考虑3D中的两个几何对象:
以下是在C++中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;
};
我想编写一个函数来检测立方体和圆锥体的交点是否为空:
// 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
}
这是问题的图解(图解是2D，但我的问题是3D):
如何有效地做到这一点(我正在寻找一种算法，因此答案可以是C，C++或Python)？
注意:这里的交集定义为:它存在于立方体和圆锥体中的非零3D体积(如果立方体在圆锥体内部，或者如果圆锥体在立方体内部，则它们相交)。
最佳答案
对于代码
该答案将比您的问题更笼统(例如，我考虑使用盒子而不是立方体)。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，让我们计算线段和圆锥之间的交点。请注意，我将在圆锥空间中进行计算。另请注意，我将Z轴设为垂直轴。将其更改为Y，留给读者练习。假定该线位于圆锥空间中。段方向未标准化；相反，该段的长度是方向 vector 的长度，从
P点开始:/*
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-圆锥形相交
现在，我们可以检查平面图的矩形部分是否与圆锥相交(这将用于检查立方体的面是否与圆锥相交)。仍处于圆锥空间中。该计划以对我们有帮助的方式通过:2个 vector 和一个点。不对 vector 进行归一化，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;
}
盒子-圆锥形相交
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]);
}
对于数学
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中线的参数方程为:
x = u + at
y = v + bt
z = w + ct
方向 vector 为(a，b，c)，点(u，v，w)位于线上。
现在，让我们将这些方程放在一起:
(u + at)² + (v + bt)² = (H - w - ct)² * tan²(alpha)
然后，在开发并重新分解了该方程后，我们得到以下结果:
At² + Bt + C = 0
其中A，B和C在第一个交集函数中显示。只需解决这个问题，然后检查z和t上的边界条件。