Archived posting to the Leica Users Group, 2004/07/07
[Author Prev] [Author Next] [Thread Prev] [Thread Next] [Author Index] [Topic Index] [Home] [Search]On 7/7/04 6:14 PM, "Henning Wulff" <henningw@archiphoto.com> typed: > At 11:25 PM +0200 7/7/04, Jean Louchet wrote: >> Hi all, >> >> Beware, the calculation maniac is back on line :-) >> >> Apart the Descartes formula, the second important formula is the "Setala >> formula" which allows to calculate the depth of field, the diameter of the >> blur circle etc. (According to the "photo club de Bievres", Mr Setala was >> a Leica User from Finland, and the inventor of the "depth of focus" scales >> one now finds on almost all lenses) >> >> >> Its parameters are: >> the diameter of the blur circle: Delta >> the diameter of the diaphragm: Diaph >> the focal distance: f >> the distance on which the lens is focussed: z >> the actual object's distance: d >> >> Setala's formula is: >> Delta = Diaph * f * abs(1/z - 1/d) >> >> It is only true when z is much greater than f (always the case except in >> macrophotography) >> >> Let's take as an example a 35mm lens on a leica. As the focal length f >> (35mm) is about equal to the width of the image (36mm), if we want to >> photograph a vertical wall at distance d, the part of the wall which >> projects onto the side of the picture will be at an actual distance z >> (measured by the rangefinder before turning the camera) such that >> >> d^2 = z^2 + e^2 where e is half the width of the part of the wall that >> will be visible on the picture (sorry I can't make a drawing here, but >> this is Pythagore's theorem), and e=z/2 (because we have a 35mm lens and a >> 18mm half-width of the picture). Thus we can calculate 1/z - 1/d, >> eliminating d (d = z * (sqrt(5)/2 = z*1.118) gives >> 1/z - 1/d = 1/z *(1 - 2/sqrt(5)) = 0.105/z >> >> At aperture f/2 we have (in millimeters) >> Diaph = 17.5mm >> and >> Delta = 17.5 * 35 * 0.105 * (1/z) (z also in millimeters!) >> Delta = 64.3/z >> >> This is the diameter of the "blurring" on the picture of the wall if we >> rely on the "focus then reframe" technique. >> >> At infinity (z = inf) the picture will be (in theory) perfectly sharp. >> At z = 10 metres we have Delta = 64.3/10000 = 6.4 microns = perfectly >> sharp in practice. >> At z = 5 metres, Delta = 13 microns = very sharp (would still resolve 80 >> lines/mm). >> At z = 2 metres, Delta = 32 microns = usually considered "good" or >> "acceptable" (30 lines/mm). >> At z - 1 metre, Delta = 64 microns :-( it is worth to make the focussing >> correction. >> >> The correction coefficient on the edges is 1.118 (say 12%). If one uses >> the rangefinder on an object and wants this object to be sharp when it >> gets on the side of the picture, he will have to SUBTRACT 12% to the >> distance given by the rangefinder. >> >> The same formulas with aperture 1.4 would give: >> At z = 10 metres Delta = 9 microns "very sharp" >> At z = 5 metres Delta = 18 microns "sharp" >> At z = 2 metres Delta = 45 microns "fair" :-| >> >> TO SUMMARISE: >> with a 35mm, at aperture 2 "focus then reframe" is OK from 2 metres >> at aperture 1.4 "FTR" is OK from 3 metres. >> >> At shorter distances, one has to reduce the focussing by, say, 10 - 12%. >> >> I did not consider focussing on an object which will finally be right in a >> corner (no practical interest). >> >> OF COURSE we luggers will take much, much better pictures now :-) >> >> If anyone is interested in having the equivalent results with a 50mm, let >> me know (private e-mail), I may well open another page on my web site. >> >> Jean > > While these calculations and formulae are correct, the point is still > moot due to the almost universal field curvature, especially of fast > lenses, at shorter distances. > > After these calculations, you won't be closer to the truth or focus, > _and_ your subject will be gone to sleep or just plain gone. This might as well be computer code. IF you were a tree would you know you were in a forest? Mark Rabiner Photography Portland Oregon http://rabinergroup.com/