Question

就像问题一样。答案最好是伪代码并引用。正确的答案应该重视速度而不是简单。

Answer 1

见Intersections of Rays, Segments, Planes and Triangles in 3D。您可以找到对多边形进行三角测量的方法。

如果你真的需要光线/多边形交叉，它是Real-Time Rendering的16.9（第二版为13.8）。

我们首先计算交叉点光线和光线之间的平面 ploygon] pie_p，很容易完成用光线替换x。

 n_p DOT (o + td) + d_p = 0 <=> t = (-d_p - n_p DOT o) / (n_p DOT d)

如果分母|n_p DOT d| < epsilon，其中epsilon非常小数字，然后考虑光线平行于多边形平面，没有交叉发生。否则，光线和光线的交点p 计算多边形平面：p = o + td。此后，问题决定p是否在...内多边形从三个减少到两个 dimentions ...

有关详细信息，请参阅本书。

Answer 2

struct point
{
    float x
    float y
    float z
}

struct ray
{
    point R1
    point R2
}

struct polygon
{
    point P[]
    int count
}

float dotProduct(point A, point B)
{
    return A.x*B.x + A.y*B.y + A.z*B.z
}

point crossProduct(point A, point B)
{
    return point(A.y*B.z-A.z*B.y, A.z*B.x-A.x*B.z, A.x*B.y-A.y*B.x)
}

point vectorSub(point A, point B)
{
    return point(A.x-B.x, A.y-B.y, A.z-B.z) 
}

point scalarMult(float a, Point B)
{
    return point(a*B.x, a*B.y, a*B.z)
}

bool findIntersection(ray Ray, polygon Poly, point& Answer)
{
    point plane_normal = crossProduct(vectorSub(Poly.P[1], Poly.P[0]), vectorSub(Poly.P[2], Poly.P[0]))

    float denominator = dotProduct(vectorSub(Ray.R2, Poly.P[0]), plane_normal)

    if (denominator == 0) { return FALSE } // ray is parallel to the polygon

    float ray_scalar = dotProduct(vectorSub(Poly.P[0], Ray.R1), plane_normal)

    Answer = vectorAdd(Ray.R1, scalarMult(ray_scalar, Ray.R2))

    // verify that the point falls inside the polygon

    point test_line = vectorSub(Answer, Poly.P[0])
    point test_axis = crossProduct(plane_normal, test_line)

    bool point_is_inside = FALSE

    point test_point = vectorSub(Poly.P[1], Answer)
    bool prev_point_ahead = (dotProduct(test_line, test_point) > 0)
    bool prev_point_above = (dotProduct(test_axis, test_point) > 0)

    bool this_point_ahead
    bool this_point_above

    int index = 2;
    while (index < Poly.count)
    {
        test_point = vectorSub(Poly.P[index], Answer)
        this_point_ahead = (dotProduct(test_line, test_point) > 0)

        if (prev_point_ahead OR this_point_ahead)
        {
            this_point_above = (dotProduct(test_axis, test_point) > 0)

            if (prev_point_above XOR this_point_above)
            {
                point_is_inside = !point_is_inside
            }
        }

        prev_point_ahead = this_point_ahead
        prev_point_above = this_point_above
        index++
    }

    return point_is_inside
}

Answer 3

全书章节一直致力于这个特殊要求 - 在这里描述一个合适的算法太长了。我建议阅读计算机图形学中的任意数量的参考书，特别是：

Ray Tracing简介，编辑。 Andrew S. Glassner，ISBN 0122861604

Answer 4

function Collision(PlaneOrigin,PlaneDirection,RayOrigin,RayDirection)
    return RayOrigin-RayDirection*Dot(PlaneDirection,RayOrigin-PlaneOrigin)/Dot(PlaneDirection,RayDirection)
end

（PlaneDirection是垂直于平面的单位矢量）

找到光线和多边形之间交点的最快方法是什么？

4 个答案: