped - 21 lines] > Calculate k = [n(x - m)] / [x(n - m)] > (should give 0 <= k <= 1 for m <= x <= n) it seems correct, but can you provide a mathmatic prove? > Now if the distance from m to n on the barrel is 1, the distance from > m to x is approximately k. [quoted text clipped - 6 lines] > > k = [(n - f)(x - m)] / [(x - f)(n - m)] this looks more complex than the one above, and it due a reasoning. Matthew, i found the contraction. you said, half way between n and m is , (2nm)/(n+m) --- (1) ( m < n ) and, you also said, for a x, if m < x < n, then x is on the k% away from m towards n, where k is, (n( x - m ))/(x(n - m) --- (2) so, let x = (2nm)/(n+m) in (2), we should expect the result is 1/2, but it is not. this proved that either (1) or (2) is not correct. - woody > Matthew, > > i found the contraction. you said, I wasn't the one who said it. > half way between n and m is , > (2nm)/(n+m) --- (1) ( m < n ) [quoted text clipped - 5 lines] > so, let x = (2nm)/(n+m) in (2), we should expect the result is 1/2, but > it is not. this proved that either (1) or (2) is not correct. Sure they are: [ n (2nm/(n+m) - m) ] / [ (2nm/(n+m)) (n - m) ] Multiply top and bottom by (n+m): [ n (2nm - m(n+m)) ] / [ 2nm (n - m) ] Multiply out the brackets: [ 2n²m - n²m - nm² ] / [ 2n²m - 2nm² ] = [ n²m - nm² ] / [ 2 (n²m - nm²) ] Divide top and bottom by (n²m - nm²): 1/2 Alternatively you can work backwards from k = 1/2 and find x: [ n (x - m) ] / [ x (n - m) ] = 1/2 Multiply out: 2nx - 2nm = xn - xm Then: xn + xm = 2nm x (n + m) = 2nm x = 2nm / (n+m) SignatureMatthew Winn [If replying by mail remove the "r" from "urk"] > > Matthew, > > [quoted text clipped - 47 lines] > Matthew Winn > [If replying by mail remove the "r" from "urk"] sorry, you are right. i made error in calculation. Does anybody really think the markings on a lens are accurate? Just consider, ANSI standards allow a 3% variation in focal length and I don't believe the manufacturer engraves focusing distances individually to allow for them. Recently, Bob G <bobjames27@sbcglobal.net> posted: > Does anybody really think the markings on a lens are accurate? That depends on the lens. The focusing scale on my 480mm APO Nikkor is over 18" long (mounted parallel to the film plane), and has separate scale markings for apertures from f 4.5 through f 32. It's *very* accurate. Of course, you can't hang this lens on a MF camera... ;-) > Just consider, ANSI standards allow a 3% variation in focal length and > I don't believe the manufacturer engraves focusing distances > individually to allow for them. I agree that the markings are just for estimations, and while these calculations may be intriguing, they shouldn't be considered to be practical for critical focus. Neil > Does anybody really think the markings on a lens are accurate? > > Just consider, ANSI standards allow a 3% variation in focal length and > I don't believe the manufacturer engraves focusing distances > individually to allow for them. The accuracy of lens markings is sometimes very good. There are several ways of dealing with normal variations in lenses. One way is to make several slightly different barrels and put the lens in the one which matches most closely. So Leitz made separate barrels for a 50/2 lens for lenses of actual focal length of 51.6mm, 51.9mm, and 52.2mm. Another way is to adjust the element spacing on the lens to adjust the focal length. This is probably suboptimal for optical reasons, but is probably less expensive. This seems to have been common for scale focussing cameras using triplet lenses. Check how accurate your lens markings are at 1 metre. Shoot wide open at a distance measured with a tape measure and check the result, The distance is usually from the film plane to the object, but on some cameras (such as my super-8 camera) it can be from the front of the lens. There is often a mark on a camera consisting of a circle with a line through it showing where the focal plane is. Such a mark should indicate that the manufacturer intends markings to be accurate enough to use. If 1 metre and the infinity marks are correct, the marks in between should also be correct. Peter. Signaturepirwin@ktb.net > > Assuming I haven't made any errors, and assuming that rotation of the > > lens barrel is proportional to the movement of the lens, the mark for [quoted text clipped - 17 lines] > > this looks more complex than the one above, and it due a reasoning. For both, start off with the basic lens equation for subject distance u, image distance v, and focal length f: 1/u + 1/v = 1/f Now consider subjects and images as follows, where the image of subject m is shown as m' (usual caveats for ASCII art apply): ^ n x m | n' x' m' ------------------------------------------------|------------- | v If the lens focusing mechanism has rotation of the lens barrel proportional to the distance moved then the ratio of distances m'-n' and m'-x' will be equal to the ratio measured on the focusing scale. Therefore we want to calculate the ratio (x'-m')/(n'-m') to find the position of the mark for distance x. We know that 1/m + 1/m' = 1/x + 1/x' = 1/n + 1/n' = 1/f from the lens equation given above. We can rewrite this to give: m' = mf / (m - f) x' = xf / (x - f) n' = nf / (n - f) Substituting these into (x'-m')/(n'-m') is left as an exercise for the reader. This results in the second equation I originally gave. The first is a simplification for the case where f is small compared with the other distances involved. SignatureMatthew Winn [If replying by mail remove the "r" from "urk"] > think of a lens, which gets distance scale on the focuing knob/ring, > there is a distance mark on it, say m feet, next to the marke there is [quoted text clipped - 6 lines] > - > woody The scales are logarithmic if that helps. |
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