我試圖導出一個CATransform3D,它將帶有4個角點的四邊形映射到具有4個新角點的另一個四邊形。我花了一點時間研究這個問題,看起來這些步驟涉及將原始的Quad轉換爲Square,然後將該Square轉換爲新的Quad。我的方法是這樣的(從here借來的代碼):返回CATransform3D將四邊形映射到四邊形
- (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;
}
計算兩個矩陣,它們相乘在一起,並在問題分配給視圖後,我結束了一個轉變的看法,但它是瘋狂不正確。事實上,無論我做什麼,它似乎都像平行四邊形剪切一樣。我錯過了什麼?
UPDATE 12年2月1日
看來我遇到問題的原因可能是,我需要適應FOV和焦距到模型視圖矩陣(這是唯一的矩陣我可以直接在Quartz中修改)。我沒有找到任何有關如何計算正確矩陣的文檔的運氣。
更新:我試過現在移植一些算法,並最終得到一個平行四邊形每一次。我必須忘記一些事情,但我很茫然。 – sevenflow 2012-02-01 18:07:17
下面是實現此目的的其他方法:http://stackoverflow.com/questions/9470493/transforming-a-rectangle-image-into-a-quadrilateral-using-a-catransform3d讓我知道這是否有幫助。 – MonsieurDart 2012-03-27 08:31:15