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Author Topic: attention color whizes: non-typical sRGB/RGB/ProPhoto question  (Read 95440 times)
gongtrip
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« on: December 30, 2010, 09:53:30 PM »
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Ok so I understand the fundamental differences of srgb, adobe rgb and prophoto rgb and I understand that working in adobe rgb or prophoto rgb theoretically gives you a wider color gamut to work with or in the case of prophoto also a greater bit depth, but what I don't understand is why would you do this if your monitor can only display the srgb gamut?  If you can't see the wider spectrum of these color spaces what is the point of using them.  Frankly, making adjustments to an image that I can't necessarily see would make me a little nervous.  What am I missing here?

 

I guess the one situation I can undersand is if you are doing extremely intensive effects work and want to use the greater bit depth of prophoto to minimize possible color artifacts but beyond that are these color spaces really a benefit?

 

Say a printer can print the entire adobe rgb gamut, again what is the point of working in this space unless you own a specialized monitor that can display the entire rgb gamut?  In effect, wouldn't your srgb monitor just be clipping the most saturated colors such that you don't really know how colors beyond the srgb gamut limit will look in the final print?

 

Another thought, since adobe rgb is also an 8 bit space like srgb, but has a wider gamut, the discreet steps between color values would be less discreet would they not?  So it seems to me that working in this space would gain you a greater range of super saturated colors but would be at the expense of losing some amount of subtlety in the overall tonal values.  Am I envisioning this correctly?
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« Reply #1 on: December 30, 2010, 10:15:14 PM »
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Answered you on the Adobe Forums...as did other people. You might want to ask a question on one forum and wait for an answer before broadcasting your question to multiple forums. And no, you DON'T "understand the fundamental differences of srgb, adobe rgb and prophoto rgb"...read up and get back to us.
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gongtrip
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« Reply #2 on: December 31, 2010, 05:41:13 AM »
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Indeed, I posted this question in a couple of other forums--I like to adopt strategies geared towards absorbing information as quickly as possible...

Thanks, to yours an others' responses the threads contain some very useful insights.  I've included hyper-links below for posterity's sake.



http://forums.adobe.com/thread/771605

http://forums.dpreview.com/forums/readflat.asp?forum=1006&message=37331953
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Steve Weldon
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« Reply #3 on: December 31, 2010, 07:15:16 AM »
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Indeed, I posted this question in a couple of other forums--I like to adopt strategies geared towards absorbing information as quickly as possible...

I suppose that's nice for you, but have you considered reasons why many people can't be bothered answering forum posts? 

Allow me to tell you what rubbed me the wrong way with your post:

"Re:  attention color whizes: xxxxxxxxxx"

a.  Gratuitous and misspelled. 

b.  That which you claimed to know you didn't.

c.  Your response portrays you as entirely thoughtless and self-serving.

Google is where one finds information to absorb.  "People" are real live humans taking the time to answer your questions.  They deserve respect and consideration.  Their time is valuable. 

I won't answer a forum post if:

a.  The OP has a history of not doing a reasonable amount of research prior to posting.  Often this 'history' can be determined in a single post.

b.  If I get even a hint that instead of trying to understand, they're 'polling' as if taking a vote.

c.  I'm not fond of gratuitous salutations, especially if not spelled correctly.

I could understand posting on multiple forums all at once if looking for a rare something or the other, or even trying to sell something.  But to gather a bulk of information to 'absorb' when its apparent they haven't searched and read past threads, Googled the topic, or maybe even read their owners manuals.. is a tad boorish no?
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« Reply #4 on: December 31, 2010, 08:56:59 AM »
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Ok so I understand the fundamental differences of srgb, adobe rgb and prophoto rgb and I understand that working in adobe rgb or prophoto rgb theoretically gives you a wider color gamut to work with or in the case of prophoto also a greater bit depth, but what I don't understand is why would you do this if your monitor can only display the srgb gamut? 

ProPhoto by itself does not have a greater bit depth (you want to be working in a wider depth as a working space gamut becomes larger). See:http://www.adobe.com/digitalimag/pdfs/phscs2ip_colspace.pdf
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joofa
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« Reply #5 on: December 31, 2010, 09:25:44 AM »
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It does not seem like that Prophoto RGB is always wider than Adobe RGB. It seems that saturated colors around blue Adobe primary might clip in Prophoto RGB in standard specification of these spaces. For more information please see:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=37330104

Joofa
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« Reply #6 on: December 31, 2010, 11:59:35 AM »
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You wrote: "It appears that when the third dimension is taken into account it is seen that the area near the saturated unit stimulus blue region of Adobe RGB requires more than unity stimulus from blue ProPhoto RGB primary, i.e., there might be colors that clip for unit stimulus blue primary of ProPhoto RGB but are within the unit stimulus of Adobe RGB."

