how to find the sharp turn in a 2d-line (curve)?

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Hi all,
I am trying to find the sharp turn in a 2d-line (curve). Line is constructed with two vectors, X and Y. In following link you can find a sample line with realized point at which there is sharp turn (red solid point).
I appreciate if you could help me out with this.
Thanks, Payam

Accepted Answer

Roger Stafford
Roger Stafford on 22 Dec 2012
One measure of a "sharp turn" is the amount of curvature between three adjacent points on your curve. Let (x1,y1), (x2,y2), and (x3,y3) be three such adjacent points. By a well-known formula, the curvature of a circle drawn through them is simply four times the area of the triangle formed by the three points divided by the product of its three sides. Using the coordinates of the points this is given by:
K = 2*abs((x2-x1).*(y3-y1)-(x3-x1).*(y2-y1)) ./ ...
Roger Stafford
Image Analyst
Image Analyst on 31 May 2019
Maybe there are not enough points to make a determination. If you have a recent version of MATLAB you might also try findchangepts().

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More Answers (3)

Jan on 22 Dec 2012
"Sharp" is relative. There is always a zoom level, which let a curve look smooth.
If you do not have a curve defined by a function, but a piecewise defined line, you are looking for neighboring elements with and included angle above a certain limit. But when such a piece has a length of 1e-200, while the others have a length of 1.0, can this have a "sharp turn"?!
But let's imagine, that you can control this fundamental problem by inventing some meaningful thresholds. Then this determines the angle between two lines:
angle = atan2(norm(cross(N1, N2)), dot(N1, N2))

Image Analyst
Image Analyst on 21 Dec 2012
Well for that example, just do
yAtTurn = min(y);
xAtTurn = find(y == yAtTurn);
If you need something more general, flexible, and robust, then you need to say how other curves might look different than the one example you supplied.
Image Analyst
Image Analyst on 30 May 2014
You can use bwboundaries() to get a list of (x,y) points.

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tafteh on 21 Dec 2012
Hi again, I have updated the image to make it more clearer regarding my problem explained in the first post. Please find the link below:

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