objective-c - 返回 CATransform3D 将四边形映射到四边形

Question

我正在尝试派生一个 CATransform3D，它将一个具有 4 个角点的四边形映射到另一个具有 4 个新角点的四边形。我花了一点时间研究这个，似乎这些步骤涉及将原始四边形转换为正方形，然后将该正方形转换为新的四边形。我的方法如下所示(代码借自 here ):

- (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;

}

在计算完两个矩阵，将它们相乘并分配给相关 View 后，我最终得到了一个转换后的 View ，但这是非常不正确的。事实上，无论我做什么，它似乎都像平行四边形一样被剪切。我错过了什么？

更新 2/1/12

看来我遇到问题的原因可能是我需要将 FOV 和焦距适应模型 View 矩阵(这是我可以在 Quartz 中直接更改的唯一矩阵。)我没有任何不过，幸运的是在网上找到了有关如何计算正确矩阵的文档。

Answer 1

我能够通过移植和组合来自这两个 URL 的四边形变形和单应性代码来实现这一点:

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

更新:我已经开源了一个小类来执行此操作:https://github.com/dominikhofmann/DHWarpView