BezierCurve is different from BezierFunction

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I am constructing Naca type profiles with Bezier curves.

controlPoints={{1, 0.}, {0.863924,0.00448168}, {0.78316,-0.019}, {0.444, -0.019}, 
  {0.269064,-0.019}, {0,-0.014478}, {0, 0}, {0, 0.017794}, {0.236028, 0.041}, 
  {0.442,0.041}, {0.616096,0.041}, {0.70006,0.0378152}, {1,0.}};

bezProfile = BezierFunction[controlPoints];

Show[Graphics[{Orange, BezierCurve[controlPoints], Red, 
   Point[controlPoints], Green, Line[controlPoints]},
   Axes -> True], ParametricPlot[bezProfile[t], {t, 0, 1}]]

Result

The BezierFunction gives a very different results over the BezierCurve which is wrong !!

Any explanation ??

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  • 3

    Please do not use the bugs tag when posting questions. See the tag description. If you do suspect a bug, always mention your Mathematica version.
    – Szabolcs
    Nov 29 at 8:40

  • 2

    use BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)]?
    – kglr
    Nov 29 at 8:42

up vote
5
down vote

favorite

1

I am constructing Naca type profiles with Bezier curves.

controlPoints={{1, 0.}, {0.863924,0.00448168}, {0.78316,-0.019}, {0.444, -0.019}, 
  {0.269064,-0.019}, {0,-0.014478}, {0, 0}, {0, 0.017794}, {0.236028, 0.041}, 
  {0.442,0.041}, {0.616096,0.041}, {0.70006,0.0378152}, {1,0.}};

bezProfile = BezierFunction[controlPoints];

Show[Graphics[{Orange, BezierCurve[controlPoints], Red, 
   Point[controlPoints], Green, Line[controlPoints]},
   Axes -> True], ParametricPlot[bezProfile[t], {t, 0, 1}]]

Result

The BezierFunction gives a very different results over the BezierCurve which is wrong !!

Any explanation ??

share|improve this question

  • 3

    Please do not use the bugs tag when posting questions. See the tag description. If you do suspect a bug, always mention your Mathematica version.
    – Szabolcs
    Nov 29 at 8:40

  • 2

    use BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)]?
    – kglr
    Nov 29 at 8:42

up vote
5
down vote

favorite

1

up vote
5
down vote

favorite

1
1

I am constructing Naca type profiles with Bezier curves.

controlPoints={{1, 0.}, {0.863924,0.00448168}, {0.78316,-0.019}, {0.444, -0.019}, 
  {0.269064,-0.019}, {0,-0.014478}, {0, 0}, {0, 0.017794}, {0.236028, 0.041}, 
  {0.442,0.041}, {0.616096,0.041}, {0.70006,0.0378152}, {1,0.}};

bezProfile = BezierFunction[controlPoints];

Show[Graphics[{Orange, BezierCurve[controlPoints], Red, 
   Point[controlPoints], Green, Line[controlPoints]},
   Axes -> True], ParametricPlot[bezProfile[t], {t, 0, 1}]]

Result

The BezierFunction gives a very different results over the BezierCurve which is wrong !!

Any explanation ??

share|improve this question

I am constructing Naca type profiles with Bezier curves.

controlPoints={{1, 0.}, {0.863924,0.00448168}, {0.78316,-0.019}, {0.444, -0.019}, 
  {0.269064,-0.019}, {0,-0.014478}, {0, 0}, {0, 0.017794}, {0.236028, 0.041}, 
  {0.442,0.041}, {0.616096,0.041}, {0.70006,0.0378152}, {1,0.}};

bezProfile = BezierFunction[controlPoints];

Show[Graphics[{Orange, BezierCurve[controlPoints], Red, 
   Point[controlPoints], Green, Line[controlPoints]},
   Axes -> True], ParametricPlot[bezProfile[t], {t, 0, 1}]]

Result

The BezierFunction gives a very different results over the BezierCurve which is wrong !!

