Question

我试图找到一条与圆相交的线。

我编写代码，但似乎该代码存在一些问题。

private Point2d[] IntersectionPoint(Point2d p1, Point2d p2, Point2d sc, double r)
    {
        Point2d[] sect = null;
        double a, b, c;
        double bb4ac;
        double mu1, mu2;
        Point2d dp;

        dp = p2 - p1;

        a = dp.X * dp.X + dp.Y * dp.Y;
        b = 2 * (dp.X * (p1.X - sc.X) + dp.Y * (p1.Y - sc.Y));
        c = sc.X * sc.X + sc.Y * sc.Y;
        c += p1.X * p1.X + p1.Y * p1.Y;
        c -= 2 * (sc.X * p1.X + sc.Y * p1.Y);
        c -= r * r;
        bb4ac = b * b - 4 * a * c;

        if (Math.Abs(a) < Double.Epsilon || bb4ac < 0)
        {
            return new Point2d[0];
        }

        mu1 = (-b + Math.Sqrt(bb4ac)) / (2 * a);
        mu2 = (-b - Math.Sqrt(bb4ac)) / (2 * a);

        // no intersection
        if ((mu1 < 0 || mu1 > 1) && (mu2 < 0 || mu2 > 1))
        {
            sect = new Point2d[0];
        }
        // one point on mu1
        else if (mu1 > 0 && mu1 < 1 && (mu2 < 0 || mu2 > 1))
        {
            sect = new Point2d[1];
            sect[0] = p1 + ((p2 - p1) * mu1);
        }
        // one point on mu2
        else if (mu2 > 0 && mu2 < 1 && (mu1 < 0 || mu1 > 1))
        {
            sect = new Point2d[1];
            sect[0] = p1 + ((p2 - p1) * mu2);
        }
        //  one or two points
        else if (mu1 > 0 && mu1 < 1 && mu2 > 0 && mu2 < 1)
        {
            //  tangential
            if (mu1 == mu2)
            {
                sect = new Point2d[1];
                sect[0] = p1 + ((p2 - p1) * mu1);
            }
            //  two points
            else
            {
                sect = new Point2d[2];
                sect[0] = p1 + ((p2 - p1) * mu1);
                sect[1] = p1 + ((p2 - p1) * mu2);
            }
        }
        else
        {
            //  should NEVER get here
            sect = new Point2d[0];
        }
        return sect;
    }

并将此功能称为

Point ptOld = points[oldPoint];
Point ptNew = points[newPoint];
Point2d p1 = new Point2d((float)ptOld.latitude, (float)ptOld.longitude);
Point2d p2 = new Point2d((float)ptNew.latitude, (float)ptNew.longitude);
Point2d sc = new Point2d((float)loc.latitude, (float)loc.longitude);

当我尝试使用这些坐标时，它失败了

30，-30

80，-40

10

https://www.dropbox.com/s/38r9eylt2p4xfvw/graph.png

Answer 1

你可以做一些线性代数：

将线条表示为原点P1和标准化方向向量N
将圆圈的中心C投影到该行：PC = P1 + dot(C - P1, N) * N
计算点dSquared和C之间的距离平方PC。
如果与radiusSquared相等（有一些小容差），则PC位于圆圈上并且是单个交叉点。
如果大于radiusSquared，则没有交集。
否则，两个交叉点由
给出
1. offset = sqrt(radiusSquared - dSquared)。
2. 交点= PC +/- offset * N。

Answer 2

有我的交叉代码：

public static Vector3? IntersectRayCircle(Vector3 rayStart, Vector3 rayPoint, Vector3 circlePosition, float circleRadiusSquared)
{
    if (rayStart == rayPoint || circleRadiusSquared <= 0)
    {
        return null;
    }

    Vector3 nearest = GetNearestPoint(circlePosition, rayStart, rayPoint, false, false);
    float distanceSquared = Vector3.DistanceSquared(nearest, circlePosition);

    if (distanceSquared > circleRadiusSquared)
    {
        return null;
    }

    Vector3 offset = Vector3.Normalize(rayPoint - rayStart) * (float)Math.Sqrt(circleRadiusSquared - distanceSquared);

    if (Vector3.DistanceSquared(circlePosition, rayStart) < circleRadiusSquared)
    {
        return nearest + offset;
    }
    else
    {
        return nearest - offset;
    }
}

public static Vector3 GetNearestPoint(Vector3 location, Vector3 segmentStart, Vector3 segmentEnd, bool trimStart, bool trimEnd)
{
    if (segmentStart == segmentEnd)
    {
        throw new ArgumentException("segmentStart cannot be equal to segmentEnd.");
    }

    Vector3 AP = location - segmentStart;
    Vector3 AB = segmentEnd - segmentStart;

    float magnitudeAB = AB.LengthSquared();
    float ABAPproduct = Vector3.Dot(AP, AB);
    float distance = ABAPproduct / magnitudeAB;

    return (distance < 0 && trimStart) ? segmentStart : (distance > 1 && trimEnd) ? segmentEnd : segmentStart + AB * distance;
}

找到圆和线的交点

2 个答案: