Lenses and sharpening

On 10/1/2014 3:39 PM, nospam wrote:
In article , PeterN
wrote:

Absent
such proof I trust Dan Margulies's opinion, more than yours. p\Peroid.

dan marguilis is wrong. period.

the link you keep referring to proves he is wrong, which means you
didn't read it.

it also means you are wrong (no surprise there) and you once again
demonstrate how much of a blithering idiot you are.

I would have to be one to accept your "proof."

you're not the final arbiter of what is true or not.

the fact is that rgb-lab-rgb is a lossy operation. period.

mathematics is not going to change because you have not accepted the
proof, one of which was in the link you refuse to read.



Well deserved response.

--
PeterN

On Thu, 18 Sep 2014 19:14:58 -0400, nospam
wrote:

In article , Eric Stevens
wrote:

I have found that using high pass on the luminiscence layer in LAB
tends to minimize halos.

Actually it is a good idea to do any/all/most sharpening on a
luminosity layer, LAB or not.

not always, since the conversion to lab and back is not lossless.

Not strictly correct:

it is completely correct.

we went through this about six months ago, and apparently will again.

https://www.ledet.com/margulis/ACT_p...LAB-damage.htm

" I have always thought that moving from either CMYK or RGB to Lab
and back was a damage free process, that is, you would end up with
the same color co-ordinates when you arrived back from Lab mode.

"RGBLABRGB is damage free, but CMYKLABCMYK is not. The damage
isn't all that great, so in many images it pays to come out of CMYK
so as to take advantage of LAB's strengths; sharpening, however, is
not one of these cases.
....
Dan Margulis"

you clearly don't understand what you're reading, since that link
agrees with what i said!

as the other posts in your link clearly show, dan margulis is wrong (as
he is about a lot of things).

read the *very* next post, from chris murphy,

Converting to and from Lab has never been a damage free process.

and the one after that,

RGBLABRGB is damage free, but CMYKLABCMYK is not.
I disagree. If you start out with all of the same spaces for RGB and
CMYK, and use only those spaces - then convert to and from Lab, you
will get some quantization errors with both.

and andrew rodney's post:
RGBLABRGB is damage free
You1re not serious are you Dan?
Take an RGB file. Duplicate it. Do an RGB to LAB to RGB conversion
and subtract the two. You can turn on or off the 8 bit dither. When
you subtract the two and create a new document and look at the
Histogram in Levels, you will see there certainly is data loss and a
change. Move the sliders of the Levels Histogram over and you1ll see
the effects of what differences between the two files you produced.
Are you saying this isn1t data loss?

that test is trivial to do. try it yourself.

This one continues to bother me. I am still inclined to agree with Dan
Margulis. I'm not quite sure what procedure Andrew Rodney is proposing
to prove his point so, using Photoshop CC, I have carried out my own
test as follows:

1. Find a JPG with a suitable range of colors. This one came from my
wife's collection:
https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg
I saved a copy as a PSD (see below for the reason).

2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved
to PSD. See
https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg

3. I then loaded the two PSD files into a new file as separate layers.
(1) above was the background layer and (2) was the next. I subtracted
the 2nd layer from the first with the result shown in
https://dl.dropboxusercontent.com/u/...Difference.jpg
That's right: solid black.

4. To confirm the point I took a screen shot. See
https://dl.dropboxusercontent.com/u/...t%20Screen.jpg
Note the histogram. All of the pixels appear to be down at the zero
end of the scale: that is, jet black.

The only conclusion I can reach is that there is no difference between
a PSD created from a RGB file and a PSD created from the same image
when it has first been converted from RGB to Lab.

I'm not wedded to the perfection of the method I have used and I would
be interested to hear from anyone who has a meaningful criticism.
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:

not always, since the conversion to lab and back is not lossless.

Not strictly correct:

it is completely correct.

we went through this about six months ago, and apparently will again.


https://www.ledet.com/margulis/ACT_p...ACT-LAB-damage.
htm

" I have always thought that moving from either CMYK or RGB to Lab
and back was a damage free process, that is, you would end up with
the same color co-ordinates when you arrived back from Lab mode.

"RGBLABRGB is damage free, but CMYKLABCMYK is not. The damage
isn't all that great, so in many images it pays to come out of CMYK
so as to take advantage of LAB's strengths; sharpening, however, is
not one of these cases.
....
Dan Margulis"

you clearly don't understand what you're reading, since that link
agrees with what i said!

as the other posts in your link clearly show, dan margulis is wrong (as
he is about a lot of things).

read the *very* next post, from chris murphy,

Converting to and from Lab has never been a damage free process.

and the one after that,

RGBLABRGB is damage free, but CMYKLABCMYK is not.
I disagree. If you start out with all of the same spaces for RGB and
CMYK, and use only those spaces - then convert to and from Lab, you
will get some quantization errors with both.

and andrew rodney's post:
RGBLABRGB is damage free
You1re not serious are you Dan?
Take an RGB file. Duplicate it. Do an RGB to LAB to RGB conversion
and subtract the two. You can turn on or off the 8 bit dither. When
you subtract the two and create a new document and look at the
Histogram in Levels, you will see there certainly is data loss and a
change. Move the sliders of the Levels Histogram over and you1ll see
the effects of what differences between the two files you produced.
Are you saying this isn1t data loss?

that test is trivial to do. try it yourself.

This one continues to bother me. I am still inclined to agree with Dan
Margulis.

don't. he's widely regarded as a quack. he is constantly proven wrong
by numerous people, including andrew rodney, bruce fraser, chris murphy
and many others.

I'm not quite sure what procedure Andrew Rodney is proposing
to prove his point so,

it's very straightforward. take an image, do an lab conversion and back
and subtract the two. the result is whatever changed.

using Photoshop CC, I have carried out my own
test as follows:

1. Find a JPG with a suitable range of colors. This one came from my
wife's collection:
https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg
I saved a copy as a PSD (see below for the reason).

2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved
to PSD. See
https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg

3. I then loaded the two PSD files into a new file as separate layers.
(1) above was the background layer and (2) was the next. I subtracted
the 2nd layer from the first with the result shown in
https://dl.dropboxusercontent.com/u/...Difference.jpg
That's right: solid black.

it may look solid black but it isn't.

4. To confirm the point I took a screen shot. See
https://dl.dropboxusercontent.com/u/...t%20Screen.jpg
Note the histogram. All of the pixels appear to be down at the zero
end of the scale: that is, jet black.

notice the differences at the left end of the histogram.

however, this is about round-tripping from rgb to lab and then back.
you only did half.

this is all explained in the link you gave. try reading it.

The only conclusion I can reach is that there is no difference between
a PSD created from a RGB file and a PSD created from the same image
when it has first been converted from RGB to Lab.

there is. it may not be a huge difference, but there is a difference.

compare a high quality jpeg with the original and you'll see black as
you did above, but there are definitely differences (and actually, less
of a difference than the rgb-lab conversion).

do you see people arguing to edit jpegs? of course not.

On Fri, 03 Oct 2014 23:42:50 -0400, nospam
wrote:

In article , Eric Stevens
wrote:

not always, since the conversion to lab and back is not lossless.

Not strictly correct:

it is completely correct.

we went through this about six months ago, and apparently will again.


https://www.ledet.com/margulis/ACT_p...ACT-LAB-damage.
htm

" I have always thought that moving from either CMYK or RGB to Lab
and back was a damage free process, that is, you would end up with
the same color co-ordinates when you arrived back from Lab mode.

"RGBLABRGB is damage free, but CMYKLABCMYK is not. The damage
isn't all that great, so in many images it pays to come out of CMYK
so as to take advantage of LAB's strengths; sharpening, however, is
not one of these cases.
....
Dan Margulis"

you clearly don't understand what you're reading, since that link
agrees with what i said!

as the other posts in your link clearly show, dan margulis is wrong (as
he is about a lot of things).

read the *very* next post, from chris murphy,

Converting to and from Lab has never been a damage free process.

and the one after that,

RGBLABRGB is damage free, but CMYKLABCMYK is not.
I disagree. If you start out with all of the same spaces for RGB and
CMYK, and use only those spaces - then convert to and from Lab, you
will get some quantization errors with both.

and andrew rodney's post:
RGBLABRGB is damage free
You1re not serious are you Dan?
Take an RGB file. Duplicate it. Do an RGB to LAB to RGB conversion
and subtract the two. You can turn on or off the 8 bit dither. When
you subtract the two and create a new document and look at the
Histogram in Levels, you will see there certainly is data loss and a
change. Move the sliders of the Levels Histogram over and you1ll see
the effects of what differences between the two files you produced.
Are you saying this isn1t data loss?

that test is trivial to do. try it yourself.

This one continues to bother me. I am still inclined to agree with Dan
Margulis.

don't. he's widely regarded as a quack. he is constantly proven wrong
by numerous people, including andrew rodney, bruce fraser, chris murphy
and many others.

