Question

我有三个GPS位置，如双拉和lng。我有三个半径对应于每个lat和lng值。半径形成围绕位置的圆圈。我想确定所有三个圆重叠的一个点。

我的出发点：

(x−lat_1)^2+(y−lng_1)^2=r_1^2
(x−lat_2)^2+(y−lng_2)^2=r_2^2
(x−lat_3)^2+(y−lng_3)^2=r_3^2

但在这里我被困 - 不仅是方程系统过度确定，还不清楚，如何将度数，分钟和秒与半径以米为单位混合。

一个函数（伪代码就足够了）看起来像接收三个位置和三个半径并输出一个显示重叠的坐标。

说到这一点，需要有一些容差，因为半径和位置都不太精确。

Answer 1

看看这个问题： Find intersecting point of three circles programmatically

我在这里发布了你需要的代码：

private static final double EPSILON = 0.000001;

private boolean calculateThreeCircleIntersection(double x0, double y0, double r0,
                                                 double x1, double y1, double r1,
                                                 double x2, double y2, double r2)
{
    double a, dx, dy, d, h, rx, ry;
    double point2_x, point2_y;

    /* dx and dy are the vertical and horizontal distances between
    * the circle centers.
    */
    dx = x1 - x0;
    dy = y1 - y0;

    /* Determine the straight-line distance between the centers. */
    d = Math.sqrt((dy*dy) + (dx*dx));

    /* Check for solvability. */
    if (d > (r0 + r1))
    {
        /* no solution. circles do not intersect. */
        return false;
    }
    if (d < Math.abs(r0 - r1))
    {
        /* no solution. one circle is contained in the other */
        return false;
    }

    /* 'point 2' is the point where the line through the circle
    * intersection points crosses the line between the circle
    * centers.
    */

    /* Determine the distance from point 0 to point 2. */
    a = ((r0*r0) - (r1*r1) + (d*d)) / (2.0 * d) ;

    /* Determine the coordinates of point 2. */
    point2_x = x0 + (dx * a/d);
    point2_y = y0 + (dy * a/d);

    /* Determine the distance from point 2 to either of the
    * intersection points.
    */
    h = Math.sqrt((r0*r0) - (a*a));

    /* Now determine the offsets of the intersection points from
    * point 2.
    */
    rx = -dy * (h/d);
    ry = dx * (h/d);

    /* Determine the absolute intersection points. */
    double intersectionPoint1_x = point2_x + rx;
    double intersectionPoint2_x = point2_x - rx;
    double intersectionPoint1_y = point2_y + ry;
    double intersectionPoint2_y = point2_y - ry;

    Log.d("INTERSECTION Circle1 AND Circle2:", "(" + intersectionPoint1_x + "," + intersectionPoint1_y + ")" + " AND (" + intersectionPoint2_x + "," + intersectionPoint2_y + ")");

    /* Lets determine if circle 3 intersects at either of the above intersection points. */
    dx = intersectionPoint1_x - x2;
    dy = intersectionPoint1_y - y2;
    double d1 = Math.sqrt((dy*dy) + (dx*dx));

    dx = intersectionPoint2_x - x2;
    dy = intersectionPoint2_y - y2;
    double d2 = Math.sqrt((dy*dy) + (dx*dx));

    if(Math.abs(d1 - r2) < EPSILON) {
        Log.d("INTERSECTION Circle1 AND Circle2 AND Circle3:", "(" + intersectionPoint1_x + "," + intersectionPoint1_y + ")");
    }
    else if(Math.abs(d2 - r2) < EPSILON) {
        Log.d("INTERSECTION Circle1 AND Circle2 AND Circle3:", "(" + intersectionPoint2_x + "," + intersectionPoint2_y + ")"); //here was an error
    }
    else {
        Log.d("INTERSECTION Circle1 AND Circle2 AND Circle3:", "NONE");
    }
    return true;
}

用法：

calculateThreeCircleIntersection(-2.0, 0.0, 2.0, // circle 1 (center_x, center_y, radius)
                                  1.0, 0.0, 1.0, // circle 2 (center_x, center_y, radius)
                                  0.0, 4.0, 4.0);// circle 3 (center_x, center_y, radius)

正如你所说，你可能需要在这里做一些单位转换。有一些复杂的公式可以计算两个地理位置之间的距离，因此您需要将其反转以从基于弧度的距离获取米。

在这里，您可以找到此计算的实现并尝试将其反转：

Calculate distance between two latitude-longitude points? (Haversine formula)

一个函数看起来像三个位置和三个半径，并输出一个显示重叠的坐标？

1 个答案: