BezierCurve is different from BezierFunction
5
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}]]
The BezierFunction gives a very different results over the BezierCurve which is wrong !!
Any explanation ??
splines
edited Nov 29 at 8:55
kglr
176k9197403
asked Nov 29 at 8:36
Maarten Mostert
362
add a comment |
5
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}]]
The BezierFunction gives a very different results over the BezierCurve which is wrong !!
Any explanation ??
splines
edited Nov 29 at 8:55
kglr
176k9197403
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
useBezierCurve[controlPoints, SplineDegree -> (Length@controlPoints - 1)]?
– kglr
Nov 29 at 8:42
add a comment |
5
5
5
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}]]
The BezierFunction gives a very different results over the BezierCurve which is wrong !!
Any explanation ??
splines
edited Nov 29 at 8:55
kglr
176k9197403
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}]]
The BezierFunction gives a very different results over the BezierCurve which is wrong !!
Any explanation ??
splines
splines
edited Nov 29 at 8:55
kglr
176k9197403
asked Nov 29 at 8:36
Maarten Mostert
362
edited Nov 29 at 8:55
kglr
176k9197403
asked Nov 29 at 8:36
Maarten Mostert
362
edited Nov 29 at 8:55
kglr
176k9197403
edited Nov 29 at 8:55
kglr
176k9197403
edited Nov 29 at 8:55
kglr
176k9197403
176k9197403
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
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
useBezierCurve[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
useBezierCurve[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
add a comment |
1 Answer
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5
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]
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]
answered Nov 29 at 8:50
kglr
176k9197403
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 isBSplineFunction
– Coolwater
Nov 29 at 11:02
@MaartenMostert, can you useBSplineCurveandBSplineFunctioninstead ofBezierCurveandBezierFunction?
– kglr
Nov 29 at 11:11
add a comment |
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5
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]
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]
answered Nov 29 at 8:50
kglr
176k9197403
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 isBSplineFunction
– Coolwater
Nov 29 at 11:02
@MaartenMostert, can you useBSplineCurveandBSplineFunctioninstead ofBezierCurveandBezierFunction?
– kglr
Nov 29 at 11:11
add a comment |
5
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]
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]
answered Nov 29 at 8:50
kglr
176k9197403
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 isBSplineFunction
– Coolwater
Nov 29 at 11:02
@MaartenMostert, can you useBSplineCurveandBSplineFunctioninstead ofBezierCurveandBezierFunction?
– kglr
Nov 29 at 11:11
add a comment |
5
5
5
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]
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]
answered Nov 29 at 8:50
kglr
176k9197403
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]
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]
answered Nov 29 at 8:50
kglr
176k9197403
edited Nov 29 at 8:56
answered Nov 29 at 8:50
kglr
176k9197403
answered Nov 29 at 8:50
kglr
176k9197403
answered Nov 29 at 8:50
kglr
176k9197403
176k9197403
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 isBSplineFunction
– Coolwater
Nov 29 at 11:02
@MaartenMostert, can you useBSplineCurveandBSplineFunctioninstead ofBezierCurveandBezierFunction?
– 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 isBSplineFunction
– Coolwater
Nov 29 at 11:02
@MaartenMostert, can you useBSplineCurveandBSplineFunctioninstead ofBezierCurveandBezierFunction?
– 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
add a comment |
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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