1 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Collision detection Static interference tests Exacxt collision prediction Conservative tests and acceleration techniques Updated January 30, 2016January 30, 2016January 30, 2016 Collision detection is one of the most challenging and most important problems in computer animation! 2 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Types of animations Camera moves in static environment Rigid motion: Objects are rigid, they only rotate and translate Articulated bodies: Parameterized linkage of rigid bodies A childs pose (forearm) is obtained from a parents pose (upper-arm) by an arbitrary rotation (elbow angle) and a fixed translation (length of upper-arm). Deformations: Each shape deforms with time Deformation may be simulated using nave physics or computed to interpolate between two shapes (3D morphing) Reactive: Responds to user actions Shape moves and deforms when approached or poked by the user All must detect collisions and adjust behavior 3 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, The collision detection problem Shapes A and B evolve (move, deform) with time. Notation for their instance at time t>0: A(t) and B(t). Find the smallest t such that A(t) B(t) . 4 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Interference A and B interfere when they share at least one common point. A B sets are disjoint sets touch sets interfere 5 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Static interference tests for collision A and B evolve (move, deform) with time: A(t) and B(t). Test and then assume A(0) B(0)= t=0; repeat { if (A(t+dt) B(t+dt) ) { repeat 5 times { if (A(t+dt/2) B(t+dt/2) ) dt=dt/2; else t+=dt/2; }; return(t); }; } return(-1);}// no collision found t redgreen 6 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, How to test for interference Often (not always) we are using the bundaries of shapes to test for interference. bundaries are disjoint bundaries overlap bundaries cross 7 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Testing boundaries A and B are disjoint if and only if their boundaries are disjoint True or false? If true, prove or justify If false, provide a counterexample and propose a correct test using the bundaries: A and B are disjoint iff their bundaries bundaries are disjoint bundaries overlap bundaries cross sets are disjoint bundaries touch sets interfere 8 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Distinguish touch and interfere How can you use tests on boundries to distinguish disjoint touching interfering bundaries are disjoint bundaries overlap bundaries cross sets are disjoint bundaries touch sets interfere 9 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Detecting collisions between balls For simplicity, we assume that objects collide when their enclosing balls collide Compute enclosing ball of center c and radius r as follows: c := ( (x min +x max )/2, (y min +y max )/2, (z min +z max )/2 ) r := max (||cv||) for all vertices v of the object Interference detection: After each time step, check whether any pair of objects interfere Ball(c 1,r 1 ) and Ball(c 2,r 2 ) interfere when Need very small steps not to miss a chock Collision prediction: Express the relative motion of object 2 in the CS of object 1 For motions with fixed velocities, v 1 and v 2, you get v 1 and v 2, v = v 2 v 1. Shrink object 2 to a point c 2 and expand object 1 to Ball(c 1,r 1 +r 2 ) Check whether curve swept by c 2 intersects Ball(c 1,r 1 +r 2 ) Constant velocity motion: find smallest time t>0 when (c 2 +tv) 2 =(r 1 +r 2 ) 2 ||c 1 c 2 || < r 1 +r 2 10 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Collision between disks Assume that disk D 1 =disk(C 1,r 1 ) will travel with constant velocity V 1. Similarly, disk D 2 =disk(C 2,r 2 ) will travel with constant velocity V 2. Assume that they are initially disjoint. How would you compute the time t when they will collide? Solve ((C 2 +tV 2 )(C 1 +tV 1 )) 2 =(r 1 +r 2 ) 2 for t (C 1 C 2 +t(V 2 V 1 )) 2 =(r 1 +r 2 ) 2 (V 2 V 1 ) 2 t 2 + 2C 1 C 2 (V 2 V 1 ) t + C 1 C 2 2 (r 1 +r 2 ) 2 = 0 return the smallest positive value of t if one exists We can reduce this problem to the one of line/circle intersection: D 2 is stationary, D 1 moves by V 1 V 2 D 2 has radius r 1 +r 2 (inflated by r 1 ), D 1 is a point (deflated by r 1 ) when will D 1 (the point) hit the inflated D 2 ? 11 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Elastic shock between two disks Assume that disk D 1 =disk(C 1,r) traveling with constant velocity V 1 and disk D 2 =disk(C 2,r) traveling with constant velocity V 2 have just collided. What should their velocities V 1 and V 2 be? (We assume that they have the same mass.) We must exchange their normal velocities N = C 1 C 2.unit (normal direction to both at contact point) U 1 = (V 1 N) N (normal components of velocities) U 2 = (V 2 N) N V 1 = V 1 U 1 + U 2 (cancel U 1 and add U 2 ) V 2 = V 2 U 2 + U 1 ( to exchange their normal velocities) 12 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Elastic shock with a fixed disk Assume that disk D 1 =disk(C 1,r), traveling with constant velocity V, has just collided with a static disk D 2 =disk(C 2,r). What should its velocity V after the bounce? (We assume that D 2 has infinite mass.) We must negate (reverse) the normal component of V N = C 1 C 2.unit (normal direction to both at contact point) U = (V N) N (normal components of V) V= V 1 2U (swap direction of U) 13 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Elastic collision between balls Tangential velocities are maintained, normal velocities are computed using (8) Masses: a and b Velocities in the normal direction: s and t before and s and t after the shock Conservation of momentum: as+bt=as+bt (1) Conservation of energy: as 2 +bt 2 =as 2 +bt 2 (2) Regrouping in (1): a(s-s)=b(t-t) (3) Regrouping in (2) and using x 2 -y 2 =(x+y)(x-y): a(s-s)(s+s)=b(t-t)(t+t) (4) Combining (3) and (4) : s+s=t+t (5) Substituting t=s+s-t obtained from (5) in (1) yields: as+bt=as+bs+bs-bt (6) Reorganizing (6) : (a+b)s=(a-b)s+2bt (7) Reorganizing and swapping (s,a) and (t,b): s=s+2b(t-s)/(a+b) and t=t+2a(s-t)/(a+b) (8) Note that when a=b, then s=t and t=s (exchange of normal velocities) When a>>b: s=s (not affected) and t=2s-t (reverse speed of b if a is static) s t s t a b chock 14 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, 2D interference between triangles Two triangles interfere if a vertex of one lies inside the other True or false? Justify of provide a counterexample and a correct test Modify your test to distinguish disjoint, interfere, and touch 15 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Test for polygon/polygon intersection in 2D? A and B are polygons They are connected sets in 2D, but may have holes Write the high-level test for checking whether they interfere For each connected component of A do, if the first vertex of that component is in or on B, return TRUE For each connected component of B do, if the first vertex of that component is in or on A, return TRUE For each edge Ea of A do, for each edge Eb of B do, if Ea intersects Eb, return TRUE Otherwise return FALSE 16 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Details Point-in-polygon Edge/edge intersection 17 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Point-in-polygon test Algorithm for testing whether point q is inside polygon P inP:=false; for each edge (a,b) of P {if (inTriangle(q,a,b,o)) then inP := !inP; return(inP); P o q ab 18 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Point-in-triangle test in 2D Is point q in tri(a,b,c) ? q c a b ab.left()aq bc.left()bq ca.left()cq all have the same sign 19 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Edge/edge intersection in 2D Write a geometric expression that returns TRUE when edges (a,b) and (c,d) intersect a b d c ( (ab.left()ac>0) != (ab.left()ad>0) ) && ( (cd.left()ca>0) != (cd.left()cb>0) ) 20 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, A and B are disjoint when the following conditions are all true V B V A = , Each verted of A is different from each vertex of B Test for equality of coordinates (use distance threshold?) V A E B = , V B E A = , Test whether any vertex of one lies on any edge of the other Use PinE(point,edge) test in 3D (see next slides) V A F B = ,V B F A = , Test whether any vertex of one lies on any face of the other Use PinT(point,triangle) test in 3D (see next slides) E A E B = , Test whether any edge of one intersects any edge of the other Compare edge-edge distance, EEdist(edge,edge), to threshold E A F B = , E B F A = , Test whether any edge of one intersects any triangle of the other Use edge-triangle intersection test, EinT(edge,triangle), in 3D Why is it unnecessary to test F A F B = ? Is the last condition sufficient? How to distinguish between touch and interfere? Intersection between triangle meshes 21 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, PinE(p, edge(a,b)) := ||ab ap|| 0 AND babp>0 Let q be the projection of p onto the line through a and b The distance ||pq|| from p to q is ||ab ap|| / ||ab|| It needs to be less than the threshold We also want q to be inside edge(a,b) abap>0 AND babp>0 PinE(point,edge): Point-in-edge test ab p q ab p q a b p q 22 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, PinT(point,triangle): Point-in-triangle test PinT(p,tri(a,b,c)) := (nap)< ||n|| AND SameSign ( (n ab)ap,(n bc)bp,(n ca)cp) ), where n=ab ac Let q be the projection of p onto the plane through a, b and c n=ab ac is orthogonal to the plane though a, b, c The distance ||pq|| is (nap)/||n|| It needs to be less than the threshold We also want q to be inside triangle(a,b,c) SameSign( (n ab)ap, (n bc)bp, (n ca)cp ) p c a b p c a b q 23 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, EintE(edge,edge): Edge-edge intersection EintE(edge(a,b),edge(c,d)):= nac< ||n|| && SameSign((ab n)ac,(ab n)da) && SameSign((cd n)ca,(cd n)cb) With n=ab ac being the common normal to both lines Let p and q be the points that minimize the distance between the two lines. The distance between the lines is ||pq|| = nac / ||n|| But we also want to ensure that it occurs inside the edges SameSign((ab n)ac,(ab n)da) AND SameSign((cd n)ca,(cd n)cb) Points c and d must be on opposite sides of plane through a with normal n Points a and b must be on opposite sides of plane through a with normal n a b d c q p 24 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, EintT(edge,triangle): edge-triangle intersection Edge(P,Q) intersects Triangle(A,B,C) if the following five expressions are all TRUE or all FALSE: s(p, a, b, c) s(p, a, b, q) s(p, b, c, q) s(p, c, a, q) !s(q, a, b, c) s(a,b,c,d) := (ab ac)ad > 0 Need not worry about cases where the edge hits the boundary of the triangle. These were handled before. p b a q c 25 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Interfering solids Assume that two triangle meshes A and B are disjoint, what else do you need to test to be able to conclude that the solids they bound are disjoint? 26 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Summary on static interference tests Simulate motion/deformation step-by-step and test for static interference between objects at each key-frame Stop when interference is detected Search for correct collision time Binary split of current time-step Static interference detection is difficult to code and slow O(n 2 ) static interference detections between pairs of objects Each checks whether an edge of one stabs the face of another Expensive (O(n 2 ) per time step) Can easily miss collisions t1t1 t2t2 t3t3 27 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Interference tests are not safe Testing collision via interfrence tests at discrete time steps is not safe. found collision missed collision 28 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Interference prediction Instead of testing interferences for discrete time steps, it would be better to predict when the collision will happen, if at all. Do not attempt this at home for complex animations. This is only practical for simple motions of rigid objects. We address it only for rigid objects Express motion of one in the CS of the other 29 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Colliding polyhedra Assume solids A and B are initially disjoint polyhedra Assume B is static and A moves by rigid-body motion M(t) At time t when first collision occurs: a vertex of is in a face of B or a vertex of B is in a face of or an edge of intersect an edge of B Use these conditions to formulate equations in t Find roots and test whether they correspond to intersections From Boyse79 30 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Case of costant velocity Assume A moves with constant velocity M(t)=tv and B is fixed If both move, compute collision in local coordinate system of B Collision may occur between A vertex p of A and (the closure of) a triangle T of B Intersect Ray(p,v) with T The closure of a triangle T of A with a vertex p of B Intersect Ray(p,-v) with T An edge (a,b) of A with an edge (c,d) of B Check when the volume of tetrahedron (a+tv,b+tv,c,d) becomes zero Make sure that, at that time, the two edges intersect b p a H v c t a b c d v 31 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Vertex hits triangle We will use: boolean s(u,v,v) {return (u v)w > 0 } A vertex starting at P traveling with velocity V follows Ray(p,v) 1) 1)Test whether Ray(p,v) hits triangle(a,b,c) Test whether triangle (a,b,c) is front facing: s(pa,pb,pc) If not, swap a and b (in the next test) Test whether triangle is stabbed by ray s(pa,pb,v) && s(pb,pc,v) && s(pc,pa,v) 2) 2)If so, compute collision time and place Collision time t = (ab ac)pa / (ab ac)v Collision point h = p + tV a p b h v c t 32 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Edge/edge collision solve (cd (ca+tv)) (cb+tv)=0 for t (cd (ca+tv)) cb + (cd (ca+tv)) tv)=0 (cd ca+t(cd v)) cb + (cd ca+t(cd v)) tv)=0 (cd ca) cb +t(cd v) cb + (cd ca) tv +t 2 (cd v) v = 0 (cd ca) cb +t(cd v) cb + (cd ca) tv = 0, because (cd v) v = 0 t = (ca cd) cb / ((cd v) cb - (cd v) ca) t = (ca cd) cb / (ab cd) v An edge (a,b) of A with an edge (c,d) of B The volume of tetrahedron (a+tv,b+tv,c,d) is zero solve (cd (ca+tv)) (cb+tv)=0 for t t = (ca cd) cb / (ab cd) v Make sure that, at time t, the two edges intersect a b c d v 33 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, General motions (with B.M. Kim) For each pair of objects A and B do Approximate relative motion by polyscrew Insert intermediate poses as needed adaptively For each screw motion segment do Use conservative rejection test to identify collision-free situations If collision may not be discarded, then do For each vertex of A(resp. B) and each triangle of B (resp. A) do If collision cannot be discarded using bounds Then find time of first collision (if one occurs) For each edge of B and each edge of A do If collision cannot be discarded using bounds Then find time of first collision (if one occurs) Stop if collision found and report the smallest time of first collision 34 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Conservative tests A collision test T is either correct or wrong T is conservative when: collision T boolean T {return(false);} is conservative by not accurate We want conservative tests that are rarelywrong Separating half-space Distances Space partition Swept regions Voxels Hardware 35 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Separating half-spaces If there is a half-space H such that A H && H B= then it is safe to concludeABAB HAB AB H H H 36 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Bounding balls Tight enclosing balls are cheap to compute (center of min/max box, radius min distance) trivial to rotate lead to inexpensive interference test no collision possible collision 37 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Bunding balls may not be tight They may generate too many false positivesA B 38 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Linear half-spaces may be betterA B H Find the normal N H, such that the convex hulls of the projections of the vertices of A and of the vertices of B onto some line parallel to N H are disjoint NHNHNHNH Candidates for N H : principal axes,principal axes, diagonal axes,diagonal axes, normals to faces of A and B,normals to faces of A and B, vector between centers,vector between centers, cross-products of edge tangents (one of A, one of B). cross-products of edge tangents (one of A, one of B). 39 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Generalization Static interference detections are not safe Collision predition is too hard for general motions or when the objects evolve So, what should we do? 40 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Extrude the problem to space-time domain 2D space domain to 3D space-time domain 3D space domain to 4D space-time domain Instances at given time t are the cross-sections Compute interference in 4D? 41 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Volume swept during screw motion Computing and visualizing pose-interpolating 3D motions Jarek R. Rossignac and Jay J. Kim (Hanyang University, Seoul, Korea), CAD, 33(4)279:291, April SweepTrimmer: Boundaries of regions swept by sculptured solids during a pose-interpolating screw motion Jarek R. Rossignac and Jay J. Kim 42 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, Using sweeps If the volumes swept by the two objects during a short time interval T do not interere, then it is save to conclude that the objects do not collide during T. The swept volume is the projection of the 4D animation onto the 3D space. 43 Georgia Tech, IIC, GVU, 2006 AnimationAnimationRossignac, How to generate swept volume bounds? Sweeps of enclosing ball (cylinder, torus, helix) Minkowski morph (of bunding box)