Maybe I need to see a shrink, but if you saw me mentioning the focal length, I suggest you re-read my message and keep your imagination in check.
I was talking about *dimensions*: d has dimensions of *length* (m, yards, leagues) c has dimensions of *length* (mm, inches, miles, a.e) m is dimensionless (magnification) f is either dimensionless (Mark, me), or has dimensions of aperture, that is, diameter, i.e. length (cm, feet, parsec) (m+1)/m^2 is dimensionless. f*c MUST have dimensions of length, to be consistent with the fact that d ~ f*c* [dimensionless constant] therefore, f *must* be dimensionless, in other words, it is an F-stop, not the actual aperture diameter. that is unless you measure DOF in square feet or magnification in inches. but that would be a totally different subject <vbg>. is that clear *now*? best, Mishka > copied and pasted from Mark's post: > > > d = 2fc*(m+1)/m^2 > > > > where d=dof, f = f stop, c = circle of confusion > > size, and m = magnification. > > Now, if anyone can see a value for focal length in > the above formula, he needs to talk to a good shrink > about his over active imagination. > > > My reply was change "f" to "a" (aperture-diameter) > to correct the formula. > (Though in the copy I have f = aperture, which is > the same as using a). > > Us people who know everything are getting real tired > of you people who think you know everything <GRIN>. > > > Ciao, > Graywolf ----- Original Message ----- From: "mishka" <[EMAIL PROTECTED]> > if you look at the formula, as Mark and I wrote it, > the DOF has dimensions of length. if you change f- > stop to the actual diameter, the dimension will > become length^2, which cannot be. > > best, > Mishka
