float MeshFillHole::findClosestPoint(const SbLine& ray, const TBoundary& polygon,
unsigned long& vertex_index, SbVec3f& closestPoint) const
{
// now check which vertex of the polygon is closest to the ray
float minDist = FLT_MAX;
vertex_index = ULONG_MAX;
const MeshCore::MeshKernel & rMesh = myMesh->Mesh.getValue().getKernel();
const MeshCore::MeshPointArray& pts = rMesh.GetPoints();
for (TBoundary::const_iterator it = polygon.begin(); it != polygon.end(); ++it) {
SbVec3f vertex;
const Base::Vector3f& v = pts[*it];
vertex.setValue(v.x,v.y,v.z);
SbVec3f point = ray.getClosestPoint(vertex);
float distance = (vertex-point).sqrLength();
if (distance < minDist) {
minDist = distance;
vertex_index = *it;
closestPoint = vertex;
}
}
return minDist;
}
/*!
Returns the two closest points on the lines. If the lines are
parallel, all points are equally close and we return \c FALSE. If
the lines are not parallel, the point positions will be stored in \a
ptOnThis and \a ptOnLine2, and we'll return \c TRUE.
Note that both SbLine instances are considered to be infinite in
both directions from their definition points, as far as this
function is concerned.
\sa getClosestPoint().
*/
SbBool
SbLine::getClosestPoints(const SbLine& line2,
SbVec3f& ptOnThis, SbVec3f& ptOnLine2) const
{
#if 1
// new optimized version based on formulas from from Boyko Bantchev
// p1 = point on line 1
// p2 = point on line 2
// d1 = line 1 direction
// d2 = line 2 direction
// q1 = closest point on line 1
// q2 = closest point on line 2
// The solution (q1 and q2) must be on their respective
// lines:
//
// q1 = p1 + t1 * d1 (0)
// q2 = p2 + t2 * d2
//
// we set u = p2 - p1, and get:
//
// q2 - q1 = u + t2*d2 - t1*d1 (1)
//
// the solution line q2 - q1 is orthogonal to d1 and d2
// (or a null vector if the lines intersect), which yields:
//
// (u + t2*d2 - t1*d1) · d1 = 0 (2)
// (u + t2*d2 - t1*d1) · d2 = 0
//
// we know |d1| and |d2| == 1, and set d1 · d2 = t
//
// t1 - t*t2 = u · d1
// t*t1 - t2 = u · d2
//
// Solve for t1, and find q1 using (0):
//
// t1 = (u·d1 - t * (u·d2))/ (1 - t^2)
//
// just find q2 by using line2.getClosestPoint(q1)
SbVec3f p1 = this->pos;
SbVec3f p2 = line2.pos;
SbVec3f d1 = this->dir;
SbVec3f d2 = line2.dir;
SbVec3f u = p2-p1;
float t = d1.dot(d2);
const float eps = 1.0e-06f;
if (t < -1.0f + eps || t > 1.0f-eps) {
// lines are parallel
return FALSE;
}
t = (u.dot(d1) - t * u.dot(d2)) / (1-t*t);
ptOnThis = p1 + t * d1;
ptOnLine2 = line2.getClosestPoint(ptOnThis);
return TRUE;
#else // old version
#if COIN_DEBUG
if(this->getDirection().length() == 0.0)
SoDebugError::postWarning("SbLine::getClosestPoints",
"This line has no direction (zero vector).");
if(line2.getDirection().length() == 0.0)
SoDebugError::postWarning("SbLine::getClosestPoints",
"argument line has no direction (zero vector).");
#endif // COIN_DEBUG
// Check if the lines are parallel.
