Evenly spaced points on boundary of polygon

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10
down vote

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I have a polygon and I would like to generate $n$ evenly spaced points along the boundary.

edited Aug 30 at 21:01

David G. Stork

21.3k11646

asked Aug 30 at 20:25

SidTheSloth

512

3

Welcome! If you have already tried, please share your attempt with us, and it will be easier for us to help
â€“Â Johu
Aug 30 at 21:07

add a commentÂ |Â

up vote
10
down vote

favorite

I have a polygon and I would like to generate $n$ evenly spaced points along the boundary.

edited Aug 30 at 21:01

David G. Stork

21.3k11646

asked Aug 30 at 20:25

SidTheSloth

512

3

Welcome! If you have already tried, please share your attempt with us, and it will be easier for us to help
â€“Â Johu
Aug 30 at 21:07

add a commentÂ |Â

up vote
10
down vote

favorite

I have a polygon and I would like to generate $n$ evenly spaced points along the boundary.

edited Aug 30 at 21:01

David G. Stork

21.3k11646

asked Aug 30 at 20:25

SidTheSloth

512

I have a polygon and I would like to generate $n$ evenly spaced points along the boundary.

edited Aug 30 at 21:01

David G. Stork

21.3k11646

asked Aug 30 at 20:25

SidTheSloth

512

edited Aug 30 at 21:01

David G. Stork

21.3k11646

edited Aug 30 at 21:01

David G. Stork

21.3k11646

edited Aug 30 at 21:01

David G. Stork

21.3k11646

asked Aug 30 at 20:25

SidTheSloth

512

asked Aug 30 at 20:25

SidTheSloth

512

asked Aug 30 at 20:25

SidTheSloth

512

3

Welcome! If you have already tried, please share your attempt with us, and it will be easier for us to help
â€“Â Johu
Aug 30 at 21:07

add a commentÂ |Â

3

Welcome! If you have already tried, please share your attempt with us, and it will be easier for us to help
â€“Â Johu
Aug 30 at 21:07

Welcome! If you have already tried, please share your attempt with us, and it will be easier for us to help
â€“Â Johu
Aug 30 at 21:07

add a commentÂ |Â

5 Answers
5

active

oldest

votes

up vote
7
down vote

You can use piecewise linear interpolation:

Let's generate a random polygon:

R = ConvexHullMesh[RandomPoint[Disk, 100]]
polygon = Append[#, #[[1]]] &@MeshCoordinates[R][[MeshCells[R, 2][[1, 1]]]];

enter image description here

Creating a piecewise-linear interpolation which is parameterized by arc length.

t = Prepend[Accumulate[Norm /@ Differences[polygon]], 0.];
ÃŽÂ³ = Interpolation[Transpose[t, polygon], 
 InterpolationOrder -> 1, 
 PeriodicInterpolation -> True
 ];

Sampling it uniformly:

s = Subdivide[ÃŽÂ³[[1, 1, 1]], ÃŽÂ³[[1, 1, 2]], 100];
newpts = ÃŽÂ³[s];

And plotting the generated points:

Show[
 R,
 ListPlot[newpts]
 ]

enter image description here

answered Aug 30 at 21:07

Henrik Schumacher

36.8k249103

add a commentÂ |Â

up vote
6
down vote

If you just want to generate a graphic, then you could use an Arrow with a custom Arrowheads directive. For example, here is a Polygon object:

poly = Polygon[
 7.361093855790543, 0.2777667989114292, 8.37955832193947, 4.837661706721057,
 8.064449639390588, 8.016729782194783, 3.518023168010062, 8.74578450698846, 
 0.8343834304870779, 2.2170380309534554, 4.807772222882807, 0.09815872747936005, 
 7.361093855790543, 0.2777667989114292
];

Graphics[RegionBoundary @ poly]

enter image description here

Show 10 evenly spaced points:

Graphics[
 Arrowheads @ Append[
 ConstantArray[
 .1, Automatic, Graphics[PointSize[Large], Point[0,0]],
 10
 ],
 0
 ],
 Arrow @@ RegionBoundary[poly]
]

enter image description here

answered Aug 30 at 20:52

Carl Woll

56k272146

add a commentÂ |Â

up vote
5
down vote

You can use BSplineFunction and ParametricPlot with ArcLength as option value for the option MeshFunctions:

