我一直在努力研究线段 - 圆筒交叉口例程,我的大脑已经融化。试图优化线条与圆柱体相交
/// Line segment VS <cylinder>
// - cylinder (A, B, r) (start point, end point, radius)
// - line has starting point (x0, y0, z0) and ending point (x0+ux, y0+uy, z0+uz) ((ux, uy, uz) is "direction")
// => start = (x0, y0, z0)
// dir = (ux, uy, uz)
// A
// B
// r
// optimize? (= don't care for t > 1)
// <= t = "time" of intersection
// norm = surface normal of intersection point
void CollisionExecuter::cylinderVSline(const Ogre::Vector3& start, const Ogre::Vector3& dir, const Ogre::Vector3& A, const Ogre::Vector3& B, const double r,
const bool optimize, double& t, Ogre::Vector3& normal) {
t = NaN;
// Solution : http://www.gamedev.net/community/forums/topic.asp?topic_id=467789
double cxmin, cymin, czmin, cxmax, cymax, czmax;
if (A.z < B.z) { czmin = A.z - r; czmax = B.z + r; } else { czmin = B.z - r; czmax = A.z + r; }
if (A.y < B.y) { cymin = A.y - r; cymax = B.y + r; } else { cymin = B.y - r; cymax = A.y + r; }
if (A.x < B.x) { cxmin = A.x - r; cxmax = B.x + r; } else { cxmin = B.x - r; cxmax = A.x + r; }
if (optimize) {
if (start.z >= czmax && (start.z + dir.z) > czmax) return;
if (start.z <= czmin && (start.z + dir.z) < czmin) return;
if (start.y >= cymax && (start.y + dir.y) > cymax) return;
if (start.y <= cymin && (start.y + dir.y) < cymin) return;
if (start.x >= cxmax && (start.x + dir.x) > cxmax) return;
if (start.x <= cxmin && (start.x + dir.x) < cxmin) return;
}
Ogre::Vector3 AB = B - A;
Ogre::Vector3 AO = start - A;
Ogre::Vector3 AOxAB = AO.crossProduct(AB);
Ogre::Vector3 VxAB = dir.crossProduct(AB);
double ab2 = AB.dotProduct(AB);
double a = VxAB.dotProduct(VxAB);
double b = 2 * VxAB.dotProduct(AOxAB);
double c = AOxAB.dotProduct(AOxAB) - (r*r * ab2);
double d = b * b - 4 * a * c;
if (d < 0) return;
double time = (-b - sqrt(d))/(2 * a);
if (time < 0) return;
Ogre::Vector3 intersection = start + dir * time; /// intersection point
Ogre::Vector3 projection = A + (AB.dotProduct(intersection - A)/ab2) * AB; /// intersection projected onto cylinder axis
if ((projection - A).length() + (B - projection).length() > AB.length()) return; /// THIS IS THE SLOW SAFE WAY
//if (projection.z > czmax - r || projection.z < czmin + r ||
// projection.y > cymax - r || projection.y < cymin + r ||
// projection.x > cxmax - r || projection.x < cxmin + r) return; /// THIS IS THE FASTER BUGGY WAY
normal = (intersection - projection);
normal.normalise();
t = time; /// at last
}
我曾想过这样的方式来加速发现交点投影是否位于圆柱体长度内。然而,它不起作用,我无法真正了解它,因为它看起来很合逻辑: 如果投影点的x,y或z坐标不在圆柱体的极限范围内,则应该在外部考虑。看来虽然这在实践中并不奏效。
任何帮助将不胜感激!
干杯,
比尔Kotsias
编辑:看来,问题上升与边界的情况下,即当气缸是平行于轴线的一个。舍入误差进入公式,“优化”停止正常工作。
if (projection.z > czmax - r + 0.001 || projection.z < czmin + r - 0.001 || ... etc...