电脑
Mathematica（8.0以上版本）
莫比乌斯带的参数方程是：x=cost(3+r*cos(t/2))，y=sint(3+r*cos(t/2))，z=r*sin(t/2)。其中，r代表的是莫比乌斯带的宽度。 用Mathematica画出图形，可以使用ParametricPlot3D：ParametricPlot3D[{Cos[t] (3 + r Cos[t/2]), Sin[t] (3 + r Cos[t/2]), r Sin[t/2]}, {r, -1, 1}, {t, 0, 2 Pi}
把坐标轴、外框、网格线都去掉，代码如下：ParametricPlot3D[{Cos[t] (3 + r Cos[t/2]), Sin[t] (3 + r Cos[t/2]), r Sin[t/2]}, {r, -1, 1}, {t, 0, 2 Pi}, Axes -> False, Mesh -> None, Boxed -> False
我一直好奇，如果莫比乌斯带很宽，当年Mobius还有可能扭出这个单侧曲面吗？让我们来看看r很大的时候，莫比乌斯带长什么模样！ParametricPlot3D[{Cos[t] (3 + r Cos[t/2]), Sin[t] (3 + r Cos[t/2]), r Sin[t/2]}, {r, -2,2}, {t, 0, 2 Pi}, Axes -> False, Mesh -> None, Boxed -> False 和ParametricPlot3D[{Cos[t] (3 + r Cos[t/2]), Sin[t] (3 + r Cos[t/2]), r Sin[t/2]}, {r, -6,6}, {t, 0, 2 Pi}, Axes -> False, Mesh -> None, Boxed -> False
把上面的过程，变成互动模型：Clear[a];Manipulate[ ParametricPlot3D[{Cos[t] (3 + r Cos[t/2]), Sin[t] (3 + r Cos[t/2]), r Sin[t/2]}, {r, -a, a}, {t, 0, 2 Pi}, Axes -> False, Mesh -> None, Boxed -> False], {a, 1, 30} 或者导出为动态图：Export['C:\\Users\\Administrator\\Desktop\\a.gif', Table[ParametricPlot3D[{Cos[t] (3 + r Cos[t/2]), Sin[t] (3 + r Cos[t/2]), r Sin[t/2]}, {r, -a, a}, {t, 0, 2 Pi}, Axes -> False, Mesh -> None, Boxed -> False, ], {a, 1, 30}]
用ColorFunction改变颜色：Manipulate[ ParametricPlot3D[{Cos[t] (3 + r Cos[t/2]), Sin[t] (3 + r Cos[t/2]), r Sin[t/2]}, {r, -a, a}, {t, 0, 2 Pi}, Axes -> False, Mesh -> None, Boxed -> False, ColorFunction -> Function[{t, r}, Hue[t]]], {a, 1, 30}]
把第五步的互动效果输出为动态图，代码如下：Export['C:\\Users\\Administrator\\Desktop\\a.gif', Table[ParametricPlot3D[{Cos[t] (3 + r Cos[t/2]), Sin[t] (3 + r Cos[t/2]), r Sin[t/2]}, {r, -a, a}, {t, 0, 2 Pi}, Axes -> False, Mesh -> None, Boxed -> False, ColorFunction -> Function[{t, r}, Hue[t]]], {a, 1, 30}]]
用旋转的杆来扭出一个莫比乌斯带，并生成互动效果，代码有点长：Manipulate[ c1 = {Yellow, Cylinder[{{-1, 0, 0}, {1, 0, 0}}, diac]}; s1 = Sphere[{-1, 0, 0}, dias]; s2 = {RGBColor[0.5, 1, 0.3], Sphere[{1, 0, 0}, dias]}; m1 = Table[ Rotate[Translate[ Rotate[{c1, s1, s2}, i fr Pi/n, {0, 1, 0}], {2, 0, 0}], i 2 Pi/n, {0, 0, 1}], {i, n}]; tor = {RGBColor[0.3, 0.5, 1], ParametricPlot3D[{(2 + diat Cos[u]) Cos[v], (2 + diat Cos[u]) Sin[ v], diat Sin[u]}, {u, 0, 2 Pi}, {v, 0, 2 Pi}, Mesh -> None, MaxRecursion -> 1][[1]]}; Graphics3D[{m1, tor}, ViewAngle -> \[Pi]/15, SphericalRegion -> True, Boxed -> False, ViewPoint -> {0, 0, 5}, ImageSize -> {380, 380}], {{n, 50, '杆的数量'}, 2, 80, 1}, {{fr, 1, '扭几下'}, 1, 8, 1, RadioButton}, {{dias, 0.17, '两端大小'}, 0.02, 1, ImageSize -> Tiny, ControlPlacement -> Left}, {{diac, 0.1, '粗细'}, 0.02, 1, ImageSize -> Tiny, ControlPlacement -> Left}, {{diat, 0.1, '中间连线大小'}, 0.02, 1, ImageSize -> Tiny, ControlPlacement -> Left}, TrackedSymbols -> Manipulate
莫比乌斯带能够膨胀成为一个“轮胎”，当然前提是，莫比乌斯带本身没有自相交：makeShape[vl_List, c1_Integer, c2_Integer] := Block[{l = vl, l1 = RotateLeft /@ vl, mesh}, mesh = {l, l1, RotateLeft[l1], RotateLeft[l]}; If[c1 == 1, mesh = Map[Drop[#, -1] &, mesh, {1}] ]; If[c2 == 1, mesh = Map[Drop[#, -1] &, mesh, {2}] ]; Polygon /@ Transpose[ Map[Flatten[#, 1] &, mesh] ] (*]*) ] /; TensorRank[vl] >= 2;Manipulate[ Graphics3D[ makeShape[ Table[{Cos[u], Sin[u], 0} + 0.5 Cos[v] { Cos[u/2] Cos[u], Cos[u/2] Sin[u], Sin[u/2]} + .5 a Sin[v] { -Sin[u/2] Cos[u], -Sin[u/2] Sin[u], Cos[u/2]}, {u, 0., 2 Pi, 2 Pi/30}, {v, 0, 2 Pi, 2 Pi/18}], 1, 1], Boxed -> False, ImageSize -> {550, 400}, SphericalRegion -> True] , {{a, 0.1, 'transform'}, 0.0, 1.0, 0.1, Appearance -> 'Labeled'}, SaveDefinitions -> True, ControlPlacement -> Top
这说明，圆环无论怎么扭，都不会变成“克莱因瓶”。最后，画个克莱因瓶看看！Manipulate[ With[{bsc = Take[{{0, 0, 0}, {0, 0, 14}, {0, 0, 20}, {0, 0, 25}, {1.7, 0, 30}, {7, 0, 32}, {10, 0, 31.5}, {13, 0, 30}, {15, 0, 26}, {13, 0, 20}, {10, 0, 17.5}, {4, 0, 13.5}, {2.5, 0, 11}, {0.33, 0, 7}, {0.2, 0, 2.5}, {0, 0, 0}}, t + 2], sizes = Take[{6.5, 14, 4, 2.3, 2.2, 2.2, 2.2, 2.2, 2.2, 2.2, 2.2, 2.6, 3.3, 3.6, 4.3, 6.5}, t + 2]}, Graphics3D[{color, CapForm[None], Opacity[opacity], Tube[BSplineCurve[bsc], sizes]}, Boxed -> False, PlotRange -> {{-15, 17}, {-15, 15}, {0, 35}}, ViewPoint -> {0, -5, 0}, SphericalRegion -> True, ImageSize -> {500, 500}]], {{t, 1, 'times'}, 1, 14, 1}, {{opacity, 0.7}, 0.1, 1}, {{color, Blue}, ColorSlider}
上面的代码，我都删了个别符号，肯定不能正常运行，但是熟悉Mathematica的人，应该很快就能纠正过来。无心交流经验和不想学习Mathematica的人，不必进！
克莱因瓶是四维空间才可能构造出来的图形，所以，我们看到的“克莱因瓶”只是个“投影”！