In the following graphic of a solid torus, how can I show in the plane z = 0 just the outer "equatorial circle", and not the smaller circle inside, without parameterizing that equatorial circle?
With[{R = 2, r = 0.5},
Show[{RegionPlot3D[(R - Sqrt[x^2 + y^2])^2 + z^2 <= r^2, {x, -3,
3}, {y, -3, 3}, {z, -1, 1}, MaxRecursion -> 5, PlotPoints -> 60,
PlotStyle -> Opacity[0.2], Mesh -> None, AxesOrigin -> {0, 0, 0},
Ticks -> None, Boxed -> False, BoxRatios -> {1, 1, 0.5}],
SliceContourPlot3D[(R - Sqrt[x^2 + y^2]) +
r, {"ZStackedPlanes", {0}}, {x, -3, 3}, {y, -3, 3}, {z, -1, 1},
Contours -> 1, ContourShading -> None,
ContourStyle -> Directive[Thick, Red], AxesOrigin -> {0, 0, 0},
Ticks -> None, Boxed -> False]}, ImageSize -> Scaled[0.7]]
]
ResourceFunction["Circle3D"]
in aGraphics3D
$\endgroup$BoundaryStyle -> None, RegionFunction -> Function[{x, y, z, f}, Sqrt[x^2 + y^2] > r]
intoSliceContourPlot3D
. $\endgroup$RegionFunction
is closest in spirit to my original. TheResourceFunction["Circle3D"]
seems awkward in that default has circle in plane perpendicular to x-axis. The solution usingTorus
sort of defeats my purpose in actually using equation of the torus, and its equatorial circle becomes a tube. $\endgroup$plot /. {dir_, l1_Line, l2_Line} :> {dir, l1}
to remove the offending line? $\endgroup$