以编程方式找到三个圆的相交点 [英] Find intersecting point of three circles programmatically

问题描述

正如标题所述， 我有 3个圈子.

As the title says, I have 3 Circle.

每个人都有不同的半径.我知道每个圆的半径.

Each one have different radius. I know the radius of each circles.

也知道每个圆的中心点.

现在我需要知道如何以编程方式计算三个圆的相交点，是否有公式或其他东西?

Now I need to know how I can calculate the intersecting point of three circles programmatically, is there is any formula or something?

如下图所示

推荐答案

您可以从此获取帮助C代码.将其移植到JAVA并不难.解释是此处.搜索/滚动到:两个圆的交点

You could get help from this C code. Porting it to JAVA should not be challenging. Explanation is here. Search for/scroll to: intersection of two circles

使用此方法，找到任意两个圆的交点.说(x,y).现在，只有第三个圆的center与点x,y之间的距离等于r时，第三个圆才会在点x,y处相交.

Using this method, find the intersection of any two circles.. lets say (x,y). Now the third circle will intersect at point x,y only if distance between its center and point x,y is equal to r.

case 1)如果为distance(center,point) == r，则x,y为交点.

case 2)如果为distance(center,point) != r，则不存在该点.

case 2) If distance(center,point) != r, then no such point exists.

代码(从[这里！全部版权归原始作者移植]:

Code (ported from [here! all credits to original author):

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;
}

按以下方式调用此方法:

Call this method as follows:

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)

此外，将EPSILON定义为适合您的应用程序要求的较小值

Also, define EPSILON to a small value that is acceptable for your application requirements

private static final double EPSILON = 0.000001;

注意:也许有人应该测试并验证结果是否正确.对于尝试过的基本情况，我找不到任何简便的方法..

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