I'm not quite sure what procedure Andrew Rodney is proposing
to prove his point so,

it's very straightforward. take an image, do an lab conversion and back
and subtract the two. the result is whatever changed.

using Photoshop CC, I have carried out my own
test as follows:

1. Find a JPG with a suitable range of colors. This one came from my
wife's collection:
https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg
I saved a copy as a PSD (see below for the reason).

2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved
to PSD. See
https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg

3. I then loaded the two PSD files into a new file as separate layers.
(1) above was the background layer and (2) was the next. I subtracted
the 2nd layer from the first with the result shown in
https://dl.dropboxusercontent.com/u/...Difference.jpg
That's right: solid black.

it may look solid black but it isn't.

4. To confirm the point I took a screen shot. See
https://dl.dropboxusercontent.com/u/...t%20Screen.jpg
Note the histogram. All of the pixels appear to be down at the zero
end of the scale: that is, jet black.

notice the differences at the left end of the histogram.

however, this is about round-tripping from rgb to lab and then back.
you only did half.

Fir comment. I've just compared the original JPG with a copy -- Lab
-- JPG again. JPGs are RGB are they not? Anyway I still got an
apparently all-black screen and here is the screen shot showing the
histogram:
https://dl.dropboxusercontent.com/u/...screen%202.jpg

An even tighter all-black bar than previously.

this is all explained in the link you gave. try reading it.

You don't have to be rude. Try reading it yourself and then explain
step by step what you think he is proposing.

The only conclusion I can reach is that there is no difference between
a PSD created from a RGB file and a PSD created from the same image
when it has first been converted from RGB to Lab.

there is. it may not be a huge difference, but there is a difference.

As soon as you do anything in Photoshop there is a difference due to
rounding errors (quantization) but is this all you are objecting to?

compare a high quality jpeg with the original and you'll see black as
you did above, but there are definitely differences (and actually, less
of a difference than the rgb-lab conversion).

What is the difference with rgb-Lab-rgb conversions and what causes
them?

do you see people arguing to edit jpegs? of course not.

What exactly do you mean by that?
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:

4. To confirm the point I took a screen shot. See
https://dl.dropboxusercontent.com/u/...t%20Screen.jpg
Note the histogram. All of the pixels appear to be down at the zero
end of the scale: that is, jet black.

notice the differences at the left end of the histogram.

however, this is about round-tripping from rgb to lab and then back.
you only did half.

Fir comment. I've just compared the original JPG with a copy -- Lab
-- JPG again. JPGs are RGB are they not?

usually but not always

Anyway I still got an
apparently all-black screen and here is the screen shot showing the
histogram:
https://dl.dropboxusercontent.com/u/...screen%202.jpg

An even tighter all-black bar than previously.

this is all explained in the link you gave. try reading it.

You don't have to be rude. Try reading it yourself and then explain
step by step what you think he is proposing.

i'm not trying to be rude. the answers really are in the link and i've
said this many times already.

The only conclusion I can reach is that there is no difference between
a PSD created from a RGB file and a PSD created from the same image
when it has first been converted from RGB to Lab.

there is. it may not be a huge difference, but there is a difference.

As soon as you do anything in Photoshop there is a difference due to
rounding errors (quantization) but is this all you are objecting to?

you do realize that adds up, right?

compare a high quality jpeg with the original and you'll see black as
you did above, but there are definitely differences (and actually, less
of a difference than the rgb-lab conversion).

What is the difference with rgb-Lab-rgb conversions and what causes
them?

read the link and pay attention to andrew rodney.

ignore marguilis, not just in that link but in general. he has claimed
that 16 bit editing was a waste, which it absolutely is not. i dunno if
he still claims it but he probably does.

do you see people arguing to edit jpegs? of course not.

What exactly do you mean by that?

you say you can't see a difference in an rgb-lab-rgb conversion and you
subtracted them and saw all black, therefore, you have deemed them to
be equivalent.

if you do the same for jpeg, you will also not see a difference, and if
you subtract, you'll also see all black. therefore, a jpeg should be
equivalent to an original raw.

the reality is that there *is* a difference. you might not consider the
difference to be significant (and indeed it is is very small), but
there *is* a difference, therefore it is *not* lossless.

bottom line: rgb-lab-rgb offers no benefit (other than possibly
contrived edge cases nobody will ever encounter).

