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java - 获取旋转矩形的角

转载 作者:塔克拉玛干 更新时间:2023-11-03 05:13:57 25 4
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我有一个围绕它的中间旋转的矩形,我有另一个矩形我想连接到旋转矩形的右上角。问题是我不知道如何得到角,以便第二个矩形始终粘在那个角上。

这是我的示例代码。现在第二个矩形将一直在同一个地方,这不是我想要的结果。

package Test;

import java.awt.*;
import java.awt.event.*;
import java.awt.geom.*;
import javax.swing.*;

class Test{

public static void main(String[] args){
new Test();
}

public Test(){
EventQueue.invokeLater(new Runnable() {
@Override
public void run() {
JFrame frame = new JFrame("Test");
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.setLayout(new BorderLayout());
frame.add(new Graphic());
frame.setSize(1000,700);
frame.setLocationRelativeTo(null);
frame.setVisible(true);
}
});
}
}

class Graphic extends JPanel{
private int x, y, windowW, windowH;
private double angle;
private Rectangle rect1, rect2;
private Path2D path;
private Timer timer;
private AffineTransform rotation;

public Graphic(){
windowW = (int) Toolkit.getDefaultToolkit().getScreenSize().getWidth();
windowH = (int) Toolkit.getDefaultToolkit().getScreenSize().getHeight();
path = new Path2D.Double();
rotation = new AffineTransform();
angle = 0;
x = windowW / 2;
y = windowH / 2;
timer = new Timer(100, new ActionListener(){
@Override
public void actionPerformed(ActionEvent e){
angle += .1;
if(angle > 360) angle -= 360;
repaint();
}
});
timer.start();
}

@Override
public void paintComponent(Graphics g){
super.paintComponent(g);
Graphics2D g2d = (Graphics2D) g;
rotation.setToTranslation(500, 200);
rotation.rotate(angle, 32, 32);
rect1 = new Rectangle(0, 0, 64, 64);
path = new Path2D.Double(rect1, rotation);
rect2 = new Rectangle(path.getBounds().x, path.getBounds().y, 10, 50);
g2d.fill(path);
g2d.fill(rect2);
}
}

最佳答案

数学解:)

public void paintComponent(Graphics g) {
super.paintComponent(g);
Graphics2D g2d = (Graphics2D) g;
rotation.setToTranslation(500, 200);
rotation.rotate(angle, 32, 32);
rect1 = new Rectangle(0, 0, 64, 64);
path = new Path2D.Double(rect1, rotation);
double r = 32.0 * Math.sqrt(2);
// (532, 232) - coordinates of rectangle center |
// you can change position of second rectangle by this V substraction (all you need to know is that the full circle corresponds to 2Pi)
int x2 = (int) Math.round(532 + r * Math.cos(angle - Math.PI / 4));
int y2 = (int) Math.round(232 + r * Math.sin(angle - Math.PI / 4));
rect2 = new Rectangle(x2, y2, 10, 50);
g2d.fill(path);
g2d.fill(rect2);
}

当然,有些常量应该是类字段,而不是方法变量。

关于java - 获取旋转矩形的角,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16777742/

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