旋转矩形的碰撞检测
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我正在创建一个乒乓球游戏。但是，在我的游戏中，桨板可以绕其中心旋转。
这些桨由angle2D对象表示。现在，这些矩形应该会碰到它们。球由一个circle2D对象表示，当球撞击桨叶时（通过使用矩形的相交方法完成），球会反转方向。当不旋转桨时，这很好用，但是当桨旋转时，相交方法不起作用。当我使用以下语句时：
paddle2.intersects(xLocation, yLocation, diameter, diameter)
g2.rotate(Math.toRadians(angle), xLocation+(width/2), yLocation+(height/2));
没有找到相关结果
已邀请:
2 个回复
古擅坛犯
/** Rectangle To Point. */ boolean testRectangleToPoint(double rectWidth, double rectHeight, double rectRotation, double rectCenterX, double rectCenterY, double pointX, double pointY) { if(rectRotation == 0) // Higher Efficiency for Rectangles with 0 rotation. return Math.abs(rectCenterX-pointX) < rectWidth/2 && Math.abs(rectCenterY-pointY) < rectHeight/2; double tx = Math.cos(rectRotation)*pointX - Math.sin(rectRotation)*pointY; double ty = Math.cos(rectRotation)*pointY + Math.sin(rectRotation)*pointX; double cx = Math.cos(rectRotation)*rectCenterX - Math.sin(rectRotation)*rectCenterY; double cy = Math.cos(rectRotation)*rectCenterY + Math.sin(rectRotation)*rectCenterX; return Math.abs(cx-tx) < rectWidth/2 && Math.abs(cy-ty) < rectHeight/2; } /** Circle To Segment. */ boolean testCircleToSegment(double circleCenterX, double circleCenterY, double circleRadius, double lineAX, double lineAY, double lineBX, double lineBY) { double lineSize = Math.sqrt(Math.pow(lineAX-lineBX, 2) + Math.pow(lineAY-lineBY, 2)); double distance; if (lineSize == 0) { distance = Math.sqrt(Math.pow(circleCenterX-lineAX, 2) + Math.pow(circleCenterY-lineAY, 2)); return distance < circleRadius; } double u = ((circleCenterX - lineAX) * (lineBX - lineAX) + (circleCenterY - lineAY) * (lineBY - lineAY)) / (lineSize * lineSize); if (u < 0) { distance = Math.sqrt(Math.pow(circleCenterX-lineAX, 2) + Math.pow(circleCenterY-lineAY, 2)); } else if (u > 1) { distance = Math.sqrt(Math.pow(circleCenterX-lineBX, 2) + Math.pow(circleCenterY-lineBY, 2)); } else { double ix = lineAX + u * (lineBX - lineAX); double iy = lineAY + u * (lineBY - lineAY); distance = Math.sqrt(Math.pow(circleCenterX-ix, 2) + Math.pow(circleCenterY-iy, 2)); } return distance < circleRadius; } /** Rectangle To Circle. */ boolean testRectangleToCircle(double rectWidth, double rectHeight, double rectRotation, double rectCenterX, double rectCenterY, double circleCenterX, double circleCenterY, double circleRadius) { double tx, ty, cx, cy; if(rectRotation == 0) { // Higher Efficiency for Rectangles with 0 rotation. tx = circleCenterX; ty = circleCenterY; cx = rectCenterX; cy = rectCenterY; } else { tx = Math.cos(rectRotation)*circleCenterX - Math.sin(rectRotation)*circleCenterY; ty = Math.cos(rectRotation)*circleCenterY + Math.sin(rectRotation)*circleCenterX; cx = Math.cos(rectRotation)*rectCenterX - Math.sin(rectRotation)*rectCenterY; cy = Math.cos(rectRotation)*rectCenterY + Math.sin(rectRotation)*rectCenterX; } return testRectangleToPoint(rectWidth, rectHeight, rectRotation, rectCenterX, rectCenterY, circleCenterX, circleCenterY) || testCircleToSegment(tx, ty, circleRadius, cx-rectWidth/2, cy+rectHeight/2, cx+rectWidth/2, cy+rectHeight/2) || testCircleToSegment(tx, ty, circleRadius, cx+rectWidth/2, cy+rectHeight/2, cx+rectWidth/2, cy-rectHeight/2) || testCircleToSegment(tx, ty, circleRadius, cx+rectWidth/2, cy-rectHeight/2, cx-rectWidth/2, cy-rectHeight/2) || testCircleToSegment(tx, ty, circleRadius, cx-rectWidth/2, cy-rectHeight/2, cx-rectWidth/2, cy+rectHeight/2); }窝头菊
。您正在告诉图形上下文旋转，以便它绘制的所有内容（例如未旋转的）都被旋转绘制，但实际的实例并未被修改。 也许您可以使用代替，并按照此答案旋转它们。这样，实际实例将被旋转（因此您无需旋转实例），并且碰撞检测应该可以正常工作。