Re: radius of curvature and discrete data question
- From: "Carolin Theuring" <c.theuring@xxxxxx>
- Date: Fri, 21 Oct 2005 11:51:57 -0400
Hi,
I have currently the same question/problem, I want to calculate the
radius of curvature from x- and y-values and I don't have the toolbox
available, that you are talking about. As I am also really a beginner
in Matlab, I would be so happy about another suggestion. Or can I
just enter the function you describe? I am not really clear what the
c is, what is xc?
Thank you for your help!
Carolin Theuring
Roger Stafford wrote:
>
>
> In article <ef17b3e.0@xxxxxxxxxxxxxxxx>, TGte
<tailgate@xxxxxxxx>
> wrote:
>
>> TGte wrote:
>> >
>> > Hey,
>> >
>> > I have discrete data points (as x and y coordinates) which
> track
>> > the
>> > movement of an object over time. I need to calculate the
radius
>> > of
>> > curvature of the set (over the time of the experiment). Is
it
>> > possible to use the rcurve command from the Mapping
Toolbox for
>> > this?
>> > Is there a better way? I have ~16 coordinates for each
track,
>> > and
>> > need to calculate the radius of curvature for 50 to 100
tracks.
>> >
>> > Thanks in advance.
>>
>> I just wanted to add that the coordinate tracks range from some
>> tracks that are very well defined arcs, to tracks which are
> nearly
>> linear. I just need to calculate a radius of curvature for each
>> track...
>>
>> Thanks again.
> ------------------
> Hi,
>
> You need to decide what criterion you want for a best fit with a
> circular arc. If you want to minimize the mean square distance
> from your
> points to the arc, then probably 'lsqnonlin' of the Optimization
> Toolbox
> would be best for your purposes, with the arc center (xc,yc) and
> radius r
> as your parameters to be optimized. The function used by
> 'lsqnonlin'
> would be:
>
> F(xc,yc,r) = sqrt((x-xc).^2 + (y-yc).^2) - r .
>
> The main problem you face is making a wise choice of the initial
> estimate
> for xc, yc, and r to start 'lsqnonlin' going properly.
>
> (Remove "xyzzy" and ".invalid" to send me email.)
> Roger Stafford
>
.
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