Shel Belinkoff wrote:
You've lost me completely. I don't understand the "math of photography."
Just snap the shutter and see what comes out. Push a few buttons in
Photoshop, or adjust the light in the darkroom. If it works one way or not
another, then the answer is clear.
Well, as long as you know that 12 bits are better than 8, then thats
enough to know that what you described is equivalent, and as bad.
Thats good enuf. I hope you know about the ratio of f-stops and all
that. There's a little math there too you know.
Anyway, I don't even understand terms like "quantization," or what a
"chunkier quantization" might be, or why or how you'd amplify a
quantization..
I love how photography has become a numbers crunching exercise for some
people. Pick up the camera, focus, press the shutter, and see what
happens, see what you get.
Shel
[Original Message]
From: Gonz
This does not make sense to me. Assuming a perfect amplification and a
perfect digitization for a moment, then a shot that would have a
complete dynamic range at ISO 1600 would only go up to 1/4 the dynamic
range at ISO 400. So when you amplify this quantization (for 12 bits
this would be 2^12/4 = 1024) to the full range, you have "chunkier"
quantitization, as if you only had a 10bit sensor instead of 12. That
leads me to believe that there would be more noise associated with this.
This is similar to the arguments of keeping your image in 16 bit mode
when editing as much as possible, until the final conversion to JPG and
8 bits. Converting to 8 bits first then editing is going to cost you
alot of information.
I'm not taking into account the effects of Bayer interpolation or other
interpolation such as uprezing, etc. That just complicates the way the
information is interpreted, but it does not change the absolute
underlying numbers.
--
Someone handed me a picture and said, "This is a picture of me when I
was younger." Every picture of you is when you were younger. "...Here's
a picture of me when I'm older." Where'd you get that camera man?
- Mitch Hedberg
