Week 13 - Wednesday CS361

Last time What did we talk about last time? Intersection testing Bounding volumes Sphere AABB OBB k-DOP SharpDX tools

Questions?

Project 4

Getting Bounding Spheres from SharpDX Models

SharpDX Picker

Intersection Methods

Ray/sphere intersection We can write the implicit sphere equation as f(p) = ||p – c|| – r = 0 p is any point on the surface c is the center r is the radius By substituting in r(t) for p, we can eventually get the equation t2 + 2tb + c = 0, where b = d • (o – c) and c = (o – c) •(o – c) – r2 If the discriminant is negative, the ray does not hit the sphere, otherwise, we can compute the location(s) where it does

Optimized ray/sphere Looking at it geometrically, we can optimize the test Find the vector from the ray origin to the center of the sphere l = c – 0 Find the squared length l2 = l • l If l2 < r2, then o is in the sphere, intersect! If not, project l onto d: s = l • d If s < 0, then the ray points away from the sphere, reject Otherwise, use the Pythagorean theorem to find the squared distance from the sphere center to the projection: m2 = l2 – s2 If m2 > r2, the ray will miss, otherwise it hits

Ray/box intersection Ray box intersection is a key element to have in your arsenal Bounding boxes are a very common form of bounding volume A ray/box intersection is often the first test you will use before going down deeper There are a couple of methods Slabs method Line segment/box overlap test

Slabs method First find the t value where the ray intersects each plane The box is made up of 3 slabs Find the min t and max t for each slab The final tmin is the max of all the tmin values The final tmax is the min of all the tmax values If tmin ≤ tmax, the ray intersects the box, otherwise it does not The idea can be extended to frustums and k-DOPs

Separating axis test For two arbitary, convex, disjoint polyhedra A and B, there exists a separating axis where the projections of the polyhedra are also disjoint Furthermore, there is an axis that is orthogonal to (making the separating plane parallel to) A face of A or A face of B or An edge from each polyhedron (take the cross product) This definition of polyhedra is general enough to include triangles and line segments

Line segment/box overlap test This method uses the separating axis test and only works for AABBs and line segments The AABB has its center at (0,0,0) and size half vector h The line segment is defined by center c and half vector w If |ci| > wi + hi for any i x,y,z then disjoint There is another test for each axis that is the cross product of the x, y, and z axis and w If any test passes, then disjoint Only if all tests fail, then overlap

Triangle representation One way to represent a triangle is with barycentric coordinates For triangles, barycentric coordinates are weights that describe where in the triangle you are, relative to the three vertices These weights are commonly labeled u, v, and w and have the following properties u ≥ 0, v ≥ 0, w ≥ 0 and u + v + w ≤ 1

Ray triangle intersection We represent a point f(u,v) on a triangle with the following explicit formula f(u,v) = (1 – u – v)p0 + up1 + vp2 Then, setting the ray equal to this equation gives o + td = (1 – u – v)p0 + up1 + vp2 This is simply a vector representation of three equations with three unknowns If the solution has a positive t, and u and v between 0 and 1, it's an intersection

Ray/polygon intersection First we compute the intersection of the ray and the plane of the polygon Then we determine if that point is inside the polygon (in 2D) The plane of the polygon is np • x + dp = 0 np is the plane normal dp is the distance along the normal from the origin to the plane Intersection is found by: np • (o + td) + dp = 0 Then project onto xy, xz, or yz plane (whichever maximizes polygon area) and see if the point is in the polygon

Crossings test If you want to see if a point is inside a polygon, you shoot a ray from that point along the positive x axis If it intersects edges of the p0lygon an even number of times, it is outside, otherwise it is inside Problems can happen if a ray intersects a vertex, so we treat all vertices with y ≥ 0 as being strictly above the x axis

Plane/box intersection Simplest idea: Plug all the vertices of the box into the plane equation n • x + d = 0 If you get positive and negative values, then the box is above and below the plane, intersection! There are more efficient ways that can be done by projecting the box onto the plane

Quiz

Upcoming

Next time… Finish intersection test methods

Reminders Keep working on Project 4 Keep reading Chapter 16