Maths Help Matrices / projection transformation

    Hey guys
    Thought i would put this one out there as you lot seem like a bunch of clever people.

    I am having some trouble with a project i am working on and the maths is killing me

    basically i have a projector which is pointed at different angles to a screen and i need to correct the skew - keystone effects via software

    i can get the corner coordinates of the polygon that appears on the screen and i can get x.y.z roll pitch, yaw of the projector

    i have looked and projection matrices look like the way but i have no idea what so ever on how to make them work

    if anyone has any ideas i would be much much appreciative



    Original Poster

    yer me too

    and thats the problem i am supposed to know this


    whoosh, that went over my head so fast, I fell off my chair,lol

    yeah me too just readin it does my head in :?

    First you need to calculate a homography that will map between your two planes - the physical and the projected image plane. I use the OpenCV software to do this which has a function cvGetPerspectiveTransform which you feed a set of corresponding points in each plane (i.e. a square in your image and the points it projects to in the real world. You can then use the calculated homography to correct your image so that is square when projected.

    From the source code: Calculates coefficients of perspective transformation
    which maps (xi,yi) to (ui,vi), (i=1,2,3,4):

    c00*xi + c01*yi + c02
    ui = ---------------------
    c20*xi + c21*yi + c22

    c10*xi + c11*yi + c12
    vi = ---------------------
    c20*xi + c21*yi + c22

    Coefficients are calculated by solving linear system:
    / x0 y0 1 0 0 0 -x0*u0 -y0*u0 \ /c00\ /u0\
    | x1 y1 1 0 0 0 -x1*u1 -y1*u1 | |c01| |u1|
    | x2 y2 1 0 0 0 -x2*u2 -y2*u2 | |c02| |u2|
    | x3 y3 1 0 0 0 -x3*u3 -y3*u3 |.|c10|=|u3|,
    | 0 0 0 x0 y0 1 -x0*v0 -y0*v0 | |c11| |v0|
    | 0 0 0 x1 y1 1 -x1*v1 -y1*v1 | |c12| |v1|
    | 0 0 0 x2 y2 1 -x2*v2 -y2*v2 | |c20| |v2|
    \ 0 0 0 x3 y3 1 -x3*v3 -y3*v3 / \c21/ \v3/

    cij - matrix coefficients, c22 = 1

    jonsend tats just making it worse cant handle it gotta goto another thread

    If you need more detail or some code I will put some up later but a bit busy atm.

    Original Poster

    Jonsend firstly thanks for the post its really appreciated

    my problem is i am doing it on a mobile device using python so i wont have access to openCV or openGL or any lovely utilities

    so from what you posted if i calculate the matrix with the values I obtain it just works it out

    also i am new to matrix miltiplication and obviously extrapolating that to code is quite a task

    so to get this working what values would i need to know
    and what is x,y and u,v
    the coeficents ci how are they calcualted in this as they arent the end product
    / x0 y0 1 0 0 0 -x0*u0 -y0*u0 \ /c00\ /u0\
    | x1 y1 1 0 0 0 -x1*u1 -y1*u1 | |c01| |u1|
    | x2 y2 1 0 0 0 -x2*u2 -y2*u2 | |c02| |u2|
    | x3 y3 1 0 0 0 -x3*u3 -y3*u3 | .|c10| |u3|,
    | 0 0 0 x0 y0 1 -x0*v0 -y0*v0 | |c11| = |v0|
    | 0 0 0 x1 y1 1 -x1*v1 -y1*v1 | |c12| |v1|
    | 0 0 0 x2 y2 1 -x2*v2 -y2*v2 | |c20| |v2|
    \ 0 0 0 x3 y3 1 -x3*v3 -y3*v3 / \c21/ \v3/

    again apologies if this is simple stuff i really am in the dark with this area

    thanks again

    Original Poster

    Jonsend just read your other message

    that would be great

    I will pm you my email address if thats ok to save the code loosing its integrity

    thanks again

    ok np
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