考虑 3D 中的两个几何对象:

• 与坐标轴对齐并由其中心位置和范围(边长)定义的立方体
• 一个未与轴对齐的圆锥体，由其顶点位置、其底边的中心位置和顶点处的半角定义

下面是用 C++ 定义这些对象的小代码:

// Preprocessor
#include <iostream>
#include <cmath>
#include <array>

// 3D cube from the position of its center and the side extent
class cube
{ 
    public:
        cube(const std::array<double, 3>& pos, const double ext)
        : _position(pos), _extent(ext) 
        {;}
        double center(const unsigned int idim) 
            {return _position[idim];}
        double min(const unsigned int idim)
            {return _position[idim]-_extent/2;}
        double max(const unsigned int idim)
            {return _position[idim]+_extent/2;}
        double extent()
            {return _extent;}
        double volume()
            {return std::pow(_extent, 3);}
    protected:
        std::array<double, 3> _position;
        double _extent;
};

// 3d cone from the position of its vertex, the base center, and the angle
class cone
{
    public:
        cone(const std::array<double, 3>& vert, 
             const std::array<double, 3>& bas, 
             const double ang)
        : _vertex(vert), _base(bas), _angle(ang)
        {;}
        double vertex(const unsigned int idim)
            {return _vertex[idim];}
        double base(const unsigned int idim)
            {return _base[idim];}
        double angle()
            {return _angle;}
        double height()
            {return std::sqrt(std::pow(_vertex[0]-_base[0], 2)+std::pow(
            _vertex[1]-_base[1], 2)+std::pow(_vertex[2]-_base[2], 2));}
        double radius()
            {return std::tan(_angle)*height();}
        double circle()
            {return 4*std::atan(1)*std::pow(radius(), 2);}
        double volume()
            {return circle()*height()/3;}
    protected:
        std::array<double, 3> _vertex;
        std::array<double, 3> _base;
        double _angle;
};

我想写一个函数来检测立方体和圆锥的交点是否为空:

// Detect whether the intersection between a 3d cube and a 3d cone is not null
bool intersection(const cube& x, const cone& y)
{
    // Function that returns false if the intersection of x and y is empty
    // and true otherwise
}

这是问题的图示(插图是 2D 的，但我的问题是 3D 的):

如何有效地做到这一点(我正在寻找一种算法，所以答案可以是 C、C++ 或 Python)？

注意:这里的交集定义为:它存在一个非空的 3D 体积，它在立方体和锥体中(如果立方体在锥体内部，或者如果锥体在立方体内部，它们相交)。

最佳答案

代码

这个答案会比你的问题更笼统(例如，我考虑的是一个盒子而不是一个立方体)。适应您的情况应该非常简单。

一些定义

/*
    Here is the cone in cone space:

            +        ^
           /|\       |
          /*| \      | H
         /  |  \     |
        /       \    |
       +---------+   v

    * = alpha (angle from edge to axis)
*/
struct Cone // In cone space (important)
{
    double H;
    double alpha;
};

/*
    A 3d plane
      v
      ^----------+
      |          |
      |          |
      +----------> u
      P
*/
struct Plane
{
    double u;
    double v;
    Vector3D P;
};

// Now, a box.
// It is assumed that the values are coherent (that's only for this answer).
// On each plane, the coordinates are between 0 and 1 to be inside the face.
struct Box
{
    Plane faces[6];
};

线-锥相交

现在，让我们计算一个线段和我们的圆锥之间的交点。请注意，我将在锥形空间中进行计算。另请注意，我将 Z 轴视为垂直轴。将其更改为 Y 留给读者作为练习。假定该线位于圆锥空间中。分段方向未归一化；相反，该段具有方向 vector 的长度，并从点 P:

/*
    The segment is points M where PM = P + t * dir, and 0 <= t <= 1
    For the cone, we have 0 <= Z <= cone.H
*/
bool intersect(Cone cone, Vector3D dir, Vector3D P)
{
    // Beware, indigest formulaes !
    double sqTA = tan(cone.alpha) * tan(cone.alpha);
    double A = dir.X * dir.X + dir.Y * dir.Y - dir.Z * dir.Z * sqTA;
    double B = 2 * P.X * dir.X +2 * P.Y * dir.Y - 2 * (cone.H - P.Z) * dir.Z * sqTA;
    double C = P.X * P.X + P.Y * P.Y - (cone.H - P.Z) * (cone.H - P.Z) * sqTA;

    // Now, we solve the polynom At² + Bt + C = 0
    double delta = B * B - 4 * A * C;
    if(delta < 0)
        return false; // No intersection between the cone and the line
    else if(A != 0)
    {
        // Check the two solutions (there might be only one, but that does not change a lot of things)
        double t1 = (-B + sqrt(delta)) / (2 * A);
        double z1 = P.Z + t1 * dir.Z;
        bool t1_intersect = (t1 >= 0 && t1 <= 1 && z1 >= 0 && z1 <= cone.H);

        double t2 = (-B - sqrt(delta)) / (2 * A);
        double z2 = P.Z + t2 * dir.Z;
        bool t2_intersect = (t2 >= 0 && t2 <= 1 && z2 >= 0 && z2 <= cone.H);

        return t1_intersect || t2_intersect;
    }
    else if(B != 0)
    {
        double t = -C / B;
        double z = P.Z + t * dir.Z;
        return t >= 0 && t <= 1 && z >= 0 && z <= cone.H;
    }
    else return C == 0;
}

矩形-圆锥相交

现在，我们可以检查平面的矩形部分是否与圆锥相交(这将用于检查立方体的一个面是否与圆锥相交)。仍在锥形空间中。该计划以一种对我们有帮助的方式通过:2 个 vector 和一个点。 vector 没有归一化，以简化计算。

/*
    A point M in the plan 'rect' is defined by:
        M = rect.P + a * rect.u + b * rect.v, where (a, b) are in [0;1]²
*/
bool intersect(Cone cone, Plane rect)
{
    bool intersection = intersect(cone, rect.u, rect.P)
                     || intersect(cone, rect.u, rect.P + rect.v)
                     || intersect(cone, rect.v, rect.P)
                     || intersect(cone, rect.v, rect.P + rect.u);

    if(!intersection)
    {
        // It is possible that either the part of the plan lie
        // entirely in the cone, or the inverse. We need to check.
        Vector3D center = P + (u + v) / 2;

        // Is the face inside the cone (<=> center is inside the cone) ?
        if(center.Z >= 0 && center.Z <= cone.H)
        {
            double r = (H - center.Z) * tan(cone.alpha);
            if(center.X * center.X + center.Y * center.Y <= r)
                intersection = true;
        }

        // Is the cone inside the face (this one is more tricky) ?
        // It can be resolved by finding whether the axis of the cone crosses the face.
        // First, find the plane coefficient (descartes equation)
        Vector3D n = rect.u.crossProduct(rect.v);
        double d = -(rect.P.X * n.X + rect.P.Y * n.Y + rect.P.Z * n.Z);

        // Now, being in the face (ie, coordinates in (u, v) are between 0 and 1)
        // can be verified through scalar product
        if(n.Z != 0)
        {
            Vector3D M(0, 0, -d/n.Z);
            Vector3D MP = M - rect.P;
            if(MP.scalar(rect.u) >= 0
               || MP.scalar(rect.u) <= 1
               || MP.scalar(rect.v) >= 0
               || MP.scalar(rect.v) <= 1)
                intersection = true;
        }
    }
    return intersection;
}

方框 - 圆锥相交

现在，最后一部分:整个立方体:

bool intersect(Cone cone, Box box)
{
    return intersect(cone, box.faces[0])
        || intersect(cone, box.faces[1])
        || intersect(cone, box.faces[2])
        || intersect(cone, box.faces[3])
        || intersect(cone, box.faces[4])
        || intersect(cone, box.faces[5]);
}

为了数学

仍然在圆锥空间中，圆锥方程为:

// 0 is the base, the vertex is at z = H
x² + y² = (H - z)² * tan²(alpha)
0 <= z <= H

现在，3D 直线的参数方程为:

x = u + at
y = v + bt
z = w + ct

方向 vector 为(a,b,c)，点(u,v,w)在直线上。

现在，让我们把方程式放在一起:

(u + at)² + (v + bt)² = (H - w - ct)² * tan²(alpha)

然后在开发和重新分解这个方程之后，我们得到以下结果:

At² + Bt + C = 0

其中 A、B 和 C 显示在第一个交集函数中。只需解决此问题并检查 z 和 t 上的边界条件。

关于python - 检测立方体和圆锥体是否相交？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/22023977/