我目前正在研究3垫式台球游戏项目。到目前为止，我已经在桌上增加了两个球。我正在尝试移动其中一个球，但是这样做很难。我应该使用计时器吗？如果是这样，那么您能告诉我在代码上使用计时器的有效方法，以便我可以移动球吗？
非常感谢您的帮助。 提前致谢。 法罕·哈桑
我试图为类球创建一个移动函数。但是我不确定应该在函数中放什么，我已经添加了xSpeed和ySpeed。 xLocation和yLocation随xSpeed和ySpeed而变化。
public class Balls
{
private Color ballFillColor;
private Color ballBorderColor;
private int ballX = 0;
private int ballY = 0;
private int xSpeed = 5;
private int ySpeed = 0;
private int ballWidth = 0;
private int ballHeight = 0;
Timer t;
public boolean fillBall = false;
private static Balls ballArray[]; //Required for drawMultipleBalls
Balls(){ //Constructor
ballBorderColor = Color.black;
}
Balls(int ballX, int ballY, int ballWidth, int ballHeight, Color ballBorderColor, JFrame window){ //Constructor
// X , Y , Width, Height, Border Colour, container
this.setBallBorderColor(ballBorderColor);
this.setBallWidth(ballWidth);
this.setBallHeight(ballHeight);
this.setBallX(ballX);
this.setBallY(ballY);
this.drawBall(window);
}
//Here is the move function. I am not really sure what to do here.
public void move()
{
if(this.ballX < 1000 - this.ballWidth)
{
this.ballX += this.xSpeed;
}
try
{
Thread.sleep(1);
}
catch(Exception e)
{
}
}
//GET AND SET FUNCTIONS HERE
//HERE ARE THE FUNCTIONS WHICH ARE RESPONSIBLE FOR DRAWING MY BALLS IN JFRAME
public void drawBall(JFrame frame)
{
frame.getContentPane().add(new MyComponent());
}
public void drawMultipleBalls(JFrame frame, Balls[] balls)
{
ballArray = balls;
frame.getContentPane().add(new MyComponent2());
}
private class MyComponent extends JComponent{
public void paintComponent(Graphics g){
if (fillBall) //Fill first, and then draw outline.
{
g.setColor(ballFillColor);
g.fillOval(getBallX(),getBallY(), getBallHeight(),getBallWidth());
}
g.setColor(getBallBorderColor());
g.drawOval(getBallX(),getBallY(), getBallHeight(),getBallWidth());
}
}
private class MyComponent2 extends JComponent{
public void paintComponent(Graphics g){
for (int i = 0; i < ballArray.length; i++)
{
if (ballArray[i].fillBall) //Fill first, and then draw outline.
{
g.setColor(ballArray[i].ballFillColor);
g.fillOval(ballArray[i].getBallX(),ballArray[i].getBallY(), ballArray[i].getBallHeight(),ballArray[i].getBallWidth());
}
g.setColor(ballArray[i].getBallBorderColor());
g.drawOval(ballArray[i].getBallX(),ballArray[i].getBallY(), ballArray[i].getBallHeight(),ballArray[i].getBallWidth());
}
}
}
希望我能有两个可移动的球，在击打屏幕边缘时弹回，并且随着时间的推移它们应该会变慢。为此，我正在考虑使用阻尼器（我将xSpeed和ySpeed乘以小于1的数字，最终会减慢球的速度）
答案 0 :(得分：0)
这是我想到的一个简单示例，展示了一个球在边缘移动并弹起。
方向根据边界改变。左边缘和上边缘仅需检查0。下边缘和右边缘需要包括球的直径。
x和y的增量是独立的。并且这些数量与计时器一起可以改变机芯。但是请注意，由于轨迹的角度等原因，使对象彼此反弹（如在台球游戏中）更加复杂。反弹的距离将根据摩擦值随着时间而变化并且变慢。其他所有内容都记录在Java API中。
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
public class MovementDemo extends JPanel implements ActionListener {
JFrame frame = new JFrame("Movement Demo");
int size = 500;
int x = 50;
int y = 200;
int diameter = 50;
int yinc = 2;
int xinc = 2;
int xdirection = 1;
int ydirection = 1;
public MovementDemo() {
setPreferredSize(new Dimension(size, size));
frame.add(this);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.pack();
frame.setLocationRelativeTo(null);
frame.setVisible(true);
}
public static void main(String[] args) {
SwingUtilities.invokeLater(() -> new MovementDemo().start());
}
public void start() {
Timer timer = new Timer(100, this);
timer.setDelay(5);
timer.start();
}
public void actionPerformed(ActionEvent ae) {
if (x < 0) {
xdirection = 1;
}
else if (x > size - diameter) {
xdirection = -1;
}
if (y < 0) {
ydirection = 1;
}
else if (y > size - diameter) {
ydirection = -1;
}
x = x + xdirection * xinc;
y = y + ydirection * yinc;
repaint();
}
public void paintComponent(Graphics g) {
super.paintComponent(g);
Graphics2D g2d = (Graphics2D) g.create();
g2d.setColor(Color.BLUE);
g2d.setRenderingHint(RenderingHints.KEY_ANTIALIASING,
RenderingHints.VALUE_ANTIALIAS_ON);
g2d.fillOval(x, y, diameter, diameter);
}
}
答案 1 :(得分：0)
通常看来，您需要弄清楚一些事情：
下面是一些示例代码，将其中的一些代码存根了。您可以首先将当前球的位置与所有其他球的位置进行比较（当然不包括当前球）。如果位置相等，则与球碰撞。如果球在窗口边界处，即撞到墙壁，请与墙壁碰撞。否则，只需根据其当前速度计算其新位置即可。
过程冲突部分只是将物理力学应用到您需要的任何复杂程度。建议的一项一般更改是更新球的速度，然后将其应用于之后的位置。您可以根据需要应用特定的速度变化计算，并且可以想象它会涉及到很多，这就是为什么我建议使用单独的方法以及可能的速度子类，而不是管理球本身中速度矢量的每个部分。因此，我将墙用作对象。物体碰撞的成分，重量，速度等都会影响最终的碰撞，但您希望处理的复杂程度取决于您。
对不起，我不是物理学专家，但我希望这能给您正确的编码方向！此外，这可能有助于您可能要使用的特定计算： https://www.khanacademy.org/science/physics/one-dimensional-motion/displacement-velocity-time/v/calculating-average-velocity-or-speed
public void move()
{
// check if balls are on same position not including this ball
for(Ball b: ballArray){
if (this.position == b.position && this != b){
processCollision(this, b, null);
} else{
// if the ball hasn't collided with anything process its movement based on speed
// this assumes a 1000 x 1000 window for keeping objects inside it
if(this.ballX < 1000 - this.ballWidth && this.ballY < 1000 - this.ballHeight){
this.ballX += this.xSpeed;
this.ballY += this.ySpeed;
}else {
processCollision(this, null, new Wall());
}
}
}
try
{
Thread.sleep(1);
}
catch(Exception e)
{
}
}
public void processCollision(Ball b1, Ball b2, Wall w){
// if ball hasn't collided with a wall, process a ball - ball collision
if(w == null){
// apply physics mechanics according the complexity desired for ball collisions
b1.xSpeed -= b2.xSpeed;
b1.ySpeed -= b2.ySpeed;
// ball 2 would end up slowing down
b2.xSpeed -= b1.xSpeed;
b2.ySpeed -= b1.ySpeed;
}
// if ball hasn't collided with a ball, process a ball - wall collision
if(b2 == null){
// apply physics mechanics for hitting a wall
// e.g as below: just send ball in opposite direction
b1.xSpeed = b1.xSpeed * -1;
b1.ySpeed = b1.ySpeed * -1;
}
// either way, process ball's new position based on its new speed
b1.ballX += b1.xSpeed;
b1.ballY += b1.ySpeed;
b2.ballX += b2.xSpeed;
b2.ballY += b2.ySpeed;
}