如何计算3D变换矩阵为矩形四边形(How to calculate 3D transformatio

希望有人能帮助，我试图找出如何从矩形变换的图像与给定x四边形，Y屏幕坐标每个角落。

到目前为止，我已经把在图像上的CALayer但需要制定出CATransform3D翘曲矩形所需四边形。什么，我想实现下面是一个例子（从A到B）。

例如矩形四路图像

如果我错了，不能使用CATransform3D做的是有这可以用一个例子来实现任何其他方式请。

我认为KennyTM的答案是接近我从这里需要..

iPhone图像拉伸（歪斜）

我已经尝试过了，并没有多少运气，他确实提到“你可能需要一个转置”，但如果是这样的话我不知道该怎么办。

Answer 1:

CATransform 3D绝对可以做到你想作什么用途。我测试了您链接到的代码和它完美地为我工作。请记住，变换矩阵一样，只定义到一个规模，因为它是齐次坐标。一旦你生成他的方程的矩阵除以右下角元素的每个元素。我能想到的，你会需要一个转，肯定是因为变换他给出的唯一原因是行优先顺序。如果你是填充柱主要变换矩阵（我相信CATransform3D是）你需要它移调您填写后。

这里是我用来测试它的代码，它采用矩阵类从OpenCV的和在C ++中，但应证明这一点

cv::Matx41d rect_tl(-10,-10,0,1);
cv::Matx41d rect_tr(10,-10,0,1);
cv::Matx41d rect_bl(-10,10,0,1);
cv::Matx41d rect_br(10,10,0,1);

cv::Matx41d quad_tl(2,2,0,1);
cv::Matx41d quad_tr(4,6,0,1);
cv::Matx41d quad_bl(2,-1,0,1);
cv::Matx41d quad_br(3,5,0,1);


double X = rect_tl(0);
double Y = rect_tl(0);
double W = 20;
double H = 20;

double x1a = quad_tl(0);
double y1a = quad_tl(1);

double x2a = quad_tr(0);
double y2a = quad_tr(1);

double x3a = quad_bl(0);
double y3a = quad_bl(1);

double x4a = quad_br(0);
double y4a = quad_br(1);



double y21 = y2a - y1a,
y32 = y3a - y2a,
y43 = y4a - y3a,
y14 = y1a - y4a,
y31 = y3a - y1a,
y42 = y4a - y2a;

double a = -H*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42);
double b = W*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43);
double c = H*X*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42) - H*W*x1a*(x4a*y32 - x3a*y42 + x2a*y43) - W*Y*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43);

double d = H*(-x4a*y21*y3a + x2a*y1a*y43 - x1a*y2a*y43 - x3a*y1a*y4a + x3a*y2a*y4a);
double e = W*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42);
double f = -(W*(x4a*(Y*y2a*y31 + H*y1a*y32) - x3a*(H + Y)*y1a*y42 + H*x2a*y1a*y43 + x2a*Y*(y1a - y3a)*y4a + x1a*Y*y3a*(-y2a + y4a)) - H*X*(x4a*y21*y3a - x2a*y1a*y43 + x3a*(y1a - y2a)*y4a + x1a*y2a*(-y3a + y4a)));

double g = H*(x3a*y21 - x4a*y21 + (-x1a + x2a)*y43);
double h = W*(-x2a*y31 + x4a*y31 + (x1a - x3a)*y42);
double i = W*Y*(x2a*y31 - x4a*y31 - x1a*y42 + x3a*y42) + H*(X*(-(x3a*y21) + x4a*y21 + x1a*y43 - x2a*y43) + W*(-(x3a*y2a) + x4a*y2a + x2a*y3a - x4a*y3a - x2a*y4a + x3a*y4a));

cv::Matx44d matrix(a,b,0,c
                   ,d,e,0,f
                   ,0,0,1,0
                   ,g,h,0,i);
matrix = matrix*(1/matrix(15));
//You may need a transpose here

cv::Matx41d test_tl = matrix*rect_tl;
test_tl *= (1/test_tl(3));
cv::Matx41d test_tr = matrix*rect_tr;
test_tr *= (1/test_tr(3));
cv::Matx41d test_bl = matrix*rect_bl;
test_bl *= (1/test_bl(3));
cv::Matx41d test_br = matrix*rect_br;
test_br *= (1/test_br(3));

执行后，在底部的所有测试变量完美匹配的四同行。希望这将清除的东西了。

Answer 2:

由于锤答案，我能得到这一切的工作，需要转置，发现如何矩阵转置的这个伟大的博客...

