javascript – 旋转矩形中的点

我在一个矩形中有一个点,我需要旋转任意度数并找到该点的x y.我怎么能用javascript做到这一点.

在x下方,y将是1,3之类,在我将90传递给方法之后,它将返回3,1.

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基本上我正在寻找这种方法的胆量

function Rotate(pointX,pointY,rectWidth,rectHeight,angle){
   /*magic*/    
   return {newX:x,newY:y};
}

解决方法:

这应该这样做：

function Rotate(pointX, pointY, rectWidth, rectHeight, angle) {
  // convert angle to radians
  angle = angle * Math.PI / 180.0
  // calculate center of rectangle
  var centerX = rectWidth / 2.0;
  var centerY = rectHeight / 2.0;
  // get coordinates relative to center
  var dx = pointX - centerX;
  var dy = pointY - centerY;
  // calculate angle and distance
  var a = Math.atan2(dy, dx);
  var dist = Math.sqrt(dx * dx + dy * dy);
  // calculate new angle
  var a2 = a + angle;
  // calculate new coordinates
  var dx2 = Math.cos(a2) * dist;
  var dy2 = Math.sin(a2) * dist;
  // return coordinates relative to top left corner
  return { newX: dx2 + centerX, newY: dy2 + centerY };
}

标签：javascript,rotation,trigonometry
来源： https://codeday.me/bug/20191005/1854682.html