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Thread started 25 Sep 2008 (Thursday) 17:17
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Why don't I get more light levels in RAW by using 16 bit instead of 8?

 
not_this_punk
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Sep 25, 2008 17:17 |  #1

Hello.

I understand that 8bit/channel means that each RGB channel has 256 luminance levels (don't know how else to call them) associated to it. I know that my 30D captures 12 bit/channel RAW and if I select 16bit/channel in ACR I may use the full 4096 luminance levels for every channel. OK, that being said, shouldn't I notice a difference in shadow/highlight clipping?

Here's what I think: Let's assume that you have 4 light levels in your picture (2 bit encoding). The first and the 2nd are clipped (value=0), the 3rd and 4th are not.
Now if you'd have 8 light levels (3 bit encoding) the 5th, 6th, 7th and 8th levels would not be clipped, obviously. What about the first 4? Well, maybe just the first 3 levels would be clipped, but the 4th wouldn't be. This could happen because "half" of the 2nd level in the previous 2 bit encoding would have been clipped, but because this was digital encoding, the device couldn't have "known".

So, based on my example, it's safe to say that increasing the number of bits used to encode the image, you'd have a smaller and smaller part of your image being clipped (so a higher dynamic range).

So what am I doing wrong? I see absolutely no difference in ACR between 8bit and 16 bit(12 bit) with both shadow and highlight clipping warnings. And considering the fact that 8 bit means 256 levels and 12 bit means 4096 levels, I'm pretty sure I should have seen a difference. It's absolutely impossible to be no difference in clipped areas because if the encoding used would be of n bits, where n is infinite, than the infinite sized picture would have approximately 0 clipped pixels.

Phew, sorry for such a boring question and for my broken English, but an answer would be very, very appreciated:D

EDIT: I just now realized that the RGB indicator beneath the histogram in ACR shows only 256 levels...So that means it still shows 8 bit even if 16 bit is selected...so...uh...I​'m lost...


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tim
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Sep 25, 2008 17:58 |  #2

ACR shows the rendered RAW using all the bits available, it's only when you take it into PS you MIGHT notice a difference. Take an image in in 8 bit, then open in ACR again and open 16 bit. Copy and paste one onto the other and change the blend mode to difference. Anything not black is different, but it will be VERY subtle. You might notice more if you adjusted the color of each layer just a tiny bit, as the skewing of the color will be more noticeable in 8 bit.


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Grimes
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Sep 25, 2008 18:03 |  #3

Not sure I understand your question. It's key to remember that when editing in 16bit, you are not gaining any dynamic range vs 8bit, you are just getting more slices of the same pie.

I'm not surprised that you don't see a difference between 16 and 8 bit. The advantage to 16 bit is that there is less rounding during computation as you mentioned, leading to less chance of posterization and other types of image degradation.


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René ­ Damkot
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Sep 26, 2008 00:43 |  #4

Grimes wrote in post #6381003 (external link)
Not sure I understand your question. It's key to remember that when editing in 16bit, you are not gaining any dynamic range vs 8bit, you are just getting more slices of the same pie.

Yep, it's not more "reach", just "finer steps".


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Sep 26, 2008 01:15 |  #5

Yes, as said above, it's like having your money in thousand dollar bills or in pennies. You still have the same amount.

Even 8 bits' 256 levels is more than the human eye can distinguish, which is around 200 levels, so you wouldn't see a difference in greater bit depths. However, as the image is edited levels are lost and the more extreme or complex the editing, the more levels are lost. 8 bit can soon drop down to that 200 levels danger point. In 16 bit you can do the editing and even when the image is eventually converted to 8 bit for monitor or print, you will still have close to 256 levels,


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not_this_punk
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Sep 26, 2008 02:53 |  #6

OK, but here's another one:
8 bit means each pixel has a value between 0 and 255.
12 bit means each pixel has a value between 0 and 4095.
16 bit means each pixel has a value between 0 and 65535.

So why on Earth the RGB indicator beneath the histogram shows a maximum level of 255 on both 8 bit and 16 bit modes? It should display a value between 0 and 65535 in 16 bit mode, right?

About the dynamic range issue, yeah, you don't have a necessarily higher DR, but look at it this way: in 8 bit, 0 and 255 are clipped. In 12 bit, 0 and 4095, so one step in 8 bit equals 16 in 12 bit. In 8 bit, a pixel may be clipped because it's closer to 0 than it is to 1. But this issue could be solved by having more steps available. So the same pixel, in 12 bit, clearly has a value anywhere between 0 and 7 (1/2 of 15), but being less than half, the analog to digital conversion made it 0 and not 1 in 8 bit mode. But in 12 bit you may have a value of 3 or something...


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Damo77
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Sep 26, 2008 05:23 |  #7

not_this_punk wrote in post #6383442 (external link)
OK, but here's another one:
8 bit means each pixel has a value between 0 and 255.
12 bit means each pixel has a value between 0 and 4095.
16 bit means each pixel has a value between 0 and 65535.

So why on Earth the RGB indicator beneath the histogram shows a maximum level of 255 on both 8 bit and 16 bit modes? It should display a value between 0 and 65535 in 16 bit mode, right?

