The surface show below is very beautiful; however, I don't know its function either as an implicit function or in parametric form.
Anyone have an idea about it and how to draw it with Mathematica?
$\begingroup$
$\endgroup$
17
-
3$\begingroup$ One way might be to browse on-line the Mathematica GUIDE book for graphics by Trott at google books books.google.com/… If anyone done something like the above, it will be in that book. Many more amazing plots there. $\endgroup$– NasserCommented Nov 24, 2013 at 9:15
-
5$\begingroup$ Do you have any affiliation with this? translate.google.com/… $\endgroup$– Dr. belisariusCommented Nov 24, 2013 at 16:45
-
$\begingroup$ Can you please give an answer to @belisarius. You have had 4 days to respond and you are back here again today. $\endgroup$– Mike HoneychurchCommented Nov 28, 2013 at 1:07
-
$\begingroup$ @belisarius I have no access to this due to the network being blocked $\endgroup$– LCFactorizationCommented Nov 28, 2013 at 1:48
-
$\begingroup$ Strange ... at least @MikeHoneychurch, four others and me aren't suffering any blocking. Someone is playing dirty with your network access. $\endgroup$– Dr. belisariusCommented Nov 28, 2013 at 1:51
|
Show 12 more comments
2 Answers
$\begingroup$
$\endgroup$
12
Consider this:
ParametricPlot3D[
RotationTransform[a, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}],
{a, 0, 2 Pi}, Evaluated -> True]
Now rotate this around a circle, while rotating it at the same time around its' origin:
ParametricPlot3D[
RotationTransform[b, {0, 0, 1}][{6, 0, 0} +
RotationTransform[a + 3 b, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}]],
{a, 0, 2 Pi}, {b, 0, 2 Pi}, PlotPoints -> 40, Evaluated -> True]
EDIT:
A color function, omitting surface mesh, fixing direction of rotation and adding a hint of transparency, like the original:
ParametricPlot3D[
RotationTransform[b, {0, 0, 1}][{6, 0, 0} +
RotationTransform[a - 3 b + Pi, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}]],
{a, 0, 2 Pi}, {b, 0, 2 Pi}, PlotPoints -> 40,
ColorFunction -> (RGBColor[#, 0, 1 - #, 4/5] &[1/2 + {1, -1}.{#1, #2}/2] &),
Mesh -> False, Evaluated -> True]
This might be slightly more intuitive way to write ColorFunction using Blend and Opacity in PlotStyle:
ParametricPlot3D[
RotationTransform[b, {0, 0, 1}][{6, 0, 0} +
RotationTransform[a - 3 b + Pi, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}]],
{a, 0, 2 Pi}, {b, 0, 2 Pi},
PlotPoints -> 40,
PlotStyle -> Opacity[4/5],
ColorFunction -> (Blend[{Red, Blue}, 1/2 + {1, -1}.{#1, #2}/2] &),
Mesh -> False, Evaluated -> True]
-
2
-
2$\begingroup$ Very nice use of
RotationTransform$\endgroup$ Commented Nov 24, 2013 at 13:11 -
3$\begingroup$ @cormullion, I love the imaginary upvote. +Sqrt[2I] :-) $\endgroup$ Commented Nov 24, 2013 at 13:12
-
1$\begingroup$ @SimonWoods Now we just need to figure out how to
Abs[]them for a rep boost :D $\endgroup$– rm -rf ♦Commented Nov 24, 2013 at 15:07 -
1$\begingroup$ Congratulations on an elusive Guru badge. :-) (Elusive to everyone but Leonid, that is.) $\endgroup$ Commented Jul 28, 2014 at 13:02
$\begingroup$
$\endgroup$
3
I'm adding this answer to put on record an answer to the second part the question, "what is the parametric equation?".
The parametric equation is implicit in Kirma's RotationTransform expression. To extract it, one need simply write something like
Clear[a, b]
quoit[a_, b_] :=
Evaluate @ RotationTransform[b, {0, 0, 1}][{6, 0, 0} +
RotationTransform[a - 3 b + Pi, {0, 1, 0}][{0, 0, Sin[3 a] + 5/4}]]
The function defined by the above expression, looks like this
Definition @ quoit
quoit[a_, b_] := { Cos[b] (6 - (5/4 + Sin[3 a]) Sin[a - 3 b]), (6 - (5/4 + Sin[3 a]) Sin[a - 3 b]) Sin[b], -Cos[a - 3 b] (5/4 + Sin[3 a]) }
answered Nov 24, 2013 at 15:11
-
$\begingroup$ thank you very much ! very useful $\endgroup$ Commented Nov 24, 2013 at 15:47
-
$\begingroup$ Indeed; maybe I left this part a bit too much implied. $\endgroup$– kirmaCommented Nov 24, 2013 at 17:23
-
4$\begingroup$ Also the order of wrapping rotation matrix is critical. When the foil rotates around the circle (outmost transform function), the Z axis is fixed. While it spins about its center, the attached Z axis changes direction. Thats why the transform with
{0,1,0}is inside the first transform. $\endgroup$– yshkCommented Nov 25, 2013 at 2:16