Linear, gamma and profiles (oh my!)

[kostache]

Thanks! I got it!

[marchermitte]

Did you try to correct your textures using 1 divided 2.2 (= 0.4546) which makes your textures look correct when redering to linear space?

[OctaneWannabe]

Did you try to correct your textures using 1 divided 2.2 (= 0.4546) which makes your textures look correct when redering to linear space?

sRGB gamma is a curve. The .4546 won't really remove it correctly - it's just and approximation.

Octane's developers should provide a node to remove it using the proper math, or it really won't look 'perfect'.

Other than that, I understand Photoshop will do a good job of converting textures.

[marchermitte]

Well, in photoshop (as I'm not sure about gamma correcting acuity in Lw) if you divide 1 (which is the linear space standard) by 2.2 (which is your screen gama and also your texture as most of the time this is the defaut gama) you do inverse the curve and get accurate gama correction for linear space image output. (the result beeing a gama of 0.4546)
That's what I do in Modo for instance on all my textures (there is an option to setup each texture gama directly)

[OctaneWannabe]

Well, in photoshop (as I'm not sure about gamma correcting acuity in Lw) if you divide 1 (which is the linear space standard) by 2.2 (which is your screen gama and also your texture as most of the time this is the defaut gama) you do inverse the curve and get accurate gama correction for linear space image output. (the result beeing a gama of 0.4546)
That's what I do in Modo for instance on all my textures (there is an option to setup each texture gama directly)

sRGB isn't really 2.2 gamma. It's a gamma curve that's close to that. http://en.wikipedia.org/wiki/SRGB

If you google a bit, you can see heated arguments on the subject by color purists.

The conversions look like this for sRGB to Linear and Linear to sRGB

def s2lin(x):
a = 0.055
return where(x <= 0.04045,
x * (1.0 / 12.92),
pow((x + a) * (1.0 / (1 + a)), 2.4))

def lin2s(x):
a = 0.055
return where(x <= 0.0031308,
x * 12.92,
(1 + a) * pow(x, 1 / 2.4) - a)

If you save your properly identified image (Photoshop must know exactly what it is to begin with - or it can't properly strip it out) as an EXR, I ? think ? that Photoshop will use the proper math to strip it out and save it as linear.

[marchermitte]

Ok, thank you for the link.
Modo does'nt require anything else than the /2.2 operation, so I'm used to the simple way and I'm confortable with It as long as It looks good. But really, I understand your request for a specific node, the whole operation is a bit tedious if you want to do It precisely.