Question

我正在使用this文章研究triange-triangle检测算法。我编写了这段代码，但不幸的是，当三角形之间存在交集时，该方法返回false。

private boolean checkTriangleCollision(Triangle triangle1, Triangle triangle2) {
    Vector3f n1 = getN(triangle1);
    Vector3f n2 = getN(triangle2);
    Vector3f v = null;
    v = Vector3f.cross(n1, n2, v);

    Vector3f p = new Vector3f();

    float pxy = n1.z * triangle1.vertex1.position.z * n1.x * n1.y;
    float px = n1.z * triangle1.vertex1.position.z * n1.x * n1.y * triangle1.vertex1.position.y;
    float py = n1.z * triangle1.vertex1.position.z * n1.x * triangle1.vertex1.position.x * n1.y;
    float p0 = n1.z * triangle1.vertex1.position.z * n1.x * triangle1.vertex1.position.x * n1.y * triangle1.vertex1.position.y;

    p.x = (p0/(px/pxy))/pxy;
    p.y = (p0/(py/pxy))/pxy;

    Vector3f x = null;
    x = Vector3f.add(p, v, x);

    Vector3f xq1 = null;
    xq1 = Vector3f.sub(x, triangle1.vertex1.position, xq1);
    float i1 = Vector3f.dot(xq1, n1);

    Vector3f xq2 = null;
    xq2 = Vector3f.sub(x, triangle2.vertex1.position, xq2);
    float i2 = Vector3f.dot(xq2, n2);

    if (i1 == 0 && i2 == 0) {
        return true;
    }
    return false;
}

private Vector3f getN(Triangle triangle) {
    Vector3f vn1 = null;
    vn1 = Vector3f.sub(triangle.vertex2.position, triangle.vertex1.position, vn1);
    Vector3f vn2 = null;
    vn2 = Vector3f.sub(triangle.vertex3.position, triangle.vertex1.position, vn2);
    Vector3f n = null;
    n = Vector3f.cross(vn1, vn2, n);
    return n;
}

我不知道为什么，但我想问题是计算p。如果是，我怎么能通过代码解决这个等式？

N1·(P0-Q1) = 0 
and
N2·(P0-R1) = 0

Answer 1

对我来说，你做错了点积。

试试这个：

    Vector3f n1 = getN(triangle1);
    Vector3f n2 = getN(triangle2);
    Vector3f v = null;
    v = Vector3f.cross(n1, n2, v);

    Vector3f p = new Vector3f();

//P0 calcuiation

    float C1 = n1.x * triangle1.vertex1.position.x + n1.y * triangle1.vertex1.position.y + n1.z * triangle1.vertex1.position.z;

    float C2 = n2.x * triangle2.vertex1.position.x + n2.y * triangle2.vertex1.position.y + n2.z * triangle2.vertex1.position.z;

    float Kn = n2.x / n1.x;

    p.y = (C2 - C1 * Kn) / (n2.y - n1.y * Kn);

    p.x = (C1 - n1.y * p.y) / n2.x;

p.z = 0;

//P0 calcuiation ends

    Vector3f x = null;
    x = Vector3f.add(p, v, x);

以下是我的计算： http://i.stack.imgur.com/3USpZ.png

三角形 - 三角形交叉检测

1 个答案: