确定矩形上垂直或水平墙上的命中

Question

目前我正在以经典的2D几何图形弹跳一个墙球。

墙壁可以在垂直和水平两侧被击中，反射的结果取决于球击中的哪一侧。

我尝试了一些不同的解决方案，但它们让我更加困惑。

如何判断球是否撞击墙壁的垂直或水平方向？

PseudoCode概述：

iterate through each wall
if (collision between ball and wall)
  determine if vertical/horizontal hit
  calculate new velocity for ball

我使用此代码进行碰撞检测，它就像一个魅力： 来源：Circle-Rectangle collision detection (intersection)

var isCollision = function (_projectile) {
    if(direction != undefined){
        var circleDistance = {};
        circleDistance.x = Math.abs(_projectile.getCenter().x - centerX);
        circleDistance.y = Math.abs(_projectile.getCenter().y - centerY);

        if (circleDistance.x > (width/2 + _projectile.getRadius())) { return false; }
        if (circleDistance.y > (height/2 + _projectile.getRadius())) { return false; }

        if (circleDistance.x <= (width/2)) { return true; }
        if (circleDistance.y <= (height/2)) { return true; }

        var cornerDistance_sq = square(circleDistance.x - width/2) + square(circleDistance.y - height/2);

        return (cornerDistance_sq <= (square(_projectile.getRadius())));
    }
    return false;
};

var square = function(_value){
    return _value * _value;
};

谢谢！

Answer 1

update after question update

Asuming the ball has a direction/velocity vector public void MinimizeApp(string parameter) { if (parameter == "-minimized") { this.WindowState = FormWindowState.Minimized; notifyIcon1.Visible = true; notifyIcon1.BalloonTipText = "Program is started and running in the background..."; notifyIcon1.ShowBalloonTip(500); Hide(); } } and has a radius of dx,dy (r, ball.x are the ball positions of the ball center) do sth like this (have used similar in a billiard game, it is basic geometry and physics):

ball.y

do similar for opposite walls if needed, since velocity vector changes, the new ball position after bouncing (or anyway) will be if (ball.x+ball.dx+r > wallV.x) ball.dx = -ball.dx // ball bounces off the vertical wall if ( ball.y+ball.dy+r > wallH.y ) ball.dy = -ball.dy // ball bounces off horizontal wall

Bonus if you add some friction factor (meaning the magnitudes of ball.x += ball.dx; ball.y += ball.dy; dx eventually fade away to zero, the ball eventually stops after following a path)

Answer 2

要正确解决碰撞问题，您无法在X =＆gt;上进行测试如果碰撞解决并返回那么y =＆gt;的测试如果碰撞解决了y。
你必须首先检查哪个 ax（x OR y）碰撞，当球同时撞到两面墙时也是如此。
所以你不能只用空间来推理：你必须处理时间，并根据球的速度计算所有墙壁的ETA估计时间到达（假设墙壁仍然是）。

伪代码：

minETA = a too big number;
colliderList = [];
for (each wall in walls) {
   thisWallETA = computeETA(ball, wall);
   if (thisWallETA > minETA) continue;
   if (thisWallETA < minETA) {
      minETA = thisWallETA;
      colliderList.length = 0;
      colliderList.push(wall);
      continue;
   } 
   if (thisWallETA == minETA) {
      colliderList.push(wall);
      continue;          
   }
}

然后当colliderList只有一个项目=＆gt;解决相应的问题。如果它有两个项目=＆gt;在两个轴上解决。
然后用minETA增加当前时间，再试一次，直到找到的currentTime + minETA为＆gt;到thisCycleTime + thisCycleDt。

如果您有兴趣，我可能会澄清一些事情。

祝你好运！