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PopPhoto and white balance color correction in Lightroom

Doug Kerr

Well-known member
In the latest print issue of Popular Photography, an article by postprocessing wizard Debbie Grossman, "Face Facts", gives hints on editing portraits in Lightroom. Step 1 is crop to suit. Step 2 is "Adjust your white balance". It suggests using the whites of the model's eyes as a neutral target. It suggests moving the eyedropper tool over that area, watching the report of the color as R,G,B, until a spot is found where the three values are nearly equal, and then clicking on that spot.

Now I may be handicapped by not ever having used Lightroom, but I am baffled by this. It would seem to me that when we click the eyedropper on a certain spot in the image, we are telling the program, "Shift the chromaticities in the image so this spot has the chromaticity of the color space white point".

But, if we have followed the directions, we have (if possible) clicked on a spot that already has almost the chromaticity of the white point (R=G=B). So nothing would be changed - that is, we have in fact ascertained that the white balance is already correct.

Now, if it is the white balance is not already correct, then presumably (since we assume that the whites of the model's eyes are everywhere very close to neutral) we would not find a place where the eyedropper reads approximately equal R, G, and B values. Then what?

So what am I missing here? Or has this "procedure" just been scrambled in the retelling?

Best regards,

Doug
 

Asher Kelman

OPF Owner/Editor-in-Chief
Doug,

It so happens, that absent a grey card, one can hunt for the whites of the eyes or even teeth and sometimes the black pupil to get white balance adjusted. A block or column of concrete, a black dress, (watch out for many dresses are actually black-purple as they have weird synthetic dyes), or even unpolished stainless steel or aluminum.

On has to hunt and make a judgement against one's mental references as to what is desirable. The secret of color correction is mostly to avoid it unless one has a grey card shot or calibrated/profiled monitor.

The software adjusts the actual RGB values to remove any contaminating hue. If one has a calibrated monitor, then one can observe whether or not sampling the white, black or grey areas that should have been neutral, gives a "better", more "natural", pleasant or agreeable color.

My rule is not to ever correct color until one has a monitor calibrate or a printer to check that color as what a monitor shows that seems off could be spot on and vice vace

After correction, of course, the area smiled will become registered as R=G=B as expected.

Asher.
 

Doug Kerr

Well-known member
Hi, Asher,
It so happens, that absent a grey card, one can hunt for the whites of the eyes or even teeth and sometimes the black pupil to get white balance adjusted. A block or column of concrete, a black dress, (watch out for many dresses are actually black-purple as they have weird synthetic dyes), or even unpolished stainless steel or aluminum.

Indeed.

On has to hunt and make a judgement against one's mental references as to what is desirable. The secret of color correction is mostly to avoid it unless one has a grey card shot or calibrated/profiled monitor.

Sounds like wise advice!

The software adjusts the actual RGB values to remove any contaminating hue. If one has a calibrated monitor, then one can observe whether or not sampling the white, black or grey areas that should have been neutral, gives a "better", more "natural", pleasant or agreeable color.

My rule is not to ever correct color until one has a monitor calibrate or a printer to check that color as what a monitor shows that seems off could be spot on and vice vace

Again, sounds like wise advice.

After correction, of course, the area smiled will become registered as R=G=B as expected.

Yes, of course.

But of course if the "neutral target" is already R=G=B in the image as taken, then eyedroppering it will not cause any change at all.

And, conversely, if the chromaticity of the illumination is not equal to the white point of the color space (so that a color adjustment is needed), then there will be no place on the target for which R=G=B.

So I have no idea what that article was trying to say!

Thanks.

Best regards,

Doug
 

Asher Kelman

OPF Owner/Editor-in-Chief
............
And, conversely, if the chromaticity of the illumination is not equal to the white point of the color space (so that a color adjustment is needed), then there will be no place on the target for which R=G=B.

...........


Doug,

If the light that illuminated the grey card shifted the R=G=B values that would be measured on the card at a temp of 5,000 degrees in the color space one chooses to work in, then the values will be offset according to the contaminating chromaticites that fall outside of the expected set of reflections.

The software will detect all these contaminating hues and create a correction curve so that all colors in that scene are decontaminated by the same amount as id the scene had been illuminated at 5,000 degrees quality light.

The obvious finding would be that if there is a point in the image one could discover, identical to the grey card in color, it would, after correction have R=G=B and all other different colors would be corrected to remove the color hue detected in that sampled grey point, according to the curve that Photoshop creates.

Asher
 

Doug Kerr

Well-known member
Hi, Asher,
Doug,

If the light that illuminated the grey card shifted the R=G=B values that would be measured on the card at a temp of 5,000 degrees in the color space one chooses to work in, then the values will be offset according to the contaminating chromaticites that fall outside of the expected set of reflections.

The software will detect all these contaminating hues and create a correction curve so that all colors in that scene are decontaminated by the same amount as id the scene had been illuminated at 5,000 degrees quality light.

Actually, as if the scene had been illuminated by light whose correlated color temperature is that of the color space in which we are working. For sRGB, that would be 6500 K. (But there is a complication, which I won't belabor here.)

The obvious finding would be that if there is a point in the image one could discover, identical to the grey card in color, . . .

Here we must mean "reflective color". But we cannot discover such a point by examination of the taken image, since the chromaticity of the light from each spot in the scene is a function of both the "reflective color" of the object at that point and the chromaticity of the illumination. We must know of an object in the scene whose reflective color is known to be essentially neutral (such as a "gray card", or a calibrated white pendant, or the whites of a typical human eye.)

it would, after correction have R=G=B . . .

