Archived posting to the Leica Users Group, 2011/11/03
[Author Prev] [Author Next] [Thread Prev] [Thread Next] [Author Index] [Topic Index] [Home] [Search]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.