返回 CATransform3D 以将四边形映射到四边形 [英] Return CATransform3D to map quadrilateral to quadrilateral
问题描述
我正在尝试导出一个 CATransform3D，它将一个带有 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，似乎这些步骤涉及将原始 Quad 转换为 Square，然后将该 Square 转换为新的 Quad.我的方法看起来像这样(从这里借用的代码):
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):
- (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;
}
在计算两个矩阵，将它们相乘并分配给有问题的视图后，我最终得到了一个转换后的视图，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?
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?
更新 2/1/12
似乎我遇到问题的原因可能是我需要将 FOV 和焦距纳入模型视图矩阵(这是我可以在 Quartz 中直接更改的唯一矩阵.)我没有任何不过，很幸运在网上找到了有关如何计算正确矩阵的文档.
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.
推荐答案
我能够通过移植和组合来自这两个 URL 的 quad warping 和 homography 代码来实现这一点:
I was able to achieve this by porting and combining the quad warping and homography code from these two URLs:
- 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，这样做:https://github.com/dominikhofmann/DHWarp视图
UPDATE: I've open sourced a small class that does this: https://github.com/dominikhofmann/DHWarpView
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