python - Detect if a cube and a cone intersect each other?

Question

Consider two geometrical objects in 3D:

• a cube aligned with the axes and defined by the position of its center and its extent (edge length)
• a cone not aligned with the axes and defined by the position of its vertex, the position of the center of its base, and the half-angle at the vertex

Here is a small code to define these objects in C++:

// 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;
};

I would like to write a function to detect whether the intersection of a cube and a cone is empty or not:

// 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
}

Here is an illustration of the problem (the illustration is in 2D, but my problem is in 3D):

How to do that efficiently (I am searching for an algorithm, so the answer can be in C, C++ or Python) ?

Note: Here intersection is defined as: it exists a non-null 3D volume that is in the cube and in the cone (if the cube is inside the cone, or if the cone is inside the cube, they intersect).

Answer 1

For the code

This answer will be slightly more general than your problem (I consider a box instead of a cube for example). Adapting to your case should be really straightforward.

Some definitions

/*
    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];
};

Line - cone intersection

Now, let's compute the intersection between a segment and our cone. Note that I will do calculations in cone space. Note also that I take the Z axis to be the vertical one. Changing it to the Y one is left as an exercise to the reader. The line is assumed to be in cone space. The segment direction is not normalized; instead, the segment is of the length of the direction vector, and starts at the point 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;
}

Rect - cone intersection

Now, we can check whether a rectangular part of a plan intersect the cone (this will be used to check whether a face of the cube intersects the cone). Still in cone space. The plan is passed in a way that will help us: 2 vectors and a point. The vectors are not normalized, to simplify the computations.

/*
    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;
}

Box - cone intersection

Now, the final part : the whole cube:

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]);
}

For the maths

Still in cone space, the cone equations are:

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

Now, the parametric equation of a line in 3D is:

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

The direction vector is (a, b, c), and the point (u, v, w) lie on the line.

Now, let's put the equations together:

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

Then after developping and re-factorizing this equation, we get the following:

At2 + Bt + C = 0

where A, B and C are shown in the first intersection function. Simply resolve this and check the boundary conditions on z and t.