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 ??

edited Nov 29 at 8:55

kglr

175k9197402

asked Nov 29 at 8:36

Maarten Mostert

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

add a comment |

up vote
5
down vote

favorite

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 ??

edited Nov 29 at 8:55

kglr

175k9197402

asked Nov 29 at 8:36

Maarten Mostert

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

add a comment |

up vote
5
down vote

favorite

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 ??

edited Nov 29 at 8:55

kglr

175k9197402

asked Nov 29 at 8:36

Maarten Mostert

362

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

edited Nov 29 at 8:55

kglr

175k9197402

asked Nov 29 at 8:36

Maarten Mostert

362

edited Nov 29 at 8:55

kglr

175k9197402

asked Nov 29 at 8:36

Maarten Mostert

362

edited Nov 29 at 8:55

kglr

175k9197402

edited Nov 29 at 8:55

kglr

175k9197402

edited Nov 29 at 8:55

kglr

175k9197402

asked Nov 29 at 8:36

Maarten Mostert

362

asked Nov 29 at 8:36

Maarten Mostert

362

asked Nov 29 at 8:36

Maarten Mostert

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

add a comment |

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

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

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

add a comment |

1 Answer
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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

edited Nov 29 at 8:56

answered Nov 29 at 8:50

kglr

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

add a comment |

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1 Answer
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active

oldest

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1 Answer
1

active

oldest

votes

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

edited Nov 29 at 8:56

answered Nov 29 at 8:50

kglr

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

add a comment |

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

edited Nov 29 at 8:56

answered Nov 29 at 8:50

kglr

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

add a comment |

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

edited Nov 29 at 8:56

answered Nov 29 at 8:50

kglr

175k9197402

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

edited Nov 29 at 8:56

answered Nov 29 at 8:50

kglr

175k9197402

edited Nov 29 at 8:56

answered Nov 29 at 8:50

kglr

175k9197402

answered Nov 29 at 8:50

kglr

175k9197402

answered Nov 29 at 8:50

kglr

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

add a comment |

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

add a comment |

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