我试图通过射线和圆环的解析方程相交而不用三角化圆环来跟踪光线。我用以下代码做到了:
void circularTorusIntersectFunc(const CircularTorus* circularToruses, RTCRay& ray, size_t item)
{
const CircularTorus& torus = circularToruses[item];
Vec3fa O = ray.org /*- sphere.p*/;
Vec3fa Dir = ray.dir;
O.w = 1.0f;
Dir.w = 0.0f;
O = torus.inv_transform.mult(O);
Dir = torus.inv_transform.mult(Dir);
// r1: cross section of torus
// r2: the ring's radius
// _____ ____
// / r1 \------->r2<--------/ \
// \_____/ \____/
float r2 = sqr(torus.r1);
float R2 = sqr(torus.r2);
double a4 = sqr(dot(Dir, Dir));
double a3 = 4 * dot(Dir, Dir) * dot(O, Dir);
double a2 = 4 * sqr(dot(O, Dir)) + 2 * dot(Dir, Dir) * (dot(O, O) - r2 - R2) + 4 * R2 * sqr(Dir.z);
double a1 = 4 * dot(O, Dir) * (dot(O, O) - r2 - R2) + 8 * R2 * O.z * Dir.z;
double a0 = sqr(dot(O, O) - r2 - R2) + 4 * R2 * sqr(O.z) - 4 * R2 * r2;
a3 /= a4; a2 /= a4; a1 /= a4; a0 /= a4;
double roots[4];
int n_real_roots;
n_real_roots = SolveP4(roots, a3, a2, a1, a0);
if (n_real_roots == 0) return;
Vec3fa intersect_point;
for (int i = 0; i < n_real_roots; i++)
{
float root = static_cast<float>(roots[i]);
intersect_point = root * Dir + O;
if ((ray.tnear <= root) && (root <= ray.tfar)) {
ray.u = 0.0f;
ray.v = 0.0f;
ray.tfar = root;
ray.geomID = torus.geomID;
ray.primID = item;
Vec3fa normal(
4.0 * intersect_point.x * (sqr(intersect_point.x) + sqr(intersect_point.y) + sqr(intersect_point.z) - r2 - R2),
4.0 * intersect_point.y * (sqr(intersect_point.x) + sqr(intersect_point.y) + sqr(intersect_point.z) - r2 - R2),
4.0 * intersect_point.z * (sqr(intersect_point.x) + sqr(intersect_point.y) + sqr(intersect_point.z) - r2 - R2) + 8 * R2*intersect_point.z,
0.0f
);
ray.Ng = normalize(torus.transform.mult(normal));
}
}
}
求解
SolveP4
函数方程的代码取自Solution of cubic and quatric functions。问题是当我们仔细观察圆环时,它的工作原理如下:
但是,当我缩小相机时,相机正看着远处的圆环,它突然变得嘈杂,并且形状识别不清。我尝试对每个像素使用1个以上的样本,但是仍然遇到相同的问题。如下:
看来我正面临一个数字问题,但我不知道如何解决。有人可以帮我吗?
另外,值得一提的是,我正在与Intel的Embree Lib交流。
更新(单色):
最佳答案
我认为很多问题是使用单精度浮点数而不是双精度数。
定义两个功能
double dsqr(double x) { return x*x; }
double ddot(const Vec3fa &a,Vec3fa &b) {
double x1 = a.x, y1 = a.y, z1 = a.z;
double x2 = b.x, y2 = b.y, z2 = b.z;
return x1*x2 + y1*y2 + z1*z2;
}
找到平方和点积,但使用双精度。更改r2 R2 a4 a3 a2 a1和a0的计算以使用这些
double r2 = dsqr(torus.r1);
double R2 = dsqr(torus.r2);
double a4 = dsqr(ddot(Dir, Dir));
double a3 = 4 * ddot(Dir, Dir) * ddot(O, Dir);
double a2 = 4 * dsqr(ddot(O, Dir)) + 2 * ddot(Dir, Dir) * (ddot(O, O) - r2 - R2)
+ 4 * R2 * dsqr(Dir.z);
double a1 = 4 * ddot(O, Dir) * (ddot(O, O) - r2 - R2) + 8 * R2 * O.z * Dir.z;
double a0 = dsqr(ddot(O, O) - r2 - R2) + 4 * R2 * dsqr(O.z) - 4 * R2 * r2;
其余所有代码都相同。在我的测试中,这使看起来模糊的图像看起来非常清晰。