kookoo121 - 1 year ago 293

C++ Question

I did use the

`findcontours()`

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Answer Source

For me curvature is:

where `t`

is the position inside the contour and `x(t)`

resp. `y(t)`

return the related `x`

resp. `y`

value. See here.

So, according to my definition of curvature, one can implement it this way:

```
std::vector< cv::Point2f > 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;
}
```

It works on non-closed pointlists as well. For closed contours, you may would like to change the boundary behavior (for the first iterations).

**UPDATE:**

Explanation for the derivatives:

A derivative for a continuous 1 dimensional function `f(t)`

is:

But we are in a discrete space and have two discrete functions `f_x(t)`

and `f_y(t)`

where the smallest step for `t`

is one.

The second derivative is the derivative of the first derivative:

Using the approximation of the first derivative, it yields to:

There are other approximations for the derivatives, if you google it, you will find a lot.

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