Yes, you are correct. It doesn't. One of the problems of this thread is that Jason was never talking about the dynamic range of the actual original scene that might be photographed, but everybody presumed he was, including me at first.
Yep, that was the topic of the original poster. Things got on a small tangent as they tend to do. The poster's original question was answered in other parts of the thread, so no big loss to him I hope.
I've got nothing against using that word to describe it... we just need to be clear what we're talking about, and understand that there are a dozen points along the path of digital photography to measure dynamic range. As you know I like a different word to describe it, but I can accomodate.(see, I didn't say " Dynamic Range"
Agreed, I made 3 and 5 bit samples files above to show this.
Correct.
Sorry about that, it was easy to miss. Here's my first post, where this all got started:
Luckily, the manufacturers tend** to use a format that efficiently matches their camera (so far). For instance, we've yet to see Canon come out with 16bit RAW files for the G3. Why? No need, just a waste of space with that sensor. They'd still have less true dynamic range than say a 1D (any way you measure it) We can't say the same for the $29 scanners obviously.
Well, let me constrain myself to saying that you got this one wrong to. A pH of 0 would mean that the input (A) to log10 was 1, so no formula is blown!
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--I wouldn't say I was polite, but I was polite for me.
I believe you've accurately described a big part of our differences in views. However, believe me, I don't have any misconceptions about the topic. I'll try to describe below.
What you've done here, is sampling "digital" data. There was no analog to digital conversion required. What this gives you, is "perfect" data.
Once again, I feel I should point out, that the difference (ratio) between black and white in a file format is infinite (actually undefined because you're dividing by zero). Does this mean that the scene had infinite dynamic range? Nope, because there is sampling error (which I think you've referred to) This sampling error happens to go back to the Dmin and Dmax values, but we should avoid that for now.
Yes, the luminosity to be seen by our eyes absolutely requires a digital to analog conversion somewhere. As I mentioned, since I'm talking about the digital format, this conversion doesn't apply. I hate to repeat myself, but I feel this is valid for the same reason that Phil's signal to noise ratio is a valid measurement (although that's not the only reason)
No, no, no. That's not what I am said. If you input a 1 into that function, you get a PH of 0. No big deal. However, you can't input a zero into that function, or it blows up. It blows up for precisely the same reason you can not divide by zero.
Yeah, a bit stubborn (Irish) but I know this stuff.
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There is a concise answer to a concise question. I'm not sure you've even narrowed down your question to a particular camera? Digital photography is a wide field.
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I don't know if this is the same concept but some CCD manufacturer was talking about making a sensor that had groups of pixels with different sensitivities. You could have a pixel set to ISO 100, and two pixels next to it set to 400 and 1600. After exposure, the image is built from the pixels that were properly exposed and the rest are thrown away. This was intended for situations where resolution wasn't as important as having proper exposure for a wide range of lighting situations (like a security camera).
