TITLE: Cutting a Möbius strip down the middle
QUESTION [21 upvotes]: Why does the result of cutting a Möbius strip down the middle lengthwise have two full twists in it? I can account for one full twist--the identification of the top left corner with the bottom right is a half twist; similarly, the top right corner and bottom left identification contributes another half twist. But where does the second full twist come from? 
Explanations with examples or analogies drawn from real life much appreciated.
edit: I'm pasting J.M.'s Mathematica code here (see his answer), modified for version 5.2.
twist[{f_, g_}, a_, b_, u_] := {Cos[u] (a + f Cos[b u] - g Sin[b u]), 
      Sin[u] (a + f Cos[b u] - g Sin[b u]), g Cos[b u] + f Sin[b u]};

With[{a = 3, b = 1/2, f = 1/2},
Block[{$DisplayFunction = Identity}, 
g1 = ParametricPlot3D[Evaluate[Append[twist[{f - v, 0}, a, b, u],
  {EdgeForm[], FaceForm[SurfaceColor[Red], SurfaceColor[Blue]]}]],
  {u, 0, 2 Pi}, {v, 0, 2 f}, Axes -> None, Boxed -> False];
g2 = ParametricPlot3D[Evaluate[Append[twist[{f - v, 0}, a, b, u],
  EdgeForm[]]], {u, 0, 4 Pi}, {v, 0, 2 f/3},
  Axes -> None, Boxed -> False];
g3 = ParametricPlot3D[Evaluate[Append[twist[{f - v, 0}, a, b, u],
  {EdgeForm[], FaceForm[SurfaceColor[Red], SurfaceColor[Blue]]}]],
  {u, 0, 2 Pi}, {v, 2 f/3, 4 f/3}, Axes -> None, Boxed -> False,
 PlotPoints -> 105]];
GraphicsArray[{{g1, Show[g2, g3]}}]];

REPLY [15 votes]: I seem to be terribly late (I'm quite slow as a typist), but hopefully this answer is complementary rather than redundant.

That the Möbius strip has a single "twist" is apparent in the color clash that appears when one attempts to color one "face" red and the opposite face blue. This color clash does not occur in the cut strip, as one requires two twists to finish a circuit through the surface.
Mathematically, consider the following parametrization of the Möbius strip:
$$\begin{align*}x&=\left(r+(2-v)\cos\frac{u}{2}\right)\cos\,u\\y&=\left(r+(2-v)\cos\frac{u}{2}\right)\sin\,u\\z&=(2-v)\sin\frac{u}{2}\end{align*}$$
where $0 \leq v \leq 2$ and $0 \leq u \leq 2\pi$. The normal vector to this surface is
$$\begin{pmatrix} \frac12\left(r\sin\frac{u}{2}-r\sin\frac{3u}{2}-2(v-2)\sin\,u\sin^2\frac{u}{2}\right) \\ \frac12\left((v-2)(\sin^2 u+\cos\,u)-2r\sin\frac{u}{2} \sin\,u\right) \\ \left(r-(v-2)\cos\frac{u}{2}\right)\cos\frac{u}{2} \end{pmatrix}$$
One can verify that the normal vector expressions at $u=0$ and $u=2\pi$ are not equal, but the normal vector expressions at $u=0$ and $u=4\pi$ do agree. This means that if you trace a pencil through the surface of the usual Möbius strip (that is, fix $v$ and vary $u$), after exactly one turn through the surface, your pencil should end up at the spot exactly under your original starting point. For the "cut" Möbius strip (same parametric equation, but $0 \leq v \leq \frac23$ and $0 \leq u \leq 4\pi$), two turns ($2\times2\pi$) will return your pencil to its starting point.

Here is sundry Mathematica code:
twist[{f_, g_}, a_, b_, u_] := {Cos[u] (a + f Cos[b u] - g Sin[b u]), 
  Sin[u] (a + f Cos[b u] - g Sin[b u]), g Cos[b u] + f Sin[b u]}

With[{a = 3, b = 1/2, f = 1/2},
 g1 = ParametricPlot3D[
   twist[{f - v, 0}, a, b, u], {u, 0, 2 Pi}, {v, 0, 2 f}, 
   Axes -> None, Boxed -> False, Mesh -> None, 
   PerformanceGoal -> "Quality", PlotStyle -> FaceForm[Red, Blue]];
 g2 = ParametricPlot3D[
   twist[{f - v, 0}, a, b, u], {u, 0, 4 Pi}, {v, 0, 2 f/3}, 
   Axes -> None, Boxed -> False, Mesh -> None, 
   PerformanceGoal -> "Quality", PlotPoints -> 85, 
   PlotStyle -> FaceForm[Red, Blue]];
 g3 = ParametricPlot3D[
   twist[{f - v, 0}, a, b, u], {u, 0, 2 Pi}, {v, 2 f/3, 4 f/3}, 
   Axes -> None, Boxed -> False, Mesh -> None, 
   PerformanceGoal -> "Quality", PlotPoints -> 85, 
   PlotStyle -> Opacity[1/10, Blue]];
 GraphicsGrid[{{g1, Show[g2, g3]}}]]