Originally posted by jpipg The rightmost bits contain less information than the even the single least significant of the leftmost bits. In the case that you have a subject with narrow dynamic range, placing the values on the left side of the register gives far larger tonal resolution because those bits contain more data.

I'm hoping I'm completely misunderstanding this, because I'm thinking binary. Individual binary bits have a value of 1 or 0 (whereas decimal digits have a value from 0 to 9) and the leftmost bits can contain no more information than that. Their numerical significance may

*represent* a decimal value of 64, or 128, or 256 etc but it's either that or zero.

Here's the number 167 in 14-bit binary:– 00000000

**1**0

**1**00

**111** (14-bit dynamic range is what we have to play with, at best)

Shift the bits left by 4 positions and you get:– 00

**1**0

**1**00

**111**0000 which is 2672 in decimal, a much bigger number, so a pixel which originally was a brightness value of 167 jumps up to 2672 by the 4-position shift operation. Great, much brighter!

What happens to a pixel whose original value was zero; which registered no brightness? It goes from 00000000000000 (zero) to 00000000000000. Which is to say, zero. No data is added, no increased brightness is recorded. Damn.

I hope I'm making this clear. It has nothing to do with ETTR, which is about fitting as much of the brightness range of the subject as possible into the actual dynamic range of the camera.

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Last edited by StiffLegged; 11-02-2019 at 03:14 PM.
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