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The specular color in metalness/roughness workflow is usually defined as following:

float3 specColor = lerp(0.03f, albedoColor, metallic);

The Cook-Torrance BRDF is given with the following formula:

Cook-Torrance BRDF

The final color will be:

Ci = (diffColor*(n.l)+Ks*specColor*cook);

My question is - won't specular be too dim for non-metals? Cook-Torrance term will be multiplied by 0.03 in case of dielectrics - this will make the specular component virtually non-existent. This doesn't seem realistic to me, because smooth plastic reflects a lot of light, almost as much as metals:

enter image description here

What am I missing?

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I think I got my answer - Beckmann distribution term from the Cook-Torrance equation becomes really high when the roughness is low: For roughness = 0.01 the Beckmann term becomes 3183; for 0.001, it becomes 318310.

The roughness has to be much lower on non-metals to make the specular highlight noticeable.

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