我无法绘制由3个点描述的最小弧:弧中心,一个“锚定”终点,以及第二个点通过确定半径来给出弧的另一端。我使用余弦定律来确定弧的长度,并尝试使用atan作为起始角度,但弧的起始位置已关闭。2点之间的Java绘制弧
我设法电弧锁定在定位点(X1,Y1),当它在第二象限,但只会工作,当它在象限2
解决方案,我可以看到所有有一堆if语句来确定2点相对于彼此的位置,但我很好奇,如果我忽略了简单的事情。任何帮助将不胜感激。
SSCCE:
import javax.swing.JComponent;
import javax.swing.JFrame;
import java.awt.event.MouseEvent;
import java.awt.event.MouseListener;
import java.awt.geom.*;
import java.awt.*;
import java.util.*;
class Canvas extends JComponent {
float circleX, circleY, x1, y1, x2, y2, dx, dy, dx2, dy2, radius, radius2;
Random random = new Random();
public Canvas() {
//Setup.
x1 = random.nextInt(250);
y1 = random.nextInt(250);
//Cant have x2 == circleX
while (x1 == 150 || y1 == 150)
{
x1 = random.nextInt(250);
y1 = random.nextInt(250);
}
circleX = 150; //circle center is always dead center.
circleY = 150;
//Radius between the 2 points must be equal.
dx = Math.abs(circleX-x1);
dy = Math.abs(circleY-y1);
//c^2 = a^2 + b^2 to solve for the radius
radius = (float) Math.sqrt((float)Math.pow(dx, 2) + (float)Math.pow(dy, 2));
//2nd random point
x2 = random.nextInt(250);
y2 = random.nextInt(250);
//I need to push it out to radius length, because the radius is equal for both points.
dx2 = Math.abs(circleX-x2);
dy2 = Math.abs(circleY-y2);
radius2 = (float) Math.sqrt((float)Math.pow(dx2, 2) + (float)Math.pow(dy2, 2));
dx2 *= radius/radius2;
dy2 *= radius/radius2;
y2 = circleY+dy2;
x2 = circleX+dx2;
//Radius now equal for both points.
}
public void paintComponent(Graphics g2) {
Graphics2D g = (Graphics2D) g2;
g.setRenderingHint(RenderingHints.KEY_ANTIALIASING,
RenderingHints.VALUE_ANTIALIAS_ON);
g.setStroke(new BasicStroke(2.0f, BasicStroke.CAP_BUTT,
BasicStroke.JOIN_BEVEL));
Arc2D.Float centerPoint = new Arc2D.Float(150-2,150-2,4,4, 0, 360, Arc2D.OPEN);
Arc2D.Float point1 = new Arc2D.Float(x1-2, y1-2, 4, 4, 0, 360, Arc2D.OPEN);
Arc2D.Float point2 = new Arc2D.Float(x2-2, y2-2, 4, 4, 0, 360, Arc2D.OPEN);
//3 points drawn in black
g.setColor(Color.BLACK);
g.draw(centerPoint);
g.draw(point1);
g.draw(point2);
float start = 0;
float distance;
//Form a right triangle to find the length of the hypotenuse.
distance = (float) Math.sqrt(Math.pow(Math.abs(x2-x1),2) + Math.pow(Math.abs(y2-y1), 2));
//Law of cosines to determine the internal angle between the 2 points.
distance = (float) (Math.acos(((radius*radius) + (radius*radius) - (distance*distance))/(2*radius*radius)) * 180/Math.PI);
float deltaY = circleY - y1;
float deltaX = circleX - x1;
float deltaY2 = circleY - y2;
float deltaX2 = circleX - x2;
float angleInDegrees = (float) ((float) Math.atan((float) (deltaY/deltaX)) * 180/Math.PI);
float angleInDegrees2 = (float) ((float) Math.atan((float) (deltaY2/deltaX2)) * 180/Math.PI);
start = angleInDegrees;
//Q2 works.
if (x1 < circleX)
{
if (y1 < circleY)
{
start*=-1;
start+=180;
} else if (y2 > circleX) {
start+=180;
start+=distance;
}
}
//System.out.println("Start: " + start);
//Arc drawn in blue
g.setColor(Color.BLUE);
Arc2D.Float arc = new Arc2D.Float(circleX-radius, //Center x
circleY-radius, //Center y Rotates around this point.
radius*2,
radius*2,
start, //start degree
distance, //distance to travel
Arc2D.OPEN); //Type of arc.
g.draw(arc);
}
}
public class Angle implements MouseListener {
Canvas view;
JFrame window;
public Angle() {
window = new JFrame();
view = new Canvas();
view.addMouseListener(this);
window.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
window.setBounds(30, 30, 400, 400);
window.getContentPane().add(view);
window.setVisible(true);
}
public static void main(String[] a) {
new Angle();
}
@Override
public void mouseClicked(MouseEvent arg0) {
window.getContentPane().remove(view);
view = new Canvas();
window.getContentPane().add(view);
view.addMouseListener(this);
view.revalidate();
view.repaint();
}
@Override
public void mouseEntered(MouseEvent arg0) {
// TODO Auto-generated method stub
}
@Override
public void mouseExited(MouseEvent arg0) {
// TODO Auto-generated method stub
}
@Override
public void mousePressed(MouseEvent arg0) {
// TODO Auto-generated method stub
}
@Override
public void mouseReleased(MouseEvent arg0) {
// TODO Auto-generated method stub
}
}
我不明白你的问题;你问问圆弧长度是圆的圆周上的两个点,给定一个任意的中心点? – cabbagery
号我问如何在2个点之间绘制最小的弧,给定一个圆的中心和2个点在圆上。我真的在寻找起点 - 从哪里开始绘制弧线,以及是否应该切换弧线的方向(顺时针/逆时针)。 –
我仍然在努力 - 乍一看看起来非常简单,但在进一步的评论中显得非常复杂,但我无法改变最终实际上相当简单的感觉。无论如何,我已经确定了一些会让事情变得棘手的场景。如果我能以可普遍的方式解决它们,我可能还没有一个可行的答案。 – cabbagery