We’re using V-Ray Sun and Sky with physical camera. I can turn off shadows in the render stats of the object, but light linking to transform1/VRayGeoSun doesn’t work at all - for shading group or geometry. I’m wondering if the link with VRaySky needs to be broken, but it doesn’t show up as a light. I tried using a vrayobjectproperties set but that doesn’t have any way to effect direct lighting (the shadows are direct / in the lighting render element). Is there a way to do this, and if not, is it in the works? I can make a separate render layer, but that’s a definite compromise - I feel this is essential for sun and sky. Thanks,
We reported that issue to the developers.
For now as a workaround you can duplicate the objects VRayMtl and set the opacity of the duplicated one to black. Then open the shading group of the original VRayMtl, go to Attribute>V-Ray>Material Override and set the duplicated VRayMtl in the Shadow material slot.
Thank you for this workaround, although it doesn’t seem to solve the problem when you need an object to cast shadows from a light but not the sun. From my quick test this seems to do the same thing as turning off the ‘cast shadows’ render stat.
I believe you’re already working on a fix, as what I’m asking about is not shadow linking. Let me explain my request with an example. I have cube A, cube B, a VRayRectLight and VRaySun/sky. Currently using the relationship editor to set light linking for VRaySun/sky does not work. No matter what, I can’t ‘turn off’ VraySun or sky for cube B. So in this scenario I’d like cube A to be affected by the sun/sky and the VRayRectLight, and cube B affected by only the VRayRectLight. Let’s say VraySun is hitting cube B and casting an ugly shadow onto cube A. If I could unlink VraySun from cube B, it would no longer be illuminated from the sun, and it would also not cast a shadow from the sun. Cube B would still be receiving light and casting shadow form the VRayRectLight, though. The only way around this currently is to make a separate render layer for cube B. Thanks.