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Splitting specular_color between dielectric tint and metallic edge color #78
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That sharing is a legacy of Standard Surface. I guess the idea was that the base substrate is a mixture of metal and dielectric, both of which produce a specular lobe (due to the metallic or dielectric Fresnel, respectively). I do agree that it's potentially confusing that Arguably this overloading of meaning is reasonable to sacrifice some precision/expressivity for simplification of the parameter set. We also do this overloading for Your example of a mix of metal and dielectric would come up practically in cases where one is blending from a metallic region to a dielectric region, e.g. modelling flaky metallic paint on shiny plastic. Then if the plastic has the default white
In #66, Brecht notes that these more non-physical / artist-driven effects are referred to as "tints" in Blender, to distinguish from physical parameters. So a tentative proposal is to have:
This then:
but at the expense of adding two new parameters. Though note the F0 of metals will depend on both |
Probably not doing this, suggest to close. |
Currently, the
specular_color
parameter serves two purposes: It acts as a multiplier on top of the Fresnel term for the dielectric reflection, and it acts as the F82 parameter for the metallic component.I'm not sure if this is a good idea, since these two purposes seem quite different to me:
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