       Re: Uniform arc length basis curve fitting

• To: mathgroup at smc.vnet.net
• Subject: [mg67401] Re: Uniform arc length basis curve fitting
• From: "Narasimham" <mathma18 at hotmail.com>
• Date: Thu, 22 Jun 2006 06:21:14 -0400 (EDT)
• References: <e7581l\$l3b\$1@smc.vnet.net><e7842v\$fm6\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```Parameter change method to arc length is clear.However what functions
f1[t] ,f2[t] describe the above example curve from beginning to end?

Narasimham

dh wrote:
> Hi Narasimham,
> If I understand right, you are asking for the derivative with repsect to
> curve lengthof a curve that is arbitrarily parametrized (not with length).
> Assume f[t]={f1[t],f2[t]} is the parametric curve, then the differential:
> df= {D[f1[t],t],D[f2[t],t]} dt
> and the length:
> ds= Sqrt[D[f1[t],t]^2+D[f2[t],t]^2] dt
> and we get the searched for derivative by:
> df/ds= {D[f1[t],t],D[f2[t],t]} / Sqrt[D[f1[t],t]^2+D[f2[t],t]^2]
>
> Daniel
>
>
> Narasimham wrote:
> > How to find slopes, curvature etc. of cubic splines as a function of
> > arc length? With this example from Help,attempted to find piecewise
> > derivatives, but it cannot be right as the given pts need not be spaced
> > evenly on the arc. TIA.
> >
> > << NumericalMath`SplineFit`
> > pts = {{0,0},{1,2},{-1,3},{0,1},{3,0} };
> > spline = SplineFit[pts, Cubic] ;
> > plspl=ParametricPlot[spline[u], {u, 0, 4}, PlotRange -> All, Compiled
> > -> False];
> > "derivative components"
> > der[x_]:=( spline[x+10^-10]-spline[x-10^-10] ) /( 2 10^-10);
> > dxu[x_]:=der[x].{1,0}; dxv[x_]:=der[x].{0,1};
> > Plot[{dxu[v],dxv[v]},{v,0,4}];
> >  ">>> slopes >>>"
> > Plot[ArcTan[dxu[v],dxv[v]] ,{v,0,4}] ;
> > plder=ParametricPlot[der[v],{v,0,4}] ;
> > der
> > Show[plspl,plder];
> >

```

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