Java 几何问题：圆弧段的周长

Question

Seq 是一种给定直线的方法：直线的点 a 和斜率 b returns 圆弧段的周长。我用圆和线的方程对x求解，然后轻松得到两个点。但是我得到的不是 (3.5,1.9) 而是 (6.528,0.0832000000000006)..

public class Circle {
Point center;
double r;

public double seq(Point a, double b) {
    double n = -b*a.getX() + a.getY();
    System.out.println("n: "+n);
    if (r*r*(b*b+1)-Math.pow(b*center.getX()-center.getY()+n, 2) > 0) {
        double temp = (1+b*b);
        System.out.println("coeficient x^2: "+temp);
        double temp1 = 2*((-1)*center.getX()+b*n-b*center.getY());
        System.out.println("coeficient x: "+ temp);
        double free_term = center.getX()*center.getX() + n*n -2*n*center.getY() + center.getY()*center.getY() - r*r;
        System.out.println("free term: "+free_term);
        double D = Math.sqrt(temp1*temp1-4*temp*free_term);
        System.out.println(temp1*temp1-4*temp*free_term);
        double x1 =  ((-1)*temp1+Math.sqrt(temp1*temp1-4*temp*free_term))/2*temp;
        double x2 = ((-1)*temp1-Math.sqrt(temp1*temp1-4*temp*free_term))/2*temp;
        double y1 = b*x1 - b*a.getX() + a.getY();
        double y2 = b*x2 - b*a.getX() + a.getY();
        System.out.println("("+x1+","+y1+")"+", ("+x2+","+y2+")");
        Point A = new Point(x1,y1);
        Point B = new Point(x2,y2);
        double d = A.dist(B);
        System.out.println("d: "+d);
        double angle = Math.acos((2*r*r-d*d)/2*r*r);
        System.out.println("angle "+ alfa);
        double l = r*Math.PI*angle/Math.toRadians(180);
        return l+d;
    } else return 0;
}
}

主要：

    Circle k = new Circle();
    Point c = new Point(0,0);
    k.center = c;
    k.r = 4;
    Point a = new Point(0,4);
    System.out.println(k.seq(a, -3.0/5.0));

控制台：

n: 4.0
coeficient x^2: 1.3599999999999999
coeficient x:  -4.8
free term: 0.0
23.04
(6.528,0.0832000000000006), (0.0,4.0)
d: 7.612890793910023
angleNaN
NaN

Answer 1

你的问题出在二次公式的实现上。您可能认为 A/2*B 意味着 A/(2*B)，但实际上它是 (A/2)*B。所以在除数周围添加括号：

double x1 = ((-1)*temp1+Math.sqrt(temp1*temp1-4*temp*free_term))/(2*temp);
double x2 = ((-1)*temp1-Math.sqrt(temp1*temp1-4*temp*free_term))/(2*temp);

Java 几何问题：圆弧段的周长

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