Actually, I only recently found out that digital did not expose like film, when I began to look around for the reasons my test exposures looked the way they did after buying my first DSLR.
I knew how film exposed and how to optimize my results, but not the math or math terminology. That stuff bores me, and beyond that it does not help me more than the simple explanation like "digital has less and less range as exposure drops into the shadows". -Aaron -----Original Message----- From: "Anthony Farr" <[EMAIL PROTECTED]> Subj: RE: Holy Crap -- Pentax 10MP body Date: Wed Aug 16, 2006 2:11 pm Size: 2K To: "'Pentax-Discuss Mail List'" <pdml@pdml.net> Aaron, Fair comment. I don't understand the math either, not enough to express it accurately, but I do understand the concept behind it and how it differs from the concept of digital exposure. For anyone who's still in the dark about why digital exposure is different to film exposure, here comes my attempt to explain. With negative film, every extra stop of exposure ideally gives the same increase in density to the developed film. Thus a stop in the shadows will, ideally, have the same density range as a stop in the highlights. However, when the film was exposed, every extra stop doubled the total quantity of light given up to the next lower stop. On a 12 stop scale, if it took n lumens to expose the first stop of density then it would take 2,048n lumens to expose to the 11th stop of density, and the 12th stop would require an additional 2,048n lumens for a total of 4,096n lumens. AND YET, and this is the point, the darkest stop that only represented n lumens will raise the negative density by the same amount as the brightest stop that represents an increase of 2,048n lumens. That is because a linear density increase is achieved by a log Exposure increase. But digital exposure is linear. Working with 12 bits, the darkest stop is from level 0 to level 1, and the brightest stop is from level 2,047 to level 4,095. The brightest stop gets you half of all the available brightness levels of the output file, the darkest stop gets only 1/2,048 as much. Is it any wonder that the shadows of a digital shot can be lost in background noise? It's easy to imagine that a non-linear A/D conversion could be better, even if it wasn't fully logarithmic. BTW, my guess is that non-linearity would be a function of the A/D conversion, not a characteristic of the sensor. None of which means that Pentax or anyone else is about to announce anything of the sort ;-) Regards, Anthony Farr > -----Original Message----- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Aaron > Reynolds > Sent: Thursday, 17 August 2006 2:18 AM > To: Pentax-Discuss Mail List > Subject: RE: Holy Crap -- Pentax 10MP body > > Why would I need to understand the math behind it to run a lab? > > -Aaron > > -----Original Message----- > > From: "Anthony Farr" <[EMAIL PROTECTED]> > Subj: RE: Holy Crap -- Pentax 10MP body > Date: Wed Aug 16, 2006 11:51 am > Size: 1K > To: "'Pentax-Discuss Mail List'" <pdml@pdml.net> > > You don't understand logE? You do/did operate a lab, did you not? > > Regards, > Anthony Farr > -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net -- PDML Pentax-Discuss Mail List PDML@pdml.net http://pdml.net/mailman/listinfo/pdml_pdml.net