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Re: Is RGB to Lab lossy? -...

Alan Browne
SubjectRe: Is RGB to Lab lossy? - was(Re: Lenses and sharpening)
FromAlan Browne
Date10/06/2014 19:03 (10/06/2014 13:03)
Message-ID<K-mdncWK3-zrVq_JnZ2dnUU7-LednZ2d@giganews.com>
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Newsgroupsrec.photo.digital
FollowsMartin Brown

On 2014.10.06, 11:33 , Martin Brown wrote:

Martin Brown
On 06/10/2014 14:19, PeterN wrote:

PeterN
On 10/5/2014 10:37 PM, Alan Browne wrote:

Alan Browne
On 2014.10.05, 20:55 , PeterN wrote:

PeterN
On 10/5/2014 6:57 PM, Alan Browne wrote:

Alan Browne
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.

PeterN
I would use the terem "color change." anstead of loss.

Alan Browne
Any change is a quality loss. Whether that is colour difference, tone, brightness, sharpness ... whatever, it's a loss.

PeterN
Then you are using a different definition of quality.

Alan Browne
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.

PeterN
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.

Martin Brown
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:

Excellent point.

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