## curvature of a discrete function

on 16 Jan 2013

### Jan Simon (view profile)

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

### Matt J (view profile)

on 16 Jan 2013

Your function looks 1D to me.

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### Jan Simon (view profile)

on 16 Jan 2013
Edited by Jan Simon

### Jan Simon (view profile)

on 16 Jan 2013

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

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

Matt J

### Matt J (view profile)

on 16 Jan 2013

CONV might be better then. GRADIENT is kind of slow

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

### Jan Simon (view profile)

on 16 Jan 2013

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 Simon

### Jan Simon (view profile)

on 16 Jan 2013
```x = rand(1, 1e6);
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

### Roger Stafford (view profile)

on 16 Jan 2013

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 Simon

on 16 Jan 2013

on 16 Jan 2013
Edited by Matt J

### Matt J (view profile)

on 16 Jan 2013

```diff(y,2)./0.5^2
```

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