Java如何定义三个圆_java – 以编程方式查找三个圆的交叉点
你可以從
C code獲得幫助.把它移植到JAVA不應該是具有挑戰性的.說明是
here.搜索/滾動到：兩個圓的交點
使用此方法,找到任意兩個圓的交點..讓我們說(x,y).現在,只有當中心與點x,y之間的距離等于r時,第三個圓將在點x,y處相交.
情況1)如果距離(中心,點)== r,則x,y是交點.
情況2)如果距離(中心,點)！= r,則不存在這樣的點.
代碼(從[這里！所有學分轉移到原作者)：
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;
}
調用此方法如下：
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,將EPSILON定義為適合您的應用程序要求的小值
private static final double EPSILON = 0.000001;
注意：也許有人應該測試并驗證結果是否正確.我找不到任何簡單的方法.我嘗試過的基本案例的工作
總結
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