Archived posting to the Leica Users Group, 2011/11/03
[Author Prev] [Author Next] [Thread Prev] [Thread Next] [Author Index] [Topic Index] [Home] [Search]Now some manufacturers - Pentax to name just one - had 50mm 1.4 and 55mm 1.8 on their catalogues: which of these offered better bokeh or faster takes? I'm so happy to have a full auto mode on my camera right now. Philippe looking for aspirin and about to call Doctor Ted Le 3 nov. 11 ? 21:40, Herbert Kanner a ?crit : > After this, the dead horse should stay dead. > > While your mathematics is correct, I differ with your conclusions. > It relates to a distinction between "f numbers" and "stops". Agreed: > reciprocal of the square of the f number is proportional to the > amount of light getting through. But the question was: "What > fraction of a stop is the difference between two f numbers?". > > Now, the stops on a lens aperture ring are an arbitrary choice, > general;y meaning a factor of two in exposure. So going, say three > stops toward smaller f numbers doubles the exposure three times, or > 2 to the power 3. Now, let's think about this example of three > stops: f/4, f/5.6, and f/11, pretend we don't know that those three > numbers are engraved on the aperture ring, and ask: "How many stops > are the distance between f/4 and f/11. First, realize that the > conventional set of stops aren't exactly a factor 2 in exposure > because the lens manufacturers in their wisdom wanted the f numbers > to not have more than two digits; who would like 3.99762 engraved on > their aperture ring? So, f/4 to f/11 is actually a change in > exposure of 7.5625, close enough to 8. > > Next let's go backward, again pretending ignorance of aperture ring > numbers, and ask how many "stops" change is f/4 to f/11. take 11/4 > and square it, getting, as before, 7.5625. Take the logarithm (base > 2) of 7.5625, and you get 2.918, which is the actual theoretical > number of stops, and close enough to the three notches on the ring. > > That is the way I calculated the number of stops between 1.8 and 1.4. > > How do you calculate log(base 2) of something. Just divide its > log(base 10) by log of 2 (base 10) > > Herb > > > >> Time to beat a dead horse! If anyone is interested: >> >> It is all geometry, namely the area of a circle. >> >> For each f-stop, we have double the light. The f-stop is related to >> the size of the aperture, which is approximated as a circle. The >> amount of light coming is proportional to the circle's area, which >> you may recall is pi times the radius squared, pi*r^2. We use the >> f# for the equivalent radius. >> >> Thus, starting with f1, and r = 1, the area is pi*r^2 = pi*1*1 = pi, >> as the relative amount of light. >> >> For an amount of light 2*pi (next f-stop, double the light), pi*r^2 >> = 2 pi. Divide both sides by pi, and you get r^2 =2. r = the >> square root of 2, or 1.414? f1.4 is the next stop. >> >> This is where the 1.4 factor George mentioned comes from; the square >> root of 2 is 1.414. >> >> Next f-stop, double the light again: pi*r^2 = 2*2 pi. r^2 = 4, f2 >> is the next stop. >> >> So, if you want fractional stops: >> >> 1/3 stop: Square root of 1.333 = 1.15456 >> 1/2 stop: Square root of 1.5 = 1.22474 >> 2/3 stop: Square root of 1.667 = 1.29112 >> >> So going back to Mark's f1.4 example: >> 1/3 stop slower = 1.414 * 1.15456 = f1.633 or f1.6 >> 1/2 stop slower = 1.414 * 1.22474 = f1.732 or f 1.7 >> 2/3 stop slower = 1.414 * 1.29112 = f1.828 or f1.8 >> 1 stop slower = 1.414 * 1.414 = f2.0 >> >> >> Matt >> >> >> >> >> >> >> >> On Nov 2, 2011, at 5:32 PM, Mark Rabiner wrote: >> >> > I looked up f 1.8 vs. 1.4 thinking it was between a half and a >> quarter of a >> > stop and they are saying its 2/3rds!?!?! Anybody know that that's >> true? >> > >> > Where is there a photo calculator that tells you these things?!?!? >> >> I always thought the basic math for 1 f stop revolved around a >> factor of 1.4. >> 1.4 x 1.4 = 1.96 >> 1.8 / 1.4 = 1.29 >> >> so - yes - 2/3 would seem close enough for? >> what? I'm not sure. >> >> Regards, >> George Lottermoser >> george at imagist.com >> http://www.imagist.com >> http://www.imagist.com/blog >> http://www.linkedin.com/in/imagist >> >> >> >> _______________________________________________ >> Leica Users Group. >> See http://leica-users.org/mailman/listinfo/lug for more information > > -- > Herbert Kanner > kanner at acm.org > 650-326-8204 > > Question authority and the authorities will question you. > > _______________________________________________ > Leica Users Group. > See http://leica-users.org/mailman/listinfo/lug for more information