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, ϕ, 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[ϕ/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, ϕ1, ϕ2, 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}];
ϕ1 = ArcTan @@ (p3 - c);
ϕ2 = ArcTan @@ (p1 - c);
c = Append[c, h];
If[OptionValue[arc] // TrueQ,
MapAt[Function[{p}, rot[q1, q2, q3].p] /@ # &, splineCircle[c, r, {ϕ1, ϕ2}], {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, ϕ1, ϕ2, ϕ3},
{c, r} = List @@ Circumsphere[{q1, q2, q3}];
If[OptionValue[arc] // TrueQ,
ϕ1 = ArcTan @@ (q1 - c);
ϕ2 = ArcTan @@ (q2 - c);
ϕ3 = ArcTan @@ (q3 - c);
{ϕ1, ϕ3} = Sort[{ϕ1, ϕ3}];

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

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.