curvature of a discrete function

Hello,

I need to compute a curvature of a simple 2D discrete function like this one:

x=1:0.5:20;
y=exp(x);

can anybody help how to do that? thanks

[Matt J]

Your function looks 1D to me.

Answer 1

[Jan]

Your function seems to be a 1D function.

Are you looking for the 2nd derivative? While diff calculates the one-sided differential quotient, gradient uses the two-sided inside the interval:

gradient(gradient(y))

If you mean the curvature as reciprocal radius of the local fitting circle:

dx  = gradient(x);
ddx = gradient(dx);
dy  = gradient(y);
ddy = gradient(dy);

num   = dx .* ddy - ddx .* dy;
denom = dx .* dx + dy .* dy;
denom = sqrt(denom) .^ 3;
curvature = num ./ denom;
curvature(denom < 0) = NaN;

Please test this, because I'm not sure if I remember the formulas correctly.

[Matt J]

CONV might be better then. GRADIENT is kind of slow

conv(x,[.25 0 -.5 0 .25],'same')

[Jan]

Therefore I'm using an efficient C-Mex function: FEX: DGradient, which is 10 to 20 times faster and handles unevenly spaced data more accurate.

[Jan]

x = rand(1, 1e6);
tic; ddx = gradient(gradient(x)); toc
tic; ddx = DGradient(DGradient(x)); toc
tic; ddx = conv(x,[.25 0 -.5 0 .25],'same'); toc

 Elapsed time is 0.251547 seconds.
 Elapsed time is 0.025728 seconds.
 Elapsed time is 0.028208 seconds

Matlab 2009a/64, Core2Duo, Win7

Answer 2

[Roger Stafford]

Let (x1,y1), (x2,y2), and (x3,y3) be three successive points on your curve. 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));

You can consider this as an approximation to the curve's curvature at the middle point of the three points.

[Jan]

+1, I've waited for this answer. See http://www.mathworks.com/matlabcentral/newsreader/view_thread/152405.

[Moreno, M.]

https://uk.mathworks.com/matlabcentral/fileexchange/107929-discrete-curvature-normals-and-evolute

Answer 3

[Matt J]

diff(y,2)./0.5^2