获取RotateTransform的结果大小

Question

我有以下代码，用于在C#中旋转图像：

private Bitmap RotateImage(Bitmap b, float angle)
{
    //create a new empty bitmap to hold rotated image
    Bitmap returnBitmap = new Bitmap(b.Width, b.Height);

    //make a graphics object from the empty bitmap
    Graphics g = Graphics.FromImage(returnBitmap);
    //move rotation point to center of image
    g.TranslateTransform((float)returnBitmap.Width / 2, (float)returnBitmap.Height / 2);

    //rotate
    g.RotateTransform(angle);
    //move image back
    g.TranslateTransform(-(float)b.Width / 2, -(float)b.Height / 2);
    //draw passed in image onto graphics object
    g.DrawImage(b, new Rectangle(new Point(0, 0), new Size(b.Width, b.Height)));
    return returnBitmap;
}

它的工作非常好，除了它剪辑的结果，当它超过了原来的界限。

据我所知，在旋转后，我必须将返回位图的大小设置为图像的大小。但是，如何找到结果有多大，以相应地设置新位图的大小？

Answer 1

您需要旋转原始图像的四个角，并计算新坐标的包围框：

    private static Bitmap RotateImage(Image b, float angle)
    {
        var corners = new[]
            {new PointF(0, 0), new Point(b.Width, 0), new PointF(0, b.Height), new PointF(b.Width, b.Height)};

        var xc = corners.Select(p => Rotate(p, angle).X);
        var yc = corners.Select(p => Rotate(p, angle).Y);

        //create a new empty bitmap to hold rotated image
        Bitmap returnBitmap = new Bitmap((int)Math.Abs(xc.Max() - xc.Min()), (int)Math.Abs(yc.Max() - yc.Min()));
        ...
    }

    /// <summary>
    /// Rotates a point around the origin (0,0)
    /// </summary>
    private static PointF Rotate(PointF p, float angle)
    {
        // convert from angle to radians
        var theta = Math.PI*angle/180;
        return new PointF(
            (float) (Math.Cos(theta)*(p.X) - Math.Sin(theta)*(p.Y)),
            (float) (Math.Sin(theta)*(p.X) + Math.Cos(theta)*(p.Y)));
    }

Answer 2

毕达哥拉斯。它是在90/270角度从原始到平方(w^2+ h^2)的任意位置。我敢打赌，它是由正弦(最大值在90)决定的。