Question

我正在尝试派生一个CATransform3D，它将带有4个角点的四边形映射到另一个带有4个新角点的四边形。我花了一点时间研究这个，似乎步骤涉及将原始Quad转换为Square，然后将Square转换为新的Quad。我的方法看起来像这样（从here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;

}

在计算了两个矩阵，将它们相乘并分配给相关视图后，我最终得到了一个转换后的视图，但它完全不正确。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

似乎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（这是我可以直接在Quartz中改变的唯一矩阵。）我没有任何幸运的是，在线找到关于如何计算适当矩阵的文档。

Answer 1

我能够通过移植和组合来自这两个网址的四元变形和单应码来实现这一目标：

更新：我已经开源了一个小课程：https://github.com/dominikhofmann/DHWarpView

返回CATransform3D以将四边形映射到四边形

1 个答案: