CS361Week 13 - Wednesday
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)1. A face of A or2. A face of B or3. 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