Java 如何在椭圆上找到扫掠给定区域的点?
我正在研究把一个椭圆分成大小相等的段的问题。有人问过这个问题,但答案是数值积分,所以我知道我在尝试什么。此代码会使扇区短路,因此集成本身的覆盖度不应超过90度。积分本身是通过将中间三角形的面积相加来完成的。下面是我尝试过的代码,但在某些情况下,它的范围超过了90度Java 如何在椭圆上找到扫掠给定区域的点?,java,geometry,ellipse,Java,Geometry,Ellipse,我正在研究把一个椭圆分成大小相等的段的问题。有人问过这个问题,但答案是数值积分,所以我知道我在尝试什么。此代码会使扇区短路,因此集成本身的覆盖度不应超过90度。积分本身是通过将中间三角形的面积相加来完成的。下面是我尝试过的代码,但在某些情况下,它的范围超过了90度 public class EllipseModel { protected double r_x; protected double r_y; private double a,a2; privat
public class EllipseModel {
protected double r_x;
protected double r_y;
private double a,a2;
private double b,b2;
boolean flip;
double area;
double sector_area;
double radstep;
double rot;
int xp,yp;
double deviation;
public EllipseModel(double r_x, double r_y, double deviation)
{
this.r_x = r_x;
this.r_y = r_y;
this.deviation = deviation;
if (r_x < r_y) {
flip = true;
a = r_y;
b = r_x;
xp = 1;
yp = 0;
rot = Math.PI/2d;
} else {
flip = false;
xp = 0;
yp = 1;
a = r_x;
b = r_y;
rot = 0d;
}
a2 = a * a;
b2 = b * b;
area = Math.PI * r_x * r_y;
sector_area = area / 4d;
radstep = (2d * deviation) / a;
}
public double getArea() {
return area;
}
public double[] getSweep(double sweep_area)
{
System.out.println(String.format("getSweep(%f) a = %f b = %f deviation = %f",sweep_area,a,b,deviation));
double[] ret = new double[2];
double[] next = new double[2];
double t_base, t_height, swept,x_mid,y_mid;
double t_area;
sweep_area = sweep_area % area;
if (sweep_area < 0d) {
sweep_area = area + sweep_area;
}
if (sweep_area == 0d) {
ret[0] = r_x;
ret[1] = 0d;
return ret;
}
double sector = Math.floor(sweep_area/sector_area);
double theta = Math.PI * sector/2d;
double theta_last = theta;
System.out.println(String.format("- Theta start = %f",Math.toDegrees(theta)));
ret[xp] = a * Math.cos(theta + rot);
ret[yp] = (1 + (((theta / Math.PI) % 2d) * -2d)) * Math.sqrt((1 - ( (ret[xp] * ret[xp])/a2)) * b2);
next[0] = ret[0];
next[1] = ret[1];
swept = sector * sector_area;
System.out.println(String.format("- Sweeping for %f sector_area=%f",sweep_area-swept,sector_area));
int c = 0;
while(swept < sweep_area) {
c++;
ret[0] = next[0];
ret[1] = next[1];
theta_last = theta;
theta += radstep;
// calculate next point
next[xp] = a * Math.cos(theta + rot);
next[yp] = (1 + (((theta / Math.PI) % 2d) * -2d)) * // selects +/- sqrt
Math.sqrt((1 - ( (ret[xp] * ret[xp])/a2)) * b2);
// calculate midpoint
x_mid = (ret[xp] + next[xp]) / 2d;
y_mid = (ret[yp] + next[yp]) / 2d;
// calculate triangle metrics
t_base = Math.sqrt( ( (ret[0] - next[0]) * (ret[0] - next[0]) ) + ( (ret[1] - next[1]) * (ret[1] - next[1])));
t_height = Math.sqrt((x_mid * x_mid) + (y_mid * y_mid));
