Sorry if this seems impertinent, but this discussion seems to have got bogged down in wood and trees.
David Kay wrote: > Now, divide this number by a factor of 2 and keep halving the remainder. > Each division represents one bit or one half the light, until you reach the > noise floor. (It's here where good data is polluted by random "noise" from > the electronic circuits and/or sources of heat within the device.) > > Surely it's obvious a 16-bit device can record more f-stops of subject > information than an 8-bit device? But, surely, as photographers, it should be only too easy to understand that if an 8 bit device stops down in 1 stop increments, then a 16 bit device is stopping down in half stop increments. i.e. the number of actual range (in f stops) stays the same... it is just the divisions that are finer. We have twice as many tones but they fall within the same range. Instead of black, grey, white, (3 steps) we have black, nearly black, dark grey, light grey, nearly white, white (6 steps... but black is still black and white is still white (whatever 'black' and 'white' happen to be for that particular physical system)). If we convert an image in photoshop from 8 to 16 bit, we do not get a load more tonal range, we just get more subtlety within the same range. The same goes for capture surely (except that we might squeeze ad extra half stop of information at either end of the range that would have been 'lumped in' with a cruder 8 bit system). I hesitate slightly to send this, because even though this seems blindingly obvious to me, the debate has gone on so long I that I am beginning to wonder whether i am, infact, deluded/bonkers. Giles Stokoe photographer/photojournalist. See some images at http://www.stokoe.co.uk =============================================================== GO TO http://www.prodig.org for ~ GUIDELINES ~ un/SUBSCRIBING ~ ITEMS for SALE
