What is p here? Is it the center of the sphere? Or is it any point in the sphere
VioIchigo
@mnicoletti15 I think p can be any surface position, based on the last slide.
kevinliu64
My understanding for why the value is Lpi instead of L2pi is because of the angle that each ray from the light source is hitting our point. Since it comes in at different angles, the energy will not be as much as if it was directly hitting it straight on. Thus, instead of it being the expected 2pi steradians, it's instead reduced because each ray is coming in at an angle.
michaeltu1
More precisely, I think p is the hit_point on the object where all the sampled light rays intersect the object.
Staffirisli
Yes, always remember Lambert's cosine law! In the hypothetical scenario where Lambert's cosine law didn't apply, then the answer would just be L∗2π
What is p here? Is it the center of the sphere? Or is it any point in the sphere
@mnicoletti15 I think p can be any surface position, based on the last slide.
My understanding for why the value is Lpi instead of L2pi is because of the angle that each ray from the light source is hitting our point. Since it comes in at different angles, the energy will not be as much as if it was directly hitting it straight on. Thus, instead of it being the expected 2pi steradians, it's instead reduced because each ray is coming in at an angle.
More precisely, I think
pis thehit_pointon the object where all the sampled light rays intersect the object.Yes, always remember Lambert's cosine law! In the hypothetical scenario where Lambert's cosine law didn't apply, then the answer would just be L∗2π