The ray misses the box if tmin is grater than tmax in some dimension.
yinxudeng
In lecture, it was mentioned that if tmin is greater than tmax in some dimension, then the ray misses the box. However, lets say we have an array starting from o pointing to the point a little above the intersection between x0 and y0. tmin is greater than tmax along the y planes, but the ray still intersects the box. Can someone please explain how we know when the ray misses the box?
CarneAsadaFry
@yinxudeng It shouldn't be possible for tmin to be greater than tmax along the planes in any single direction by definition. In the situation you described, I believe we would define tmin to be the intersection with the line y1 and tmax to be the intersection with y0, since this is the "order" the ray hits the two lines in. Then, along the x planes, tmin would be the intersection with x0 and tmax would be the intersection with x1, and then you'll see that the intersection of these intervals is non-empty (meaning the global tmin and tmax have tmin<tmax) as expected.
In summary, I believe we define tmin and tmax so tmin < tmax along each direction. But then to check if the ray is in the slab, we intersect all these intervals, and if the interval is empty, then the ray did not intersection (corresponding to the fact that there is no interval of time when the ray alingns with the box in along all axes). So, I think @bronyayang's description is not quite right.
cchendyc
I don't quite understand this slide...if we have x and y planes are trying to measure the intersection of planes and the ray, what does this y plane mean? how can we combine the two results?
melodysifry
To answer the question above, we combine the two results by making our final tmin be the greater of the two tmins we found for the x and y intersections, and tmax being the smaller of the two
The ray misses the box if tmin is grater than tmax in some dimension.
In lecture, it was mentioned that if tmin is greater than tmax in some dimension, then the ray misses the box. However, lets say we have an array starting from o pointing to the point a little above the intersection between x0 and y0. tmin is greater than tmax along the y planes, but the ray still intersects the box. Can someone please explain how we know when the ray misses the box?
@yinxudeng It shouldn't be possible for tmin to be greater than tmax along the planes in any single direction by definition. In the situation you described, I believe we would define tmin to be the intersection with the line y1 and tmax to be the intersection with y0, since this is the "order" the ray hits the two lines in. Then, along the x planes, tmin would be the intersection with x0 and tmax would be the intersection with x1, and then you'll see that the intersection of these intervals is non-empty (meaning the global tmin and tmax have tmin<tmax) as expected.
In summary, I believe we define tmin and tmax so tmin < tmax along each direction. But then to check if the ray is in the slab, we intersect all these intervals, and if the interval is empty, then the ray did not intersection (corresponding to the fact that there is no interval of time when the ray alingns with the box in along all axes). So, I think @bronyayang's description is not quite right.
I don't quite understand this slide...if we have x and y planes are trying to measure the intersection of planes and the ray, what does this y plane mean? how can we combine the two results?
To answer the question above, we combine the two results by making our final tmin be the greater of the two tmins we found for the x and y intersections, and tmax being the smaller of the two