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Find length of edges of Moebius strip formed by cutting along length and re-joining from a circular cylinder segment:

$( a \cos \theta, a \sin \theta, z), ( \theta, 0, 2 \pi ),( z,0,b) $ after half rotation.

Since the Moebius strip has negative Gauss curvature its edges must be longer than $ 2 \pi a $.

What is its non-isometric mapping or parametrization that includes $a$ and $b$?

The question is not about its topology but its geometry.

EDIT1:

 Moebius Band from Mathworld.. 
 a = 1;
 ParametricPlot3D[{ (a + v  Cos [u/2])  Cos[u], ( a + v  Cos [u/2])  Sin[u],
 v  Sin [u/2] }, {v, -4, 3}, {u, 0, 2 Pi},
 PlotStyle -> {Yellow, Opacity[0.75]},
 PlotLabel -> Moebius_Band _MathWorld]

EDIT2:

The Band shown below with given parametrization has self intersections!

MoebiusBandMathWorld

Narasimham
  • 40,495

1 Answers1

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(I write $\phi$ instead of $\theta$.)

Imagine the Moebius strip $M$ being produced in the following way: A stick of length $2b$ moves along a circle of radius $a$ in the $(x,y)$-plane, whereby the center of the stick is always on the circle and the stick is vertical at $\phi=0$. At the same time the stick is "screwed" such that it is always in a plane with the $z$-axis and makes a half turn during its trip.

A parametrization implementing this idea is given by

$$M:\quad(\phi,t)\mapsto(a\cos\phi,a\sin\phi,0)+ t\left(\sin{\phi\over2}\cos\phi,\ \sin{\phi\over2}\sin\phi,\ \cos{\phi\over2}\right)$$ with $0\leq\phi\leq 2\pi$, $\>-b\leq t\leq b$.

The length of the boundary curve can now be computed by the standard formula; but I'm afraid that the resulting integral is not elementary.

Christian Blatter
  • 226,825
  • An expression for length is given. Has it been updated? if $ z' $ is included,would it then simplify? http://mathworld.wolfram.com/MoebiusStrip.html – Narasimham Nov 30 '14 at 07:47
  • It's a typo; it should read $\sqrt{x'^2+y'^2+z'^2}$ there. But I have not checked the resulting expression. – Christian Blatter Nov 30 '14 at 08:53
  • Is there some sort of standard parametrization? or from topological view point no standardization is required? Even self-intersection of Mathworld Moebius Band is not disconcerting! – Narasimham Nov 30 '14 at 12:50