Any explanation ??

splines

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edited Nov 29 at 8:55

kglr

175k9197402

175k9197402

asked Nov 29 at 8:36

Maarten Mostert

362

362

  • 3

    Please do not use the bugs tag when posting questions. See the tag description. If you do suspect a bug, always mention your Mathematica version.
    – Szabolcs
    Nov 29 at 8:40

  • 2

    use BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)]?
    – kglr
    Nov 29 at 8:42

  • 3

    Please do not use the bugs tag when posting questions. See the tag description. If you do suspect a bug, always mention your Mathematica version.
    – Szabolcs
    Nov 29 at 8:40

  • 2

    use BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)]?
    – kglr
    Nov 29 at 8:42

3

3

Please do not use the bugs tag when posting questions. See the tag description. If you do suspect a bug, always mention your Mathematica version.
– Szabolcs
Nov 29 at 8:40

Please do not use the bugs tag when posting questions. See the tag description. If you do suspect a bug, always mention your Mathematica version.
– Szabolcs
Nov 29 at 8:40

2

2

use BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)]?
– kglr
Nov 29 at 8:42

use BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)]?
– kglr
Nov 29 at 8:42

1 Answer
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up vote
5
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Use the option SplineDegree -> (Length@controlPoints - 1) with BezierCurve:

Show[Graphics[{Orange, Thick, 
    BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)], 
    Red, Point[controlPoints], Green, Line[controlPoints]}, Axes -> True], 
 ParametricPlot[bezProfile[t], {t, 0, 1}, 
  PlotStyle -> Directive[Thickness[.01], Opacity[.5]]], 
 AspectRatio -> 1/GoldenRatio]

enter image description here

BezierCurve >> Details and Options:

BezierCurve by default represents a composite cubic Bézier curve.

Graphics[{Orange, Thick, BezierCurve[controlPoints], 
  Thickness[.01], Opacity[.3], Blue, 
  BezierCurve[controlPoints, SplineDegree -> 3]},
Axes -> True, AspectRatio -> 1/GoldenRatio]

enter image description here

share|improve this answer

  • Thank you for swift reply. The BezierFunction does not have SplineDegree as an option only the BezierCurve has Why ? The correct curve is the last one you plotted with y=0 for x=0 and x=1. Should I split the curve in upper and lower one or are there other solutions.
    – Maarten Mostert
    Nov 29 at 10:51

  • @MaartenMostert For composite bezier functions there is BSplineFunction
    – Coolwater
    Nov 29 at 11:02

  • @MaartenMostert, can you use BSplineCurve and BSplineFunction instead of BezierCurve and BezierFunction?
    – kglr
    Nov 29 at 11:11

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up vote
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Use the option SplineDegree -> (Length@controlPoints - 1) with BezierCurve:

Show[Graphics[{Orange, Thick, 
    BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)], 
    Red, Point[controlPoints], Green, Line[controlPoints]}, Axes -> True], 
 ParametricPlot[bezProfile[t], {t, 0, 1}, 
  PlotStyle -> Directive[Thickness[.01], Opacity[.5]]], 
 AspectRatio -> 1/GoldenRatio]

enter image description here

BezierCurve >> Details and Options:

BezierCurve by default represents a composite cubic Bézier curve.

Graphics[{Orange, Thick, BezierCurve[controlPoints], 
  Thickness[.01], Opacity[.3], Blue, 
  BezierCurve[controlPoints, SplineDegree -> 3]},
Axes -> True, AspectRatio -> 1/GoldenRatio]

enter image description here

share|improve this answer

  • Thank you for swift reply. The BezierFunction does not have SplineDegree as an option only the BezierCurve has Why ? The correct curve is the last one you plotted with y=0 for x=0 and x=1. Should I split the curve in upper and lower one or are there other solutions.
    – Maarten Mostert
    Nov 29 at 10:51

  • @MaartenMostert For composite bezier functions there is BSplineFunction
    – Coolwater
    Nov 29 at 11:02

  • @MaartenMostert, can you use BSplineCurve and BSplineFunction instead of BezierCurve and BezierFunction?
    – kglr
    Nov 29 at 11:11

up vote
5
down vote

Use the option SplineDegree -> (Length@controlPoints - 1) with BezierCurve:

Show[Graphics[{Orange, Thick, 
    BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)], 
    Red, Point[controlPoints], Green, Line[controlPoints]}, Axes -> True], 
 ParametricPlot[bezProfile[t], {t, 0, 1}, 
  PlotStyle -> Directive[Thickness[.01], Opacity[.5]]], 
 AspectRatio -> 1/GoldenRatio]

enter image description here

BezierCurve >> Details and Options:

BezierCurve by default represents a composite cubic Bézier curve.

