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I have some control points and I could just plot the Bezier curve, but I also need to have the function! How can I get the exact function of the plotted Bezier curve?

Alexey Popkov
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sara
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3 Answers3

12

In the documentation for BezierCurve, under Properties & Relations, it is written that a Bezier curve can be constructed from a sum of Bernstein polynomials. They give the following example:

pts = {{0, -1}, {1, 1}, {2, -1}, {3, 1}};
Graphics[BezierCurve[pts, SplineDegree -> 3], Frame -> True]

Mathematica graphics

f[t_] := Sum[pts[[i + 1]] BernsteinBasis[3, i, t], {i, 0, 3}]
ParametricPlot[f[t], {t, 0, 1}, Frame -> True, Axes -> False]

Mathematica graphics

You can read about this relationship on MathWorld.

C. E.
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2

In fact, you can just write f = BezierFunction[pts]. See BezierFunction.

MarcoB
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Zhou Rui
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  • Please note that BezierCurve is different from BezierFunction, and by default produces different result: (1), (2). – Alexey Popkov Aug 06 '22 at 16:39
  • Get parametric polynomial expression for x and y with PiecewiseExpand g[t_] = f[t] // PiecewiseExpand // Simplify[#, 0 <= t <= 1] & yields {3 t, (-1 + 2 t)^3} . ParametricPlot[g[t], {t, 0, 1}, Frame -> True, Axes -> False, GridLines -> Automatic] Get y as a function of x Solve[Thread[{x, y} == {3 t, (-1 + 2 t)^3}], y, {t}] yields {{y -> 1/27 (-3 + 2 x)^3}} – Akku14 Aug 07 '22 at 04:23
2

I've made a function that can convert any BezierCurve of any degree into a set of curves, each parameterized from 0 to 1. First, convert the BezierCurve points to a new set of points:

compositeBezierPoints[bezierCurvePoints_, degree_Integer] :=
Select[
  Partition[bezierCurvePoints, UpTo[degree + 1], 1][[1 ;; -1 ;; degree]],
  Length[#] > 1 &]

compositeBezierPoints[bezierCurvePoints_, 1] :=
Select[
  Partition[bezierCurvePoints, UpTo[2], 1],
  Length[#] > 1 &]

If you input the points of the BezierCurve, and the degree (Automatic is degree 3), then compositeBezierPoints breaks up the points into a new set of points expected by BezierFunction. You would map BezierFunction over this list to get a list of functions that are each parameterized from 0 to 1. Here's a Manipulate showing the equivalence for a range of degrees:

pts = Table[{i, RandomReal[{-10, 10}]}, {i, 15}];
Manipulate[
  Show[
    Graphics[{Red, Point[pts]}], 
    ParametricPlot[
      Evaluate[Through[(BezierFunction /@ compositeBezier[pts, d])[t]]],
      {t, 0, 1}], 
    Graphics[{Black, Dashed, BezierCurve[pts, SplineDegree -> d]}]],
  {d, 1, 6, 1}]

You can re-parameterize the composite parts into a Piecewise function that itself runs from 0 to 1. I use BezierSymbolicFunction (found somewhere on MSE) to get the arc lengths in order to scale the segments:

BezierSymbolicFunction[pts_?MatrixQ] :=
Function[Evaluate[Sum[pts[[i+1]] BernsteinBasis[Length[pts]-1, i, #], {i, 0, Length[pts]-1}]]]

makeBezierPiecewise[bezierCurvePoints_, degree_Integer] :=
Module[{symbolicSegments, bezierSegments, lengths, ends},
  symbolicSegments = BezierSymbolicFunction /@ compositeBezier[bezierCurvePoints, degree];
  bezierSegments = BezierFunction /@ compositeBezier[bezierCurvePoints, degree];
  lengths = ArcLength[#[t], {t, 0, 1}]& /@ symbolicSegments;
  ends = Prepend[Accumulate[lengths/Total[lengths]], 0.];
  Function[Evaluate[
    Piecewise[
      Table[
        {
          bezierSegments[[i]][(#-ends[[i]])/(ends[[i+1]] - ends[[i]])],
          ends[[i]] < # <= ends[[i+1]]},
        {i, 1, Length[ends]-1}]]]]]
kevind
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