Extruding a weird 2D shape to make a prismatic 3D shape
Michael Seifert's answer is the easiest for curves that can be plotted parametrically, but there is a slightly more general method that can be used to construct an extrusion of any curve that can be plotted in 2D.
First, note that one can always extract the points from a 2D plotted curve, because Mathematica never forgets. For instance, with the curve provided in the question:
R[θ_] := (1 + 0.5 Sin[2 θ]);
shape1 = PolarPlot[R[θ], {θ, 0, 2 π},
Axes -> False,
PlotStyle -> {Orange, Thickness[0.02]}
];
the points are located via
points = (Flatten @ shape1[[1]])[[2, 1]]
Other information about the curve can be found similarly, and using that info and list manipulations, we can use Polygon
s to construct a surface. Here is an extrusion function that does what is necessary:
Options[Extrude] = Join[Options[Graphics3D], {Closed -> True, Capped -> True}];
Extrude[curve_, {zmin_, zmax_}, opts : OptionsPattern[]] :=
Module[{info, points, color, tube, caps},
info = Flatten @ {curve[[1]]};
points = Select[info, Head[#] === Line &][[1, 1]];
If[OptionValue[Closed], points = points ~Join~ {points[[1]]}];
color = Select[info, Head[#] === Directive &];
If[Length[color] == 0, color = Orange, color = First @ Select[color[[1]], ColorQ]];
tube = Polygon[
Partition[
Flatten[
Transpose[points /. {x_, y_} -> {x, y, #} & /@ {zmin, zmax}], 1], 3, 1]
];
If[OptionValue[Closed] && OptionValue[Capped],
caps = Polygon[points /. {x_, y_} -> {x, y, #}] & /@ {zmin, zmax};
tube = Flatten@{tube, caps},
tube = {tube}
];
Graphics3D[
Flatten @ {EdgeForm[None], color, #} & /@ tube,
FilterRules[{opts}, Options[Graphics3D]]
]
];
For the case in hand, we get
Extrude[shape1, {-2, 5}, Boxed -> False]
This is really a lot of work for the same result that Michael's answer gives more easily, but we can use this to close and extrude any plotted 2D curve:
shape2 = Plot[x^2, {x, -2, 2}, Axes -> False]
Extrude[shape2, {-2, 5}, Boxed -> False]
This will not work with Graphics
primitives, as they do not provide a list of points that can be extracted (well, it will work with a Graphics[Line[...]]
). Also, to close a non-closed shape, it simply connects the first and last points, which might not be the behavior always desired. Lastly, note that one can leave the caps off:
shape3 = Graphics[
Line[{{0, 0}, {1, 1}, {2, -1}, {3, 0}, {4, -2}, {5, 1}, {-1, 2}, {0, 0}}]]
Extrude[shape3, {-2, 5}, Capped -> False, Boxed -> False]
You could create parametric plots for the endcaps as well:
tube = ParametricPlot3D[{R6[θ] Cos[θ], R6[θ] Sin[θ], z}, {θ, 0, 2 π}, {z, -2, 5},
Axes -> False, Boxed -> False, Mesh -> None]
endcap1 = ParametricPlot3D[{r R6[θ] Cos[θ], r R6[θ] Sin[θ], 5}, {θ, 0, 2 π}, {r, 0, 1},
Mesh -> False];
endcap2 = ParametricPlot3D[{r R6[θ] Cos[θ], r R6[θ] Sin[θ], -2}, {θ, 0, 2 π}, {r, 0, 1},
Mesh -> False];
Show[tube, endcap1, endcap2]
This answer was written to demonstrate that you can use BSplineSurface[]
to render custom prisms, so that you won't need to carry as many polygons around as the other answers. To wit,
(* Lee's method, http://dx.doi.org/10.1016/0010-4485(89)90003-1 *)
parametrizeCurve[pts_List, a : (_?NumericQ) : 1/2] :=
FoldList[Plus, 0, Normalize[(Norm /@ Differences[pts])^a, Total]] /;
MatrixQ[pts, NumericQ]
shape1 = First[Cases[Normal[PolarPlot[1 + 0.5 Sin[2 θ], {θ, 0, 2 π},
MaxRecursion -> 1, PlotPoints -> 45]],
Line[l_] :> l, ∞]];
tvals = parametrizeCurve[shape1, 1]; (* chord-length parametrization *)
knots = Join[{0, 0}, ArrayPad[tvals, -1], {1, 1}];
{zmin, zmax} = {-1, 2};
prispts = Outer[Append, shape1, {zmin, zmax}, 1];
Graphics3D[{EdgeForm[], Polygon[Transpose[prispts]],
BSplineSurface[prispts,
SplineDegree -> 1, SplineKnots -> {knots, {0, 0, 1, 1}}]},
Boxed -> False]
Of course, you can omit the Polygon[]
caps: