An efficient circular arc primitive for Graphics3D

Question

As many people have noted, the 2D graphics primitive Circle doesn't work in a Graphics3D environment. Several solutions to this problem have been proposed, both on this site and on StackOverflow.

They all have the disadvantage that they result in either rather ugly circles or highly inefficient ones because these circles were generated by a polygon with several hundreds of facets, making interactive graphics incredibly slow. Other alternatives involve the use of ParametricPlot which doesn't generate efficient graphics either or yield a primitive that can't be used with GeometricTransformation.

I would like to have a more elegant solution that creates a smooth circular arc in 3D without requiring zillions of coordinates. The resulting arc should be usable in combination with Tube and can be used with GeometricTransformation.

Answer 1

In principle, Non-uniform rational B-splines (NURBS) can be used to represent conic sections. The difficulty is finding the correct set of control points and knot weights. The following function does this.

---

EDIT : Better handling of cases where end angle < start angle

---

ClearAll[splineCircle];
splineCircle[m_List, r_, angles_List: {0, 2 \[Pi]}] :=
 Module[{seg, \[Phi], start, end, pts, w, k},
   {start, end} = Mod[angles // N, 2 \[Pi]];
   If[ end <= start, end += 2 \[Pi]];
   seg = Quotient[end - start // N, \[Pi]/2];
   \[Phi] = Mod[end - start // N, \[Pi]/2];
   If[seg == 4, seg = 3; \[Phi] = \[Pi]/2];
   pts = r RotationMatrix[start ].# & /@ 
     Join[Take[{{1, 0}, {1, 1}, {0, 1}, {-1, 1}, {-1,0}, {-1, -1}, {0, -1}}, 2 seg + 1], 
      RotationMatrix[seg \[Pi]/2 ].# & /@ {{1, Tan[\[Phi]/2]}, {Cos[ \[Phi]], Sin[ \[Phi]]}}];
   If[Length[m] == 2, 
    pts = m + # & /@ pts, 
    pts = m + # & /@ Transpose[Append[Transpose[pts], ConstantArray[0, Length[pts]]]]
   ];
   w = Join[
        Take[{1, 1/Sqrt[2], 1, 1/Sqrt[2], 1, 1/Sqrt[2], 1}, 2 seg + 1], 
        {Cos[\[Phi]/2 ], 1}
       ];
   k = Join[{0, 0, 0}, Riffle[#, #] &@Range[seg + 1], {seg + 1}];
   BSplineCurve[pts, SplineDegree -> 2, SplineKnots -> k, SplineWeights -> w]
 ] /; Length[m] == 2 || Length[m] == 3

This looks rather complex, and it is. However, the output (the only thing that ends up in the final graphics) is clean and simple:

splineCircle[{0, 0}, 1, {0, 3/2 \[Pi]}]

Just a single BSplineCurve with a few control points.

It can be used both in 2D and 3D Graphics (the dimensionality of the center point location is used to select this):

DynamicModule[{sc},
 Manipulate[
  Graphics[
    {FaceForm[], EdgeForm[Black], 
     Rectangle[{-1, -1}, {1, 1}], Circle[], 
      {Thickness[0.02], Blue, 
       sc = splineCircle[m, r, {start Degree, end Degree}]
      }, 
      Green, Line[sc[[1]]], Red, PointSize[0.02], Point[sc[[1]]]
    }
  ],
  {{m, {0, 0}}, {-1, -1}, {1, 1}},
  {{r, 1}, 0.5, 2},
  {{start, 45}, 0, 360},
  {{end, 180}, 0, 360}
  ]
 ] 

Manipulate[
 Graphics3D[{FaceForm[], EdgeForm[Black], 
   Cuboid[{-1, -1, -1}, {1, 1, 1}], Blue, 
   sc = splineCircle[{x, y, z}, r, {start Degree, end Degree}], Green,
    Line[sc[[1]]], Red, PointSize[0.02], Point[sc[[1]]]}, 
  Boxed -> False],
 {{x, 0}, -1, 1},
 {{y, 0}, -1, 1},
 {{z, 0}, -1, 1},
 {{r, 1}, 0.5, 2},
 {{start, 45}, 0, 360},
 {{end, 180}, 0, 360}
 ]

With Tube and various transformation functions:

Graphics3D[
  Table[
   {
    Hue@Random[],
    GeometricTransformation[
     Tube[splineCircle[{0, 0, 0}, RandomReal[{0.5, 4}], 
       RandomReal[{\[Pi]/2, 2 \[Pi]}, 2]], RandomReal[{0.2, 1}]], 
     TranslationTransform[RandomReal[{-10, 10}, 3]].RotationTransform[
       RandomReal[{0, 2 \[Pi]}], {0, 0, 1}].RotationTransform[
       RandomReal[{0, 2 \[Pi]}], {0, 1, 0}]]
    },
   {50}
   ], Boxed -> False
  ]

---

Additional uses

I used this code to make the partial disk with annular hole asked for in this question.

Answer 2

Might as well... what follows is a routine that isn't as general as the routine Sjoerd gave, but gives simpler results in some cases. This is based on work by Piegl and Tiller (see their nice book as well).

arc[center_?VectorQ, {start_?VectorQ, end_?VectorQ}] := Module[{ang, co, r},
  ang = VectorAngle[start - center, end - center];
  co = Cos[ang/2]; r = EuclideanDistance[center, start];
  BSplineCurve[{start, center + r/co Normalize[(start + end)/2 - center], end}, 
   SplineDegree -> 2, SplineKnots -> {0, 0, 0, 1, 1, 1}, SplineWeights -> {1, co, 1}]]

For example:

{Graphics[arc[{0, 0}, {{1, 1}, {-1, 1}}]],
 Graphics3D[arc[{0, 0, 0}, {{1, 1, 1}, {-1, 1, 1}}]]} // GraphicsRow

This routine works as long as the angle determined by the arc lies in the open interval $(0,\pi)$ (an inherent limitation of the simple method), and that EuclideanDistance[center, start] == EuclideanDistance[center, end] (otherwise, it draws an elliptical arc). For reflex angles (that is, angles in the interval $(\pi,2\pi)$), you will have to stitch together two of these arc[]s properly.

(A little note: though Piegl and Tiller show in their work that one can use negative weights to generate an arc corresponding to a reflex angle, BSplineCurve[] handles negative weights poorly:

Graphics[BSplineCurve[{{-1/Sqrt[2], 1/Sqrt[2]}, {0, Sqrt[2]}, {1/Sqrt[2], 1/Sqrt[2]}},
 SplineDegree -> 2, SplineKnots -> {0, 0, 0, 1, 1, 1}, SplineWeights -> {1, -1/Sqrt[2], 1}], 
 PlotRange -> {{-1, 1}, {-1, 1}}]

but maybe it might get fixed in a future version of Mathematica.

)

Finally, here's how to render a unit semicircle with BSplineCurve[] (the generalization to the three-dimensional case is left to the reader):

Graphics[BSplineCurve[{{1, 0}, {1, 1}, {-1, 1}, {-1, 0}}, 
  SplineDegree -> 2, SplineKnots -> {0, 0, 0, 1/2, 1, 1, 1}, 
  SplineWeights -> {1, 1/2, 1/2, 1}]]

Again, see Piegl and Tiller's work if you want to learn more about these things.