Archived posting to the Leica Users Group, 2011/11/03
[Author Prev] [Author Next] [Thread Prev] [Thread Next] [Author Index] [Topic Index] [Home] [Search]Well its not as if there are that many pictures which cant be taken with an 1.8 that can be taken with a 1.4. But with all this conversation I saw that I only have one 1.4 lens and that's my 28. So I put that on tonight and it weighs a full pound and kept thinking I'd bought apples or bananas and had them also in my bag and had forgotten to take them out. It was really a lead weight. Tomorrow I put on my 50 1.8 which weights a third of the 28 1.4. -- Mark R. http://gallery.leica-users.org/v/lugalrabs/ > From: Jim Nichols <jhnichols at lighttube.net> > Reply-To: Leica Users Group <lug at leica-users.org> > Date: Thu, 3 Nov 2011 16:49:05 -0500 > To: Leica Users Group <lug at leica-users.org> > Subject: Re: [Leica] 1.8 vs. 1.4!?!? > > For sharper images, use the 55mm 1.8. I have them both. I find that > sharpness trumps speed and bokeh in producing images that I am happy with. > > Jim Nichols > Tullahoma, TN USA > ----- Original Message ----- > From: "philippe.amard" <philippe.amard at sfr.fr> > To: "Leica Users Group" <lug at leica-users.org> > Sent: Thursday, November 03, 2011 4:07 PM > Subject: Re: [Leica] 1.8 vs. 1.4!?!? > > >> 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 >> >> _______________________________________________ >> Leica Users Group. >> See http://leica-users.org/mailman/listinfo/lug for more information >> >> > > > > _______________________________________________ > Leica Users Group. > See http://leica-users.org/mailman/listinfo/lug for more information