java - 增加距离后计算点的 x y

Question

我试图在增加距离 r 后获取点的 x 和 y 值。也许还有更好的方法来计算角度 phi，这样我就不需要检查该点在哪个象限。 0 点位于窗口宽度和高度的一半处。这是我的尝试:

package test;

import java.awt.Color;
import java.awt.Graphics;
import java.awt.Point;

public final class Laser extends java.applet.Applet implements Runnable{

private static final long serialVersionUID = -7566644836595581327L;

Thread runner;

int width = 800;
int height = 600;

Point point = new Point(405,100);
Point point1 = new Point(405,100);

public void calc(){

    int x = getWidth()/2;
    int y = getHeight()/2;
    int px = point.x;
    int py = point.y;
    int px1 = point1.x;
    int py1 = point1.y;
    double r = 0;
    double phi = 0;

    // Point is in:
    // Quadrant 1
    if(px > x && py < y){
        r = Math.hypot(px1-x, y-py1);
        phi = Math.acos((px1-x)/r)*(180/Math.PI);
    }/*
    // Quadrant 2
    else if(px < x && py < y){
        r = Math.hypot(x-px, y-py);
        phi = Math.acos((px-x)/r)*(180/Math.PI);
    }
    // Quadrant 3
    else if(px < x && py > y){
        r = Math.hypot(x-px, py-y);
        phi = Math.acos((px-x)/r)*(180/Math.PI)+180;
    }
    // Quadrant 4
    else if(px > x && py > y){
        r = Math.hypot(px-x, py-y);
        phi = Math.acos((px-x)/r)*(180/Math.PI)+180;
    }*/
    r += 1;
    point1.x = (int) (r*Math.cos(phi));
    point1.y = (int) (r*Math.sin(phi));
    System.out.println(r+";"+point1.x+";"+point1.y);

}

public void paint(Graphics g) {

    g.setColor(Color.ORANGE);
    calc(); 
    g.drawLine(point.x, point.y, point1.x, point1.y);

    int h = getHeight();
    int w = getWidth();
    g.setColor(Color.GREEN);
    g.drawLine(0, h/2, w, h/2);
    g.drawLine(w/2, 0, w/2, h);

}
/*
public void initPoints(){

    for(int i = 0; i < pointsStart.length; i++){
        int x = (int)(Math.random()*getWidth());
        int y = (int)(Math.random()*getHeight());
        pointsStart[i] = pointsEnd[i] = new Point(x,y);
    }

}
*/
public void start() {

    if (runner == null) {

        runner = new Thread(this);
        setBackground(Color.black);
        setSize(width, height);
        //initPoints();
        runner.start();

    }

}

@SuppressWarnings("deprecation")
public void stop() {

    if (runner != null) {

        runner.stop();

        runner = null;

    }

}


public void run() {

    while (true) {

        repaint();

        try { Thread.sleep(700); }

        catch (InterruptedException e) { }

    }

}

public void update(Graphics g) {

    paint(g);

}

}

Answer 1

您正在将 (x,y) 从其他点更改为 r，而它之前距该点有一定距离 r'，对吗？那么为什么不避免三角函数，而只是将每个分量从该点缩放 r/r' 呢？

编辑:好的，沿着较长的分量(x 或 y)迭代像素(假设它是 y)；对于 (0..x) 中的每个 xi，yi = xi*(y/x)，然后绘制 (xi,yi)。