Subject | Re: Is RGB to Lab lossy? - was(Re: Lenses and sharpening) |
From | Alan Browne |
Date | 10/06/2014 19:03 (10/06/2014 13:03) |
Message-ID | <K-mdncWK3-zrVq_JnZ2dnUU7-LednZ2d@giganews.com> |
Client | |
Newsgroups | rec.photo.digital |
Follows | Martin Brown |
Martin BrownExcellent point.
On 06/10/2014 14:19, PeterN wrote:PeterNMartin Brown
On 10/5/2014 10:37 PM, Alan Browne wrote:Alan BrownePeterN
On 2014.10.05, 20:55 , PeterN wrote:PeterNAlan Browne
On 10/5/2014 6:57 PM, Alan Browne wrote:Alan BrownePeterN
On 2014.10.05, 14:42 , PeterN wrote:We went through all this some many months ago. I demonstrated clearly that the amount of 'loss' was negligible in practical terms.PeterNAlan Browne
I would use the terem "color change." anstead of loss.
Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss.
Then you are using a different definition of quality.
Not at all. A non lossy process would have:
RGB-A -->X-format -->RGB-B
with RGB-A identical to RGB-B
But - the fact is that with Lab
RGB-A -->Lab -->RGB-B
RGB-A =/= RGB-B, therefore there was quality loss.
It seems to me that the assumption in that logic is: the quality of RGB-A >quality of RGB-B. LAB has a larger color gamut than RGB. If there is no processing in LAB I would think that there would be no need for interpolation on the return trip.
The problem is not one of interpolation but that there are unavoidable minor rounding errors in the nonlinear transform from RGB to CIELAB and also on the way back due to the finite representation of the results. See: