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2.4. PRIMITIVE TESTS -CLOSEST POINT
Closest point forms of intersection detection
Closest point forms of intersection detection
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:
• 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).
Forms of closest point test
Consult the recommended course text for details of the
following tests:
•
•
•
•
•
•
•
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•
•
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.
Closest point between an OBB and a Point
Closest point on an OBB to a Point
Consider an OBB with
centre point C, orthogonal
unit vectors (u0, u1, u2) and
half-width distances (e0, e1,
e2) and a point P.
Closest point on an OBB to a Point
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
Closest point on an OBB to a Point
1. Transform P into the OBB
P
●
coordinate system:
The transformed point Q is
found from:
P-C
●
Qx = (P – C)● uo
Qy = (P – C)● u1
Qz = (P – C)● u2
(P-C)● uo
●
i.e. the distance vector of the
point from the centre of the OBB
as projected onto each defined
OBB axis
C
u0
Closest point on an OBB to a Point
2. Computing OBB point closest
to the transformed point.
If the transformed point u0, u1, u2
distances are respectively longer
than e0, e1, e2 then clamp to halfwidth extents
Qi = (P–C)● ui
if(Qi>ei) Qi=ei
if(Qi<-ei) Qi=-ei
3. Transform the point back into
world coordinates
Qi = C + Qiui
P
●
●
u1
●
C
Q
e1
u0
e2
Closest point on an OBB to a Point
Expressed as code:
ClosestPointToOBB(Point point, OBB box, out Point closestPt)
{
Vector distance = point- box.centre; Initially set closest point as box centre
closestPt = box.centre;
for (int i = 0; i < 3; i++) { Iterate over box axis Determine projection
float length = Dot(distance, box.axis[i]);along axis
Clamp if needed
if (length > box.extent[i]) length = box.extent[i];
if (length < -box.extent[i]) length = -box.extent[i];
closestPt += length * box.axis[i]; Build world location component
by moving determine distance
}
along OBB axis
}
Closest point on an OBB to a Point
The separation (actual or squared)
can then be determined simply as:
float distance =
(closestPt – point).length()
Closest point between a Triangle and a Point
Closest point on a triangle to a point
Given a triangle defined by
points ABC and a point P,
assume Q defines the point of
ABC closest to P.
A
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.
C
Voronoi region
of edge CB
B
A
Closest point on a triangle to a point
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 (BA) and the other with normal (CA).
A
A
C
B
A
Closest point on a triangle to a point
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:
n =
rab
rbc
rca
abc
u =
(B-A)×(C-A)
= n ● ((A-P)×(B-P))
= n ● ((B-P)×(C-P))
= n ● ((C-P)×(A-P))
= rab + rbc + rca
rbc/abc, v = rca/abc, w = rab/abc
C
Voronoi region
of edge CB
B
Closest point on a triangle to a point
To determine if P is within a triangle edge Voronoi
region the following conditions must hold:
A
1. Calculated area of the barycentric region
must be <= 0 (i.e. for AB, rab <= 0)
C
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
B
B
Closest point on a Triangle to a Point
Points ABC defined the triangle, point P is the source point, point Q is the closest triangle point to P
ClosestPointToTriangle(
Point a, Point b, Point c, Point p, out Point q
{
Build edge vectors
Vector ab = b – a, ac = c – a, bc = c - b;
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
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; Vertex Voronoi
region hit early out
Parametric position u for P projected onto BC
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;
Closest point on a Triangle to a Point
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.
Vector n = Cross(b - a, c - a);
float vc = Dot(n, Cross(a - p, b - p));
If P is outside of AB (signed area <= 0) and within Voronoi feature region,
then return projection of P onto AB
if (vc <= 0.0f && snom >= 0.0f && sdenom >= 0.0f)
return a + snom / (snom + sdenom) * ab;
Repeat the same test for P onto BC
float va = Dot(n, Cross(b
if (va <= 0.0f && unom >=
return b + unom / (unom +
Repeat the same test for P onto CA
float vb = Dot(n, Cross(c
if (vb <= 0.0f && tnom >=
return a + tnom / (tnom +
- p, c - p));
0.0f && udenom >= 0.0f)
udenom) * bc;
- p, a - p));
0.0f && tdenom >= 0.0f)
tdenom) * ac;
P must project onto inside face. Find closest point using the barycentric coordinates
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;
}
Directed mathematical reading
Directed reading
• Read Chapter 5 (pp125-155)
of Real Time Collision
Detection
• Related papers can be
found from:
http://realtimecollisiondetection.net/books/rtcd/references/
Summary
Today we
explored:
 Notion of
closest point
forms of
intersection
testing.
 Closest point
to an OBB or
triangle

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