Werfen Sie einen Blick auf diese Frage: Find intersecting point of three circles programmatically
Ich bin hier, um den Code zu veröffentlichen, das tut, was Sie brauchen:
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;
}
Verbrauch:
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)
Wie Sie sagten, Wahrscheinlich müssen Sie hier eine Einheitenumrechnung vornehmen. Es gibt eine komplizierte Formel, die den Abstand zwischen zwei geolocations berechnet, also müssen Sie es umkehren, um Meter von der radianbasierten Entfernung zu erhalten.
Hier können Sie Implementierungen dieser Berechnung finden und versuchen, sie rückgängig zu machen:
Calculate distance between two latitude-longitude points? (Haversine formula)
Haben Sie zumindest diese SO Beitrag finden: http://stackoverflow.com/questions/19723641/find-intersecting -Punkt-von-drei-Kreisen-programmgesteuert –