actionscript-3 - 基于矩阵计算新的 BitmapData 的大小

Question

我正在尝试将 BitmapData 的内容绘制到另一个尚未创建的内容中。
但在绘制之前，我需要缩放和旋转图像，然后绘制它。
我的问题是我不知道转换后 BitmapData 的大小，所以我无法创建新的来绘制它。

这个方法显示了我的意思:

public function getTransformedBitmapData(origin:BitmapData):BitmapData
{
    var matrix:Matrix = new Matrix();

    // ajusting the anchor point and rotating
    matrix.translate(-origin.width / 2, -origin.height / 2);
    matrix.rotate(Math.PI / 4); // 45 deg
    matrix.translate(origin.width / 2, origin.height / 2);

    // scaling
    matrix.scale(1.5, 1.5);

    // Calculating the size of the new BitmapData
    var width:Number = 0; // I don't know this value!
    var height:Number = 0; // I don't know this value!

    // Creating and drawing (with transformation)
    var result:BitmapData = new BitmapData(width, height, true, 0);
    result.draw(origin, matrix);

    return result;
}

有人知道我应该怎么做才能找出(计算)转换后该图像的大小？

这张图片说明了旋转的 Action ，以及我想了解的内容:

Answer 1

好的，使用@ansiart 答案作为起点，我设法以这种方式计算尺寸:

public function getTransformedBitmapData(origin:BitmapData):BitmapData
{
    var matrix:Matrix = new Matrix();

    // ajusting the anchor point and rotating
    matrix.translate(-origin.width / 2, -origin.height / 2);
    matrix.rotate(Math.PI / 4); // 45 deg
    matrix.translate(origin.width / 2, origin.height / 2);

    // scaling
    matrix.scale(1.5, 1.5);

    // Finding the four corners of the bounfing box after transformation
    var topLeft:Point = matrix.transformPoint(new Point(0, 0));
    var topRight:Point = matrix.transformPoint(new Point(origin.width, 0));
    var bottomLeft:Point = matrix.transformPoint(new Point(0, origin.height));
    var bottomRight:Point = matrix.transformPoint(new Point(origin.width, origin.height));

    // Calculating "who" is "where"
    var top:Number = Math.min(topLeft.y, topRight.y, bottomLeft.y, bottomRight.y);
    var bottom:Number = Math.max(topLeft.y, topRight.y, bottomLeft.y, bottomRight.y);
    var left:Number = Math.min(topLeft.x, topRight.x, bottomLeft.x, bottomRight.x);
    var right:Number = Math.max(topLeft.x, topRight.x, bottomLeft.x, bottomRight.x);

    // Ajusting final position
    matrix.translate(-left, -top);

    // Calculating the size of the new BitmapData
    var width:Number = right - left;
    var height:Number = bottom - top;

    // Creating and drawing (with transformation)
    var result:BitmapData = new BitmapData(width, height, false, 0);
    result.draw(origin, matrix);

    return result;
}

我认为这可能有点矫枉过正，但确实有效。