To state the obvious, 16-bit integer offers more precision than 8-bit
integer:
*There are 255 tonal steps from 0 to 255 for 8-bit integer precision.
*There are 65535 tonal steps from 0 to 65535 for 16-bit integer precision.
*65535 steps divided by 255 steps is 257. So for every tonal step in an
8-bit image there are 257 steps in a 16-bit image.
I've read, and it makes sense because with floating point you have to
share the available precision with the numbers on both sides of the
decimal place, that 16-bit integer is more precise than 16-bit floating
point. And 32-bit integer is more precise than 32-bit floating point.
My question is, for Gimp from git, is 32-bit floating point more precise
than 16-bit integer? If so, by how much, and does it depend on the
machine and/or things like the math instructions of the processor
(whatever that means)? And if not, how much less precise is it?
To restate the question, in decimal notation, 1 divided by 65535 is
0.00001525878906250000. So 16-bit integer precision requires 16 decimal
places (lop off the four trailing zeros) in floating point to express
the floating point equivalent of 1 16-bit integer tonal step, yes? no?
The Gimp eyedropper displays 6 decimal places for RGB values.
0.00001525878906250000 rounded to 6 places is 0.000015.
0.000015 times 65535 is 0.983025.
0.000016 times 65535 is 1.04856.
How many decimal places does Gimp 32-bit floating point actually provide?
Elle
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