How do I prevent clipping when rotating an image i

Question

I just went through a bunch of stuff trying to figure out how to get the image to even rotate. That works but now it's clipping and I'm not really sure how to make it stop... I'm using this rotateImage method:

public static Image RotateImage(Image img, float rotationAngle)
    {
        //create an empty Bitmap image
        Bitmap bmp = new Bitmap(img.Width, img.Height);

        //turn the Bitmap into a Graphics object
        Graphics gfx = Graphics.FromImage(bmp);

        //now we set the rotation point to the center of our image
        gfx.TranslateTransform((float)bmp.Width / 2, (float)bmp.Height / 2);

        //now rotate the image
        gfx.RotateTransform(rotationAngle);

        gfx.TranslateTransform(-(float)bmp.Width / 2, -(float)bmp.Height / 2);

        //set the InterpolationMode to HighQualityBicubic so to ensure a high
        //quality image once it is transformed to the specified size
        gfx.InterpolationMode = InterpolationMode.HighQualityBicubic;

        //now draw our new image onto the graphics object
        gfx.DrawImage(img, new System.Drawing.Point(0, 0));

        //dispose of our Graphics object
        gfx.Dispose();

        //return the image
        return bmp;
    }

I tried making the empty bitmap larger but that only works for one side since the image is pinned to the upper left corner of the bitmap. Any ideas would be appreciated!

Answer 1

I found some help from another website. Here's what I ended up doing for those of you who want to know:

// Rotates the input image by theta degrees around center.
    public static Bitmap RotateImage(Bitmap bmpSrc, float theta)
    {
        Matrix mRotate = new Matrix();
        mRotate.Translate(bmpSrc.Width / -2, bmpSrc.Height / -2, MatrixOrder.Append);
        mRotate.RotateAt(theta, new System.Drawing.Point(0, 0), MatrixOrder.Append);
        using (GraphicsPath gp = new GraphicsPath())
        {  // transform image points by rotation matrix
            gp.AddPolygon(new System.Drawing.Point[] { new System.Drawing.Point(0, 0), new System.Drawing.Point(bmpSrc.Width, 0), new System.Drawing.Point(0, bmpSrc.Height) });
            gp.Transform(mRotate);
            System.Drawing.PointF[] pts = gp.PathPoints;

            // create destination bitmap sized to contain rotated source image
            Rectangle bbox = boundingBox(bmpSrc, mRotate);
            Bitmap bmpDest = new Bitmap(bbox.Width, bbox.Height);

            using (Graphics gDest = Graphics.FromImage(bmpDest))
            {  // draw source into dest
                Matrix mDest = new Matrix();
                mDest.Translate(bmpDest.Width / 2, bmpDest.Height / 2, MatrixOrder.Append);
                gDest.Transform = mDest;
                gDest.DrawImage(bmpSrc, pts);
                return bmpDest;
            }
        }
    }

    private static Rectangle boundingBox(Image img, Matrix matrix)
    {
        GraphicsUnit gu = new GraphicsUnit();
        Rectangle rImg = Rectangle.Round(img.GetBounds(ref gu));

        // Transform the four points of the image, to get the resized bounding box.
        System.Drawing.Point topLeft = new System.Drawing.Point(rImg.Left, rImg.Top);
        System.Drawing.Point topRight = new System.Drawing.Point(rImg.Right, rImg.Top);
        System.Drawing.Point bottomRight = new System.Drawing.Point(rImg.Right, rImg.Bottom);
        System.Drawing.Point bottomLeft = new System.Drawing.Point(rImg.Left, rImg.Bottom);
        System.Drawing.Point[] points = new System.Drawing.Point[] { topLeft, topRight, bottomRight, bottomLeft };
        GraphicsPath gp = new GraphicsPath(points,
                                                            new byte[] { (byte)PathPointType.Start, (byte)PathPointType.Line, (byte)PathPointType.Line, (byte)PathPointType.Line });
        gp.Transform(matrix);
        return Rectangle.Round(gp.GetBounds());
    }

Answer 2

A rotated image may require a larger bitmap to be contained without clipping. A fairly simple way to compute the bounds is to apply the transformation matrix to the corners of the original bounding rectangle. Since the new bitmap is larger, the original image must be drawn such that it is centered, rather than at (0,0).

This is demonstrated in the following revision to your code:

public static Image RotateImage(Image img, float rotationAngle)
{
    int minx = int.MaxValue, maxx = int.MinValue, miny = int.MaxValue, maxy = int.MinValue;

    using (Bitmap bmp = new Bitmap(1, 1)) // Dummy bitmap, so we can use TransformPoints to figure out the correct size.
    {
        using (Graphics g = Graphics.FromImage(bmp))
        {
            g.TranslateTransform((float)img.Width / 2, (float)img.Height / 2);
            g.RotateTransform(rotationAngle);
            g.TranslateTransform(-(float)img.Width / 2, -(float)img.Height / 2);

            Point[] pts = new Point[4];
            pts[0] = new Point(0, 0);
            pts[1] = new Point(img.Width, 0);
            pts[2] = new Point(img.Width, img.Height);
            pts[3] = new Point(0, img.Height);
            g.TransformPoints(CoordinateSpace.Device, CoordinateSpace.World, pts);

            foreach (Point pt in pts)
            {
                minx = Math.Min(minx, pt.X);
                maxx = Math.Max(maxx, pt.X);
                miny = Math.Min(miny, pt.Y);
                maxy = Math.Max(maxy, pt.Y);
            }
        }
    }

    Bitmap bmp2 = new Bitmap(maxx - minx, maxy - miny);
    using (Graphics g = Graphics.FromImage(bmp2))
    {
        g.TranslateTransform((float)bmp2.Width / 2, (float)bmp2.Height / 2);
        g.RotateTransform(rotationAngle);
        g.TranslateTransform(-(float)bmp2.Width / 2, -(float)bmp2.Height / 2);
        g.InterpolationMode = InterpolationMode.HighQualityBicubic;
        g.DrawImage(img, bmp2.Width / 2 - img.Width / 2, bmp2.Height / 2 - img.Height / 2);
    }

    return bmp2;
}

Answer 3

follow these steps:

1. Create empty destination image whose width & height = bounding box of your rotated image
2. Draw your source image onto the destination image. (0,0) of source will map to (ox,oy) of destination
3. Now do the steps to rotate the destination image.

Details to accomplish above steps:

W,H = width & height of destination

w,h = width & height of source

c=|cos(theta)|

s=|sin(theta)|

w * c + h * s = W

w * s + h * c = H

ox = ( W - w ) / 2

oy = ( H - h ) / 2

Answer 4

public Bitmap rotateImage(Bitmap b, float angle)
    {
        if (angle > 0)
        {
            int l = b.Width;
            int h = b.Height;
            double an = angle * Math.PI / 180;
            double cos = Math.Abs(Math.Cos(an));
            double sin = Math.Abs(Math.Sin(an));
            int nl = (int)(l * cos + h * sin);
            int nh = (int)(l * sin + h * cos);
            Bitmap returnBitmap = new Bitmap(nl, nh);
            Graphics g = Graphics.FromImage(returnBitmap);
            g.TranslateTransform((float)(nl-l) / 2, (float)(nh-h) / 2);
            g.TranslateTransform((float)b.Width / 2, (float)b.Height / 2);
            g.RotateTransform(angle);
            g.TranslateTransform(-(float)b.Width / 2, -(float)b.Height / 2);
            g.DrawImage(b, new Point(0, 0));
            return returnBitmap;
        }
        else return b;
    }