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[2] > > Show[plspl,plder]; > >