返回CATransform3D，将四边形映射为四边形

I'm trying to derive a CATransform3D that will map a quad with 4 corner points to another quad with 4 new corner points. I've spent a little bit of time researching this and it seems the steps involve converting the original Quad to a Square, and then converting that Square to the new Quad. My methods look like this (code borrowed from here):

我正在尝试推导一个CATransform3D，它将一个有4个角点的四轴飞行器映射到另一个有4个新角点的四轴飞行器。我花了一点时间研究这个，看起来步骤包括把原来的四边形变成正方形，然后把正方形变成新的四边形。我的方法是这样的(从这里借来的代码):

- (CATransform3D)quadFromSquare_x0:(float)x0 y0:(float)y0 x1:(float)x1 y1:(float)y1 x2:(float)x2 y2:(float)y2 x3:(float)x3 y3:(float)y3 {

    float dx1 = x1 - x2,    dy1 = y1 - y2;
    float dx2 = x3 - x2,    dy2 = y3 - y2;
    float sx = x0 - x1 + x2 - x3;
    float sy = y0 - y1 + y2 - y3;
    float g = (sx * dy2 - dx2 * sy) / (dx1 * dy2 - dx2 * dy1);
    float h = (dx1 * sy - sx * dy1) / (dx1 * dy2 - dx2 * dy1);
    float a = x1 - x0 + g * x1;
    float b = x3 - x0 + h * x3;
    float c = x0;
    float d = y1 - y0 + g * y1;
    float e = y3 - y0 + h * y3;
    float f = y0;

    CATransform3D mat;

    mat.m11 = a;
    mat.m12 = b;
    mat.m13 = 0;
    mat.m14 = c;

    mat.m21 = d;
    mat.m22 = e;
    mat.m23 = 0;
    mat.m24 = f;

    mat.m31 = 0;
    mat.m32 = 0;
    mat.m33 = 1;
    mat.m34 = 0;

    mat.m41 = g;
    mat.m42 = h;
    mat.m43 = 0;
    mat.m44 = 1;

    return mat;

}

- (CATransform3D)squareFromQuad_x0:(float)x0 y0:(float)y0 x1:(float)x1 y1:(float)y1 x2:(float)x2 y2:(float)y2 x3:(float)x3 y3:(float)y3 {

    CATransform3D mat = [self quadFromSquare_x0:x0 y0:y0 x1:x1 y1:y1 x2:x2 y2:y2 x3:x3 y3:y3];

    // invert through adjoint

    float a = mat.m11,      d = mat.m21,    /* ignore */            g = mat.m41;
    float b = mat.m12,      e = mat.m22,    /* 3rd col*/            h = mat.m42;
    /* ignore 3rd row */
    float c = mat.m14,      f = mat.m24;

    float A =     e - f * h;
    float B = c * h - b;
    float C = b * f - c * e;
    float D = f * g - d;
    float E =     a - c * g;
    float F = c * d - a * f;
    float G = d * h - e * g;
    float H = b * g - a * h;
    float I = a * e - b * d;

    // Probably unnecessary since 'I' is also scaled by the determinant,
    //   and 'I' scales the homogeneous coordinate, which, in turn,
    //   scales the X,Y coordinates.
    // Determinant  =   a * (e - f * h) + b * (f * g - d) + c * (d * h - e * g);
    float idet = 1.0f / (a * A           + b * D           + c * G);

    mat.m11 = A * idet;     mat.m21 = D * idet;     mat.m31 = 0;    mat.m41 = G * idet;
    mat.m12 = B * idet;     mat.m22 = E * idet;     mat.m32 = 0;    mat.m42 = H * idet;
    mat.m13 = 0       ;     mat.m23 = 0       ;     mat.m33 = 1;    mat.m43 = 0       ;
    mat.m14 = C * idet;     mat.m24 = F * idet;     mat.m34 = 0;    mat.m44 = I * idet;

    return mat;

}

After calculating both matrices, multiplying them together, and assigning to the view in question, I end up with a transformed view, but it is wildly incorrect. In fact, it seems to be sheared like a parallelogram no matter what I do. What am I missing?

在计算了这两个矩阵，将它们相乘，并分配给所讨论的视图之后，我最终得到了一个转换后的视图，但它是非常不正确的。事实上，它看起来就像一个平行四边形，不管我怎么做。我缺少什么?

UPDATE 2/1/12

更新2/1/12

It seems the reason I'm running into issues may be that I need to accommodate for FOV and focal length into the model view matrix (which is the only matrix I can alter directly in Quartz.) I'm not having any luck finding documentation online on how to calculate the proper matrix, though.

我遇到问题的原因可能是我需要将FOV和焦距调整到模型视图矩阵中(这是我在Quartz中唯一可以直接修改的矩阵)。不过，我在网上找不到关于如何计算适当矩阵的文档。

1 个解决方案

#1

I was able to achieve this by porting and combining the quad warping and homography code from these two URLs:

我可以通过将这两个url的四轴扭曲和同源代码移植并结合在一起来实现这一点:

http://forum.openframeworks.cc/index.php/topic,509.30.html
http://forum.openframeworks.cc/index.php/topic 509.30.html
http://forum.openframeworks.cc/index.php?topic=3121.15
http://forum.openframeworks.cc/index.php?topic=3121.15

UPDATE: I've open sourced a small class that does this: https://github.com/dominikhofmann/DHWarpView

更新:我打开了一个小类的源代码:https://github.com/dominikhofmann/DHWarpView

#1