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
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採用された回答
Roger Stafford
2012 年 12 月 22 日
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)) ./ ...
sqrt(((x2-x1).^2+(y2-y1).^2)*((x3-x1).^2+(y3-y1).^2)*((x3-x2).^2+(y3-y2).^2));
Roger Stafford
3 件のコメント
Maradona Rodrigues
2019 年 5 月 31 日
Hi I tried using the above code and resulted in some erronous results
my 2d line is list = [0 0; 4 0.5; 8 6; 6 25; 3 7; 1 1] , with the biggest curvature being at the point [6,25]. However i didnt get the second lowest at that point?
Image Analyst
2019 年 5 月 31 日
Maybe there are not enough points to make a determination. If you have a recent version of MATLAB you might also try findchangepts().
その他の回答 (3 件)
Jan
2012 年 12 月 22 日
"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))
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Image Analyst
2012 年 12 月 21 日
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.
6 件のコメント
Alessandro
2014 年 5 月 30 日
Hi Image Analyst... your code works fine for me but i have a bw image instead of a set of (x,y) points... so that, i've not an ordered set of (x,y)... how can i figure out? (my purpose is still find inflection points)
Image Analyst
2014 年 5 月 30 日
You can use bwboundaries() to get a list of (x,y) points.
Roger's Answer is in the FAQ http://matlab.wikia.com/wiki/FAQ#How_do_I_find_.22kinks.22_in_a_curve.3F
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