Aside from the terminology being a bit, odd, the fact is you are talking about theoretical colors in the extreme blue region which prolly don't exist on earth and certainly won't be able to be captured on a digital camera...there ARE however plenty of REAL colors that a camera can capture that can't be contained in Adobe RGB and some like yellow and ocher can actually be printed on an Epson 9900 (and even the 3880).

The irony is that the opening illustration of the 2D plots is actually an image I created for an Adobe white paper. I see it's been appropriated and is now on Wikipedia? BTW, the gamut of the 2200 is WAY below the gamut of the Ultrachrome K3 with vivid magenta or the 10 color HDR ink set. Maybe I should update that...
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« Reply #7 on: December 31, 2010, 01:45:21 PM »
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Aside from the terminology being a bit, odd, the fact is you are talking about theoretical colors in the extreme blue region which prolly don't exist on earth and certainly won't be able to be captured on a digital camera...there ARE however plenty of REAL colors that a camera can capture that can't be contained in Adobe RGB and some like yellow and ocher can actually be printed on an Epson 9900 (and even the 3880).

The irony is that the opening illustration of the 2D plots is actually an image I created for an Adobe white paper. I see it's been appropriated and is now on Wikipedia? BTW, the gamut of the 2200 is WAY below the gamut of the Ultrachrome K3 with vivid magenta or the 10 color HDR ink set. Maybe I should update that...

Sorry for the terminology being a little confusing. It is certainly a pleasant surprise that the Wikipedia image was created by you actually.

You are right about Adobe RGB gamut falling short of some useful colors. My article is not claiming otherwise. It just points out that it appears with the standard specification of white points for Prophoto RGB and Adobe RGB some saturated Adobe blues might not have exact representation in Prophoto space. But, you do have a point that these blue color shades may not be practically observable after all. You have a vast experience in color science so I can buy that statement.

Sincerely,

Joofa
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bjanes
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« Reply #8 on: December 31, 2010, 03:25:52 PM »
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It does not seem like that Prophoto RGB is always wider than Adobe RGB. It seems that saturated colors around blue Adobe primary might clip in Prophoto RGB in standard specification of these spaces. For more information please see:

http://forums.dpreview.com/forums/read.asp?forum=1018&message=37330104

Joofa

Joofa,

I don't quite understand how you are applying chromatic adaption in your calculations. ?Bradford

For a concrete example, please give the extreme blue in aRGB that clips in ProPhotoRGB. Then we can use Bruce Lindbloom's CIE color calculator to check the results.

Regards,

Bill
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joofa
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« Reply #9 on: December 31, 2010, 05:19:37 PM »
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Joofa,

I don't quite understand how you are applying chromatic adaption in your calculations. ?Bradford

For a concrete example, please give the extreme blue in aRGB that clips in ProPhotoRGB. Then we can use Bruce Lindbloom's CIE color calculator to check the results.

Regards,

Bill

Hi Bill,

Yes, it is Bradford in concept. I checked out the calculator on Bruce Lindbloom's site and played around a little. The Adobe blue primary with standard D65 white point has the XYZ coordinates, [0.188185   0.075274   0.991108], which using Lindbloom's calculator for conversion from D65 to D50 white point with Bradford transformation results in [0.149224 0.063220 0.744839]. You can use one of the matrices in Lindbloom's arsenal to convert this tristimulus value to Prophoto RGB with D50 white point and that results in [0.146617   0.029374   0.902605]. This situation corresponds to following case in my note:

Quote
Joofa wrote on DPReview:

Fraction of unit stimulus blue ProPhoto RGB primary needed to match unit stimulus blue Adobe RGB primary:
(3) Adobe RGB white point=D50, PropPhoto RGB white point=D50, Fraction needed=0.88

The value of 0.902605 is close to 0.88 that I mentioned, and may be attributed to the fact that a direct calculation of Adobe RGB blue primary with D50 white point results in XYZ tristimulus of [0.137826   0.055130   0.725885] (compare Lindblom's [0.149224 0.063220 0.744839]).

EDIT: For the case (1) in my note you have to do some matrix manipulation. But the starting point will be the same Adobe D65 primary of XYZ = [0.188185   0.075274   0.991108].

Sincerely,

Joofa
« Last Edit: December 31, 2010, 05:30:52 PM by joofa » Logged

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bjanes
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« Reply #10 on: January 01, 2011, 08:05:21 AM »
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Hi Bill,

Yes, it is Bradford in concept. I checked out the calculator on Bruce Lindbloom's site and played around a little. The Adobe blue primary with standard D65 white point has the XYZ coordinates, [0.188185   0.075274   0.991108], which using Lindbloom's calculator for conversion from D65 to D50 white point with Bradford transformation results in [0.149224 0.063220 0.744839]. You can use one of the matrices in Lindbloom's arsenal to convert this tristimulus value to Prophoto RGB with D50 white point and that results in [0.146617   0.029374   0.902605]. This situation corresponds to following case in my note:

The value of 0.902605 is close to 0.88 that I mentioned, and may be attributed to the fact that a direct calculation of Adobe RGB blue primary with D50 white point results in XYZ tristimulus of [0.137826   0.055130   0.725885] (compare Lindblom's [0.149224 0.063220 0.744839]).

EDIT: For the case (1) in my note you have to do some matrix manipulation. But the starting point will be the same Adobe D65 primary of XYZ = [0.188185   0.075274   0.991108].


Joofa,

Calculation of the primaries for various color spaces is an interesting exercise, but that was not the point of my original question. For simplicity and since I am not that familiar with color calculations, I prefer to do all the calculations within Lindbloom's calculator. One can start out with with a blank entry and select Adobe RGB with the most saturated possible blue (0, 0, 255) as shown in panel 1. One then calculate the XYZ values for this color as in Panel 2, change the RGB to ProPhotoRGB and calculate the Prophoto values for this XYZ.

0,0,255 in aRGB is 88.8, 35.9, 240.9 in ProPhotoRGB. My challenge is to find a saturated  blue in aRGB which can not be represented in ProPhotoRGB. My intuition is that there is no such color.

Regards,

Bill

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joofa
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« Reply #11 on: January 01, 2011, 12:28:52 PM »
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Joofa,

Calculation of the primaries for various color spaces is an interesting exercise, but that was not the point of my original question. For simplicity and since I am not that familiar with color calculations, I prefer to do all the calculations within Lindbloom's calculator. One can start out with with a blank entry and select Adobe RGB with the most saturated possible blue (0, 0, 255) as shown in panel 1. One then calculate the XYZ values for this color as in Panel 2, change the RGB to ProPhotoRGB and calculate the Prophoto values for this XYZ.

0,0,255 in aRGB is 88.8, 35.9, 240.9 in ProPhotoRGB. My challenge is to find a saturated  blue in aRGB which can not be represented in ProPhotoRGB. My intuition is that there is no such color.

Regards,

Bill




Hi Bill,

I think you selected the D50 white point for Adobe RGB, where as the standard says to use D65. I used Lindbloom's calculator with [0 0 255] for Adobe RGB with D65 white point and got the XYZ = [0.188186 0.075274 0.991109]. Now I converted that to ProPhoto RGB with D50 white point on the calculator and it gave me [99.3813 37.2770 282.3180] for gamma = 1.8, or if you use gamma = 1.0 you get [46.76 8 306.2649]. The value of blue in both, i.e., 282 and 306 is greater than 255, and so will clip for 8 bit numbers. Incidently, 306/255=1.2 for linear calculation, and that is what I reported as case 1 in my note as shown below:

Quote
Joofa wrote on DPReview:

Fraction of unit stimulus blue ProPhoto RGB primary needed to match unit stimulus blue Adobe RGB primary:

(1) Adobe RGB white point=D65, PropPhoto RGB white point=D50, Fraction needed=1.2

Sincerely,

Joofa
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« Reply #12 on: January 01, 2011, 02:11:27 PM »
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Hi Bill,

I think you selected the D50 white point for Adobe RGB, where as the standard says to use D65.
Sincerely,

Joofa

Joffa,

I don't think that I selected a D50 white point for Adobe RGB, since the calculator automatically and correctly uses a D65 white point for Adobe RGB. The D50 on the calculator refers to the white point of the reference (XYZ) space. When the RGB values of 0,0,255 in Adobe RGB are converted to ProPhotoRGB with the calculator, I get RGB values of 88.76, 35.93 and 240.89. As a check, I filled an AdobeRGB image with blue (0,0,255) in Photoshop and then converted to ProPhotoRGB and got 88, 36, 241.

I'm hardly a color expert and hope that Eric Chan or some other expert will clarify the issue. I just can't believe that a saturated blue in AdobeRGB can not be contained in ProPhotoRGB. I don't understand your calculations, but do note that the blue primary of ProPhotoRGB is imaginary--it lies outside the CIE L*a*b space.

Regards,

Bill
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« Reply #13 on: January 01, 2011, 02:31:30 PM »
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Joffa,

I don't think that I selected a D50 white point for Adobe RGB,

The gif attachment in your earlier message shows that you picked D50 for Adobe RGB. Please change that to D65 and you will get the numbers that I'm getting.

