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VRayUserColor to pass a coordinate through to a shader

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  • VRayUserColor to pass a coordinate through to a shader

    What would be the best way to read a coordinate system from a Vray Proxy mesh for use in a displacement shader in Maya?

    We've written a rest attribute generated for a mesh in houdini into a meshed alembic, and this is read by the VrayProxy - you can see the channel exists in the Proxy file in Maya ok.

    Looking through the Vray documentation I can see that a VrayUserColor can be used to read user attributes from the VrayProxy, but will this be able to read the reference coordinate system? Any knowledge on the best way to wire this all together to drive a reference coordinate system?


    Thanks in advance!





  • #2
    You can add custom V-Ray attributes to the 3D placement or 2D placement node that is connected to your displacement map, and you can specify the name of the rest attribute there (see attachments).

    Best regards,
    Vlado
    Attached Files
    I only act like I know everything, Rogers.

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    • #3
      Lovely - thanks for the reply

      Comment


      • #4
        Hey Vlado,

        Expanding on this; if I wanted to read a color set from a VrayProxy to pipe into a shader, I'm assuming it's a similar sort of setup?
        i.e., create a VrayUserColor node, specify the name of the color set in the proxy, and apply the shader to the proxy?

        I have tried to create a VrayLightMaterial and directly pipe the VrayUserColor into the color of the light material. In the VrayUserColor I've specified the color set, but I'm just getting the color swatch, not the user attribute.
        Any pointers?

        --thanks!

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        • #5
          It should be exactly the same. If you scroll down in the V-Ray proxy attributes in the attribute editor, you will see a rollout "VRayMesh file info", which lists the available UV and color sets (you can generally use those interchangeably).

          Best regards,
          Vlado
          Attached Files
          I only act like I know everything, Rogers.

          Comment

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