On 2014.10.03, 23:29 , Eric Stevens wrote:
On Thu, 18 Sep 2014 19:14:58 -0400, nospam
wrote:

In article , Eric Stevens
wrote:

I have found that using high pass on the luminiscence layer in LAB
tends to minimize halos.

Actually it is a good idea to do any/all/most sharpening on a
luminosity layer, LAB or not.

not always, since the conversion to lab and back is not lossless.

Not strictly correct:

it is completely correct.

we went through this about six months ago, and apparently will again.

https://www.ledet.com/margulis/ACT_p...LAB-damage.htm

" I have always thought that moving from either CMYK or RGB to Lab
and back was a damage free process, that is, you would end up with
the same color co-ordinates when you arrived back from Lab mode.

"RGBLABRGB is damage free, but CMYKLABCMYK is not. The damage
isn't all that great, so in many images it pays to come out of CMYK
so as to take advantage of LAB's strengths; sharpening, however, is
not one of these cases.
....
Dan Margulis"

you clearly don't understand what you're reading, since that link
agrees with what i said!

as the other posts in your link clearly show, dan margulis is wrong (as
he is about a lot of things).

read the *very* next post, from chris murphy,

Converting to and from Lab has never been a damage free process.

and the one after that,

RGBLABRGB is damage free, but CMYKLABCMYK is not.
I disagree. If you start out with all of the same spaces for RGB and
CMYK, and use only those spaces - then convert to and from Lab, you
will get some quantization errors with both.

and andrew rodney's post:
RGBLABRGB is damage free
You1re not serious are you Dan?
Take an RGB file. Duplicate it. Do an RGB to LAB to RGB conversion
and subtract the two. You can turn on or off the 8 bit dither. When
you subtract the two and create a new document and look at the
Histogram in Levels, you will see there certainly is data loss and a
change. Move the sliders of the Levels Histogram over and you1ll see
the effects of what differences between the two files you produced.
Are you saying this isn1t data loss?

that test is trivial to do. try it yourself.

This one continues to bother me. I am still inclined to agree with Dan
Margulis. I'm not quite sure what procedure Andrew Rodney is proposing
to prove his point so, using Photoshop CC, I have carried out my own
test as follows:

1. Find a JPG with a suitable range of colors. This one came from my
wife's collection:
https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg
I saved a copy as a PSD (see below for the reason).

2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved
to PSD. See
https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg

3. I then loaded the two PSD files into a new file as separate layers.
(1) above was the background layer and (2) was the next. I subtracted
the 2nd layer from the first with the result shown in
https://dl.dropboxusercontent.com/u/...Difference.jpg
That's right: solid black.

4. To confirm the point I took a screen shot. See
https://dl.dropboxusercontent.com/u/...t%20Screen.jpg
Note the histogram. All of the pixels appear to be down at the zero
end of the scale: that is, jet black.

The only conclusion I can reach is that there is no difference between
a PSD created from a RGB file and a PSD created from the same image
when it has first been converted from RGB to Lab.

I'm not wedded to the perfection of the method I have used and I would
be interested to hear from anyone who has a meaningful criticism.

We went through all this some many months ago. I demonstrated clearly
that the amount of 'loss' was negligible in practical terms.

Procedu

1. An original in JPG
2. The original converted to LAB version
3. The LAB version converted to JPG.

delta 1-2, delta 2-3.

In the deltas you will see the actual difference (hard to see unless
your screen is turned up bright) and it is not something that would be
noticeable in an actual screen or print of a shot.

--
Among Broad Outlines, conception is far more pleasurable
than “carrying [the children] to fruition.”
Sadly, “there’s a high infant mortality rate among
Broad Outlines—they often fall prey to Nonstarters.”
"Bestiary of Intelligence Writing" - CIA

On Sat, 04 Oct 2014 04:48:23 -0400, nospam
wrote:

In article , Eric Stevens
wrote:

4. To confirm the point I took a screen shot. See
https://dl.dropboxusercontent.com/u/...t%20Screen.jpg
Note the histogram. All of the pixels appear to be down at the zero
end of the scale: that is, jet black.

notice the differences at the left end of the histogram.

however, this is about round-tripping from rgb to lab and then back.
you only did half.

Fir comment. I've just compared the original JPG with a copy -- Lab
-- JPG again. JPGs are RGB are they not?

usually but not always

Then what else might they be and under what circumstances?

Anyway I still got an
apparently all-black screen and here is the screen shot showing the
histogram:
https://dl.dropboxusercontent.com/u/...screen%202.jpg

An even tighter all-black bar than previously.

this is all explained in the link you gave. try reading it.

