Luminous Landscape Forum
Raw & Post Processing, Printing => Colour Management => Topic started by: Ellis Vener on September 18, 2015, 01:16:38 pm
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I know the Whitebal is not designed to be used as an exposure target, but does anyone have a good idea of what the R G B values in ACR should be for the gr black, gray, and white target area of the target? I am using the target as a neutral to show differences in exposure ranges ( by dialing the flash power down from maximum energy on the lights and then adjusting ISO and aperture as needed) for various electronic flashes
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I know the Whitebal is not designed to be used as an exposure target
any uniform surface can be
, but does anyone have a good idea of what the R G B values in ACR should be for the gr black, gray, and white target area of the target?
you can as well use a spectrophotometer to check those spots and you will find out the difference in reflectance (in the absence of contrast reducing conditions of your actual shooting scene that is)... as for ACR - which profile and conversion parameters ?
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The back of the WhiBal card gives the Lab readouts. Establish a base exposure by trial and error by opening the test shot in ACR and checking the RGB numbers along with L* readouts that match the WhiBal numbers.
Keep in mind the RGB readouts in ACR are determined by output working space encoding (i.e. ProPhotoRGB 1.8 gamma vs AdobeRGB 2.2).
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Have you asked Michael Tapes about your "application" ?
Jpegman
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Have you asked Michael Tapes about your "application" ?
Jpegman
Yes I did, and he responded. I thought I had posted his reply, but apparently not so here is his answer:
"The L* value of the WhiBal gray is 75. So in PS it is 75 in l*a*b* mode and in Lightroom 75%"
Deriving various RGB color space values of an L* value of 75 yields the following:
Pro Photo RGB (16 bit): 170/170/170
Adobe RGB (1998) (8 bit): 184/184/184
sRGB (8 bit): 185/185/185
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Yes I did, and he responded. I thought I had posted his reply, but apparently not so here is his answer:
"The L* value of the WhiBal gray is 75. So in PS it is 75 in l*a*b* mode and in Lightroom 75%"
Deriving various RGB color space values of an L* value of 75 yields the following:
Pro Photo RGB (16 bit): 170/170/170
Adobe RGB (1998) (8 bit): 184/184/184
sRGB (8 bit): 185/185/185
Yes I did, and he responded. I thought I had posted his reply, but apparently not so here is his answer:
"The L* value of the WhiBal gray is 75. So in PS it is 75 in l*a*b* mode and in Lightroom 75%"
Deriving various RGB color space values of an L* value of 75 yields the following:
Pro Photo RGB (16 bit): 170/170/170
Adobe RGB (1998) (8 bit): 184/184/184
sRGB (8 bit): 185/185/185
I don't want to appear pedantic, but I do not think Lightroom readouts in per cent correspond to L* but rather to normalized sRGB values expressed as percentages. To normalize sRGB, one divides the 8 bit value (0..255) by 255.
This graph shows sRGB vs L* from calculated values according to the formulas for the two spaces. The red line is the identity line. The sRGB values are close to the L* over the entire range, but deviate most prominently in the mid-range.
(https://bjanes.smugmug.com/Gamma-Calculations/i-cQrFgJL/0/O/GammaVsL_Star.png)
I created files in L*a*b in Photoshop and filled them with L* values of 50 and 75 and noted the sRGB values in Photoshop. I then imported the files into Lightroom and read out the luminance values in per cent. The results are shown below:
(https://bjanes.smugmug.com/Gamma-Calculations/i-XJG6Fjp/0/O/PSTable.png)
Note that L* of 75 is 185 in sRGB, or 0.725 when normalized. The value in Lightroom is 72.6%, not 75%. For L* of 50, the sRGB values are 119 and 0.467 and LR gives 46.7%, not 50%.
Regards,
Bill
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I don't want to appear pedantic, but I do not think Lightroom readouts in per cent correspond to L* but rather to normalized sRGB values expressed as percentages.
I believe you are absolutely correct Bill. It's using a 2.2 sRGB TRC outside of soft proofing which then uses the profile selected.
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any uniform surface can be
Ellis said "designed to be used as" - so not sure that this comment is relevant ::)