Alan321
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Pondria wrote:
Alan,
I'll use your terminlogy to get my thought across better to you.
I have been maintaing the position that if you just shift the 12bit data to the right by two bits to make it a 14bit data and pad the two LSBs with zero, it would be effectively the same as the 14bit data that 1D3 produces UNLESS 1D3 has improved S/N. Because the 2 LSBs that 14bit has WITHOUT the improved S/N would be just random numbers (noise) containing no relevant information for the image.
I've been asking for the data to show us if there is any S/N improvements. Jeff is going to send the raw files.
...Show more →
Yes it would be the same as the 1D3 produces IF you multiplied the analogue data of another camera by 4 before converting to digital. However, if you do the analogue conversion to digital at 12 bit and THEN multiply those values by 4 you will still have the same level of quantisation as the original 12-bit data. There will be bigger differences between adjacent used levels but the extra levels in between will remain unused.
Consider this subset of data values:
12-bit data prior to multiplication:
....,10,11,12,13,...
would become
....,40,44,48,52,... (no pixels would have values 41, 42, 43, 45, 46, etc.)
when multiplied by 4. i.e. the increment is now 4 instead of 1.
Because the 1D3 does the A to D conversion at 14-bit we would expect to see
....,40,41,42,43,44,45,46,47,48,...
where the increment is just 1 instead of 4.
These values will have to be interpreted as meaning a quarter of their actual value but there are still 4 times as many levels to help reduce posterisation during processing.
How might this be used in practise ? Well lets suppose we underexposed and we want to brighten the shadows by one stop.
Our original 12 bit data goes from ...,10,11,12,13,... to ...,20,22,24,26,.... (increments of 2)
Our original 14 bit data goes from ...,40,41,42,43,.... to ...,80,82,84,86,... (increments of 2) and after allowing for shift of middle tone would become ...,20,20,21,21,22,... (increments of 1 after removing decimal fractions, because the data file only stores integer values)
The scaled up 12 bit data would go from ...,10,11,12,13,... to ...,40,44,48,52,... (increments of 4) to ...,80,88,96,104,... (increments of 8) and after allowing for shift of middle tone would become ...,20,22,24,26,... (increments of 2) which is just as with the standard 12-bit data and still with half the number of discrete levels as the 14-bit data.
In converting to 8-bit these would still be squeezed into only 256 possible levels but there should still be less posterisation
Here's another set of values, this time with an exposure adjustment of two stops:
12-bit 10,11,12,13 ---> 40,44,48,52 (increments of 4)
14-bit 40,41,42,43 ---> 160,164,168,172 ---> 40,41,42,43 (increments of 1)
modified 12-bit 10,11,12,13 ---> 40,44,48,52 (increments of 4) ---> 160,176,192,208 (increments of 16) ---> 40,44,48,52 (increments of 4)
And this time with a three stop exposure adjustment:
12-bit 10,11,12,13 ---> 80,88,96,104 (increments of 8)
14-bit 40,41,42,43 ---> 320,328,336,344 ---> 80,82,84,86 (increments of 2)
modified 12-bit 10,11,12,13 ---> 40,44,48,52 (increments of 4) ---> 320,352,384,408 (increments of 32) ---> 80,88,96,102 (increments of 8)
Conclusion: The 14-bit data initially has the same increment (1) between adjacent values as the 12-bit dat and pseudo-14-bit data (modified 12-bit data)(after returning it to 12-bit), but there are 4 times as many values to choose from.
When you increase the exposure level in software (not at image capture) you get smaller increments between adjacent levels in the 14-bit file by a factor of up to 4 compared with the 12-bit data (1 at nil change, 2 at +1 stop, 4 at +2 or more stops). There is no advantage to using the pseudo-14-bit data by multiplying the 12-bit data by 4 prior to manipulation and dividing by 4 afterwards, because the quantisation damage has already been done.
The significance of each value increment varies according to the data value. In very dark data it is a large portion of the total value, but at bright values it is only a small portion.
Recently obtained information shows that data below -7 stops from middle tone apparently cannot be utilised, at least by DPP software. In the interval between -6 and -7 stops there are only about 10 levels in 14-bit data and 2 or 3 in 12-bit data. Scale that up until you can see it and there will be a significant advantage in the 14-bit raw image data. However, the extent to which that advantage translates to an 8-bit data file for printing depends on the tone conversion curve. It would probably all be black unless you did a major levels increase.
A more realistic example would be data at -5 stops being adjusted to -3 stops in editing. Here the number of discrete data levels in the darkest stop visible in print would be about 40 for the 14-bit data and 10 for the 12-bit data. If you can see a tenth of a stop on a print then you will notice the improvement in the 1D3 files that have been lightened by two stops. If the data has not been lightened by two stops then you are unlikely to appreciate the difference in prints even though you may see it on a computer screen, because the data will be buried in the black area of the print.
The computer screens have 16-bit colours so you will always see the 14-bit advantage more readily on a computer than you will on a print or on an 8-bit file.
- Alan
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). Again, with that horribly small histogram dialog box (per my post in the 'DPP Comparison' thread), it's really tough to see accurately. I know that in some cases moving the black point slider affected the image before I'd gotten to the point where I thought the curve started, but I wasn't paying enough attention at the time, plus, it was kinda late. 
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