Skip to main content

graphics - An efficient circular arc primitive for Graphics3D



As many people have noted, the 2D graphics primitive Circle doesn't work in a Graphics3D environment (even in v10.0-v10.4, where many geometric regions were added). 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 using polygons with several hundreds of edges, 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



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.



UPDATE (2016-05-22): Added a convenience function to draw a circle or circular arc in 3D specified by three points (see bottom of post)


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



ClearAll[splineCircle];

splineCircle[m_List, r_, angles_List: {0, 2 π}] :=
 Module[{seg, Ï•, start, end, pts, w, k},
   {start, end} = Mod[angles // N, 2 π];
   If[ end <= start, end += 2 π];
   seg = Quotient[end - start // N, π/2];
   ϕ = Mod[end - start // N, π/2];
   If[seg == 4, seg = 3; ϕ = π/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 π/2 ].# & /@ {{1, Tan[ϕ/2]}, {Cos[ ϕ], Sin[ ϕ]}}];

   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[Ï•/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 π}]

Mathematica graphics


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}
  ]
 ]

Mathematica graphics


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}
 ]


Mathematica graphics


With Tube and various transformation functions:


Graphics3D[
  Table[
   {
    Hue@Random[],
    GeometricTransformation[
     Tube[splineCircle[{0, 0, 0}, RandomReal[{0.5, 4}], 
       RandomReal[{π/2, 2 π}, 2]], RandomReal[{0.2, 1}]], 

     TranslationTransform[RandomReal[{-10, 10}, 3]].RotationTransform[
       RandomReal[{0, 2 π}], {0, 0, 1}].RotationTransform[
       RandomReal[{0, 2 π}], {0, 1, 0}]]
    },
   {50}
   ], Boxed -> False
  ]

enter image description here




Additional uses


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



Specification of a circle or circular arc using three points


[The use of Circumsphere here was a tip by J.M.. Though it doesn't yield an arc, it can be used to obtain the parameters of an arc]


[UPDATE 2020-02-08: CircleThrough, introduced in v12, can be used instead of Circumsphere as well]


Options[circleFromPoints] = {arc -> False};

circleFromPoints[m : {q1_, q2_, q3_}, OptionsPattern[]] :=
Module[{c, r, Ï•1, Ï•2, p1, p2, p3, h, 

        rot = RotationMatrix[{{0, 0, 1}, Cross[#1 - #2, #3 - #2]}] &},
  {p1, p2, p3} = {q1, q2, q3}.rot[q1, q2, q3];
  h = p1[[3]];
  {p1, p2, p3} = {p1, p2, p3}[[All, ;; 2]];
  {c, r} = List @@ Circumsphere[{p1, p2, p3}];
  Ï•1 = ArcTan @@ (p3 - c);
  Ï•2 = ArcTan @@ (p1 - c);
  c = Append[c, h];
  If[OptionValue[arc] // TrueQ,
    MapAt[Function[{p}, rot[q1, q2, q3].p] /@ # &, splineCircle[c, r, {Ï•1, Ï•2}], {1}],

    MapAt[Function[{p}, rot[q1, q2, q3].p] /@ # &, splineCircle[c, r], {1}]
  ]
] /; MatrixQ[m, NumericQ] && Dimensions[m] == {3, 3}

Example of usage:


{q1, q2, q3} = RandomReal[{-10, 10}, {3, 3}];
Graphics3D[
 {
  Red,
  PointSize[0.02],

  Point[{q1, q2, q3}],
  Black,
  Text["1", q1, {0, -1}],
  Text["2", q2, {0, -1}],
  Text["3", q3, {0, -1}],
  Green,
  Tube@circleFromPoints[{q1, q2, q3}, arc -> True
  }
 ]


enter image description here


Similarly, one can define a 2D version:


 circleFromPoints[m : {q1_List, q2_List, q3_List}, OptionsPattern[]] :=
 Module[{c, r, Ï•1, Ï•2, Ï•3},
   {c, r} = List @@ Circumsphere[{q1, q2, q3}];
   If[OptionValue[arc] // TrueQ,
    Ï•1 = ArcTan @@ (q1 - c);
    Ï•2 = ArcTan @@ (q2 - c);
    Ï•3 = ArcTan @@ (q3 - c);
    {Ï•1, Ï•3} = Sort[{Ï•1, Ï•3}];

    splineCircle[c, r, 
     If[ϕ1 <= ϕ2 <= ϕ3, {ϕ1, ϕ3}, {ϕ3, ϕ1 + 2 π}]],
    splineCircle[c, r]
    ]
   ] /; MatrixQ[m, NumericQ] && Dimensions[m] == {3, 2}

Demo:


Manipulate[
 c = Circumsphere[{q1, q2, q3}][[1]];
 Graphics[

  {
   Black,
   Line[{{q1, c}, {q2, c}, {q3, c}}],
   Point[c],
   Text["1", q1, {0, -1}],
   Text["2", q2, {0, -1}],
   Text["3", q3, {0, -1}],
   Green,
   Thickness[thickness], Arrowheads[10 thickness],
   sp@circleFromPoints[{q1, q2, q3}, arc -> a]

   }, PlotRange -> {{-3, 3}, {-3, 3}}
  ],
 {{q1, {0, 0}}, Locator},
 {{q2, {0, 1}}, Locator},
 {{q3, {1, 0}}, Locator},
 {{a, False, "Draw arc"}, {False, True}},
 {{sp, Identity, "Graphics type"}, {Identity, Arrow}},
 {{thickness, 0.01}, 0, 0.05}
 ]


enter image description here


For versions without Circumsphere (i.e, before v10.0) one could use the following function to get the circle center (c in the code above, r would then be the EuclideanDistance between c and p1):


getCenter[{{p1x_, p1y_}, {p2x_, p2y_}, {p3x_, p3y_}}] := 
   {(1/2)*(p1x + p2x + ((-p1y + p2y)*
           ((p1x - p3x)*(p2x - p3x) + (p1y - p3y)*(p2y - p3y)))/
            (p1y*(p2x - p3x) + p2y*p3x - p2x*p3y + p1x*(-p2y + p3y))), 
    (1/2)*(p1y + p2y + ((p1x - p2x)*
            ((p1x - p3x)*(p2x - p3x) + (p1y - p3y)*(p2y - p3y)))/
            (p1y*(p2x - p3x) + p2y*p3x - p2x*p3y + p1x*(-p2y + p3y)))}

Comments

Post a Comment

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more