转置矩阵

我已创建的结果的工作方法如下...

- (CATransform3D)rectToQuad:(NSRect)rect quadTLX:(double)x1a quadTLY:(double)y1a quadTRX:(double)x2a quadTRY:(double)y2a quadBLX:(double)x3a quadBLY:(double)y3a quadBRX:(double)x4a quadBRY:(double)y4a
{
    double X = rect.origin.x;
    double Y = rect.origin.y;
    double W = rect.size.width;
    double H = rect.size.height;

    double y21 = y2a - y1a;
    double y32 = y3a - y2a;
    double y43 = y4a - y3a;
    double y14 = y1a - y4a;
    double y31 = y3a - y1a;
    double y42 = y4a - y2a;

    double a = -H*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42);
    double b = W*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43);
    double c = H*X*(x2a*x3a*y14 + x2a*x4a*y31 - x1a*x4a*y32 + x1a*x3a*y42) - H*W*x1a*(x4a*y32 - x3a*y42 + x2a*y43) - W*Y*(x2a*x3a*y14 + x3a*x4a*y21 + x1a*x4a*y32 + x1a*x2a*y43);

    double d = H*(-x4a*y21*y3a + x2a*y1a*y43 - x1a*y2a*y43 - x3a*y1a*y4a + x3a*y2a*y4a);
    double e = W*(x4a*y2a*y31 - x3a*y1a*y42 - x2a*y31*y4a + x1a*y3a*y42);
    double f = -(W*(x4a*(Y*y2a*y31 + H*y1a*y32) - x3a*(H + Y)*y1a*y42 + H*x2a*y1a*y43 + x2a*Y*(y1a - y3a)*y4a + x1a*Y*y3a*(-y2a + y4a)) - H*X*(x4a*y21*y3a - x2a*y1a*y43 + x3a*(y1a - y2a)*y4a + x1a*y2a*(-y3a + y4a)));

    double g = H*(x3a*y21 - x4a*y21 + (-x1a + x2a)*y43);
    double h = W*(-x2a*y31 + x4a*y31 + (x1a - x3a)*y42);
    double i = W*Y*(x2a*y31 - x4a*y31 - x1a*y42 + x3a*y42) + H*(X*(-(x3a*y21) + x4a*y21 + x1a*y43 - x2a*y43) + W*(-(x3a*y2a) + x4a*y2a + x2a*y3a - x4a*y3a - x2a*y4a + x3a*y4a));

    //Transposed matrix
    CATransform3D transform;
    transform.m11 = a / i;
    transform.m12 = d / i;
    transform.m13 = 0;
    transform.m14 = g / i;
    transform.m21 = b / i;
    transform.m22 = e / i;
    transform.m23 = 0;
    transform.m24 = h / i;
    transform.m31 = 0;
    transform.m32 = 0;
    transform.m33 = 1;
    transform.m34 = 0;
    transform.m41 = c / i;
    transform.m42 = f / i;
    transform.m43 = 0;
    transform.m44 = i / i;
    return transform;
}

一个例子调用此方法如下...

NSImage *image = // load a image

CALayer *layer = [CALayer layer];
[layer setContents:image];
[view setLayer:myLayer];
[view setFrame:NSMakeRect(0, 0, image.size.width, image.size.height)];

view.layer.transform = [self rectToQuad:view.frame quadTLX:0 quadTLY:0 quadTRX:image.size.width quadTRY:20 quadBLX:0 quadBLY:image.size.height quadBRX:image.size.width quadBRY:image.size.height + 90];