11 ms·
It does demonstrate that to some extent: specular light is omitted from the normal map, so it correctly maps the room behind the light. The last example also c
by gste 1mo ago
It does demonstrate that to some extent: specular light is omitted from the normal map, so it correctly maps the room behind the light.
The last example also correctly identified broken panes of glass in the background, so that the glass is a solid and the spaces have nothing there.