在旋转后获取div角点像素位置 [英] get div corners pixel positions after rotation

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问题描述

在设置旋转弧度/度之后,如何计算div的左上角,左下角,右上角,右下角像素位置?


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这解决了我在旋转后计算新像素位置的问题。我创建了一个要使用的函数,传入x / y位置,div的半宽/高度和新的旋转角度。

  function getPixelsByAngle(x,y,halfWidth,halfHeight,angle){
var bounds = [
// left left
[x +(halfWidth) - (halfHeight)* Math.sin(angle)+ halfWidth,y +(halfHeight)* Math.cos(angle)+(halfWidth)* Math.sin(angle)+ halfHeight],
// upper right
[x - (halfWidth)* Math.cos(angle) - (halfHeight)* Math.sin(angle)+ halfWidth,y +(halfHeight)* Math.cos(angle) - (halfWidth)* Math.sin (angle)+ halfHeight],
//右下角
[x - (halfWidth)* Math.cos(angle)+(halfHeight)* Math.sin(angle)+ halfWidth,y- )* Math.cos(angle) - (halfWidth)* Math.sin(angle)+ halfHeight],
//左下角
[x +(halfWidth) halfHeight)* Math.sin(angle)+ halfWidth,y - (halfHeight)* Math.cos(angle)+(halfWidth)* Math.sin(angle)+ halfHeight]
];
return bounds;
}


how can i calculate top-left, bottom-left, top-right, bottom-right pixel positions of a div after you have set a rotation radian/degree on it?

An example would be helpful.

解决方案

This solved my issue with calculating new pixel positions after rotating. I created a function to be used, passing in the x/y position, half width/height of the div and the new rotation angle.

function getPixelsByAngle(x,y,halfWidth,halfHeight,angle){
   var bounds = [
      //upper left
      [x + (halfWidth) * Math.cos(angle) - (halfHeight) * Math.sin(angle) + halfWidth, y + (halfHeight) * Math.cos(angle) + (halfWidth) * Math.sin(angle) + halfHeight],
      //upper right
      [x - (halfWidth) * Math.cos(angle) - (halfHeight) * Math.sin(angle) + halfWidth, y + (halfHeight) * Math.cos(angle) - (halfWidth) * Math.sin(angle) + halfHeight],
      //bottom right
      [x - (halfWidth) * Math.cos(angle) + (halfHeight) * Math.sin(angle) + halfWidth, y - (halfHeight) * Math.cos(angle) - (halfWidth) * Math.sin(angle) + halfHeight],
      //bottom left
      [x + (halfWidth) * Math.cos(angle) + (halfHeight) * Math.sin(angle) + halfWidth, y - (halfHeight) * Math.cos(angle) + (halfWidth) * Math.sin(angle) + halfHeight]
   ];
   return bounds;
}

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