if (tymin > tmin) tmax = tzmax; t1z = (B1z - Oz) / Dz tmin = tzmin; if (tzmin > tzmax) swap(tzmin, tzmax); hitSurface, t = surface.intersect(ray, tMin, tBest); if hitSurface is not null {tBest = t; firstSurface = hitSurface;}} ... Top face of cube receives a certain amount of light Top face of 60º rotated cube intercepts half the light In general, light per unit area is proportional to t0y = (B0y - Oy) / Dy In fact they are the same if you replace \(O\) with \(b\) and \(D\) with \(m\). For instance the line equation for the x component of the bounding volume's minimum extent can be written as:The x component of the bounding volume's maximum extent can be used in a similar way to compute \(t1x\). t0y = (B0y - Oy) / Dy if (tzmin > tmin) }if (ray.dir.x >= 0) { Considering that all the logic behind the tests we are doing later on relies on \(t0\) being always smaller than \(t1\) we need to swap the values if \(t1\) is smaller than \(t0\).There is a couple of improvement we can make to the previous code to not only make it faster but also more robust. } Figure 1: equation of a line. else { t1y = (B1y - Oy) / Dy else { if ((tmin > tymax) || (tymin > tmax)) bounds[0] = vmin;
The problem at this stage is to find which of these six values correspond to an intersection of the ray with the box (if the ray intersects the box at all. To represent an axis-aligned bounding volume, all we need are two points representing the minimum and maximum extent of the box (called bounds in the code).The bounds of the volume define a set of lines parallel to each axis of the coordinate system which we can also expressed using the line equation. Here is the final version of the ray-box intersection method:The optimised version runs on our machine approximately 1.5 faster than the first version (the speedup might depend on the compiler).The code from this lesson returns intersections with the box which are in front or behind the origin of the ray. bounds[1] = vmax;
All we need is to remember that a straight line can be defined using the following analytical equation:In mathematics, the \(m\) term is called the slope (or gradient) and is actually responsible for the orientation of the line and \(b\) corresponds to the point where the line intersects the y-axis.
This though should be handled properly by the compiler which shall return +∞. tmin = (min.x - r.orig.x) / ray.dir.x; It is a kernel operation for BVH-based ray intersection, on which modern high-performance ray-casting algorithms are built. sign[2] = (invdir.z < 0); Note than when the values for \(t\) are negative, the box is behind the ray.Finally if we apply the same technique to the y and z components we will have at the end of this process a set of six values indicating where the ray intersects the box planes parallel to the x, y and z axis.Note that if the ray is parallel to an axis it won't intersect with the bounding volume plane for this axis (in this case, the line equation for the ray is reduced to the constant \(b\) and there's no solution to equation 1). The ray first intersects the planes defined by the minimum extent of the box in two places: \(t0x\) and \(t0y\). t1x : t1yt0z = (B0z - Oz) / Dz Treat those two cases as tests, starting with the first one: a ray intersecting a cube. { tmax = (max.x - r.orig.x) / ray.dir.x; } The intersection algorithm must decide whether a given ray intersects any of the cube’s six faces or whether the ray misses the cube altogether. Show that the local_intersect function for a cube correctly identifies intersections on any face. { sign[1] = (invdir.y < 0); };t0x = (B0x - Ox) / Dx return false; if (t1z < tmax) tmax = t1zbool intersect(const Ray &r) t1y = (B1y - Oy) / Dy
public: Box3(cont Vec3f &vmin, const Vec3f &vmax) As you can see on the figure, the ray misses the box when \(t0x\) is greater than \(t1y\) and when \(t0y\) is greater than \(t1x\).Finally we can extend the technique to the 3D case by computing \(t\) values for the z component and compare them to the \(t\) values we have computed so far for the x and y components:Here is a full implementation of the method in C++ (min and max are the minimum and maximum extent of the bounding box):Note that depending on the ray direction tmin might be greater than tmax. float tymax = (max.y - r.orig.y) / r.dir.y; Ray(const Vec3f &orig, const Vec3f &dir) : orig(orig), dir(dir)
return false;
Remember that a ray can be expressed using the following parametric form: Where O represents th… float tmax = (max.x - r.orig.x) / r.dir.x;
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