Doug Schwarz <see@sig.for.address.edu> wrote in message <see1B4CC7.13383010112008@news.frontiernet.net>...
> [top posting repaired]
>
> In article <gf9sur$r89$1@fred.mathworks.com>,
> "Dave Brackett" <davebrackett@hotmail.com> wrote:
>
> > Doug Schwarz <see@sig.for.address.edu> wrote in message
> > <seeAE3628.11371810112008@news.frontiernet.net>...
> > > In article <gf9fl8$4kh$1@fred.mathworks.com>,
> > > "Dave Brackett" <davebrackett@hotmail.com> wrote:
> > >
> > > > Hi,
> > > >
> > > > I have plotted a curve which connects up some data points. The curve is
> > > > not a
> > > > straight line.
> > > >
> > > > I would like to compute a new set of data points from the plotted curve,
> > > > but
> > > > equally spaced over the length of the curve. The number of points would
> > > > be
> > > > specified which would control the resolution.
> > > >
> > > > Does anyone know how I could do this? If you need any more information
> > > > please
> > > > ask. Thanks.
> > >
> > > Here's something I plan on contributing to the FEX when I get a chance
> > > to clean it up and document it. It's not especially fast, but if you
> > > already have x and y in functional form it could be faster.
> > >
> > >
> > >  linspacearc.m 
> > > function [x2,y2] = linspacearc(x,y,n)
> > > m = length(x);
> > > t = linspace(0,1,m);
> > > ppx = spline(t,x);
> > > ppy = spline(t,y);
> > >
> > > dppx = pp_deriv(ppx);
> > > dppy = pp_deriv(ppy);
> > > integrand = @(tt) sqrt(ppval(dppx,tt).^2 + ppval(dppy,tt).^2);
> > > arc_length = quadgk(integrand,0,1);
> > > s = linspace(0,arc_length,n);
> > >
> > > inv_arc_len = @(arc,est) fzero(@(u)(quadgk(integrand,0,u))  arc,est);
> > >
> > > t2 = zeros(1,n);
> > > t2(1) = inv_arc_len(s(1),0);
> > > for i = 2:n
> > > t2(i) = inv_arc_len(s(i),t2(i1));
> > > end
> > >
> > > x2 = ppval(ppx,t2);
> > > y2 = ppval(ppy,t2);
> > >
> > >
> > > function dpp = pp_deriv(pp)
> > > % pp_deriv: derivative of piecewise polynomial (pp)
> > >
> > > dpp = pp;
> > > n = pp.order;
> > > dpp.coefs = bsxfun(@times,n1:1:1,pp.coefs(:,1:n1));
> > > dpp.order = n  1;
> > > 
> > >
> > >
> > > Simply pass in your x and y vectors and the number of desired output
> > > points and get back your new x and y vectors.
> >
> >
> > Thanks for your code Doug, however, it doesn't do what I want. Your code
> > computes points over the curve equally spaced along the x axis. What I want
> > is to compute points over the curve equally spaced along the curve.
> >
> > My curve has regions of zero gradient and regions of much steeper gradient. I
> > want the distance between each set of two points to be the same over the
> > length of the curve.
> >
> > Do you know how your code could be edited to enable this? Thanks for your
> > help so far.
>
>
> No, it does exactly what you asked for, but it is possible that your
> specific data tricks it up somehow or makes this not obvious. In
> particular, it is quite easy to be fooled if you are judging this
> visually and you haven't set the axes to have equal scaling in the X and
> Y directions (use "axis equal"). If you can't seem to get it to work
> please post or send me as small an example as possible and I would be
> glad to take a look at it.
>
> 
> Doug Schwarz
> dmschwarz&ieee,org
> Make obvious changes to get real email address.
Ah yes, you are right, sorry about that  i didn't have my axes equally scaled. once I set that the code works fine. thanks a lot for your help.