// FIXME: should probably use equals() here.
if(line2.dir == this->dir) return FALSE;
else if(line2.dir == -this->dir) return FALSE;
// From the discussion on getClosestPoint(), we know that the point
// we wish to find on a line can be expressed as:
//
// (Q1-P0)·D0
// Q0 = P0 + D0 * ----------
// |D0|
//
// ...where P0 is a point on the first line, D0 is the direction
// vector and Q1 is the "closest point" on the other line. From this
// we get two equations with two unknowns. By substituting for
// Q1 we get a new equation with a single unknown, Q0:
//
// ( (Q0 - P1)·D1 )
// (P1 + D1 * ------------ - P0) · D0
// ( |D1| )
// Q0 = P0 + D0 * ------------------------------------
// |D0|
//
// Which obviously is bloody hard (perhaps impossible?) to solve
// analytically. Damn. Back to the pen and pencil stuff.
//
// Ok, new try. Since we're looking for the minimum distance between the
// two lines, we should be able to solve it by expressing the distance
// between the points we want to find as a parametrized function and
// take the derivative:
//
// f(t0, t1) = |Q1 - Q0| = |P1+D1*t1 - (P0+D0*t0)|
//
// (t1*D1 - P0)·D0
// t0 can be expressed as --------------- which gives us
// |D0|
//
// f(t) = |P1 + D1*t - P0 - D0N * ((t*D1 - P0)·D0)|, t = t1
// D0N = D0 normalized
// _____________
// ..which is eual to f(t) = \/Þ² + ß² + ð² , where Þ, ß, and ð
// is the full expression above with the x, y, and z components
// of the vectors.
//
// Since we're looking for the minimum value of the function, we can just
// ignore the square root. We'll do the next parts of the math on a
// general components case, since it's the same for the x, y and z parts.
//
// Expanding any of the Þ, ß, or ð expressions, we get this:
// (P1[i] - D1[i]*t - P0[i] - D0N[i]*D0[x]*D1[x]*t + D0N[i]*D0[x]*P0[x]
// - D0N[i]*D0[y]*D1[y]*t + D0N[i]*D0[y]*P0[y] - D0N[i]*D0[z]*D1[z]*t
// + D0N[i]*D0[z]*P0[z])² ,
// where i=[x|y|z].
//
// Derivating this by using the chain rule (i.e. g(t)² =
// 2*g(t)*g'(t)), we'll get this equation for finding the t yielding
// the minimum distance between two points Q0 and Q1 on the lines:
//
// -(cx*dx+cy*dy+cz*dz)
// t = --------------------
// dx² + dy² + dz²
//
// di = D1[i] - D0N[i] * (D0[x]*D1[x] + D0[y]*D1[y] + D0[z]*D1[z])
// and
// ci = P1[i] - P0[i] + D0N[i] * (D0[x]*P0[x] + D0[y]*P0[y] + D0[z]*P0[z])
// where i=[x|y|z].
//
// Now we'll substitute t back in for t1 in Q1 = P1 + D1*t1, which'll
// also let us find Q0 by an invocation of getClosestPoint().
//
// That's it. I can't believe this took me 4 hours to complete. Code worked
// on the first run, though. :-)
// 19980815 mortene.
SbVec3f P0 = this->pos;
SbVec3f P1 = line2.pos;
SbVec3f D0 = this->dir;
SbVec3f D1 = line2.dir;
SbVec3f D0N = D0;
// we warn about lines with no direction above, just normalize
(void) D0N.normalize();
float c[3], d[3];
for (int i=0; i < 3; i++) {
d[i] =
D1[i] - D0N[i]*(D0[0]*D1[0] + D0[1]*D1[1] + D0[2]*D1[2]);
c[i] =
P1[i] - P0[i] + D0N[i]*(D0[0]*P0[0] + D0[1]*P0[1] + D0[2]*P0[2]);
}
float t = -(c[0]*d[0]+c[1]*d[1]+c[2]*d[2]) / (d[0]*d[0]+d[1]*d[1]+d[2]*d[2]);
ptOnLine2 = line2.pos + line2.dir * t;
ptOnThis = this->getClosestPoint(ptOnLine2);
return TRUE;
#endif
}