测地球的算法

Question

我必须用较小的均匀分布的球制作一个球体.我认为最佳方法是构建一个基于三角形的测地球体,并使用顶点作为我球的中间点.但我没有编写生成顶点的算法.用C++或伪代码回答会更好.

测地球的示例:http://i.stack.imgur.com/iNQfP.png

Answer 1

使用链接@Muckle_ewe给了我,我能够编写以下算法:在...之外 main()

class Vector3d {  // this is a pretty standard vector class
public:
    double x, y, z;
    ...
}

void subdivide(const Vector3d &v1, const Vector3d &v2, const Vector3d &v3, vector<Vector3d> &sphere_points, const unsigned int depth) {
    if(depth == 0) {
        sphere_points.push_back(v1);
        sphere_points.push_back(v2);
        sphere_points.push_back(v3);
        return;
    }
    const Vector3d v12 = (v1 + v2).norm();
    const Vector3d v23 = (v2 + v3).norm();
    const Vector3d v31 = (v3 + v1).norm();
    subdivide(v1, v12, v31, sphere_points, depth - 1);
    subdivide(v2, v23, v12, sphere_points, depth - 1);
    subdivide(v3, v31, v23, sphere_points, depth - 1);
    subdivide(v12, v23, v31, sphere_points, depth - 1);
}

void initialize_sphere(vector<Vector3d> &sphere_points, const unsigned int depth) {
    const double X = 0.525731112119133606;
    const double Z = 0.850650808352039932;
    const Vector3d vdata[12] = {
        {-X, 0.0, Z}, { X, 0.0, Z }, { -X, 0.0, -Z }, { X, 0.0, -Z },
        { 0.0, Z, X }, { 0.0, Z, -X }, { 0.0, -Z, X }, { 0.0, -Z, -X },
        { Z, X, 0.0 }, { -Z, X, 0.0 }, { Z, -X, 0.0 }, { -Z, -X, 0.0 }
    };
    int tindices[20][3] = {
        {0, 4, 1}, { 0, 9, 4 }, { 9, 5, 4 }, { 4, 5, 8 }, { 4, 8, 1 },
        { 8, 10, 1 }, { 8, 3, 10 }, { 5, 3, 8 }, { 5, 2, 3 }, { 2, 7, 3 },
        { 7, 10, 3 }, { 7, 6, 10 }, { 7, 11, 6 }, { 11, 0, 6 }, { 0, 1, 6 },
        { 6, 1, 10 }, { 9, 0, 11 }, { 9, 11, 2 }, { 9, 2, 5 }, { 7, 2, 11 }
    };
    for(int i = 0; i < 20; i++)
        subdivide(vdata[tindices[i][0]], vdata[tindices[i][1]], vdata[tindices[i][2]], sphere_points, depth);
}

然后在main():

vector<Vector3d> sphere_points;
initialize_sphere(sphere_points, DEPTH);  // where DEPTH should be the subdivision depth
for(const Vector3d &point : sphere_points)
    const Vector3d point_tmp = point * RADIUS + CENTER;  // Then for the sphere I want to draw, I  iterate over all the precomputed sphere points and with a linear function translate the sphere to its CENTER and chose the proper RADIUS

实际上,您只需要使用initialize_sphere()一次并将结果用于您想要绘制的每个球体.