圆柱交贯线可视化
Posted on 06 Jun 2015 Mathematica 添加评论
绘制曲线方程Mathematica非常有一套，这里演示下如何用Mathematica画圆柱的交贯线。
两个圆柱都要出现的话，最简单：
ContourPlot3D[{x^2 + z^2 - 1 == 0, y^2 + x^2 - 1/2 == 0}, {x, -2,
2}, {y, -2, 2}, {z, -2, 2}, ContourStyle -> Opacity[0.5],
Mesh -> None, PlotPoints -> 20, Boxed -> False, ImageSize -> Large,
PlotTheme -> "Web"]
不要一个圆柱的话，建议的做法是使用MeshFunction选项：
ContourPlot3D[x^2 + z^2 - 1 == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
ContourStyle -> Opacity[0.5], PlotPoints -> 20,
MeshFunctions -> Function[{x, y, z}, y^2 + x^2 - 1],
MeshStyle -> {Thick, Red},
Mesh ->0.5, BoundaryStyle -> None, Boxed -> False]
两个都不要的话，很简单：
ContourPlot3D[x^2 + z^2 - 1 == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
ContourStyle -> Opacity[0], PlotPoints -> 20,
MeshFunctions -> Function[{x, y, z}, y^2 + x^2 - 1],
MeshStyle -> {Thick, Red},
Mesh ->0.5, BoundaryStyle -> None, Boxed -> False]
要来个立体的交贯线的话，就改改MeshStyle:
ContourPlot3D[x^2 + z^2 - 1 == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
ContourStyle -> Opacity[0], PlotPoints -> 20,
MeshFunctions -> Function[{x, y, z}, y^2 + x^2 - 1],
MeshStyle -> {Tube[0.05]},
Mesh ->0.5, BoundaryStyle -> None, Boxed -> False
开头那个样子的图像呢，就是这样：
ContourPlot3D[x^2 + z^2 - 1 == 0, {x, -2, 2}, {y, -2, 2}, {z, -2, 2},
ContourStyle -> Opacity[0.5], PlotPoints -> 20,
MeshFunctions -> Function[{x, y, z}, y^2 + x^2 - 1],
MeshStyle -> {Tube[0.03]},
Mesh -> {Range[-1, 0, .2]},
ColorFunction -> CreateColorFunction[Paired, 4],
BoundaryStyle -> None, Boxed -> False]
怎么样，是不是很强大。同样的方法还可以画各种类型的曲面的交贯线。
