在C＃高效AABB /三角形相交 [英] Efficient AABB/triangle intersection in C#

问题描述

谁能推荐任何公共AABB /三角形相交算法的高效端口CSHARP。

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

我一直在寻找穆勒的做法，抽象描述的此处，如果我是将它移植，我可能会从的这个C ++版本。的这个C由Mike Vandelay ++库好像它也可能是一个很好的起点。

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...任何其他的车轮，可以采取Vector3类型的一个三角形，并告诉我，如果将其与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.

似乎有不同的算法，但多数似乎是用C ++编写，或者只是在白皮书中描述抽象的，我需要交流＃具体的实施为我们的应用程序。效率不是关键，但C＃是。 （虽然效率显然是很好过的课程; 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＃的选择，之前我涉水通过数学端口;）将大大AP preciated！谢谢你。

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

推荐答案

对于任何两个凸目，发现他们是否相交，则需要检查是否存在分离平面。如果是这样，他们不相交。该平面可以从任何面任一形状的，或边缘交叉的产品被拾取

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.

的平面被定义为一个正常的和从欧瑞偏移。所以，你只需要检查三个面的AABB，以及一个面三角形的。

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; }
}

---

一个很好的例子是箱（±10，±10，±10）和三角形（12,9,9），（9,12,9），（19,19,20）。没有一个面可以作为一个分离平面，但它们不相交。该分离轴线为＆lt; 1,1,0>，它是从交叉乘积得到之间＆所述; 1,0,0>和＆lt;。-3,3,0>

---

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>.

这篇关于在C＃高效AABB /三角形相交的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！