4095 and 65535 is equivalent to 255 in 8-bit. It would be very very confusing if Adobe did things the way you've described. Much better to stick to the standard.

not_this_punk wrote in post #6383442 (external link)
About the dynamic range issue, yeah, you don't have a necessarily higher DR, but look at it this way: in 8 bit, 0 and 255 are clipped. In 12 bit, 0 and 4095, so one step in 8 bit equals 16 in 12 bit. In 8 bit, a pixel may be clipped because it's closer to 0 than it is to 1. But this issue could be solved by having more steps available. So the same pixel, in 12 bit, clearly has a value anywhere between 0 and 7 (1/2 of 15), but being less than half, the analog to digital conversion made it 0 and not 1 in 8 bit mode. But in 12 bit you may have a value of 3 or something...

Come on man, go out and have a beer or something. Honestly, you're worrying about something that's SO insignificant that it practically doesn't exist.


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agedbriar
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Sep 26, 2008 11:14 |  #8

Damo77 wrote in post #6383703 (external link)
Come on man, go out and have a beer or something.

I'm afraid that wouldn't do. ;)

Some of us just like to know how things work. I understand the OP's deduction and to me his question is interesting, but unfortunately I have no answer.

The larger and smaller slices are exactly the point the OP is making. If you cut the pie into larger slices and then found out that one slice is burnt up, by cutting the burnt slice into smaller slices, you should be able to eat some of that part of the pie that would otherwise get entirely discarded.




  
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Grimes
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Sep 26, 2008 17:07 |  #9

I think I see where the confusion is - and Damo77 already pointed it out. Even if you are using 16-bit editing, the pixel values for RGB that you see will still range from 0 to 255. That doesn't change. All that changes are the way that the "internal calculations" are done.

As for rounding, you're not going to be able to see the difference between say, RGB (125, 175, 200) and (124, 174, 201). However, if you keep editing an image in 8-bit mode, there is a chance that you will spread certain pixel values far enough apart that you will start to notice a difference.


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not_this_punk
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Sep 26, 2008 17:33 |  #10

agedbriar wrote in post #6385217 (external link)
I'm afraid that wouldn't do. ;)

Some of us just like to know how things work. I understand the OP's deduction and to me his question is interesting, but unfortunately I have no answer.

The larger and smaller slices are exactly the point the OP is making. If you cut the pie into larger slices and then found out that one slice is burnt up, by cutting the burnt slice into smaller slices, you should be able to eat some of that part of the pie that would otherwise get entirely discarded.

Thank you so much, I couldn't have explained it better :D

Anyway, at least I know now that I'm not doing something wrong in ACR and this is the way everybody does it. I know that the answer to my question won't make my pictures better, but as agedbriar pointed out, I just want to know how things work:)


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Player9
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Sep 26, 2008 19:22 as a reply to  @ not_this_punk's post |  #11

The pixel is the thing

It would seems to me that a pixel is a real thing (not just a bunch of digital numbers that the camera's computer assigns to the analog signal that comes from the pixel). If the pixel is blown out (i.e., it received more photons than it could handle during the exposure), it's done. Whether the camera uses 8-bit or 16-bit to assign a number to that pixel, I wouldn't think it would matter. In that sense, the pixel is not like a burnt pie, as there is no good part to slice off.


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agedbriar
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Sep 27, 2008 04:23 |  #12

I agree, the saturation of a photosite (to distinguish from the pixel as the digital image element) is a physical state. However, level 255 is populated by pixels that correctly depict that value as well as those, which have been mapped there from the oversaturation range. Both of them are shown as clipped, even if some actually aren't.

I'd expect an increasing number of bits to reduce the number of pixels flagged as clipped, though IMO their number would not limit to zero, as the OP suggested, but rather to their correct amount, progressively excluding "true" top level pixels from the clipped group as the encoding bits rose.




  
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Damo77
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Sep 27, 2008 04:33 |  #13

Anybody who is concerned about their pixels to this level is undoubtedly going to shoot Raw, so detail can be recovered and clipping avoided to a large degree.

Sorry, I just can't take this thread seriously. I'm a nerd, but this is incredible ...


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not_this_punk
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Sep 27, 2008 08:41 |  #14

agedbriar wrote in post #6390006 (external link)
I agree, the saturation of a photosite (to distinguish from the pixel as the digital image element) is a physical state. However, level 255 is populated by pixels that correctly depict that value as well as those, which have been mapped there from the oversaturation range. Both of them are shown as clipped, even if some actually aren't.

I'd expect an increasing number of bits to reduce the number of pixels flagged as clipped, though IMO their number would not limit to zero, as the OP suggested, but rather to their correct amount, progressively excluding "true" top level pixels from the clipped group as the encoding bits rose.

No, no, I said that there will be approximately 0 clipped pixels (but not exactly 0), if the number of bits used to encode the image is infinite.I just wanted to present an extrapolation so that I could better point out what I was saying;)

But you're right, the clipped pixels are made up of truly clipped ones, that surpass the DR of the camera and will be clipped in any encoding, as well as those that are clipped because of the encoding used. I was inadvertently referring to the second category of pixels:)

Damo77 wrote in post #6390021 (external link)
Anybody who is concerned about their pixels to this level is undoubtedly going to shoot Raw, so detail can be recovered and clipping avoided to a large degree.

Sorry, I just can't take this thread seriously. I'm a nerd, but this is incredible ...

What do you know, you're not that much of a nerd:D
But, in all honesty, I kind of agree with you. After all, this is a photography forum and not an electronics one. I can't say that I like these sort of things, but studying electronics & IT at the university has that geeky effect on you and makes you want to know the most intricate detail...:lol:


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agedbriar
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Sep 27, 2008 14:22 |  #15

Well, now... where's that beer Damien volunteered?




  
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Why don't I get more light levels in RAW by using 16 bit instead of 8?
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