Yes. But of course in the scenario about which I am inquiring (the one of the PopPhoto "instructions"), the recorded color of the chosen spot is already (very near to) R=G=B. That is the conundrum at hand.

and all other different colors would be corrected to remove the color hue detected in that sampled grey point, according to the curve that Photoshop creates.

Sure.

Best regards,

Doug
 

Doug Kerr

Well-known member
I alluded in my previous message to a "complication" regarding the white point of the sRGB color space.

The white point of the color space is illuminant D65, which has a correlated color temperature (CCT) of 6500 K.

That means that in an image encoded as sRGB, a spot for which R=G=B would ideally create in a display or a print a coior whose chromaticity was D65 (and thus whose CCT was 6500 K).

But here is the complication. The sRGB standard prescribes that a scene for which an image is to be encoded in the sRGB color space (and not needing any white balance color correction) be illuminated by light of chromaticity D50, with a CCT of 5000 K.

Thus in effect there is built into the scheme of sRGB a white balance color miscorrection.

This is a "plug" put in to deal with the reality that if a printer creates "white" at the highest possible level by depositing no ink on the paper, for the printer papers that were common at the time, the "virgin" paper, seen under an illumination of D65, would look slightly bluish. (Like fluorescent dyes in laundry detergent, this is intended to make things look "whiter" on the paper.) Thus the entire image would "look a little more bluish" than it should.

So if, although the white point of sRGB has a CCT of 6500 K, we still think of illuminating the scene with illumination with a CCT of 5000 K, and ideally make no "color correction" for that discrepancy. Thus, everything in the image will be "a little redder" than it should, which will compensate for the fact that the image will be printed on "slightly bluish" paper.

And, for viewing the image on a monitor, the sRGB standard recommends illumination of the viewing environment by D50 light (5000 K). Thus the eye's chromatic adaptation to that chromaticity will shift the perception of the colors in the image toward the blue, making this all "come out even".

This is the usual mess we have in photographic standards!

This is no doubt, Asher, why you speak of the "ideal" illumination for a scene as having a CCT of 5000 K, and of color correcting the image to a "5000 K" basis.

Best regards,

Doug
 

Asher Kelman

OPF Owner/Editor-in-Chief
"This is no doubt, Asher, why you speak of the "ideal" illumination for a scene as having a CCT of 5000 K, and of color correcting the image to a "5000 K" basis."

Exactly, but in truth, I did not go through the light-engineering steps you have illuminated for us!

❤️

Thanks,

Asher
 

Doug Kerr

Well-known member
The paradox in the scheme I describe above, which I do not pretend to have fully resolved in my own mind, is this:

Suppose we have in our scene a "neutral" object, perhaps a pendant (nicely nestled in the model's ample cleavage) whose "stone" is made of a neutral-reflectance material, such as Spectralon. It is quite reasonably said to be a "white" object. (If that isn't a "white" object, I don't know what would be.) The scene is illuminated by light whose chromaticity we do not know.

Our actual end-to-end objective (rarely mentioned) is that a person viewing this image (on-screen, or as a print) will perceive the image of that object to be "white".

• For viewing on-screen, that means that the light emitted by the display for this object must have the same chromaticity as that of the light that illuminates the "viewing location".

• For viewing in a print, it means that the area on the print for this object must itself have a neutral reflectance.

Now let us consider our processing of this "shot". With the image loaded into our favorite photo processor, we place the "white balance eyedropper" on the image of the pendant stone. The program will show us the current RGB coordinates for that spot. Likely we will not see R=G=B.

We click to tell the program, "make this spot on the image "white" and adjust all the other colors in the image accordingly.

Of course "white" means the chromaticity of the white point of the color space in use, for which by definition R=G=B.

And in fact, after this has happened, with the eyedropper still on the pendant stone, we see that the color there now has R=G=B.

And, if the color space in use is sRGB, any cold code with R=G=B implies a color whose chromaticity is the "white point" defined for teh sRGB color space, namely D65 (which, among other specifications, has a correlated color temperature (CCT) of 6500 K.

Now, at the "viewing location", I have always assumed that when my display system is calibrated (based on sRGB), the objective is to have a signal with R=G=B generate light that is like D65, whose CCT is 6500 K. (And if the display proper can't do that exactly, over a range of luminances, the "profile" for the display that is used by the displaying software will compensate for that.

But if in fact, as recommended (and as is the case here), the viewing environment is illuminated by D50 light (CCT 5000 K), then to me the D65 light emitted from the display for the pendant stone will look "bluer than white".

So I am missing something about the "scheme" crafted by the developers of the sRGB standard.

Best regards,

Doug
 

Asher Kelman

OPF Owner/Editor-in-Chief
Yes.

The ample clevage.

Best.

I have never been an avid follower of cleavage, ample or otherwise as that is merely an empty space with at best some intriguing shadows. What's on either side is much more interesting and enticing.

In the natural state, there is no cleavage. When it is seen, it's just an artifact of eye-catching teasing prudery instead of open display of the female form! It's never shown by accident, but meant to be seen. The paradox of social custom is that to overtly stare at a cleavage is considered rude, but the woman displaying a generous cleavage is tantalizing!

Anyway, we men are suckers for this manipulation!

Asher
 

fahim mohammed

Well-known member
Asher, ' white ' and ' balance '...that in
And of itself is an oximyron.

Clevage could just be a distraction..perceived or
Otherwise.
 
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