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Data processing

NN146125UL
Hi folks! A simple question that I hope someone can answer! Why is it that when I view an image that I am processing the luminance values shown are always 8 bit i.e., 0 to 255, even if my original data is 12 or 14 bit?

Thanks in advance, Martin
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2 comments

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  • Paul Steunebrink
    Martin, I think two issues are involved here.

    First, a camera's raw recording bit value/rate (i.e. 12- or 14-bit) is a different parameter than a 8- or 16-bit color channel value. Raw is (in general) a linear luminance level (a pixel can't see color). Each pixel is translated in three color channels (RGB or CMYK) with 8- or 16-bits/channel on output. The process in between called demosaicing not only translates luminance to color but also applies a non-linear curve (in general). The bit rate the raw converter itself works with might (and probably will be) another, higher, value.

    Next, how does the program display levels to the user so he/she is able to master the process described above and optimize adjustments? To give the photographer an idea of the color and luminance level 8-bit values are used because it is easy to read. Open an image in Photoshop (CS3/4) and get the Info tool, select 16- or 32-bit and see what I mean. (Yes, you can set color channel readout of an 8-bit image to 16-bit and v.v.).
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  • NN146125UL
    Thanks for the explanation Paul! I am trying to understand the processes involved because I have made a huge improvement to my photographic results by shooting RAW and using Capture One 4 rather than leaving the processing to the camera's mini-lab!

    I wonder how many amateurs spend huge sums of money upgrading DSLRs but then simply accept the processed output rather than processing it themselves? The photo magazines don't help either because they always focus (ha-ha) on how many picture functions and fancy modes a DSLR has instead of viewing it as box with a sensor in it where the only important features, in my opinion, are; quality of sensor, accuracy of exposure and focus.

    Once again thanks very much for the explanation!

    Kind regards, Martin
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