Are SA(L) and SA(R) here each representing a portion that sums up to 1 or representing the surface area? The notation here is different from the Prob() used in two slide before. I wonder if there is a difference.
jgforsberg
If L and R are the two children of an internal node N, then SA(L)+SA(R) = SA(N). So, the probability is just P(L) = SA(L)/SA(N). However, since this is an approximation, I think we can say P(L) is proportional to SA(L). It avoids having to divide by SA(N) and should be proportional to the cost using probabilities. We made this assumption at lot in ML and AI.
jeshlee121
In a few slides back, C_trav was just the cost to traverse which makes sense to me, but I don’t quite understand it in this slide. Why is C_trav a ratio, or more specifically the ratio of cost to traverse to cost to intersect tri?
zehric
I think a few slides ago this was explained at a higher level of abstraction. In this more concrete setting, we need to define exactly what "cost" means; here we just compare the cost of traversal to the cost of intersecting a triangle at the current level.
GitMerlin
I agree with @jgforsberg mostly, but, correct me if I'm wrong, I don't think SA(L)+SA(R)=SA(N). First off, SA() is computing the surface area, not the area of bounding box. Also, bounding boxes of subsets do not add up to bounding box of the entire set.
I don't understand the C_trav here either. It seems that C_trav in this slide incorporates the meaning of both C_trav and C_isect in previous slides. I don't know for sure.
knguyen0811
Are 'cell' and 'node' interchangeable here? The previous few slides used 'node' as the input to the cost function but here it is 'cell' and it's tripping me up...
raghav-cs184
[deleted]
raghav-cs184
Is a uniform ray distribution the best way to go about it? It seems to me that there could be a better distribution that ignores the edges (widely spaced rays) and concentrates in the center ot increase detail in the middle. I'm curious to see the kind of effects non-uniform distributions can achieve in images.
Gilbert-Han
I'm thinking C_trav is the ratio because the cost in this slide is scaled by a factor of 1/(cost to intersect tri) compared to the previous cost fn.
Are SA(L) and SA(R) here each representing a portion that sums up to 1 or representing the surface area? The notation here is different from the Prob() used in two slide before. I wonder if there is a difference.
If L and R are the two children of an internal node N, then SA(L)+SA(R) = SA(N). So, the probability is just P(L) = SA(L)/SA(N). However, since this is an approximation, I think we can say P(L) is proportional to SA(L). It avoids having to divide by SA(N) and should be proportional to the cost using probabilities. We made this assumption at lot in ML and AI.
In a few slides back, C_trav was just the cost to traverse which makes sense to me, but I don’t quite understand it in this slide. Why is C_trav a ratio, or more specifically the ratio of cost to traverse to cost to intersect tri?
I think a few slides ago this was explained at a higher level of abstraction. In this more concrete setting, we need to define exactly what "cost" means; here we just compare the cost of traversal to the cost of intersecting a triangle at the current level.
I agree with @jgforsberg mostly, but, correct me if I'm wrong, I don't think SA(L)+SA(R)=SA(N). First off, SA() is computing the surface area, not the area of bounding box. Also, bounding boxes of subsets do not add up to bounding box of the entire set.
I don't understand the C_trav here either. It seems that C_trav in this slide incorporates the meaning of both C_trav and C_isect in previous slides. I don't know for sure.
Are 'cell' and 'node' interchangeable here? The previous few slides used 'node' as the input to the cost function but here it is 'cell' and it's tripping me up...
[deleted]
Is a uniform ray distribution the best way to go about it? It seems to me that there could be a better distribution that ignores the edges (widely spaced rays) and concentrates in the center ot increase detail in the middle. I'm curious to see the kind of effects non-uniform distributions can achieve in images.
I'm thinking C_trav is the ratio because the cost in this slide is scaled by a factor of 1/(cost to intersect tri) compared to the previous cost fn.