Question

我正在使用以下函数计算两条线的交点：

// Functions of lines as per requested:
// f(y1) = starty1 + x * d1
// f(y2) = starty2 + x * d2
// x1 and y1 are the coordinates of the first point
// x2 and y2 are the coordinates of the second point
// d1 and d2 are the deltas of the corresponding lines
private static double[] intersect(double x1, double y1, double d1, double x2, double y2, double d2) {
    double starty1 = y1 - x1 * d1;
    double starty2 = y2 - x2 * d2;
    double rx = (starty2 - starty1) / (d1 - d2);
    double ry = starty1 + d1 * rx;

    tmpRes[0] = rx;
    tmpRes[1] = ry;

    return tmpRes;
}

// This is the same function, but takes 4 points to make two lines, 
// instead of two points and two deltas.
private static double[] tmpRes = new double[2];
private static double[] intersect(double x1, double y1, double x2, double y2, double x3, double y3, double x4, double y4) {
    double d1 = (y1 - y2) / (x1 - x2);
    double d2 = (y3 - y4) / (x3 - x4);
    double starty1 = y1 - x1 * d1;
    double starty2 = y3 - x3 * d2;
    double rx = (starty2 - starty1) / (d1 - d2);
    double ry = starty1 + d1 * rx;

    tmpRes[0] = rx;
    tmpRes[1] = ry;

    return tmpRes;
}

然而，随着d1或d2变大（对于垂直线），结果变得越来越不准确。我怎样才能防止这种情况发生？

对于我的情况，我有两条线彼此垂直。如果线条旋转45度，我会得到准确的结果。如果线条处于0度或90度，我得到的结果不准确（交叉点的一个轴是正确的，另一个轴遍布整个地方。

修改

使用交叉产品：

private static double[] crTmp = new double[3];
public static double[] cross(double a, double b, double c, double a2, double b2, double c2){
    double newA = b*c2 - c*b2;
    double newB = c*a2 - a*c2;
    double newC = a*b2 - b*a2;
    crTmp[0] = newA;
    crTmp[1] = newB;
    crTmp[2] = newC;
    return crTmp;
}


public static double[] linesIntersect(double x1, double y1, double d1, double x2, double y2, double d2)
{
    double dd1 = 1.0 / d1;
    double dd2 = 1.0 / d2;

    double a1, b1, a2, b2, c1, c2;
    if (Math.abs(d1) < Math.abs(dd1)) {
        a1 = d1;
        b1 = -1.0;
        c1 = y1 - x1 * d1;
    } else {
        a1 = 1.0;
        b1 = dd1;
        c1 = -x1 - y1 * dd1;
    }
    if (Math.abs(d2) < Math.abs(dd2)) {
        a2 = d2;
        b2 = -1.0;
        c2 = y2 - x2 * d2;
    } else {
        a2 = 1.0;
        b2 = dd2;
        c2 = -x2 - y2 * dd2;
    }

    double[] v1 = {a1, b1, c1};
    double[] v2 = {a2, b2, c2};
    double[] res = cross(v1[0], v1[1], v1[2], v2[0], v2[1], v2[2]);
    tmpRes[0] = res[0] / res[2];
    tmpRes[1] = res[1] / res[2];
    return tmpRes;
}

Answer 1

如果使用同类符号，这是最简单的：

将您的行代表从y = d*x + c更改为
```
d*x - y + c = 0 = [d -1 c] . [x y 1]
```
（其中.表示内部产品）
使用此表示法，您可以将行写为两个向量：[d1 -1 y1]和[d2 -1 y2]
取这两个向量的叉积，给你一个新的向量：
```
[d1 -1 y1] x [d2 -1 y2] = [a b c]
```
（我会让你看看如何计算交叉乘积，但它只是简单的乘法）

两点的交点位于(a/c, b/c)。如果两条线不平行，则c将为非零。

请参阅：http://robotics.stanford.edu/~birch/projective/node4.html

线方程的a*x + b*y + c = 0形式的一个优点是您可以自然地表示垂直线：您不能以x = 1形式表示线y = m*x + c，因为{ {1}}是无限的，而m可以。

线与垂直线交叉的精度

1 个答案: