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I have a complicated parametric curve which takes too long to integrate, therefore I want to sample points and interpolate to speed up calculations.

The curve is:

     T2[s_, y_] := c1[s] + 0.1*(j[s]*Sin[y] - v[s]*Cos [y])

where

    y = 5 Pi*s

I start by

    Table[T2[s, y], {s, 0, 250, 0.1}]
    func = Interpolation[%]

which doesn't give any correct point on the curve.

Definitions of functions are:

    a = 5
    g[t_] := {-(a + 3*Cos[6*t])*Sin[7*t], (a + 3*Cos[6*t])*Cos[7*t], 
    3*Sin[6*t]}

    dg[t_] := If[t - 2*Pi <= 0, g'[t], g'[2*Pi]];

    tfn = NDSolveValue[{t'[s] == 1/Norm[dg[t[s]]], t[0] == 0, 
    WhenEvent[t[s] == 2*Pi, "StopIntegration"]}, 
    t, {s, 0, 2*Pi + NIntegrate[Norm[g'[t]], {t, 0, 2*Pi}]}];

    l = NIntegrate[Norm[D[g[t], t]], {t, 0, 2*Pi}]
    c1[s_] := g[tfn[s]]
    j[s_] := Normalize[Cross[D[c1[s], s], D[D[c1[s], s], s]]]
    v[s_] := Normalize[Cross[j[s], Normalize[D[c1[s], s]]]]

I think the problem is with the func part.

Thank you.

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closed as unclear what you're asking by Dr. belisarius, Yves Klett, MarcoB, dr.blochwave, bbgodfrey May 29 '15 at 14:47

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The problem, seems, lies in the definition of the functions j[s] and v[s] . as well as in the fact that c1[s] yields a list of functions, rather than a single one. One can make sure by direct plotting them. After that is fixed, in the definition of the table for the interpolation function try this: Table[{s,T2[s, y]}, {s, 0, 250, 0.1}] . $\endgroup$ – Alexei Boulbitch May 29 '15 at 12:26
  • $\begingroup$ j and v are vector functions, while when I plot c1, it seems OK to me. i.imgur.com/liwvzZI.png $\endgroup$ – thedude May 29 '15 at 12:29
  • $\begingroup$ ParametricPlot3D[Evaluate@T2[s, y], {p, 0, l}, PlotPoints -> 1000] i.imgur.com/5H9EIqn.png $\endgroup$ – thedude May 29 '15 at 13:05
  • $\begingroup$ This is your third or so question about the same problem and I believe your main issue is clarity. Think a little further about the math and the function definitions and try to make them easier to understand (for you and others). Just as an example, you don't need two Normalize[ ]s on the v[s_] definition. $\endgroup$ – Dr. belisarius May 29 '15 at 13:24
  • $\begingroup$ Also, a curve is a one-dimensional variety (mostly) so for example the reason of using two parameters to describe it should be stated $\endgroup$ – Dr. belisarius May 29 '15 at 13:34
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As I already said in the comments to your question I believe the functions aren't clear enough. Anyway perhaps something like this may help:

a = 5
g[t_] := {-(a + 3*Cos[6*t])*Sin[7*t], (a + 3*Cos[6*t])*Cos[7*t], 3*Sin[6*t]}
l = NIntegrate[Norm[g'[t]], {t, 0, 2*Pi}]
tfn = NDSolveValue[{t'[s] == 1/Norm[g'[t[s]]], t[0] == 0}, t, {s, 0, l}];
c1[s_] := g[tfn[s]]
j[s_] := Normalize[Cross[c1'[s], c1''[s]]]
v[s_] := Normalize[Cross[j[s], c1'[s]]]
T2[s_] := c1[s] + 0.1*(j[s]*Sin[5 Pi s] - v[s]*Cos[5 Pi s])
tab = Table[{s, T2[s]}, {s, 0, l, .05}];

fns = Interpolation /@ Transpose[tab /. {t_, {x_, y_, z_}} -> {{t, x}, {t, y}, {t, z}}]

ParametricPlot3D[Through[fns[x]], {x, 0, l}, PlotPoints -> 30 IntegerPart@l]

Mathematica graphics

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