SeedRandom[77]
pts = RandomReal[-1, 1, 7, 2];
coords = pts[[FindShortestTour[pts][[2]]]];
n = 10;

bsF = BSplineFunction[coords, SplineDegree -> 1];
ParametricPlot[bsF[t], t, 0, 1, Frame -> True, Axes -> False, AspectRatio -> Automatic, 
 MeshFunctions -> ArcLength, 
 Mesh -> n, 
 MeshStyle -> Directive[Red, PointSize[Large]], 
 Epilog -> Red, PointSize[Large] , Point@bsF[0]]

enter image description here

edited Aug 30 at 21:17

answered Aug 30 at 21:12

kglr

159k8183382

add a commentÂ |Â

up vote
4
down vote

We could also use mesh functions, as Vitaliy showed here:

plot = ListLinePlot[
 coords,
 Mesh -> 20,
 MeshFunctions -> "ArcLength"
 ];
pts = Cases[Normal[plot], Point[pt_] :> pt, Infinity];
pts = Prepend[pts, First[coords]];

Graphics[
 Line[coords],
 PointSize[Large], Red, Point[pts]
 ]

Mathematica graphics

I reused the polygon in Carl's answer.

edited Aug 30 at 22:47

answered Aug 30 at 22:29

C. E.

47.6k390192

add a commentÂ |Â

up vote
0
down vote

I think that the broad scheme of doing it would be as follows:

Load the Polytopes package:

<< Polytopes`

Use the vertices command (e.g. assuming that you are working with an octagon):

a=Vertices[Octagon]

Set up a function which interpolates between two points using a parameter t.

Use the Map command to calculate the interpolated co-ordinates for any given value of t on successive values of your vertices.

answered Aug 30 at 20:45

GerardF123

764

add a commentÂ |Â

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

up vote
7
down vote

You can use piecewise linear interpolation:

Let's generate a random polygon:

R = ConvexHullMesh[RandomPoint[Disk, 100]]
polygon = Append[#, #[[1]]] &@MeshCoordinates[R][[MeshCells[R, 2][[1, 1]]]];

enter image description here

Creating a piecewise-linear interpolation which is parameterized by arc length.

t = Prepend[Accumulate[Norm /@ Differences[polygon]], 0.];
ÃŽÂ³ = Interpolation[Transpose[t, polygon], 
 InterpolationOrder -> 1, 
 PeriodicInterpolation -> True
 ];

Sampling it uniformly:

s = Subdivide[ÃŽÂ³[[1, 1, 1]], ÃŽÂ³[[1, 1, 2]], 100];
newpts = ÃŽÂ³[s];

And plotting the generated points:

Show[
 R,
 ListPlot[newpts]
 ]

enter image description here

answered Aug 30 at 21:07

Henrik Schumacher

36.8k249103

add a commentÂ |Â

up vote
7
down vote

You can use piecewise linear interpolation:

Let's generate a random polygon:

R = ConvexHullMesh[RandomPoint[Disk, 100]]
polygon = Append[#, #[[1]]] &@MeshCoordinates[R][[MeshCells[R, 2][[1, 1]]]];

enter image description here

Creating a piecewise-linear interpolation which is parameterized by arc length.

t = Prepend[Accumulate[Norm /@ Differences[polygon]], 0.];
ÃŽÂ³ = Interpolation[Transpose[t, polygon], 
 InterpolationOrder -> 1, 
 PeriodicInterpolation -> True
 ];

Sampling it uniformly:

s = Subdivide[ÃŽÂ³[[1, 1, 1]], ÃŽÂ³[[1, 1, 2]], 100];
newpts = ÃŽÂ³[s];

And plotting the generated points:

Show[
 R,
 ListPlot[newpts]
 ]

enter image description here

answered Aug 30 at 21:07

Henrik Schumacher

36.8k249103

add a commentÂ |Â

up vote
7
down vote

You can use piecewise linear interpolation:

Let's generate a random polygon:

R = ConvexHullMesh[RandomPoint[Disk, 100]]
polygon = Append[#, #[[1]]] &@MeshCoordinates[R][[MeshCells[R, 2][[1, 1]]]];

enter image description here

Creating a piecewise-linear interpolation which is parameterized by arc length.

t = Prepend[Accumulate[Norm /@ Differences[polygon]], 0.];
ÃŽÂ³ = Interpolation[Transpose[t, polygon], 
 InterpolationOrder -> 1, 
 PeriodicInterpolation -> True
 ];