You don't have to be rude. Try reading it yourself and then explain
step by step what you think he is proposing.

i'm not trying to be rude. the answers really are in the link and i've
said this many times already.

Do you mean where he says:

"ANY colorspace conversion can cause these quantization errors (RGB
to RGB as an example)."

The only conclusion I can reach is that there is no difference between
a PSD created from a RGB file and a PSD created from the same image
when it has first been converted from RGB to Lab.

there is. it may not be a huge difference, but there is a difference.

As soon as you do anything in Photoshop there is a difference due to
rounding errors (quantization) but is this all you are objecting to?

you do realize that adds up, right?

Yes, and it's common to evrything you do. So why does converting to
Lab allegedly make it so much worse?

compare a high quality jpeg with the original and you'll see black as
you did above, but there are definitely differences (and actually, less
of a difference than the rgb-lab conversion).

What is the difference with rgb-Lab-rgb conversions and what causes
them?

read the link and pay attention to andrew rodney.

Do you mean where he says:

"ANY colorspace conversion can cause these quantization errors (RGB
to RGB as an example)."

ignore marguilis, not just in that link but in general. he has claimed
that 16 bit editing was a waste, which it absolutely is not. i dunno if
he still claims it but he probably does.

I bet you are quoting him out of context.

do you see people arguing to edit jpegs? of course not.

What exactly do you mean by that?

you say you can't see a difference in an rgb-lab-rgb conversion and you
subtracted them and saw all black, therefore, you have deemed them to
be equivalent.

I didn't say that. Read it all again carefully. I compared an
rgb-lab-rgb conversion to the original JPG.

if you do the same for jpeg, you will also not see a difference, and if
you subtract, you'll also see all black. therefore, a jpeg should be
equivalent to an original raw.

That is squiffy logic and it's not even a good parody of what I did.

the reality is that there *is* a difference. you might not consider the
difference to be significant (and indeed it is is very small), but
there *is* a difference, therefore it is *not* lossless.

bottom line: rgb-lab-rgb offers no benefit (other than possibly
contrived edge cases nobody will ever encounter).

You have backed off considerably from your original opinion on this
matter.
--

Regards,

Eric Stevens

On Sat, 04 Oct 2014 10:57:31 -0400, Alan Browne
wrote:

On 2014.10.03, 23:29 , Eric Stevens wrote:
On Thu, 18 Sep 2014 19:14:58 -0400, nospam
wrote:

--- snip ---

and the one after that,

RGBLABRGB is damage free, but CMYKLABCMYK is not.
I disagree. If you start out with all of the same spaces for RGB and
CMYK, and use only those spaces - then convert to and from Lab, you
will get some quantization errors with both.

and andrew rodney's post:
RGBLABRGB is damage free
You1re not serious are you Dan?
Take an RGB file. Duplicate it. Do an RGB to LAB to RGB conversion
and subtract the two. You can turn on or off the 8 bit dither. When
you subtract the two and create a new document and look at the
Histogram in Levels, you will see there certainly is data loss and a
change. Move the sliders of the Levels Histogram over and you1ll see
the effects of what differences between the two files you produced.
Are you saying this isn1t data loss?

that test is trivial to do. try it yourself.

This one continues to bother me. I am still inclined to agree with Dan
Margulis. I'm not quite sure what procedure Andrew Rodney is proposing
to prove his point so, using Photoshop CC, I have carried out my own
test as follows:

1. Find a JPG with a suitable range of colors. This one came from my
wife's collection:
https://dl.dropboxusercontent.com/u/...20IMG_2154.jpg
I saved a copy as a PSD (see below for the reason).

2. Copy and convert to Lab. I couldn't save to JPG from Lab so I saved
to PSD. See
https://dl.dropboxusercontent.com/u/...54-via-Lab.jpg

3. I then loaded the two PSD files into a new file as separate layers.
(1) above was the background layer and (2) was the next. I subtracted
the 2nd layer from the first with the result shown in
https://dl.dropboxusercontent.com/u/...Difference.jpg
That's right: solid black.

4. To confirm the point I took a screen shot. See
https://dl.dropboxusercontent.com/u/...t%20Screen.jpg
Note the histogram. All of the pixels appear to be down at the zero
end of the scale: that is, jet black.

The only conclusion I can reach is that there is no difference between
a PSD created from a RGB file and a PSD created from the same image
when it has first been converted from RGB to Lab.

