BezierCurve is different from BezierFunction

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

174k9196402

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

174k9196402

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

174k9196402

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

174k9196402

asked Nov 29 at 8:36

Maarten Mostert

362

edited Nov 29 at 8:55

kglr

174k9196402

asked Nov 29 at 8:36

Maarten Mostert

362

edited Nov 29 at 8:55

kglr

174k9196402

edited Nov 29 at 8:55

kglr

174k9196402

edited Nov 29 at 8:55

kglr

174k9196402

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
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

174k9196402

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 |

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\$","\$"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "387"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f186949%2fbeziercurve-is-different-from-bezierfunction%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

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

174k9196402

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

174k9196402

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

174k9196402

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

174k9196402

edited Nov 29 at 8:56

answered Nov 29 at 8:50

kglr

174k9196402

answered Nov 29 at 8:50

kglr

174k9196402

answered Nov 29 at 8:50

kglr

174k9196402

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 |

draft saved

draft discarded

Thanks for contributing an answer to Mathematica Stack Exchange!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Vrftjkry