I finally have time to catch on to this thread that I initated. I also see that the list mom have ordered no more talking on this subject. However, I'd still like to get an answer on it.
Based on the following post: (you wrote a lot of posts) I get to see your argument, and I thought what you wrote made sense. But then I read the post made by Giles Stokoe, and I *think* I see a step you missed out. "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))." Do you see what I see from Giles' post? I think the key here is, that: "if an 8 bit device stops down in 1 stop increments, then a 16 bit device is stopping down in half stop increments." Martin Evening should give his comments on this, don't you think? On 11/3/04 6:21 pm, "David Kay" <[EMAIL PROTECTED]> wrote: > Douglas Burns wrote:> > >> Have been following this thread for a while, and staring to get a bit >> confused. With all due respect to David Kay I just don't follow his logic. I >> agree with all the other posts that the dynamic range is simply the >> difference between the highest and lowest values recordable (hence the term >> "range"). This is something I was taught on my first day of college (and the >> reason I'm writing this post in simplistic terms with the minimum of >> jargon!). The bit rate simply determines the number of discrete steps within >> that range that the device can record. > > Am I deep in the hot cocky cack or what? > At the risk of being banished to the colonies forever... > > Can we agree that Dynamic Range is the ratio between the lightest and > darkest Tonal Values which can be detected and/or reproduced? > Any argument here? > > And can we agree that Bit Depth is the number of Tonal Values in which a > Contrast Range is divided? I certainly hope so! > > Then this leaves the issue of how a CCD responds to light reflected from the > subject. > > Electrons are generated proportionate to the intensity of light. > So when twice as many electrons are generated, up to the saturation point of > the CCD, we have one f-stop more exposure. And when half as many are > generated, we have one f-stop less exposure. Agreed? > > I'm not counting electrons here, here's an arithmetic example... > In an earlier post, I suggested that a two-stop under-exposure with an 8-bit > device would provide a range as follows: 64, 32, 16, 8, 4, 2, 1. > > Now with a 16-bit device, here's what we get with a two-stop under-exposure. > 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1 > > Now on the basis that light generates electrons proportionately, then it's > clear to me that the 16-bit device is a clear winner with 8 f-stops more > being resolved! > > I know the CCD is an analogue device and it's here where voltages are > created to represent the various light levels. However, in a capture device, > the A/D converter is also of primary importance. It must be be capable of > converting every voltage at each pixel from analogue to digital data. I have > heard that a manufacturer can save a lot of money by fitting a lower > specified A/D converter. I believe I can see the effect of this as less > "sparkle" in an image! And you can actually see this when images are > compared side by side that are taken under identical conditions! Most images > seen in isolation look reasonable until you compare two devices this way. > Some photographers don't pay more for this level of quality. > > And finally, there's the frequency at which the data is read out from the > CCD. Faster shooting speeds implies that the read-out of data is > over-clocked and therefore of lesser quality than provided by the optimum or > slower frequency. The result is more noise. > > Phew. I rest my case your Honour. > > Am I to be banished down-under forever? > > > Best regards, > > David Kay > > > =============================================================== > GO TO http://www.prodig.org for ~ GUIDELINES ~ un/SUBSCRIBING ~ ITEMS for SALE =============================================================== GO TO http://www.prodig.org for ~ GUIDELINES ~ un/SUBSCRIBING ~ ITEMS for SALE
