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As per Principled shader docs, Specular parameter controls dielectric specular reflection.

As I understand it, when I set metallic to 1., I only get the specular (tinted) reflection, and no diffuse reflection.

What does changing the specular slider do for the fully metallic material? Should I set it to 0. for a physically accurate metal?

sygi
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1 Answers1

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Example of specularity

The above image is inspired by a diagram on page 13 of Disney's "Physically-Based Shading at Disney" "Brief note" about the original principled shader.

This was generated using a principled BSDF shader with everything except Metalic and Specularity set to default. Metalic is set to 1 and specularity to the value above the sphere.

It's a very subtle effect, but the way the principled shader is designed, it is in effect using the IOR of a material to simulate the Fresnel effect. It, in fact, implements a variant of the Fresnel equation to accomplish the transition between specular and diffuse.

Marty Fouts
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  • So that's the picture, but what does it correspond to physically? Is specularity the portion of the light that gets reflected in a specular way? If so, why is the remaining one diffused if metals don't have a diffused reflection?

    From the picture, it looks like it's defining whether the Fresnel effect is applied: does it mean only 1. is physically accurate?

     – sygi Nov 21 '21 at 09:30
  • @sygi Another explanation here: https://blender.stackexchange.com/a/155424/78972 – jachym michal Nov 21 '21 at 10:37
  • CG does not use the terms 'diffuse' and 'specular' in precisely the same way as physics does and the principled shader uses 'specularity' in a very non-physics manner. At best it only approximates physics and you're close, it's defining how the Fresnel effect is applied through a formula that relates 'specularity' to an IOR value and uses that IOR value to calculate Fresnel. – Marty Fouts Nov 21 '21 at 14:57