Graphics[{Orange, Thick, BezierCurve[controlPoints], 
  Thickness[.01], Opacity[.3], Blue, 
  BezierCurve[controlPoints, SplineDegree -> 3]},
Axes -> True, AspectRatio -> 1/GoldenRatio]

enter image description here

share|improve this answer

  • Thank you for swift reply. The BezierFunction does not have SplineDegree as an option only the BezierCurve has Why ? The correct curve is the last one you plotted with y=0 for x=0 and x=1. Should I split the curve in upper and lower one or are there other solutions.
    – Maarten Mostert
    Nov 29 at 10:51

  • @MaartenMostert For composite bezier functions there is BSplineFunction
    – Coolwater
    Nov 29 at 11:02

  • @MaartenMostert, can you use BSplineCurve and BSplineFunction instead of BezierCurve and BezierFunction?
    – kglr
    Nov 29 at 11:11

up vote
5
down vote

up vote
5
down vote

Use the option SplineDegree -> (Length@controlPoints - 1) with BezierCurve:

Show[Graphics[{Orange, Thick, 
    BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)], 
    Red, Point[controlPoints], Green, Line[controlPoints]}, Axes -> True], 
 ParametricPlot[bezProfile[t], {t, 0, 1}, 
  PlotStyle -> Directive[Thickness[.01], Opacity[.5]]], 
 AspectRatio -> 1/GoldenRatio]

enter image description here

BezierCurve >> Details and Options:

BezierCurve by default represents a composite cubic Bézier curve.

Graphics[{Orange, Thick, BezierCurve[controlPoints], 
  Thickness[.01], Opacity[.3], Blue, 
  BezierCurve[controlPoints, SplineDegree -> 3]},
Axes -> True, AspectRatio -> 1/GoldenRatio]

enter image description here

share|improve this answer

Use the option SplineDegree -> (Length@controlPoints - 1) with BezierCurve:

Show[Graphics[{Orange, Thick, 
    BezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)], 
    Red, Point[controlPoints], Green, Line[controlPoints]}, Axes -> True], 
 ParametricPlot[bezProfile[t], {t, 0, 1}, 
  PlotStyle -> Directive[Thickness[.01], Opacity[.5]]], 
 AspectRatio -> 1/GoldenRatio]

enter image description here

BezierCurve >> Details and Options:

BezierCurve by default represents a composite cubic Bézier curve.

Graphics[{Orange, Thick, BezierCurve[controlPoints], 
  Thickness[.01], Opacity[.3], Blue, 
  BezierCurve[controlPoints, SplineDegree -> 3]},
Axes -> True, AspectRatio -> 1/GoldenRatio]

enter image description here

share|improve this answer

share|improve this answer

share|improve this answer

edited Nov 29 at 8:56

answered Nov 29 at 8:50

kglr

175k9197402

175k9197402

  • Thank you for swift reply. The BezierFunction does not have SplineDegree as an option only the BezierCurve has Why ? The correct curve is the last one you plotted with y=0 for x=0 and x=1. Should I split the curve in upper and lower one or are there other solutions.
    – Maarten Mostert
    Nov 29 at 10:51

  • @MaartenMostert For composite bezier functions there is BSplineFunction
    – Coolwater
    Nov 29 at 11:02

  • @MaartenMostert, can you use BSplineCurve and BSplineFunction instead of BezierCurve and BezierFunction?
    – kglr
    Nov 29 at 11:11

  • Thank you for swift reply. The BezierFunction does not have SplineDegree as an option only the BezierCurve has Why ? The correct curve is the last one you plotted with y=0 for x=0 and x=1. Should I split the curve in upper and lower one or are there other solutions.
    – Maarten Mostert
    Nov 29 at 10:51

  • @MaartenMostert For composite bezier functions there is BSplineFunction
    – Coolwater
    Nov 29 at 11:02

  • @MaartenMostert, can you use BSplineCurve and BSplineFunction instead of BezierCurve and BezierFunction?
    – kglr
    Nov 29 at 11:11

Thank you for swift reply. The BezierFunction does not have SplineDegree as an option only the BezierCurve has Why ? The correct curve is the last one you plotted with y=0 for x=0 and x=1. Should I split the curve in upper and lower one or are there other solutions.
– Maarten Mostert
Nov 29 at 10:51

Thank you for swift reply. The BezierFunction does not have SplineDegree as an option only the BezierCurve has Why ? The correct curve is the last one you plotted with y=0 for x=0 and x=1. Should I split the curve in upper and lower one or are there other solutions.
– Maarten Mostert
Nov 29 at 10:51

@MaartenMostert For composite bezier functions there is BSplineFunction
– Coolwater
Nov 29 at 11:02

@MaartenMostert For composite bezier functions there is BSplineFunction
– Coolwater
Nov 29 at 11:02

@MaartenMostert, can you use BSplineCurve and BSplineFunction instead of BezierCurve and BezierFunction?
– kglr
Nov 29 at 11:11

@MaartenMostert, can you use BSplineCurve and BSplineFunction instead of BezierCurve and BezierFunction?
– kglr
Nov 29 at 11:11

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