C# 中的高效 AABB/三角形相交

问题描述

任何人都可以向 CSharp 推荐任何公共 AABB/三角形相交算法的有效端口.

Can anyone recommend an efficient port to CSharp of any of the public AABB/triangle intersection algorithms.

我一直在研究 Moller 的方法，抽象地描述了 这里，如果我要移植它，我可能会从 这个 C++ 版本.这个由 Mike Vandelay 编写的 C++ 库 似乎这也是一个很好的起点.

I've been looking at Moller's approach, described abstractly here, and if I were to port it, I would probably start from this C++ version. This C++ library by Mike Vandelay seems like it could also be a great starting point.

...或...任何其他轮子"，可以采用 Vector3 的三角形并告诉我它是否与 AABB 相交)，相对有效.

...or... any other "wheel" that can take a triangle of Vector3's and tell me if it intersects with an AABB), relatively efficiently.

似乎有各种各样的算法，但大多数似乎都是用 c++ 编写的，或者只是在白皮书中抽象地描述了，我需要为我们的应用程序提供一个特定于 c# 的实现.效率不是关键，但 c# 是.(虽然效率当然也很不错;p)

There seem to be a variety of algorithms, but most seem to be written in c++, or just described abstractly in white papers and I need a c# specific implementation for our application. Efficiency is not key, but c# is. (though efficiency is obviously nice too of course ;p )

任何 C# 选项，在我涉足数学"端口之前 ;) 将不胜感激！谢谢.

Any C# options, before I wade through a "math" port ;) would be greatly appreciated! Thanks.

推荐答案

对于任意两个凸网格，要判断它们是否相交，需要检查是否存在分离平面.如果是这样，它们就不会相交.可以从任意形状的任何面或边缘叉积中拾取平面.

For any two convex meshes, to find whether they intersect, you need to check if there exist a separating plane. If it does, they do not intersect. The plane can be picked from any face of either shape, or the edge cross-products.

平面被定义为法线和从 Origo 的偏移.因此，您只需检查 AABB 的三个面和三角形的一个面.

The plane is defined as a normal and an offset from Origo. So, you only have to check three faces of the AABB, and one face of the triangle.

bool IsIntersecting(IAABox box, ITriangle triangle)
{
    double triangleMin, triangleMax;
    double boxMin, boxMax;

    // Test the box normals (x-, y- and z-axes)
    var boxNormals = new IVector[] {
        new Vector(1,0,0),
        new Vector(0,1,0),
        new Vector(0,0,1)
    };
    for (int i = 0; i < 3; i++)
    {
        IVector n = boxNormals[i];
        Project(triangle.Vertices, boxNormals[i], out triangleMin, out triangleMax);
        if (triangleMax < box.Start.Coords[i] || triangleMin > box.End.Coords[i])
            return false; // No intersection possible.
    }

    // Test the triangle normal
    double triangleOffset = triangle.Normal.Dot(triangle.A);
    Project(box.Vertices, triangle.Normal, out boxMin, out boxMax);
    if (boxMax < triangleOffset || boxMin > triangleOffset)
        return false; // No intersection possible.

    // Test the nine edge cross-products
    IVector[] triangleEdges = new IVector[] {
        triangle.A.Minus(triangle.B),
        triangle.B.Minus(triangle.C),
        triangle.C.Minus(triangle.A)
    };
    for (int i = 0; i < 3; i++)
    for (int j = 0; j < 3; j++)
    {
        // The box normals are the same as it's edge tangents
        IVector axis = triangleEdges[i].Cross(boxNormals[j]);
        Project(box.Vertices, axis, out boxMin, out boxMax);
        Project(triangle.Vertices, axis, out triangleMin, out triangleMax);
        if (boxMax <= triangleMin || boxMin >= triangleMax)
            return false; // No intersection possible
    }

    // No separating axis found.
    return true;
}

void Project(IEnumerable<IVector> points, IVector axis,
        out double min, out double max)
{
    double min = double.PositiveInfinity;
    double max = double.NegativeInfinity;
    foreach (var p in points)
    {
        double val = axis.Dot(p);
        if (val < min) min = val;
        if (val > max) max = val;
    }
}

interface IVector
{
    double X { get; }
    double Y { get; }
    double Z { get; }
    double[] Coords { get; }
    double Dot(IVector other);
    IVector Minus(IVector other);
    IVector Cross(IVector other);
}

interface IShape
{
    IEnumerable<IVector> Vertices { get; }
}

interface IAABox : IShape
{
    IVector Start { get; }
    IVector End { get; }
}

interface ITriangle : IShape {
    IVector Normal { get; }
    IVector A { get; }
    IVector B { get; }
    IVector C { get; }
}

<小时>

一个很好的例子是方框 (±10, ±10, ±10) 和三角形 (12,9,9),(9,12,9),(19,19,20).没有一个面可以用作分离平面，但它们不相交.分离轴为<1,1,0>，由<1,0,0>和<-3,3,0>的叉积得到.

A good example is the box (±10, ±10, ±10) and the triangle (12,9,9),(9,12,9),(19,19,20). None of the faces can be used as a separating plane, yet they do not intersect. The separating axis is <1,1,0>, which is obtained from the cross product between <1,0,0> and <-3,3,0>.

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