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I could not re-parametrize the curve

r[s_] := {-(5 + 2*Cos[2*s])*Sin[3*s], (5 + 2*Cos[2*s])*Cos[3*s], 
2*Sin[2*s]}

neither by hand nor with Mathematica. Is there any method else to parametrize the curve with uniform velocity (edit: not velocity, speed) without taking the integral with respect to parameter s?

Thank you.


UPDATE2:

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.49.3151&rep=rep1&type=pdf

http://ieeexplore.ieee.org/xpl/login.jsp?tp=&arnumber=4127017&url=http%3A%2F%2Fieeexplore.ieee.org%2Fxpls%2Fabs_all.jsp%3Farnumber%3D4127017

http://algorithmist.net/docs/arcparam.pdf

I have not read them all thoroughly but I think this is the solution. I would like to ask a question on Mathematica SE for coding these but I cannot due to the rules of the website. Does anyone know about these methods, how and which one to use?

Thanks.


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    $\begingroup$ You shouldn't be surprised. This is hardly ever possible. Curves parameterized by arc length rarely exist outside of textbooks $\endgroup$ – bubba Apr 15 '15 at 11:54
  • $\begingroup$ But it has to be possible to parametrize it by something similar, if not arclength. How can it not be? $\endgroup$ – user230541 Apr 15 '15 at 11:57
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    $\begingroup$ To compute the arc length, you find yourself integrating $\sqrt{180 \cos (2 t)+18 \cos (4 t)+259}$ which, as far as I know, as no elementary anti-derivative. The integral can easily be computed numerically, however, and the resulting arc length function can also be inverted numerically. As a result, there is an effective numerical procedure to compute the arc length parametrization. $\endgroup$ – Mark McClure Apr 15 '15 at 12:06
  • $\begingroup$ Mathematica has computed the numerical integral (from 0 to 2*Pi) in 5 minutes and got 97.8. I actually need to solve s for arclength in order to replace the parameter but not sure if software can. $\endgroup$ – user230541 Apr 15 '15 at 12:15
  • $\begingroup$ It has just calculated but it is very long, now I will try to solve. Hang on... $\endgroup$ – user230541 Apr 15 '15 at 12:16
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Your notation is a bit risky, as $s$ usually denotes the curvilinear abscissa. Here I am using $a$ for the latter.

I recommend recasting the problem as a differential equation.

You start from the expression of the curvilinear abscissa $a$ $$\frac{da}{ds}=\sqrt{\left(\frac{dx}{ds}\right)^2+\left(\frac{dy}{ds}\right)^2+\left(\frac{dz}{ds}\right)^2}=\sqrt{x'^2+y'^2+z'^2}$$ and solve for the parameter $s$ as a function of $a$: $$\frac{ds}{da}=\frac1{\sqrt{x'^2+y'^2+z'^2}}.$$ You can integrate with Runge-Kutta.


Alternatively, you can sample points on the curve and approximate by a sequence of line segments or, much better, circular arcs (taking points in triples).

Rectification of the approximating curve is straightforward.

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  • $\begingroup$ Because of my limited knowledge in mathematics, the second option attracted me more. So, I give values to s, get the sample points, group them in threes, define new arcs and add them together to construct an approximation curve. Then I just reparametrize. $\endgroup$ – user230541 Apr 15 '15 at 14:50
  • $\begingroup$ Working with line segments is very easy, but not so accurate. With arcs, the consecutive triples will share a point. The "hard" part is to find the circle equation. Reparameterization is easy as the arc lengths is proportional to the angle. Don't be afraid by the first method, Runge-Kutta isn't so difficult, look it up on Wikipedia. $\endgroup$ – Yves Daoust Apr 15 '15 at 14:55
  • $\begingroup$ I don't want to risk it, because the method -as you tell- implies the original curve is actually possible to reparametrize without approximation if I understand correctly, whereas this would remove the need of all (?) approximations. Or this curve is a special one, that can be parametrized using Runge-Kutta, but I want to try it later. $\endgroup$ – user230541 Apr 15 '15 at 15:14
  • $\begingroup$ You can reparameterize the approximation curve exactly, but not the original curve. If you insist that the reparameterization be exact, then replace the original by the approximation, keeping in mind that its curvature is discontinuous (piecewise constant). $\endgroup$ – Yves Daoust Apr 15 '15 at 15:17
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The given lines are center-lines of a standard MoebiusBand.So it would appear it is at all possible to re-parametrize the Band with respect to toroidal coordinates with some transformations to find arc length.

MoebiusBandParametrization

r[s_,v_]:={-(5+2*Cos[2*s])*Sin[3*s+v],(5+2*Cos[2*s])*Cos[3*s+v],2*Sin[2*s]}
ParametricPlot3D[r[s,t],{s,0,2Pi},{t,0,Pi/3},Mesh->{40,3},PlotStyle->{Yellow},Boxed->False,Axes->None]
r[s_,v_]:={-(5+2*Cos[2*s])*Sin[3*s+v],(5+2*Cos[2*s])*Cos[3*s+v],2*Sin[2*s]}
ParametricPlot3D[r[s,t],{s,0,2Pi},{t,0,Pi 2},Mesh->{40,6},PlotStyle->{Yellow},Boxed->False,Axes->None]
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  • $\begingroup$ The question asks for a re-parametrization yielding the same path but with constant speed. Does your answer yield a technique to find that re-parametrization? $\endgroup$ – Mark McClure Apr 15 '15 at 14:04
  • $\begingroup$ Thanks. I tried to indicate more of what it is rather than how, to be on more familiar grounds.Constant speed on Moebius Band is doable. Actually wanted to put in comments only, but size and image prompted into answer space. But still thinking of a way out. $\endgroup$ – Narasimham Apr 15 '15 at 14:14
  • $\begingroup$ Right now looking at the possibility only as at least topologically Euler characteristic is same =0, for MB and torus. Also you need a line, not a surface. $\endgroup$ – Narasimham Apr 15 '15 at 14:36

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