// add triangle area to swept
t_area = 0.5d * t_base * t_height;
swept += t_area;
}
System.out.println(String.format("- Theta end = %f (%d)",Math.toDegrees(theta_last),c));
return ret;
}
}
有没有办法修正积分公式,创建一个函数,返回扫掠给定区域的椭圆上的点?使用此代码的应用程序将总面积除以所需的段数,然后使用此代码确定每个段开始和结束的角度。不幸的是,它没有按预期工作
*编辑*
我相信上面的积分失败了,因为基数和高度公式不正确。应用仿射变换将椭圆变成圆,最好是单位圆。然后在应用逆变换之前,将其拆分为大小相等的段。变换将按相同的因子缩放所有区域(相对于长度),因此等面积转换为等面积。无需变换,请使用椭圆的参数方程
x=x0+rx*cos(a)
y=y0+ry*sin(a)
其中a=<0,2.0*M_PI>
- 若你们把椭圆除以从中心到x,y的直线,从上面的方程
- 角度a均匀地增大
- 然后这些段将具有相同的大小
double a,da,x,y,x0=0,y0=0,rx=50,ry=20; // ellipse x0,y0,rx,ry
int i,N=32; // divided to N = segments
da=2.0*M_PI/double(N);
for (a=0.0,i=0;i<N;i++,a+=da)
{
x=x0+(rx*cos(a));
y=y0+(ry*sin(a));
// draw_line(x0,y0,x,y);
}
展位:
- x、 y是重点
- 段分割段的数目
- 线段为扫掠区域(小于0,线段)
double a,da,x,y,x0=0,y0=0,rx=50,ry=20; // ellipse x0,y0,rx,ry
int i,N=32; // divided to N = segments
da=2.0*M_PI/double(N);
for (a=0.0,i=0;i<N;i++,a+=da)
{
x=x0+(rx*cos(a));
y=y0+(ry*sin(a));
// draw_line(x0,y0,x,y);
}
double x0,y0,rx,ry; // ellipse parameters
// [Edit2] sorry forgot to add these constants but they are I thin straight forward
const double pi=M_PI;
const double pi2=2.0*M_PI;
// [/Edit2]
double atanxy(double x,double y) // atan2 return < 0 , 2.0*M_PI >
{
int sx,sy;
double a;
const double _zero=1.0e-30;
sx=0; if (x<-_zero) sx=-1; if (x>+_zero) sx=+1;
sy=0; if (y<-_zero) sy=-1; if (y>+_zero) sy=+1;
if ((sy==0)&&(sx==0)) return 0;
if ((sx==0)&&(sy> 0)) return 0.5*pi;
if ((sx==0)&&(sy< 0)) return 1.5*pi;
if ((sy==0)&&(sx> 0)) return 0;
if ((sy==0)&&(sx< 0)) return pi;
a=y/x; if (a<0) a=-a;
a=atan(a);
if ((x>0)&&(y>0)) a=a;
if ((x<0)&&(y>0)) a=pi-a;
if ((x<0)&&(y<0)) a=pi+a;
if ((x>0)&&(y<0)) a=pi2-a;
return a;
}
bool is_pnt_in_segment(double x,double y,int segment,int segments)
{
double a;
a=atanxy(x-x0,y-y0); // get sweep angle
a/=2.0*M_PI; // convert angle to a = <0,1>
if (a>=1.0) a=0.0; // handle extreme case where a was = 2 Pi
a*=segments; // convert to segment index a = <0,segments)
a-=double(segment );
// return floor(a); // this is how to change this function to return points segment id
// of course header should be slightly different: int get_pnt_segment_id(double x,double y,int segments)
if (a< 0.0) return false; // is lower then segment
if (a>=1.0) return false; // is higher then segment
return true;
}
void get_edge_pnt(double &x,double &y,int segment,int segments)
{
double a;
a=2.0*M_PI/double(segments);
a*=double(segment); // this is segments start edge point
//a*=double(segment+1); // this is segments end edge point
x=x0+(rx*cos(a));
y=y0+(ry*sin(a));
}