rotating an image, by not using the defaulted upper left corner for rotation point

[jkw119]

If anyone can help me out here, i would really appreciate it. I am trying to do a relatively simple task -> Rotate a Picture. Which works fine. But the problem is, my code uses the upper left corner to rotate the image. Is it possible to change that to the middle, or the origin, so that i can rotate the picture from the center. If so, how is this done? Here is the sample code, basically it is a panel on a form, and this is it's paint event. It loads a picture from the programs bin folder. Thanks alot...

Jeff

Private Sub TextArea_Paint(ByVal sender As Object, ByVal e As System.Windows.Forms.PaintEventArgs) Handles TextArea.Paint
 
        Dim sa As System.Drawing.Size
        sa = TextArea.Size
 
        Dim middle As New Point(sa.Width / 4, sa.Height / 3)
        e.Graphics.TranslateTransform(middle.X, middle.Y)
 
        'Apply all transforms....
        e.Graphics.RotateTransform(45) 'angle to rotate
 
        Dim bmp As New Bitmap("sample.jpg")
        e.Graphics.DrawImage(bmp, 0, 0)
 
    End Sub

[Lunatic3]

Your code rotates the text area not the image. Try it on a form and you will see some intersting results, for example from resize handler goes out of your form.

Try this code I adopted from C# from here.
Dont forget to import system.math and that your image should be 'Centered' in a PictureBox.

Public Function rotate(ByVal img As Bitmap, ByVal angle As Single) As Bitmap
        Dim pi2 As Double = PI / 2.0
        Dim oldWidth As Double = img.Width
        Dim oldHeight As Double = img.Height
        Dim theta As Double = angle * PI / 180.0
        Dim locked_theta As Double = theta
        Dim newWidth, newHeight As Double
        Dim nWidth, nHeight As Integer
        Dim adjacentTop, oppositeTop As Double
        Dim adjacentBottom, oppositeBottom As Double
 
 
        If (locked_theta >= 0.0 And locked_theta < pi2) Or (locked_theta >= Math.PI And locked_theta < (Math.PI + pi2)) Then
 
            adjacentTop = Math.Abs(Math.Cos(locked_theta)) * oldWidth
            oppositeTop = Math.Abs(Math.Sin(locked_theta)) * oldWidth
 
            adjacentBottom = Math.Abs(Math.Cos(locked_theta)) * oldHeight
            oppositeBottom = Math.Abs(Math.Sin(locked_theta)) * oldHeight
 
        Else
 
            adjacentTop = Math.Abs(Math.Sin(locked_theta)) * oldHeight
            oppositeTop = Math.Abs(Math.Cos(locked_theta)) * oldHeight
 
            adjacentBottom = Math.Abs(Math.Sin(locked_theta)) * oldWidth
            oppositeBottom = Math.Abs(Math.Cos(locked_theta)) * oldWidth
        End If
        newWidth = adjacentTop + oppositeBottom
        newHeight = adjacentBottom + oppositeTop
        nWidth = Math.Ceiling(newWidth)
        nHeight = Math.Ceiling(newHeight)
        Dim rotatedBmp As Bitmap = New Bitmap(nWidth, nHeight)
        Dim g As Graphics = Graphics.FromImage(rotatedBmp)
        Dim points(2) As Point
        If locked_theta >= 0.0 And locked_theta < pi2 Then
            points(0) = New Point(CInt(oppositeBottom), 0)
            points(1) = New Point(nWidth, CInt(oppositeTop))
            points(2) = New Point(0, CInt(adjacentBottom))
 
        ElseIf locked_theta >= pi2 And locked_theta < Math.PI Then
            points(0) = New Point(nWidth, CInt(oppositeTop))
            points(1) = New Point(CInt(adjacentTop), nHeight)
            points(2) = New Point(CInt(oppositeBottom), 0)
 
        ElseIf locked_theta >= Math.PI And locked_theta < (Math.PI + pi2) Then
            points(0) = New Point(CInt(adjacentTop), nHeight)
            points(1) = New Point(0, CInt(adjacentBottom))
            points(2) = New Point(nWidth, CInt(oppositeTop))
        Else
            points(0) = New Point(0, CInt(adjacentBottom))
            points(1) = New Point(CInt(oppositeBottom), 0)
            points(2) = New Point(CInt(adjacentTop), nHeight)
 
        End If
        g.DrawImage(img, points)
 
        g.Dispose()
        Return rotatedBmp
 
    End Function

[jkw119]

thanks so much, that was exactly what i wanted to do!

Jeff