public static ArrayList<IntPoint> getCircleLineIntersectionPoint(IntPoint pointA, IntPoint pointB, IntPoint center, int radius) {
// returns a list of intersection points between a line which passes through given points,
// pointA and pointB, and a circle described by given radius and center coordinate
double disc, A, B, C, slope, c;
double x1, x2, y1, y2;
IntPoint point1, point2;
ArrayList<IntPoint> intersections = new ArrayList<IntPoint>();
try{
slope = Util.calculateSlope(pointA, pointB);
}catch (UndefinedSlopeException e){
C = Math.pow(center.y, 2) + Math.pow(pointB.x, 2) - 2 * pointB.x * center.x + Math.pow(center.x, 2) - Math.pow(radius, 2);
B = -2 * center.y;
A = 1;
disc = Math.pow(B, 2) - 4 * 1 * C;
if (disc < 0){
return intersections;
}
else{
y1 = (-B + Math.sqrt(disc)) / (2 * A);
y2 = (-B - Math.sqrt(disc)) / (2 * A);
x1 = pointB.x;
x2 = pointB.x;
}
point1 = new IntPoint((int)x1, (int)y1);
point2 = new IntPoint((int)x2, (int)y2);
if (Util.euclideanDistance(pointA, point2) > Util.euclideanDistance(pointA, point1)){
intersections.add(point1);
}
else{
intersections.add(point2);
}
return intersections;
}
if (slope == 0){
C = Math.pow(center.x, 2) + Math.pow(center.y, 2) + Math.pow(pointB.y, 2) - 2 * pointB.y * center.y - Math.pow(radius, 2);
B = -2 * center.x;
A = 1;
disc = Math.pow(B, 2) - 4 * 1 * C;
if (disc < 0){
return intersections;
}
else{
x1 = (-B + Math.sqrt(disc)) / (2*A);
x2 = (-B - Math.sqrt(disc)) / (2*A);
y1 = pointB.y;
y2 = pointB.y;
}
}
else{
c = slope * pointA.x + pointA.y;
B = (2 * center.x + 2 * center.y * slope + 2 * c * slope);
A = 1 + Math.pow(slope, 2);
C = (Math.pow(center.x, 2) + Math.pow(c, 2) + 2 * center.y * c + Math.pow(center.y, 2) - Math.pow(radius, 2));
disc = Math.pow(B, 2) - (4 * A * C);
if (disc < 0){
return intersections;
}
else{
x1 = (-B + Math.sqrt(disc)) / (2 * A);
x2 = (-B - Math.sqrt(disc)) / (2 * A);
y1 = slope * x1 - c;
y2 = slope * x2 - c;
}
}
point1 = new IntPoint((int)x1, (int)y1);
point2 = new IntPoint((int)x2, (int)y2);
if (Util.euclideanDistance(pointA, point2) > Util.euclideanDistance(pointA, point1)){
//if (Util.angleBetween(pointA, pointB, point1) < Math.PI/2){
intersections.add(point1);
//}
}
else{
//if (Util.angleBetween(pointA, pointB, point1) < Math.PI/2){
intersections.add(point2);
//}
}
return intersections;
}
我正在使用上述算法来测试圆和直线之间的交点。它有时很好，但在其他时候它失败了。代码表示从圆和线方程(x-a)^+(y-b)^2=r^2和y = mx - mx1 + y1同时求解x得到的方程。有没有人知道我在数学或其他地方出错了？
答案 0 :(得分：25)
您的计算似乎很长，我没有看到您测试的不同案例的使用情况。
无论如何，因为我发现问题很有趣，我试图自己解决它并提出以下内容。可以按double radius替换int radius并使用IntPoint s，但要注意每次投射时，如评论中所讨论的那样，不是精确整数交叉点的结果将会出错
执行计算的背景是：从A点开始，矢量AB的缩放版本指向圆上的点。该点与中心的距离半径。因此，| AC + scalingFactor * AB | = r。
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class CircleLine {
public static List<Point> getCircleLineIntersectionPoint(Point pointA,
Point pointB, Point center, double radius) {
double baX = pointB.x - pointA.x;
double baY = pointB.y - pointA.y;
double caX = center.x - pointA.x;
double caY = center.y - pointA.y;
double a = baX * baX + baY * baY;
double bBy2 = baX * caX + baY * caY;
double c = caX * caX + caY * caY - radius * radius;
double pBy2 = bBy2 / a;
double q = c / a;
double disc = pBy2 * pBy2 - q;
if (disc < 0) {
return Collections.emptyList();
}
// if disc == 0 ... dealt with later
double tmpSqrt = Math.sqrt(disc);
double abScalingFactor1 = -pBy2 + tmpSqrt;
double abScalingFactor2 = -pBy2 - tmpSqrt;
Point p1 = new Point(pointA.x - baX * abScalingFactor1, pointA.y
- baY * abScalingFactor1);
if (disc == 0) { // abScalingFactor1 == abScalingFactor2
return Collections.singletonList(p1);
}
Point p2 = new Point(pointA.x - baX * abScalingFactor2, pointA.y
- baY * abScalingFactor2);
return Arrays.asList(p1, p2);
}
static class Point {
double x, y;
public Point(double x, double y) { this.x = x; this.y = y; }
@Override
public String toString() {
return "Point [x=" + x + ", y=" + y + "]";
}
}
public static void main(String[] args) {
System.out.println(getCircleLineIntersectionPoint(new Point(-3, -3),
new Point(-3, 3), new Point(0, 0), 5));
System.out.println(getCircleLineIntersectionPoint(new Point(0, -2),
new Point(1, -2), new Point(1, 1), 5));
System.out.println(getCircleLineIntersectionPoint(new Point(1, -1),
new Point(-1, 0), new Point(-1, 1), 5));
System.out.println(getCircleLineIntersectionPoint(new Point(-3, -3),
new Point(-2, -2), new Point(0, 0), Math.sqrt(2)));
}
答案 1 :(得分：0)
这是一个import javax.vecmath.Vector2d;
static Vector2d[] circleLineIntersection1(Vector2d a, Vector2d b, Vector2d o, double radius) {
Vector2d p1 = new Vector2d(a);
Vector2d p2 = new Vector2d(b);
p1.sub(o);
p2.sub(o);
Vector2d d = new Vector2d();
d.sub(p2, p1);
double det = p1.x * p2.y - p2.x * p1.y;
double dSq = d.lengthSquared();
double discrimant = radius * radius * dSq - det * det;
if (discrimant < 0) {
return new Vector2d[0];
}
if (discrimant == 0) {
Vector2d[] t = new Vector2d[1];
t[0] = new Vector2d(det * d.y / dSq + o.x, -det * d.x / dSq + o.y);
return t;
}
double discSqrt = Math.sqrt(discrimant);
double sgn = 1;
if (d.y < 0) {
sgn = -1;
}
Vector2d[] t = new Vector2d[2];
t[0] = new Vector2d((det * d.y + sgn * d.x * discSqrt) / dSq + o.x, (-det * d.x + Math.abs(d.y) * discSqrt) / dSq + o.y);
t[1] = new Vector2d((det * d.y - sgn * d.x * discSqrt) / dSq + o.x, (-det * d.x - Math.abs(d.y) * discSqrt) / dSq + o.y);
return t;
}