Does PS use a linear shadow segment for sRGB and ProPhotoRGB?

Posted By
Feb 20, 2009
In the ACR forum, there was a recent discussion about linear segments for the shadows. These are official for sRGB and ProPhotoRGB but are not part of the aRGB specification, but a slope of less than 32 is allowed in the shadows for raw converters and is used by ACR for all three of these spaces.

Does Photoshop currently implement a linear segment for these color spaces and does it make that much difference in practical photography?

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Feb 20, 2009
Photoshop has always used a slope limiting segment in such gamma curves. It normally helps image quality. (and certainly helps the math by avoiding infinities)
Feb 21, 2009

Thanks for the authoritative answer. I am a bit confused, as the latest revision of the aRGB spec specifically states that there is no linear segment. The Annex C does state that implementations may use a linear segment in the shadows with a slope limit of 1/32. Is that what Photoshop uses?

One also hears about simplified sRGB in Photoshop. Is there such a thing?

Feb 22, 2009
The ICC profile AdobeRGB1998 doesn’t contain
a linear segment. The tone reproduction curve
is defined by a mathematical power function
with exponent 2.2.
Similarly for ProPhotoRGB, with exponent 1.8.
This can be shown by ICC Profile Inspector.

There are good reasons for using a linear slope
at the dark end, but IMO the profiles should be
re-defined. Otherwise some inaccuracies can be

ICC Profile Inspector doesn’t tell us an exponent
for sRGB, because the TRC is defined by a linear
part and a power function with exponent 2.4.
This delivers with good accuracy an average expo-
nent of 2.2.

Best regards –Gernot Hoffmann
Feb 23, 2009
Bill – yes, we use the slope limit of 1/32 even when the profile gives a single number gamma value. Not doing that would cause too many problems whenever you need to invert the gamma table. (people who write an explicit gamma table into the profile using a pure power function find this out the hard way — they can’t be correctly inverted)

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