# Drawing a cuboid with rounded corners

I want to draw a cuboid with rounded corners, like this:

RoundingRadius only works with Rectangle or Framed. I have no idea how to draw a cuboid with rounded corners. What are your ideas? Thanks.

• I think should be possible to do this using BSplineSurface and it would be the "ultimate" way, creating a precise, single-piece shape, but frankly I'm too lazy to figure it out as it's far from trivial. – Mr.Wizard Jun 8 '14 at 10:08

Two more options. These allow direct control of the box dimensions and the rounding radius.

Using ContourPlot3D:

signedDistance[p_, p0_, p1_] := Piecewise[
{{-Min[p - p0, p1 - p], And @@ Thread[p0 <= p <= p1]},
{EuclideanDistance[p, MapThread[Min[Max[#1, #2], #3] &, {p, p0, p1}]], True}}]
roundedCuboidPlot[p0 : {x0_, y0_, z0_}, p1 : {x1_, y1_, z1_}, r_, opts : OptionsPattern[]] :=
ContourPlot3D[signedDistance[{x, y, z}, p0 + r, p1 - r] == r,
{x, x0 - r, x1 + r}, {y, y0 - r, y1 + r}, {z, z0 - r, z1 + r}, opts]

roundedCuboidPlot[{0, 0, 0}, {1, 2, 3}, 1/4, BoxRatios -> Automatic, Mesh -> None]


Using Graphics3D primitives:

roundedCuboid[p0 : {x0_, y0_, z0_}, p1 : {x1_, y1_, z1_}, r_] :=
{EdgeForm[None],
Cuboid[p0 + {0, r, r}, p1 - {0, r, r}],
Cuboid[p0 + {r, 0, r}, p1 - {r, 0, r}],
Cuboid[p0 + {r, r, 0}, p1 - {r, r, 0}],
Table[Cylinder[{{x0 + r, y, z}, {x1 - r, y, z}}, r],
{y, {y0 + r, y1 - r}}, {z, {z0 + r, z1 - r}}],
Table[Cylinder[{{x, y0 + r, z}, {x, y1 - r, z}}, r],
{x, {x0 + r, x1 - r}}, {z, {z0 + r, z1 - r}}],
Table[Cylinder[{{x, y, z0 + r}, {x, y, z1 - r}}, r],
{x, {x0 + r, x1 - r}}, {y, {y0 + r, y1 - r}}],
Table[Sphere[{x, y, z}, r],
{x, {x0 + r, x1 - r}}, {y, {y0 + r, y1 - r}}, {z, {z0 + r, z1 - r}}]}

Graphics3D[{roundedCuboid[{0, 0, 0}, {1, 2, 3}, 1/4]}]


• The second way is great!Since I need to add one texture on cuboid, i just replace a Cuboid to Polygon，then I can use Texture.Thx. – Apple Jun 8 '14 at 7:35
• Nice, I was to lazyto write this ;) p.s. you can skip Spheres and replace Cylinders with Tubes. – Kuba Jun 8 '14 at 8:24
ClearAll[roundedCuboidF]
roundedCuboidF[hprof_: 10, vprof_: 10, taper_: 1][box_] :=
ChartElementDataFunction["DoubleProfileCube", "HorizontalProfile" -> hprof,
"VerticalProfile" -> vprof, "TaperRatio" -> taper][box]

Graphics3D[roundedCuboidF[][{{0, 1}, {0, 1}, {0, 1}}], Boxed -> False]


or

ContourPlot3D[Norm[{x, y, z}, 4], {x, -1, 1}, {y, -1, 1}, {z, -1, 1},
Mesh -> None, Boxed -> False, Axes -> False, Lighting -> "Neutral",
ContourStyle ->  Directive[Orange, Opacity[0.8], Specularity[White, 30]]]


• not for keypad use: roundedCuboidF[1, 1, 0.01]. Nice thing to play with and very instructive :) – eldo Jun 7 '14 at 15:41
• Thx!But I can not find ChartElementDataFunction in Reference Centre, so How should I learn this function？ – Apple Jun 7 '14 at 15:42
• @Chenminqi, try ChartElementFunction and examples there and Chart Element Schemes in the Palettes menu. – kglr Jun 7 '14 at 15:44
• @eldo, thank you! -- looks nicer with Directive[Red, Specularity[White, 30]] and Lighting->"Neutral" :) – kglr Jun 7 '14 at 15:47

Rahul's otherwise fine approach has a drawback that can be seen if you include an Opacity[] directive:

Graphics3D[{Opacity[2/3, Pink], roundedCuboid[{0, 0, 0}, {1, 2, 3}, 1/4]},
Boxed -> False]


The "ribs" may or may not be desirable in an application, so I sought an alternative that does not use too many Polygon[]s (as with the solutions based on plotting) and yet looks fine when made translucent.

