Is it possible to create such kind of GIF via Mathematica?
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7$\begingroup$ It is possible... $\endgroup$– AppleNov 30, 2014 at 15:09
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5$\begingroup$ As @Chenminqi answered, it is possible. But before anyone actually show you the ways, it would be better you show the effort you have made. So what have you tried? $\endgroup$– SilviaNov 30, 2014 at 15:13
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$\begingroup$ @Silvia Writing code.. $\endgroup$– AppleNov 30, 2014 at 15:35
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2$\begingroup$ a little bit related: Morphing a “sheet of paper” into a torus $\endgroup$– Kuba ♦Nov 30, 2014 at 18:25
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2$\begingroup$ @LCFactorization I mean you might want to include a description (even better with some code) on what you have tried in solving the problem in Mathematica, so instead of accomplishing your work from scratch, people can see the specific point where you are stuck in, so they might have a better chance giving more specific and effective answers. That would fit more in the spirit of the site, also more polite for people who are reading and trying to answer your questions. :) $\endgroup$– SilviaDec 2, 2014 at 4:35
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2 Answers
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Clear["Global`*"]
f[x_, θ_] =
RotationTransform[θ, {1, 0, 1}, {5 Pi, 0, 5 Pi}][{x,
0, -((10 Pi)/6) Sin[x] + 5 Pi}][[{1, 3}]];
p1 = ParametricPlot[{x, x}, {x, -10 Pi, 10 Pi},
PlotRange -> {{-10 Pi, 10 Pi}, {-10 Pi, 10 Pi}, {-10 Pi, 10 Pi}},
ImageSize -> 300, Axes -> True];
n = 7;
g[a_] := Evaluate[
t^(1/n) (5 a π^2)/(1 + 5 a π) + (1 - t^(1/n)) (
10 a π)/(1 + 5 a π) /.
t -> Rescale[a, {0.002, 2 Pi}, {0, 1}] // Simplify];
t = Pi;
Manipulate[
If[var < Pi + 0.0025,
Show[p1, ParametricPlot[f[x, var], {x, -10 Pi, 10 Pi}]],
Show[p1, PolarPlot[
5 Pi + 1/(var - t) -
5/3 π Sin[(2 Pi)/(2*g[var - t]/10) θ], {θ, -g[
var - t], g[var - t]},
PlotRange -> {{-10 Pi, 10 Pi}, {-10 Pi, 10 Pi}},
ImageSize -> 300] /.
Line[data___] :>
Translate[Line[data], {-(1/(var - t)), 0}]]], {var, 0,
2 Pi + Pi}]
The best way is take suitable discrete points of var artificially, not let var change uniform.
Update 1 A better solution from other people.
Manipulate[
ParametricPlot[{1 - 1/y +
Cos[θ] (2 + 1/y - Sin[(10 θ)/y]),
Sin[θ] (2 + 1/y -
Sin[(10 θ)/y])}, {θ, -π y, π y},
PlotRange -> {{-5, 5}, {-5, 5}}], {y, 0.01, 1}]
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2$\begingroup$ Why is it so slow at the end, when linking the two points to create a closed path? $\endgroup$ Dec 1, 2014 at 18:21
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2$\begingroup$ Related link, blog.csdn.net/stereohomology/article/details/41625867 $\endgroup$– chyanogDec 23, 2014 at 5:54
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Here's a start, the 2nd transformation is tricky for me.
data = Table[{i, 0.1 Sin[100 i] + 0.7, 0}, {i, 0, 1, 0.01}];
gr = Graphics3D[{Thick, Red, Line@data}, Boxed -> False];
Manipulate[Graphics3D[
{If[t < 0.1 Pi, {Dashed, Blue, Line[{{0, 0, 0}, {1, 1, 0}}]}, {}],
Arrow[{{0.5, 0, 0}, {0.5, 1, 0}}],
Arrow[{{0, 0.5, 0}, {1, 0.5, 0}}],
GeometricTransformation[{Thick, Red, Line@data},
RotationTransform[t, {1, 1, 0}]]},
PlotRange -> {{0, 1}, {0, 1}, {-1, 1}}, Boxed -> False,
SphericalRegion -> True, ViewPoint -> Top], {t, 0, 0.99 Pi}]
answered Nov 30, 2014 at 17:15
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