Quote
I get RGB values of 88.76, 35.93 and 240.89. As a check, I filled an AdobeRGB image with blue (0,0,255) in Photoshop and then converted to ProPhotoRGB and got 88, 36, 241.

The notion of "blue" in the Adobe RGB with D50 and D65 are different. Probably Photoshop is also doing the same track of calculation as you did above.

Quote
I just can't believe that a saturated blue in AdobeRGB can not be contained in ProPhotoRGB.

You can represent saturated Adobe RGB blues if the white point is D50 and you then go to ProPhoto with D50 white point. However, with the standard specification of D65 white point for Adobe RGB the resulting saturated blue is different than with D50 white point. And that blue color cannot be contained in ProPhoto primaries with standard D50 white point without using more than unity stimulus from blue Prophoto RGB. Incidently you can still represent that blue in the same ProPhoto primaries (same x,y chromacity numbers) if you change the ProPhoto white point from D50 to D65.

Quote
I don't understand your calculations,

This time I used Bruce Lindbloom's calculator in my calculations as shown in the above message.

Quote
but do note that the blue primary of ProPhotoRGB is imaginary--it lies outside the CIE L*a*b space.

Its okay to have imaginary primaries. All three XYZ primaries are imaginary as far as human vision is concerned.

Sincerely,

Joofa
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« Reply #14 on: January 01, 2011, 02:45:14 PM »
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The gif attachment in your earlier message shows that you picked D50 for Adobe RGB. Please change that to D65 and you will get the numbers that I'm getting.
Joofa

I don't think so. See the attached image with the spaces circled in yellow. Does Photoshop also make the same mistake? I've gone as far as my current knowledge permits in this discussion and will leave further comment to others. Anyway, it has been an interesting learning experience.

Regards,

Bill
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« Reply #15 on: January 01, 2011, 02:49:12 PM »
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I don't think so. See the attached image with the spaces circled in yellow. Does Photoshop also make the same mistake? I've gone as far as my current knowledge permits in this discussion and will leave further comment to others. Anyway, it has been an interesting learning experience.

Regards,

Bill

You have highlighted the incorrect parameter. See the drop-down above where you have circled. It clearly says D50. Please change that to D65 for this Adobe RGB. However, when you go to ProPhoto calculation change that back to D50.

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« Reply #16 on: January 01, 2011, 02:49:50 PM »
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I don't think so.

Neither do I.

The WP of Adobe RGB (1998) is D65. Its as simple as that. If you try to define the space using D50, it ain’t Adobe RGB (1998) anymore. Go ahead and call it joofaRGB. <G>
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« Reply #17 on: January 01, 2011, 02:52:23 PM »
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Neither do I.

The WP of Adobe RGB (1998) is D65. Its as simple as that. If you try to define the space using D50, it ain’t Adobe RGB (1998) anymore. Go ahead and call it joofaRGB. <G>

Yeah, but Bill is using the D50 white point when he thought he used the D65 white point. Please see my message above.

Joofa
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« Reply #18 on: January 01, 2011, 03:50:58 PM »
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Yeah, but Bill is using the D50 white point when he thought he used the D65 white point. Please see my message above.

I think its still correct as you can see, he circled D65, which the calculator places automatically based on the correct WP for the working space selected. Notice it is again correct when he selected ProPhoto, the WP is now D50. The PCS WP should be set to D50, again as he’s selected (and on by default).
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« Reply #19 on: January 01, 2011, 04:25:44 PM »
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I think its still correct as you can see, he circled D65, which the calculator places automatically based on the correct WP for the working space selected. Notice it is again correct when he selected ProPhoto, the WP is now D50. The PCS WP should be set to D50, again as he’s selected (and on by default).

I don't think that what you are saying is correct. In the equations sections of Bruce Lindbloom's website he has clearly listed the following matrix for transformation from Adobe RGB (D65) to XYZ

0.5767309  0.1855540  0.1881852
0.2973769  0.6273491  0.0752741
0.0270343  0.0706872  0.9911085

See the last column, which is [0.1881852 0.0752741 0.9911085], and which are the XYZ for a saturated blue in Adobe RGB with D65 white point, and this is the same set of numbers which I have mentioned above you will get by selecting the D65 in the drop down menu, which Bill did not do in his calculation. Even if you ignore the calculator, this matrix, which Lindbloom has himself provided on the same website, clearly establishes that saturated blue in this system is [0.1881852 0.0752741 0.9911085] and not [0.149224 0.063220 0.744839] as Bill has mistakenly used in his calculation.

Joofa
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