I'm not wedded to the perfection of the method I have used and I would
be interested to hear from anyone who has a meaningful criticism.

We went through all this some many months ago. I demonstrated clearly
that the amount of 'loss' was negligible in practical terms.

Procedu

1. An original in JPG
2. The original converted to LAB version
3. The LAB version converted to JPG.

delta 1-2, delta 2-3.

In the deltas you will see the actual difference (hard to see unless
your screen is turned up bright) and it is not something that would be
noticeable in an actual screen or print of a shot.

I couldn't see the difference at all, but then I didn't want to push
the screen with excessive brightness. Instead I relied on the
subtraction and the histogram to find the differences, which were
almost zero.

nospam has backed off considerably from his original views but I
expect that won't stop him from trumpeting them again in the future.
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:

4. To confirm the point I took a screen shot. See
https://dl.dropboxusercontent.com/u/...t%20Screen.jpg
Note the histogram. All of the pixels appear to be down at the zero
end of the scale: that is, jet black.

notice the differences at the left end of the histogram.

however, this is about round-tripping from rgb to lab and then back.
you only did half.

Fir comment. I've just compared the original JPG with a copy -- Lab
-- JPG again. JPGs are RGB are they not?

usually but not always

Then what else might they be and under what circumstances?

cmyk


The only conclusion I can reach is that there is no difference between
a PSD created from a RGB file and a PSD created from the same image
when it has first been converted from RGB to Lab.

there is. it may not be a huge difference, but there is a difference.

As soon as you do anything in Photoshop there is a difference due to
rounding errors (quantization) but is this all you are objecting to?

you do realize that adds up, right?

Yes, and it's common to evrything you do. So why does converting to
Lab allegedly make it so much worse?

i didn't say converting to lab was much worse.

i said that rgb-lab-rgb is not lossless. you may not care about the
loss, but it's definitely there.

margulis is wrong.

compare a high quality jpeg with the original and you'll see black as
you did above, but there are definitely differences (and actually, less
of a difference than the rgb-lab conversion).

What is the difference with rgb-Lab-rgb conversions and what causes
them?

read the link and pay attention to andrew rodney.

Do you mean where he says:

"ANY colorspace conversion can cause these quantization errors (RGB
to RGB as an example)."

that's part of it.

ignore marguilis, not just in that link but in general. he has claimed
that 16 bit editing was a waste, which it absolutely is not. i dunno if
he still claims it but he probably does.

I bet you are quoting him out of context.

nope.

http://www.brucelindbloom.com/index.html?DanMargulis.html
...If an example is presented that shows an 8-bit/16-bit difference,
a rule is immediately created, on-the-spot, that disqualifies the
image. None of Dan's original six conditions would disqualify a
ProPhoto image (you can read these conditions below in section I),
but it appears as though ProPhoto images are no longer acceptable. If
one takes this technique to its logical conclusion, Dan's 16-bit
challenge would become "When considering all images showing no 16-bit
advantage, 16-bit images show no advantage."

do you see people arguing to edit jpegs? of course not.

What exactly do you mean by that?

you say you can't see a difference in an rgb-lab-rgb conversion and you
subtracted them and saw all black, therefore, you have deemed them to
be equivalent.

I didn't say that. Read it all again carefully. I compared an
rgb-lab-rgb conversion to the original JPG.

you said you saw black when subtracting them.

if you do the same for jpeg, you will also not see a difference, and if
you subtract, you'll also see all black. therefore, a jpeg should be
equivalent to an original raw.

That is squiffy logic and it's not even a good parody of what I did.

it's *exactly* the same logic.

you're position is if you can't see it then there is no difference.

the reality is that there *is* a difference. you might not consider the
difference to be significant (and indeed it is is very small), but
there *is* a difference, therefore it is *not* lossless.

bottom line: rgb-lab-rgb offers no benefit (other than possibly
contrived edge cases nobody will ever encounter).

You have backed off considerably from your original opinion on this
matter.

no i haven't at *all*.

stop lying about what i say.

In article , Eric Stevens
wrote:

I couldn't see the difference at all, but then I didn't want to push
the screen with excessive brightness. Instead I relied on the
subtraction and the histogram to find the differences, which were
almost zero.

'almost zero' is not zero.

you are actually proving my point.

nospam has backed off considerably from his original views but I
expect that won't stop him from trumpeting them again in the future.

i have *not* done any such thing. stop lying and twisting what i say.

i have *always* said it's not lossless and it is not.

this is a fact, no matter how much you or anyone else say otherwise.