OK - let's see what we've got here:
Wilt wrote in post #10418983
False logic, Xarqi. You could represent range of full black to full white with 256 levels (8 bit), or 4096 levels (12 bits), or even 65535 levels (16 bits).
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The difference is the 'granularity' of the intermediate levels of gray, but not necessarily the magnitude of the range.
False logic, Xarqi. You could represent range of full black to full white with 256 levels (8 bit), or 4096 levels (12 bits), or even 65535 levels (16 bits).
<snip>
The difference is the 'granularity' of the intermediate levels of gray, but not necessarily the magnitude of the range.
Fair enough - I think that was the point I wanted to make, but I went a bit sideways considering mapping to a space with poorer resolution (i.e. fewer bits).
tonylong wrote in post #10419675
So, the question I have is "what does that 255 value mean?"
That is indeed the question. For the data given, 0 and 255 represent the lowest and highest levels of pixel brightness as far as photoshop is concerned, but the difference between them in real terms is unknown. We can learn nothing about the sensitivity of the sensor, nor of its range of sensitivities, that is its DR, by examining such data.
So, the question I have is "what does that 255 value mean?"
That is indeed the question. For the data given, 0 and 255 represent the lowest and highest levels of pixel brightness as far as photoshop is concerned, but the difference between them in real terms is unknown. We can learn nothing about the sensitivity of the sensor, nor of its range of sensitivities, that is its DR, by examining such data.
Wilt wrote in post #10420119
For 8 bits, 255 merely mean 'max white'. For 12 bits, 4096 similarly means 'max white'. For there to be more dynamic range, the distance between max black and max white needs to be broadened, and if that ever happens it may make sense to increase the number of levels beyond 255 or beyone 4096.
For 8 bits, 255 merely mean 'max white'. For 12 bits, 4096 similarly means 'max white'. For there to be more dynamic range, the distance between max black and max white needs to be broadened, and if that ever happens it may make sense to increase the number of levels beyond 255 or beyone 4096.
OK. That would give greater resolution of brightness within the dynamic range present, certainly, but no information about exactly how broad that range is.
DrPablo wrote in post #10420318
That doesn't make sense. A "stop" is not a physical unit. It's a statement of relativity.
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There is an infinite continuity between those extreme values. You can encode them with two values -- 0 and 1. If you do this, there will be only black and white with no intermediates.
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But the number of intermediate values is not what is meant by dynamic range. It's the ability to capture the extremes.
That doesn't make sense. A "stop" is not a physical unit. It's a statement of relativity.
<snip>
There is an infinite continuity between those extreme values. You can encode them with two values -- 0 and 1. If you do this, there will be only black and white with no intermediates.
<snip>
But the number of intermediate values is not what is meant by dynamic range. It's the ability to capture the extremes.
Agreed.
Let's see where I went wrong:
Originally Posted by DrPablo :
It seems to me that whether 8 bits, 16 bits, or 32 bits, and whether linear or curved, dynamic range is delimited by 00000000 and 11111111. You can add in an infinite number of intermediates with greater bit depth, but that doesn't make the highest value "brighter" or the lowest value "darker".
To which I replied:
If, for example, the dynamic range in question is 10 stops, at least 10 bits are required in order to uniquely represent all of the possible brightness levels in a linear space. Mapping those 10 stops of DR to a space of only 8 bits means that some compression or clipping must occur. In fact, losing those 2 bits reduces the DR by a factor of 4!
Yup - completely wrong. I've confused the precision of representation of the points between 'black' and 'white', with the measurement of the distance between them.
My apologies and thanks to those more on the ball than me!
I should have posted the graph and gone to bed as planned!
