Question

我确实使用findcontours()方法从图像中提取轮廓，但我不知道如何从一组轮廓点计算曲率。有人能帮助我吗？非常感谢你！

Answer 1

虽然Gombat的答案背后的理论是正确的，但代码和公式中都存在一些错误（分母t+n-t应该是std::vector<double> getCurvature(std::vector<cv::Point> const& vecContourPoints, int step) { std::vector< double > vecCurvature( vecContourPoints.size() ); if (vecContourPoints.size() < step) return vecCurvature; auto frontToBack = vecContourPoints.front() - vecContourPoints.back(); std::cout << CONTENT_OF(frontToBack) << std::endl; bool isClosed = ((int)std::max(std::abs(frontToBack.x), std::abs(frontToBack.y))) <= 1; cv::Point2f pplus, pminus; cv::Point2f f1stDerivative, f2ndDerivative; for (int i = 0; i < vecContourPoints.size(); i++ ) { const cv::Point2f& pos = vecContourPoints[i]; int maxStep = step; if (!isClosed) { maxStep = std::min(std::min(step, i), (int)vecContourPoints.size()-1-i); if (maxStep == 0) { vecCurvature[i] = std::numeric_limits<double>::infinity(); continue; } } int iminus = i-maxStep; int iplus = i+maxStep; pminus = vecContourPoints[iminus < 0 ? iminus + vecContourPoints.size() : iminus]; pplus = vecContourPoints[iplus > vecContourPoints.size() ? iplus - vecContourPoints.size() : iplus]; f1stDerivative.x = (pplus.x - pminus.x) / (iplus-iminus); f1stDerivative.y = (pplus.y - pminus.y) / (iplus-iminus); f2ndDerivative.x = (pplus.x - 2*pos.x + pminus.x) / ((iplus-iminus)/2*(iplus-iminus)/2); f2ndDerivative.y = (pplus.y - 2*pos.y + pminus.y) / ((iplus-iminus)/2*(iplus-iminus)/2); double curvature2D; double divisor = f1stDerivative.x*f1stDerivative.x + f1stDerivative.y*f1stDerivative.y; if ( std::abs(divisor) > 10e-8 ) { curvature2D = std::abs(f2ndDerivative.y*f1stDerivative.x - f2ndDerivative.x*f1stDerivative.y) / pow(divisor, 3.0/2.0 ) ; } else { curvature2D = std::numeric_limits<double>::infinity(); } vecCurvature[i] = curvature2D; } return vecCurvature; }）。我做了一些改变：

使用对称导数来获得更精确的曲率最大位置
允许使用步长进行微分计算（可用于减少噪声等值线的噪声）
使用闭合轮廓

修正： *如果分母为0（不为0），则返回无穷大作为曲率 *在分母中增加了平方计算 *正确检查0除数

importrange()

Answer 2

对我来说曲率是：

$|K| = \sqrt{ \frac{ \left(\frac{d^2y(t)}{dt^2}\frac{dx(t))}{dt}-\frac{d^2x(t)}{dt^2}\frac{dy(t)}{dt} \right )^2 }{ \left( \frac{d^2x(t)}{dt^2}+\frac{d^2y(t)}{dt^2} \right )^{3} } }$

其中t是轮廓内的位置，x(t)表示。 y(t)返回相关的x resp。 y值。std::vector< float > vecCurvature( vecContourPoints.size() ); cv::Point2f posOld, posOlder; cv::Point2f f1stDerivative, f2ndDerivative; for (size_t i = 0; i < vecContourPoints.size(); i++ ) { const cv::Point2f& pos = vecContourPoints[i]; if ( i == 0 ){ posOld = posOlder = pos; } f1stDerivative.x = pos.x - posOld.x; f1stDerivative.y = pos.y - posOld.y; f2ndDerivative.x = - pos.x + 2.0f * posOld.x - posOlder.x; f2ndDerivative.y = - pos.y + 2.0f * posOld.y - posOlder.y; float curvature2D = 0.0f; if ( std::abs(f2ndDerivative.x) > 10e-4 && std::abs(f2ndDerivative.y) > 10e-4 ) { curvature2D = sqrt( std::abs( pow( f2ndDerivative.y*f1stDerivative.x - f2ndDerivative.x*f1stDerivative.y, 2.0f ) / pow( f2ndDerivative.x + f2ndDerivative.y, 3.0 ) ) ); } vecCurvature[i] = curvature2D; posOlder = posOld; posOld = pos; }请参阅here。

所以，根据我对曲率的定义，可以用这种方式实现它：

f(t)

它也适用于非封闭的点列表。对于闭合轮廓，您可能希望更改边界行为（对于第一次迭代）。

<强>更新

衍生物的解释：

连续1维函数f_x(t)的导数是：

$f'(t)=\lim\limits_{n \rightarrow 0}{\frac{f(t+n)-f(t))}{t+n-x}} = \lim\limits_{n \rightarrow 0}{\frac{f(t+n)-f(t))}{n}}$

但我们在一个离散的空间中有两个离散函数f_y(t)和t，其中resources的最小步长为1。

$f_x'(t)=\lim\limits_{n \rightarrow 0}{\frac{f_x(t+n)-f_x(t))}{t+n-x}} = \lim\limits_{n \rightarrow 0}{\frac{f_x(t+n)-f_x(t))}{n}} \approx \frac{f_x(t+1)-f_x(t))}{1}$

二阶导数是一阶导数的导数：

$f_x''(t)=\lim\limits_{n \rightarrow 0}{\frac{f_x'(t+n)-f_x'(t))}{t+n-x}} = \lim\limits_{n \rightarrow 0}{\frac{f_x'(t+n)-f_x'(t))}{n}} \approx \frac{f_x'(t+1)-f_x'(t))}{1}$

使用一阶导数的近似值，得到：

$f_x''(t)=(f_x(t+2)-f_x(t+1)) - (f_x(t+1)-f_x(t)) = f_x(t+2) - 2f_x(t+1) + f_x(t)$

衍生品还有其他近似值，如果你谷歌它，你会发现很多。

如何通过opencv计算提取轮廓的曲率？

2 个答案: