1. Week 13 - Wednesday
CS361

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

3. Questions?

4. Project 4

5. Getting Bounding Spheres from SharpDX Models

6. SharpDX Picker

7. Intersection Methods

8. 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

9. 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

10. 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

11. 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

12. 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

13. 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

14. 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

15. 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

16. 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

17. 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

18. 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

19. Quiz

20. Upcoming

21. Next time…
Finish intersection test methods

22. Reminders
Keep working on Project 4
Keep reading Chapter 16