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2.4. P RIMITIVE T ESTS - C LOSEST POINT Closest point forms of intersection detection

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Finding the closest point between two objects provides a number of benefits. If positive, the objects are separated, if negative, the objects are interpenetrating (and knowledge of the closest point can be used to provide contact information). Two different approaches can be used to find the minimum point: Keeping a record of the closest points between two objects can often be incrementally maintained at a low cost (object coherency), providing fast collision testing. Formulate the test as a minimisation problem and solve it using calculus. Use the geometric properties of the objects to geometrically derive the closest point (considered here).

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Consult the recommended course text for details of the following tests: Closest Point on Plane to Point Closest Point on Line Segment to Point Closest Point on AABB to Point Closest Point on OBB to Point Closest Point on Triangle to Point Closest Point on Convex Polyhedron to Point Closest Points of Two Lines Closest Points of Two Line Segments Closest Points of a Line Segment and a Triangle Closest Points of Two Triangles Two illustrative forms of closest point test are explored next.

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Closest point between an OBB and a Point

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Consider an OBB with centre point C, orthogonal unit vectors (u 0, u 1, u 2 ) and half-width distances (e 0, e 1, e 2 ) and a point P.

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The closest point on/in the OBB to P can be determined by: 1. Transforming the point P into the local coordinate system of the OBB, 2. Computing the point on the OBB (now effectively an AABB) closest to the transformed point. 3. Transforming the resulting point back into world coordinates

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1. Transform P into the OBB coordinate system: The transformed point Q is found from: i.e. the distance vector of the point from the centre of the OBB as projected onto each defined OBB axis Q x = (P – C) ● u o Q y = (P – C) ● u 1 Q z = (P – C) ● u 2 C ● P ● P-C (P-C) ● u o ● u0u0

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2. Computing OBB point closest to the transformed point. If the transformed point u 0, u 1, u 2 distances are respectively longer than e 0, e 1, e 2 then clamp to half- width extents 3. Transform the point back into world coordinates Q i = (P–C)● u i if(Q i >e i ) Q i =e i if(Q i <-e i ) Q i =-e i C ● P ● ● u0u0 u1u1 Q e1e1 e2e2 Q i = C + Q i u i

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Expressed as code: ClosestPointToOBB(Point point, OBB box, out Point closestPt) { Vector distance = point- box.centre; closestPt = box.centre; for (int i = 0; i < 3; i++) { float length = Dot(distance, box.axis[i]); if (length > box.extent[i]) length = box.extent[i]; if (length < -box.extent[i]) length = -box.extent[i]; closestPt += length * box.axis[i]; } } Initially set closest point as box centre Iterate over box axis Determine projection along axis Clamp if needed Build world location component by moving determine distance along OBB axis

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The separation (actual or squared) can then be determined simply as: float distance = (closestPt – point).length() Somewhat obviously, if all components of the closest OBB point (expressed in OBB space) are less than the half-width extents, then the point is within the OBB.

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Closest point between a Triangle and a Point

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Given a triangle defined by points ABC and a point P, assume Q defines the point of ABC closest to P. An efficient means of computing Q is: 1. Determine which triangle Voronoi region P is within (i.e. find which feature P is closest to). 2. Orthogonally project P onto the determined closest feature. A B C A Voronoi region of vertex A Voronoi region of edge CB

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Intersection with the Voronoi region for vertex A, B and C can be determined as the intersection of the negative half-space of the two planes through that point, e.g. for A each plane passes through A, one with a normal (B- A) and the other with normal (C- A). A A Voronoi region of vertex A C B

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The Voronoi region of each triangle edge can be efficiently computed by determining the triangle barycentric coordinates of P. Assuming n is the normal of triangle ABC and R is the projection of P onto the triangle’s plane, i.e. R = P – tn some for t, the barycentric coordinates of R (R = uA + vB + wC) are given by: B C A Voronoi region of edge CB n = (B-A)×(C-A) rab = n ● ((A-P)×(B-P)) rbc = n ● ((B-P)×(C-P)) rca = n ● ((C-P)×(A-P)) abc = rab + rbc + rca u = rbc/abc, v = rca/abc, w = rab/abc Aside: the barycentric coordinates of R are given as the ratios of the signed areas of triangles RAB, RBC, and RCA to the signed area of ABC

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To determine if P is within a triangle edge Voronoi region the following conditions must hold: Aside: It is not sufficient just to test if P is outside AB, in that for a triangle with an obtuse angle at A, P could be outside AB and actually be located in the Voronoi region of edge CA (see figure). B C A B (X-A) ●(B-A) > 0 (X-B) ●(A-B) > 0 1. Calculated area of the barycentric region must be <= 0 (i.e. for AB, rab <= 0) 2. The point must be within the positive half- spaces of the planes (assuming an edge AB): (X-A) ●(B-A) = 0 and (X-B) ●(A-B) = 0

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ClosestPointToTriangle( Point a, Point b, Point c, Point p, out Point q { Vector ab = b – a, ac = c – a, bc = c - b; float snom = Dot(p - a, ab), sdenom = Dot(p - b, a - b); float tnom = Dot(p - a, ac), tdenom = Dot(p - c, a - c); if (snom <= 0.0f && tnom <= 0.0f) return a; float unom = Dot(p - b, bc), udenom = Dot(p - c, b - c); if (sdenom <= 0.0f && unom <= 0.0f) return b; if (tdenom <= 0.0f && udenom <= 0.0f) return c; Points ABC defined the triangle, point P is the source point, point Q is the closest triangle point to P Vertex Voronoi region hit early out Build edge vectors Determine the parametric position s for the projection of P onto AB (i.e. P’ = A+s*AB, where s = snom/ (snom+sdenom), and then parametric position t for P projected onto AC Aside: the projection of P onto AB from point A is: (P-A)●(B-A) and from P onto BA is: (P-B)●(A-B). Added together they sum to the length of AB Parametric position u for P projected onto BC

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Vector n = Cross(b - a, c - a); float vc = Dot(n, Cross(a - p, b - p)); if (vc = 0.0f && sdenom >= 0.0f) return a + snom / (snom + sdenom) * ab; float va = Dot(n, Cross(b - p, c - p)); if (va = 0.0f && udenom >= 0.0f) return b + unom / (unom + udenom) * bc; float vb = Dot(n, Cross(c - p, a - p)); if (vb = 0.0f && tdenom >= 0.0f) return a + tnom / (tnom + tdenom) * ac; float u = va / (va + vb + vc); float v = vb / (va + vb + vc); float w = 1.0f - u - v; // = vc / (va + vb + vc) return u * a + v * b + w * c; } Determine if P is outside (or on) edge AB by finding the area formed by vectors PA, PB and the triangle normal. A scalar triple product is used. If P is outside of AB (signed area <= 0) and within Voronoi feature region, then return projection of P onto AB Repeat the same test for P onto BC Repeat the same test for P onto CA P must project onto inside face. Find closest point using the barycentric coordinates Aside: The Lagrange identity (a × b) · (c × d) = (a · c)(b · d) − (a · d)(b · c) can be used to express the three scalar triple products in a manner that can be calculated more quickly.

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Directed mathematical reading Directed reading

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Directed reading Read Chapter 5 (pp ) of Real Time Collision Detection Related papers can be found from:

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To do: Read the directed material After reading the directed material, have a ponder if this is the type of material you would like to explore within a project. Today we explored: Notion of closest point forms of intersection testing. Closest point to an OBB or triangle

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