如何计算矩形到四边形的 3D 变换矩阵

Question

希望有人可以提供帮助，我正在尝试研究如何将图像从矩形转换为四边形，每个角都有给定的 x,y 屏幕坐标。

到目前为止，我已将图像放在 CALayer 上，但需要计算 CATransform3D 以将矩形扭曲为所需的四边形。下面是我想要实现的一个示例（从 a 到 b）。

Example Rect to Quad image

如果我错了并且无法使用 CATransform3D 完成，请问有没有其他方法可以通过示例来实现。

我认为 KennyTM 的答案与我在这里所需要的很接近..

iPhone image stretching (skew)

我已经尝试过了，但运气不佳，他确实提到“您可能需要转置”，但如果是这种情况，我不知道该怎么做。

Answer 1

CATransform 3D 绝对可以满足您的需求。我测试了code you linked to，它非常适合我。请记住，像这样的变换矩阵仅定义为一个比例，因为它位于齐次坐标中。一旦你用他的方程生成矩阵，将每个元素除以右下角的元素。我能想到你需要转置的唯一原因是因为他给出的变换是行主要顺序的。如果要填充列主要转换矩阵（我相信 CATransform3D 是），则需要在填充后对其进行转置。

这是我用来测试它的代码，它使用来自 openCV 的矩阵类并且是在 c++ 中，但应该证明这一点

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，底部的所有测试变量都与它们的四边形对应物完美匹配。希望这可以解决问题。

Answer 2

感谢 Hammers 的回答，我能够让这一切正常工作，需要转置，并找到了这篇关于如何转置矩阵的精彩博客...

Transpose Matrix

我创建的结果工作方法如下...

- (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;
}

对该方法的调用示例如下...

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];

Answer 3

感谢@Equinox2000 的帮助！

请注意，iOS CALayer 的默认锚点为 (0.5, 0.5)。如果您尝试应用所有值都相对于左上角坐标的变换，则需要将锚点更改为 (0.0, 0.0)。