Archived posting to the Leica Users Group, 2002/03/26
[Author Prev] [Author Next] [Thread Prev] [Thread Next] [Author Index] [Topic Index] [Home] [Search]> What scanner manufacturers *actually* mean by Dmax is that the A/D > converters in their machines capture so many bits of information. > > For example, a scanner which can capture 12 bits is capable of representing > 4096 distinct shades of gray (ignoring color for the moment). It's > *theoretical* dynamic range is therefore > > log10(4096) - log10(1) > > Which is 3 and a bit (sorry don't have slide rule handy) it's 3.6. For n bits your formula gives TDR = log10(2^n) = n log10(2) = 0.3 n This seems to work for the few scanners I know of. Do you know if it is a universal practice? I always assumed they worked out the number of bits needed after managing to make a device with whatever dynamic range. I'm confused about what your formula means. It says that th. dyn. range in *stops* = number of bits. This means that for n stops of range, you divide each stop into 2^n/n regions: 2-stop range (0.6 of density) ==> 2 different greys per stop. 3 (0.9) ~3 10 (3.0) 100 13 (3.9) 630 I thought that visually we needed a constant number of greys-per-stop in order not to see the steps. That would suggest a theoretical dynamic range for an n-bit number of TDR = 2^n 0.3 / #vis-greys-per-stop or n proportional log(TDR) although that just doesn't seem right. Can anyone see what I've mis-understood? (This will probably be embarassing!) Confusedly yours, Michael (With apologies to all the non-mathematicians here. Less OT than lawyers though.) - -- To unsubscribe, see http://mejac.palo-alto.ca.us/leica-users/unsub.html