数学-几何

点到直线的垂足

Point1 GetFootOfPerpendicular(const Point1 &pt, const Point1 &begin, const Point1 &end){Point1 retVal;double dx = begin.x - end.x;double dy = begin.y - end.y;double dz = begin.z - end.z;if (abs(dx) < 0.00000001 && abs(dy) < 0.00000001 && abs(dz) < 0.00000001){retVal = begin;return retVal;}double u = (pt.x - begin.x)*(begin.x - end.x) +(pt.y - begin.y)*(begin.y - end.y) + (pt.z - begin.z)*(begin.z - end.z);u = u / ((dx*dx) + (dy*dy) + (dz*dz));retVal.x = begin.x + u*dx;retVal.y = begin.y + u*dy;retVal.z = begin.z + u*dz;return retVal;}

直线与圆的交点

int CompNode(double * v, double r, double *node, double c, double d){double k, b;                                  //直线方程：y = kx + b；   //圆方程：(x - c) ^ 2 + (y - d) ^ 2 = r ^ 2;       k = (v[1] - v[3]) / (v[0] - v[2]);b = v[1] - k*v[0];if (fabs(k*c - d + b) / sqrt(k*k + b*b)){node[0] = (2 * c - 2 * k*(b - d) + sqrt(pow((2 * k*(b - d) - 2 * c), 2) - 4 * (k*k + 1)*((b - d)*(b - d) + c*c - r*r))) / (2 * k*k + 2);node[1] = k*node[0] + b;node[2] = (2 * c - 2 * k*(b - d) - sqrt(pow((2 * k*(b - d) - 2 * c), 2) - 4 * (k*k + 1)*((b - d)*(b - d) + c*c - r*r))) / (2 * k*k + 2);node[3] = k*node[2] + b;}return 0;}

直线与球的交点

int FindLineSphereIntersections(Point1 linePoint0, Point1 linePoint1, Point1 circleCenter, double circleRadius, Point1 *node){double cx = circleCenter.x;double cy = circleCenter.y;double cz = circleCenter.z;double px = linePoint0.x;double py = linePoint0.y;double pz = linePoint0.z;double vx = linePoint1.x - px;double vy = linePoint1.y - py;double vz = linePoint1.z - pz;double A = vx * vx + vy * vy + vz * vz;double B = 2.0 * (px * vx + py * vy + pz * vz - vx * cx - vy * cy - vz * cz);double C = px * px - 2 * px * cx + cx * cx + py * py - 2 * py * cy + cy * cy +pz * pz - 2 * pz * cz + cz * cz - circleRadius * circleRadius;// discriminantdouble D = B * B - 4 * A * C;if (D < 0){return -1;}double t1 = (-B - sqrt(D)) / (2.0 * A);node[0].x = linePoint0.x * (1 - t1) + t1 * linePoint1.x;node[0].y = linePoint0.y * (1 - t1) + t1 * linePoint1.y;node[0].z = linePoint0.z * (1 - t1) + t1 * linePoint1.z;if (D == 0){return 0;}double t2 = (-B + sqrt(D)) / (2.0 * A);node[1].x = linePoint0.x * (1 - t2) + t2 * linePoint1.x;node[1].y = linePoint0.y * (1 - t2) + t2 * linePoint1.y;node[1].z = linePoint0.z * (1 - t2) + t2 * linePoint1.z;// prefer a solution that's on the line segment itselfif (abs(t1 - 0.5) < abs(t2 - 0.5)){return 0;}return 0;}