6 ms·

You still have quantization noise at 20 bits. It's just less than at 8.

by wl 9y ago

chrisseaton 9y ago
Exactly - it's a real range! You always have quantisation noise. What I'm asking is why is 20 bits the correct minimum resolution? One extreme being a factor of 1,000,000 to the other seems irrelevant for determining the correct minimum resolution.
- corysama 9y ago
  You can represent a range of 0 to 1,000,000 with a single bit. It's just that you can't do much with that representation :P So, for display purposes, 8 bits (with a non-linear mapping) has been sufficient for the brightness range of TV and monitors until recently. Now we have displays going High Dynamic Range. If you tried to display a 10x wider range of brightness using the old 8-bit format, you'd start to see a lot of ugly banding (posterization) much more than you have seen before because each of those 255 flat steps would be associated with a much larger section of the brightness curve. You'll need at least a couple more bits to keep your brighter&darker image looking as smooth as the mid-range-only images did in the past. For storage purposes, the 16-bit per channel floating point OpenEXR format has been used by the film industry as a intermediate format between image processing tools for quite some time now. I think that demonstrates that 16 bit floats are sufficient for storage. For computational purposes, 16 bit floats can build up a lot of numerical error very quickly. It's OK for small bit of math (moderate real-time shaders), but when things get complicated, it's necessary to go full 32-bit float.
  - Dylan16807 9y ago
    > For computational purposes, 16 bit floats can build up a lot of numerical error very quickly. It's OK for small bit of math (moderate real-time shaders), but when things get complicated, it's necessary to go full 32-bit float. What kind of graphics calculation would build up enough error to matter? You don't exactly need to do A + 10000 - 10000 in a rendering pipeline. If you're summing huge amounts of numbers in a raytracer it's appropriate. Where else?
    - tripzilch 9y ago
      You're completely right to question that assumption. It's very much possible to quantify the amount of error, or even just ballpark it. But numerical math is hard! :) And .. even a raytracer doesn't sum that many numbers, maybe a few thousand[0]? So even then it's still a very reasonable question to ask. [0] assuming a Monte Carlo pathtracer type of thing with 5-10 bounces or so. BTW, do you think you'd ever need more than five? Unless you're rendering close-to-parallel reflective surfaces, but the appearance of those is often as confusing in real life as it is when rendered correctly.
  - tripzilch 9y ago
    > You can represent a range of 0 to 1,000,000 with a single bit. It's just that you can't do much with that representation :P Tell that to D/A converters everywhere. > If you tried to display a 10x wider range of brightness using the old 8-bit format, you'd start to see a lot of ugly banding (posterization) much more than you have seen before because-- Because you didn't dither, is why. The only thing you lose with less bits is SNR, and thanks to the wonders of noise-shaping you can pretty much decide where to put this noise. And I'm just going to make a very broad statement that ever since we managed to climb out of 320x200 resolutions, spatial resolution is where you should put it, not in banding. Maybe even one or two bits in the time resolution if you want to get fancy about it. Basic dithering can be as easy as just adding a uniform 0..1 random number just before truncation. Triangle noise (difference of two uniforms) can give a bit better visual accuracy for graphics, but has one or two snags that take attention to get right.
ttoinou 9y ago
It's less than at 8, making acceptable new previously darker parts of the image. But I agree with the "infinite scale" argument

You still have quantization noise at 20 bits. It's just less than at 8.

by wl 9y ago

You still have quantization noise at 20 bits. It's just less than at 8.

Exactly - it's a real range! You always have quantisation noise. What I'm asking is why is 20 bits the correct minimum resolution? One extreme being a factor of 1,000,000 to the other seems irrelevant for determining the correct minimum resolution.

corysama 9y ago
You can represent a range of 0 to 1,000,000 with a single bit. It's just that you can't do much with that representation :P So, for display purposes, 8 bits (with a non-linear mapping) has been sufficient for the brightness range of TV and monitors until recently. Now we have displays going High Dynamic Range. If you tried to display a 10x wider range of brightness using the old 8-bit format, you'd start to see a lot of ugly banding (posterization) much more than you have seen before because each of those 255 flat steps would be associated with a much larger section of the brightness curve. You'll need at least a couple more bits to keep your brighter&darker image looking as smooth as the mid-range-only images did in the past. For storage purposes, the 16-bit per channel floating point OpenEXR format has been used by the film industry as a intermediate format between image processing tools for quite some time now. I think that demonstrates that 16 bit floats are sufficient for storage. For computational purposes, 16 bit floats can build up a lot of numerical error very quickly. It's OK for small bit of math (moderate real-time shaders), but when things get complicated, it's necessary to go full 32-bit float.
- Dylan16807 9y ago
  > For computational purposes, 16 bit floats can build up a lot of numerical error very quickly. It's OK for small bit of math (moderate real-time shaders), but when things get complicated, it's necessary to go full 32-bit float. What kind of graphics calculation would build up enough error to matter? You don't exactly need to do A + 10000 - 10000 in a rendering pipeline. If you're summing huge amounts of numbers in a raytracer it's appropriate. Where else?
  - tripzilch 9y ago
    You're completely right to question that assumption. It's very much possible to quantify the amount of error, or even just ballpark it. But numerical math is hard! :) And .. even a raytracer doesn't sum that many numbers, maybe a few thousand[0]? So even then it's still a very reasonable question to ask. [0] assuming a Monte Carlo pathtracer type of thing with 5-10 bounces or so. BTW, do you think you'd ever need more than five? Unless you're rendering close-to-parallel reflective surfaces, but the appearance of those is often as confusing in real life as it is when rendered correctly.
- tripzilch 9y ago
  > You can represent a range of 0 to 1,000,000 with a single bit. It's just that you can't do much with that representation :P Tell that to D/A converters everywhere. > If you tried to display a 10x wider range of brightness using the old 8-bit format, you'd start to see a lot of ugly banding (posterization) much more than you have seen before because-- Because you didn't dither, is why. The only thing you lose with less bits is SNR, and thanks to the wonders of noise-shaping you can pretty much decide where to put this noise. And I'm just going to make a very broad statement that ever since we managed to climb out of 320x200 resolutions, spatial resolution is where you should put it, not in banding. Maybe even one or two bits in the time resolution if you want to get fancy about it. Basic dithering can be as easy as just adding a uniform 0..1 random number just before truncation. Triangle noise (difference of two uniforms) can give a bit better visual accuracy for graphics, but has one or two snags that take attention to get right.

ttoinou 9y ago

It's less than at 8, making acceptable new previously darker parts of the image. But I agree with the "infinite scale" argument