Translating my bitmap rotation (Zaei?)

[Janus]

I'm testing out an algorithm I found on a few C++ sites for doing a destination-source mapped bitmap rotate, and it all works, except I can't translate the rotation from 0,0 to the centerpoint. The first time I messed with it, I got it to work, but I can't figure out how I did it. I figure Zaei or someone else mathy can help me (I still haven't taken any advanced math or geometry).

Don't worry about the calls and everything, they all work fine.
RotateRect calculates the necessary Width and Height to contain the rotated image.
Resize resizes the buffer to hold it.
GetPixelAARolloff is basically GetPixel, but BiLinear.

sAngle = Angle * (c_dblPi / 180)
    duCol = Sin(-sAngle) * (1)
    dvCol = Cos(-sAngle) * (1)
    m_dblX2 = SourceImage.Width
    m_dblY2 = SourceImage.Height
    SoftFX.RotateRect m_dblX2, m_dblY2, m_dblXO, m_dblYO, CDbl(sAngle)
    DestImage.Resize m_dblX2, m_dblY2
    X = m_dblX2 / 2
    Y = m_dblY2 / 2
    DestImage.Clear F2RGB(64, 96, 128, 0)
    duRow = dvCol
    dvRow = -duCol
    startu = (SourceImage.Width / 2)
    startv = (SourceImage.Height / 2)
    rowu = startu - ((X * duCol) + (Y * duRow))
    rowv = startv - ((X * dvCol) + (Y * dvRow))
    For dY = 0 To DestImage.Height - 1
        sX = rowu
        sY = rowv
        For dX = 0 To DestImage.Width - 1
            DestImage.SetPixel dX, dY, SourceImage.GetPixelAARolloff(sX, sY)
            sX = sX + duRow
            sY = sY + dvRow
        Next dX
        rowu = rowu + duCol
        rowv = rowv + dvCol
    Next dY

[/\/\isanThr0p]

alright... I looked at this, and I think you need some major changing before this code will rotate around the center...
take a look at the rotation tutorial on www.ur.co.nz ... I am pretty sure that that's the best way to turn around the center...

http://www.ur.co.nz/urcorp/default.a...ata_article=69

you need a log in for that though.. it's free and I never got unwanted emails...

[Sastraxi]

Can you not just add or subtract 0.5(newwidth) and 0.5(newheight) to get the centrepoint? (ex. if it's rotated around the top-left corner, add both. If its around the bottom-right corner, subtract both).

Or is it rotated in some supremely odd way?

[Zaei]

Dim cA as Single
Dim sA as Single
cA = Cos(Theta*(3.14159/180))
sA = Sin(Theta*(3.14159/180))

Dim l as Single

Dim i as Long
Dim j as Long

For i = 0 to width
  For j = 0 to height
    l = sqr(i^2 + j^2)
    SetPixel(l*cA, l*sA, GetPixel(i, j))
   Next j
Next i

That should do a standard rotation, around the upper left corner.

Now, for the translation...

Dim cA as Single
Dim sA as Single
Dim cX as Long
Dim cY as Long
Dim dcX as Long
Dim dcY as Long

cA = Cos(Theta*(3.14159/180))
sA = Sin(Theta*(3.14159/180))

Dim l as Single

Dim i as Long
Dim j as Long

cX = width/2
cY = height/2
dcX = destWidth/2
dcY = destHeight/2

For i = 0 to width
  For j = 0 to height
    l = sqr((cX+i)^2 + (cY + j)^2)
    SetPixel(l*cA+dcX, l*sA+dcY, GetPixel(i, j))
   Next j
Next i

Try that... Im not sure if it will work (its off the top of my head, and I cant test it).

Z.

[Janus]

I wasn't sure this code would rotate around the center either... the annoying thing is, I know it does, because I got it to by accident once, and it does in a sample in another language. However, the sample is too simple to be used for this.

Zaei, I appreciate the try, but I have to use this algorithm. I'll see if I can find any tricks to use from that one.

Adding and subtracting doesn't seem to actually translate the rotation, it just translates the bitmap.

[Sastraxi]

Oohhh I get it then.

Well what you do is use cos() and sin() with the radius as the number of pixels away from the centre pixel...

[Janus]

Using Cos() or Sin() in the main loop would totally ruin my performance - this function's very fast.

[/\/\isanThr0p]

well the advantage of this function is that it calculates sin and cos once... and than goes through the picture at an angle with help of the x and y component (found through cos and sin).

But using sin and cos does not have to slow down your function completely...
sin(x)=cos(x-90) or sin(x)=cos(x-Pi/2) if you are using radians

so all you have to do now is precalculate a little array from one to 360.
Pseudocode (you'd need to do it for radians)
For x = 0 to 359
lsin(x) = sin(x)
next

now get cos of an angle like lsin(angle-90) and the sin lsin(angle)
this still is pretty fast.

[Zaei]

You only need compute the sine and cosine once, as you only have one angle.

Z.

[/\/\isanThr0p]

how does the algorithm look than? (I rember having seen something like that.. pretty close to the one posted here I guess... well I really do not feel like figuring it out again...)

it can be done really easy with a whole table of sin and cos... the other way must be faster though...

[Janus]

The Cos and Sin aren't the only part that's slow, the Floating Point Multiply is much slower as well. I guess I could try and integerize it.