【问题标题】:Rotating a shape vertically around the x-axis围绕 x 轴垂直旋转形状
【发布时间】:2015-01-25 23:37:03
【问题描述】:

我有一个带有 x 和 y 轴的 2d 图,我试图围绕一个轴旋转一个形状(一系列点)。这种旋转需要包含一个缩放函数。

import java.awt.*;
import java.awt.event.*;
import java.awt.geom.AffineTransform;
import java.awt.geom.Point2D;
import javax.swing.*;
import java.lang.reflect.Array;

public class test extends JPanel implements ActionListener {

    int[] p1x = {200, 200, 240, 240, 220, 220, 200};
    int[] p1y = {200, 260, 260, 240, 240, 200, 200};
    int[] p2x = {600, 600, 620, 620, 640, 640, 660, 660, 600};
    int[] p2y = {400, 420, 420, 460, 460, 420, 420, 400, 400};
    int[] p3x = {400, 400, 460, 460, 440, 440, 420, 420, 400};
    int[] p3y = {400, 460, 460, 400, 400, 440, 440, 400, 400};
    int delay = 1000;
    int dx = 0;
    int dy = 5;
    int steps = 121;
    Polygon t;
    Timer tim = new Timer(delay, this);

    public void actionPerformed(ActionEvent event) {
        for (int i = 0; i < Array.getLength(p2x); i++) {
            //p2x[i] = (int) (p2x[i]*Math.cos(Math.toRadians(1))- p2y[i]*Math.sin(Math.toRadians(1)));
            //p2y[i] = (int) (p2x[i]*Math.sin(Math.toRadians(1))+ p2y[i]*Math.cos(Math.toRadians(1)));;

            Point2D original = new Point2D.Double(p2x[i], p2y[i]);
            AffineTransform at = new AffineTransform();
            //at.setToRotation(.02, 250, 250);
            at.scale(1, -1);
            Point2D rotated = at.transform(original, null);
            p2x[i] = (int) rotated.getX();
            p2y[i] = (int) rotated.getY();
        }
        repaint();

        if (--steps == 0) {
            tim.stop();
        }
    }

    public void paintComponent(Graphics g) {
        super.paintComponent(g);

        this.setBackground(Color.white);

        g.drawLine(this.getWidth() / 2, 0, this.getWidth() / 2, this.getWidth());
        g.drawLine(0, this.getHeight() / 2, this.getHeight(), this.getHeight() / 2);

        Polygon t = new Polygon(p2x, p2y, 9);
        g.drawPolygon(t);

        Letters u = new Letters(p3x, p3y, 9);
        u.draw(g);

        Letters l = new Letters(p1x, p1y, 7);
        l.draw(g);
    }

    public static void main(String[] args) {

        JFrame frame = new JFrame("Drawing line and a moving polygon");
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

        test sl = new test();
        frame.getContentPane().add(sl);
        frame.setSize(700, 700);
        frame.setVisible(true);

        sl.tim.start();
    }
}

【问题讨论】:

  • 重新格式化的代码;如果不正确,请恢复。

标签: java rotation transformation


【解决方案1】:

没有一个明确的问题,下面显示了一个使用坐标数组的简单动画。一般来说,您可以转换图形上下文 (g2d) 或多边形 Shape 本身 (p3);该示例显示了两者。调整窗口大小以查看每个效果。

注意at 中转换的last-specified-first-applied 顺序。首先,p3 上的一个合适的点被平移到原点,然后p3 被缩放,然后p3 被平移到面板的中心。应用于p3+ 10 软糖因子是没有对称旋转点的伪影。相对于原点定义多边形可能更容易,如example 所示。

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

/** @see http://stackoverflow.com/questions/3405799 */
public class AffineTest extends JPanel implements ActionListener {

    private static final double DELTA_THETA = Math.PI / 45; // 4°
    private static final double DELTA_SCALE = 0.1;
    private int[] p1x = {200, 200, 240, 240, 220, 220, 200};
    private int[] p1y = {200, 260, 260, 240, 240, 200, 200};
    private int[] p2x = {600, 600, 620, 620, 640, 640, 660, 660, 600};
    private int[] p2y = {400, 420, 420, 460, 460, 420, 420, 400, 400};
    private int[] p3x = {400, 400, 460, 460, 440, 440, 420, 420, 400};
    private int[] p3y = {400, 460, 460, 400, 400, 440, 440, 400, 400};
    private Polygon p1 = new Polygon(p1x, p1y, p1x.length);
    private Polygon p2 = new Polygon(p2x, p2y, p2x.length);
    private Polygon p3 = new Polygon(p3x, p3y, p3x.length);
    private AffineTransform at = new AffineTransform();
    private double dt = DELTA_THETA;
    private double theta;
    private double ds = DELTA_SCALE;
    private double scale = 1;
    private Timer timer = new Timer(100, this);

    public AffineTest() {
        this.setPreferredSize(new Dimension(700, 700));
        this.setBackground(Color.white);
        p1.translate(-50, +100);
        p2.translate(-100, -100);
    }

    @Override
    public void actionPerformed(ActionEvent event) {
        theta += dt;
        scale += ds;
        if (scale < .5 || scale > 4) {
            ds = -ds;
        }
        repaint();
    }

    @Override
    public void paintComponent(Graphics g) {
        super.paintComponent(g);
        Graphics2D g2d = (Graphics2D) g;
        g2d.setRenderingHint(
            RenderingHints.KEY_ANTIALIASING,
            RenderingHints.VALUE_ANTIALIAS_ON);
        int w = this.getWidth();
        int h = this.getHeight();
        g2d.drawLine(w / 2, 0, w / 2, h);
        g2d.drawLine(0, h / 2, w, h / 2);
        g2d.rotate(theta, w / 2, h / 2);
        g2d.drawPolygon(p1);
        g2d.drawPolygon(p2);
        at.setToIdentity();
        at.translate(w / 2, h / 2);
        at.scale(scale, scale);
        at.translate(-p3x[5] + 10, -p3y[5]);
        g2d.setPaint(Color.blue);
        g2d.fill(at.createTransformedShape(p3));
    }

    public void start() {
        timer.start();
    }

    public static void main(String[] args) {
        JFrame frame = new JFrame("Affine Test");
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        AffineTest sl = new AffineTest();
        frame.add(sl);
        frame.pack();
        frame.setVisible(true);
        sl.start();
    }
}

【讨论】:

  • 谢谢你,这很有帮助。帮助解开一些困惑。我试图围绕 x 轴而不是围绕 2D 图上的原点旋转这些字母。请注意,顶点向“z 方向”(框架外)的旋转是通过缩放实现的;并且每个字母的坐标轴相对于框架的坐标轴都有一个相对位置。
  • 优秀。有一个sscce 供参考是非常宝贵的;但是,作为一名学生,我不想享受所有的乐趣。 :-) 您应该能够通过在正交方向上缩放和平移来模拟绕轴的旋转;离轴通常需要剪切。请考虑接受和/或投票赞成您收到的这个和其他答案。
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