接收三个位置和三个半径并输出一个显示重叠的坐标的函数会是什么样子？

Question

我有三个 gps 位置，分别是双纬度和经度。我有三个半径对应于每个 lat 和 lng 值。半径在位置周围形成圆圈。我想确定所有三个圆重叠的一点。

我的起点：

(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

但我被困在这里 - 不仅方程组多定，而且还不清楚如何将度、分和秒与以米为单位的半径混合。

接收三个位置和三个半径并输出一个显示重叠的坐标的函数（伪代码就足够了）会是什么样子。

说起来还是要有一定的公差，半径和位置都不能太精确

Answer 1

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

我在这里发布了满足您需要的代码：

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。有一些计算两个地理位置之间距离的复杂公式，因此您需要将其反转以从基于弧度的距离中获取米。

在这里您可能会找到此计算的实现并尝试反转它：

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

接收三个位置和三个半径并输出一个显示重叠的坐标的函数会是什么样子？

What could a function look like that receives three locations and three radius and outputs one coordinate showing the overlapping?

android

gps

triangulation