Question

我需要能够计算2个圆圈之间的交点。我确信总会有2个交叉点。不是1，不是0，不是无限，总是2.这是我想要做的图表：

这是我目前的尝试：

public static List<Vector2> intersect(Vector3 c1, Vector3 c2, float rad1, float rad2)
{
    List<Vector2> rLp = new List<Vector2>();
    float d = Vector2.Distance(c1, c2);

    if (d > (rad1 + rad2))
        return rLp;
    else if (d == 0 && rad1 == rad2)
        return rLp;
    else if ((d + Mathf.Min(rad1, rad2)) < Mathf.Max(rad1, rad2))
        return rLp;
    else
    {
        float a = (rad1 * rad1 - rad2 * rad2 + d * d) / (2 * d);
        float h = Mathf.Sqrt(rad1 * rad1 - a * a);

        Vector2 p2 = new Vector2((float)(c1.x + (a * (c2.x - c1.x)) / d), (float)(c1.y + (a * c2.y - c1.y) / d));

        Vector2 i1 = new Vector2((float)(p2.x + (h * (c2.y - c1.y)) / d), (float)(p2.y - (h * (c2.x - c1.x)) / d));
        Vector2 i2 = new Vector2((float)(p2.x - (h * (c2.y - c1.y)) / d), (float)(p2.y + (h * (c2.x - c1.x)) / d));

        if (d == (rad1 + rad2))
            rLp.Add(i1);
        else
        {
            rLp.Add(i1);
            rLp.Add(i2);
        }

        return rLp;
    }
}

它给了我以下结果：

如您所见，代表两个圆圈之间截取点的白色方块位于错误的位置。我真的可以在这个领域使用一些帮助。任何人都可以看到什么是错的吗？

Answer 1

数学代码的复制和粘贴对我来说几乎总是出错（例如，在定义中翻转signes等）。从其他人调试这样的代码只是纯粹的恐怖（当我深入数学时，调试我自己很难）。您应该尝试自己进行计算并将其转换为代码。如果出现问题，您可以使用调试器进行调试，并使用袖珍计算器进行交叉检查。

你真的应该尝试自己。拿一张纸然后算好数学。但这是我将使用的方法：

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提供代码后

更新：

您的方法与我的方法非常相似（private void intersectionTwoCircles(double c1x, double c1y, double r1, double c2x, double c2y, double r2, out double a1x, out double a1y, out double a2x, out double a2y) { /* error handling is missing complettely - left as an exercise A1 /| \ r1 / | \ r2 / | \ / |h \ /g1 | \ (g1 means angle gamma1) C1----P-----C2 d1 d2 */ double dx = c1x - c2x; double dy = c1y - c2y; double d = Math.Sqrt(dx*dx + dy*dy); // d = |C1-C2| double gamma1 = Math.Acos((r2*r2 + d*d - r1*r1)/(2*r2*d)); // law of cosines double d1 = r1*Math.Cos(gamma1); // basic math in right triangle double h = r1*Math.Sin(gamma1); double px = c1x + (c2x - c1x) / d*d1; double py = c1y + (c2y - c1y) / d*d1; // (-dy, dx)/d is (C2-C1) normalized and rotated by 90 degrees a1x = px + (-dy)/d*h; a1y = py + (+dx) / d * h; a2x = px - (-dy) / d * h; a2y = py - (+dx) / d * h; }和a=r1 cos(gamma_1)）。您只需避免直接计算h=r1 sin(gamma_1)。它更快（只有一个gamma而不是sqrt，cos和sin）。但我认为，我的更具可读性;）考虑重载载体类的操作数（+， - ，*，...）......你的代码将变得更容易阅读。

我认为我发现了你的错误......

acos

Answer 2

对于任何希望使用PointF而不是离散坐标的实现的人，我都采用了五年前Alex K提到的link的解决方案，以产生以下内容。尽管尚未对OP所述的已知两点情况进行优化，但它可以工作并容纳相交于一个，两个或无点的圆。

using System.Drawing;
using static System.Math;

public static class CircleIntersections {

    /// <summary>
    /// Gets the intersections of two circles
    /// </summary>
    /// <param name="center1">The first circle's center</param>
    /// <param name="center2">The second circle's center</param>
    /// <param name="radius1">The first circle's radius</param>
    /// <param name="radius2">The second circle's radius. If omitted, assumed to equal the first circle's radius</param>
    /// <returns>An array of intersection points. May have zero, one, or two values</returns>
    /// <remarks>Adapted from http://csharphelper.com/blog/2014/09/determine-where-two-circles-intersect-in-c/</remarks>
    public static PointF[] GetCircleIntersections(PointF center1, PointF center2, double radius1, double? radius2 = null) {

        var (r1, r2) = (radius1, radius2 ?? radius1);
        (double x1, double y1, double x2, double y2) = (center1.X, center1.Y, center2.X, center2.Y);
        // d = distance from center1 to center2
        double d = Sqrt(Pow(x1 - x2, 2) + Pow(y1 - y2, 2));
        // Return an empty array if there are no intersections
        if (!(Abs(r1 - r2) <= d && d <= r1 + r2)) { return new PointF[0]; }

        // Intersections i1 and possibly i2 exist
        var dsq = d * d;
        var (r1sq, r2sq) = (r1 * r1, r2 * r2);
        var r1sq_r2sq = r1sq - r2sq;
        var a = r1sq_r2sq / (2 * dsq);
        var c = Sqrt(2 * (r1sq + r2sq) / dsq - (r1sq_r2sq * r1sq_r2sq) / (dsq * dsq) - 1);

        var fx = (x1 + x2) / 2 + a * (x2 - x1);
        var gx = c * (y2 - y1) / 2;

        var fy = (y1 + y2) / 2 + a * (y2 - y1);
        var gy = c * (x1 - x2) / 2;

        var i1 = new PointF((float)(fx + gx), (float)(fy + gy));
        var i2 = new PointF((float)(fx - gx), (float)(fy - gy));

        return i1 == i2 ? new PointF[] { i1 } : new PointF[] { i1, i2 };
    }
}

获得2个圆圈的交点

2 个答案: