Week 13 - Friday

 What did we talk about last time?  Ray/sphere intersection  Ray/box intersection  Slabs method  Line segment/box overlap test  Ray/triangle intersection  Ray/polygon 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

 Seeing if bounding volumes collide is a fundamental part of most collision detection algorithms  Does this ship hit that asteroid?  Bounding volumes are often arranged in hierarchies of bounding volumes  Bounding volumes allow for easy reject cases

 Sphere/sphere is the easiest  Is the distance between their centers bigger than the sum of their radii?  If yes, they are disjoint  If no, they overlap c1c1 c2c2 r1r1 r2r2

 Remember that an AABB is defined by two points, a min and a max  We go through each axis (x, y, and z) and first check to see if we can reject based on distance on that axis being too large  If not, we add the squared axis's distance to the total  If you want to do a sphere/OBB intersection, transform the sphere's center into the axes of the OBB and then the OBB will be an AABB

intersect( c, r, A ) { d = 0 for i in x,y,z if( (e = c i – a i min ) < 0 ) if( e < -r ) return DISJOINT d = d + e 2 else if ( (e = c i – a i max ) > 0 ) if( e > r ) return DISJOINT d = d + e 2 if ( d > r 2 ) return DISJOINT return OVERLAP } intersect( c, r, A ) { d = 0 for i in x,y,z if( (e = c i – a i min ) < 0 ) if( e < -r ) return DISJOINT d = d + e 2 else if ( (e = c i – a i max ) > 0 ) if( e > r ) return DISJOINT d = d + e 2 if ( d > r 2 ) return DISJOINT return OVERLAP }

 We test each dimension to see if the min of one box is greater than the max of the other or vice versa  If that's ever true, they're disjoint  If it's never true, they overlap intersect(A, B ) { for i in x,y,z if(a i min > b i max or b i min > a i max ) return DISJOINT return OVERLAP } intersect(A, B ) { for i in x,y,z if(a i min > b i max or b i min > a i max ) return DISJOINT return OVERLAP }

 An AABB is a special case of a 6-DOP  Use the same test for an AABB, looking at the mins and maxes of each slab  Because the axes of a k-DOP are not necessarily orthogonal, this test is inexact (unlike for the AABB)  It is conservative: Some disjoint k-DOPs will report that they overlap, but overlapping k-DOPs will never report disjoint intersect(A, B ) { for i in 1 … k/2 if(d i A,min > d i B,max or d i B,min > d i A,max ) return DISJOINT return OVERLAP } intersect(A, B ) { for i in 1 … k/2 if(d i A,min > d i B,max or d i B,min > d i A,max ) return DISJOINT return OVERLAP }

 Again, an OBB is surprisingly complex  The fastest way found involves the separating axis test  There are 15 different axes you've got to test for overlap before you can be sure that the boxes overlap  When all the math is worked out, the test is quite fast

 Because anything visible on the screen will be in the view frustum, we can save time by ignoring objects that are not  Remember that the frustum is defined by 6 planes: near, far, left, right, top and bottom  When test the frustum against bounding volumes, we will want three answers: outside, inside, and intersect

 To test frustum intersection, it is necessary to know the plane equations for each of the six frustum planes  If the view matrix is V and the projection matrix is P, the final transform is M = PV  If m i means the i th row of M, the equations for each plane are as follows:  -(m 3 + m 0 ) (x, y, z, 1) = 0(left)  -(m 3 – m 0 ) (x, y, z, 1) = 0(right)  -(m 3 + m 1 ) (x, y, z, 1) = 0(bottom)  -(m 3 – m 1 ) (x, y, z, 1) = 0(top)  -(m 3 + m 2 ) (x, y, z, 1) = 0(near)  -(m 3 – m 2 ) (x, y, z, 1) = 0(far)

 We take the center of the sphere p and plug it into each of the plane equations, getting signed distance values  Note that the normals of the planes point outwards  If the distance to any given plane is greater than radius r, the sphere is outside the frustum  If the distances to all six planes are less than –r, the sphere is inside  Otherwise, the sphere intersects  This test is conservative: Reports some outside spheres as intersections

 We skipped over the section that says how to test a plane for intersection with a box, but it's a simple calculation  To do frustum/AABB intersection, we test the AABB against every plane of the frustum  If the box is outside any plane, we return outside  If the box is not outside any plane but intersects some plane, we return intersects  Otherwise, we return inside

 We will only look at the 2D problem, but the book has discussion of 3D lines as well  For a 2D vector (x, y), we define its perp dot product (x, y)  = (-y, x)  Thus, we can work through the equations of a line (only the path to the value of s is shown)  r 1 (s) = r 2 (t)  o 1 + sd 1 = o 2 + td 2  sd 1 d 2  = (o 2 – o 1 ) d 2   s = ((o 2 – o 1 ) d 2  ) / (d 1 d 2  )

 Collision detection

 Keep working on Project 4  Start Assignment 5  Read Chapter 17