The following routine is not quite Mr. Wizard's wish in the comments, but it is certainly built from BSplineSurface[] + Polygon[] components:

roundedCuboid[p1_?VectorQ, p2_?VectorQ, r_?NumericQ] :=
Module[{csk, csw, cv, ei, fi, ocp, osk, owt},
cv = Tuples[Transpose[{p1 + r, p2 - r}]];
ocp = {{{1, 0, 0}, {1, 1, 0}, {0, 1, 0}},
{{1, 0, 1}, {1, 1, 1}, {0, 1, 1}},
{{0, 0, 1}, {0, 0, 1}, {0, 0, 1}}};
osk = {{0, 0, 0, 1, 1, 1}, {0, 0, 0, 1, 1, 1}};
owt = {{1, 1/Sqrt[2], 1}, {1/Sqrt[2], 1/2, 1/Sqrt[2]},
{1, 1/Sqrt[2], 1}};
ei = {{{4, 8}, {2, 6}, {1, 5}, {3, 7}},
{{6, 8}, {2, 4}, {1, 3}, {5, 7}},
{{7, 8}, {3, 4}, {1, 2}, {5, 6}}};
csk = {{0, 0, 1, 1}, {0, 0, 0, 1, 1, 1}};
csw = {{1, 1/Sqrt[2], 1}, {1, 1/Sqrt[2], 1}};
fi = {{8, 6, 5, 7}, {8, 7, 3, 4}, {8, 4, 2, 6},
{4, 3, 1, 2}, {2, 1, 5, 6}, {1, 3, 7, 5}};
Flatten[{EdgeForm[], BSplineSurface3DBoxOptions ->
{Method -> {"SplinePoints" -> 35}},
MapIndexed[BSplineSurface[Map[
AffineTransform[{RotationMatrix[π Mod[#2[[1]] - 1, 4]/2,
{0, 0, 1}], #1}],
ocp.DiagonalMatrix[r {1, 1, If[Mod[#2[[1]] - 1, 8] < 4,
1, -1]}], {2}],
SplineDegree -> 2, SplineKnots -> osk, SplineWeights -> owt] &,
cv[[{8, 4, 2, 6, 7, 3, 1, 5}]]],
MapIndexed[Function[{idx, pos},
BSplineSurface[Outer[Plus, cv[[idx]],
Composition[Insert[#, 0, pos[[1]]] &,
RotationTransform[π (pos[[2]] - 1)/2]] /@
(r {{1, 0}, {1, 1}, {0, 1}}), 1], SplineDegree -> {1, 2},
SplineKnots -> csk, SplineWeights -> csw]], ei, {2}],
{fi, Join[#, -#] &[IdentityMatrix[3]]}]]}]]


Using this version instead in the first snippet yields the following picture:

Some more examples:

Graphics3D[{Yellow, roundedCuboid[{0, 0, 0}, {1, 3, 1}, 1/10],
Blue, roundedCuboid[{2, 1, 1}, {4, 2, 3}, 1/4]}, Boxed -> False]


Graphics3D[{{EdgeForm[Gray], Opacity[1/2, Green], Cuboid[{2, 1, 1}, {4, 2, 3}]},
{Pink, roundedCuboid[{2, 1, 1}, {4, 2, 3}, 1/5]}},
Boxed -> False, Lighting -> "Neutral"]


• B-spline surface was my first thought, too, but it would take me too long to figure out how to do it. Nice job. +1 – Michael E2 Mar 4 '16 at 23:19
• Thanks, Michael. I already knew how to build NURBS representations of sphere octants and cylinder quarters; organizing them to form the corner filleting was what took the most work from me. – J. M.'s technical difficulties Mar 5 '16 at 11:48
f = PolyhedronData["Cube", "RegionFunction"][x, y, z];

r = 2; u = 0.6;

RegionPlot3D[f, {x, -r, r}, {y, -r, r}, {z, -r, r}, Mesh -> False,
PlotPoints -> 55, PlotRange -> {{-u, u}, {-u, u}, {-u, u}}]


Weakness of this approach: You have to find the right number for PlotPoints (here 55) by trial and error.

• f is a cube with straight edges covering {{-0.5,0.5},{-0.5,0.5},{-0.5,0.5}}. This volume is completely inside {{-2,2},...} and {{-0.6,0.6},...} so how does this even work? Is it a side effect of RegionPlot3D not having high enough resolution? – C. E. Jun 7 '14 at 17:36
• @C.E. Yes, it is (see last sentence of my answer). – eldo Jun 7 '14 at 18:53

This approach takes advantage of a weakness in the implementation of DiscretizeRegion wherein I give it a perfectly cubic region, and it returns a poor approximation of it,

MeshRegion[
DiscretizeRegion@
ImplicitRegion[{0, 0, 0} <= {x, y, z} <= {1, 1, 1}, {x, y, z}],
`