Archived posting to the Leica Users Group, 2004/07/07
[Author Prev] [Author Next] [Thread Prev] [Thread Next] [Author Index] [Topic Index] [Home] [Search]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. -- * Henning J. Wulff /|\ Wulff Photography & Design /###\ mailto:henningw@archiphoto.com |[ ]| http://www.archiphoto.com