Sampling it uniformly:

s = Subdivide[ÃŽÂ³[[1, 1, 1]], ÃŽÂ³[[1, 1, 2]], 100];
newpts = ÃŽÂ³[s];

And plotting the generated points:

Show[
 R,
 ListPlot[newpts]
 ]

enter image description here

answered Aug 30 at 21:07

Henrik Schumacher

36.8k249103

You can use piecewise linear interpolation:

Let's generate a random polygon:

R = ConvexHullMesh[RandomPoint[Disk, 100]]
polygon = Append[#, #[[1]]] &@MeshCoordinates[R][[MeshCells[R, 2][[1, 1]]]];

enter image description here

Creating a piecewise-linear interpolation which is parameterized by arc length.

t = Prepend[Accumulate[Norm /@ Differences[polygon]], 0.];
ÃŽÂ³ = Interpolation[Transpose[t, polygon], 
 InterpolationOrder -> 1, 
 PeriodicInterpolation -> True
 ];

Sampling it uniformly:

s = Subdivide[ÃŽÂ³[[1, 1, 1]], ÃŽÂ³[[1, 1, 2]], 100];
newpts = ÃŽÂ³[s];

And plotting the generated points:

Show[
 R,
 ListPlot[newpts]
 ]

enter image description here

answered Aug 30 at 21:07

Henrik Schumacher

36.8k249103

answered Aug 30 at 21:07

Henrik Schumacher

36.8k249103

answered Aug 30 at 21:07

Henrik Schumacher

36.8k249103

answered Aug 30 at 21:07

Henrik Schumacher

36.8k249103

add a commentÂ |Â

up vote
6
down vote

If you just want to generate a graphic, then you could use an Arrow with a custom Arrowheads directive. For example, here is a Polygon object:

poly = Polygon[
 7.361093855790543, 0.2777667989114292, 8.37955832193947, 4.837661706721057,
 8.064449639390588, 8.016729782194783, 3.518023168010062, 8.74578450698846, 
 0.8343834304870779, 2.2170380309534554, 4.807772222882807, 0.09815872747936005, 
 7.361093855790543, 0.2777667989114292
];

Graphics[RegionBoundary @ poly]

enter image description here

Show 10 evenly spaced points:

Graphics[
 Arrowheads @ Append[
 ConstantArray[
 .1, Automatic, Graphics[PointSize[Large], Point[0,0]],
 10
 ],
 0
 ],
 Arrow @@ RegionBoundary[poly]
]

enter image description here

answered Aug 30 at 20:52

Carl Woll

56k272146

add a commentÂ |Â

up vote
6
down vote

If you just want to generate a graphic, then you could use an Arrow with a custom Arrowheads directive. For example, here is a Polygon object:

poly = Polygon[
 7.361093855790543, 0.2777667989114292, 8.37955832193947, 4.837661706721057,
 8.064449639390588, 8.016729782194783, 3.518023168010062, 8.74578450698846, 
 0.8343834304870779, 2.2170380309534554, 4.807772222882807, 0.09815872747936005, 
 7.361093855790543, 0.2777667989114292
];

Graphics[RegionBoundary @ poly]

enter image description here

Show 10 evenly spaced points:

Graphics[
 Arrowheads @ Append[
 ConstantArray[
 .1, Automatic, Graphics[PointSize[Large], Point[0,0]],
 10
 ],
 0
 ],
 Arrow @@ RegionBoundary[poly]
]

enter image description here

answered Aug 30 at 20:52

Carl Woll

56k272146

add a commentÂ |Â

up vote
6
down vote

If you just want to generate a graphic, then you could use an Arrow with a custom Arrowheads directive. For example, here is a Polygon object:

poly = Polygon[
 7.361093855790543, 0.2777667989114292, 8.37955832193947, 4.837661706721057,
 8.064449639390588, 8.016729782194783, 3.518023168010062, 8.74578450698846, 
 0.8343834304870779, 2.2170380309534554, 4.807772222882807, 0.09815872747936005, 
 7.361093855790543, 0.2777667989114292
];

Graphics[RegionBoundary @ poly]

enter image description here

Show 10 evenly spaced points:

Graphics[
 Arrowheads @ Append[
 ConstantArray[
 .1, Automatic, Graphics[PointSize[Large], Point[0,0]],
 10
 ],
 0
 ],
 Arrow @@ RegionBoundary[poly]
]

enter image description here

answered Aug 30 at 20:52

Carl Woll

56k272146

If you just want to generate a graphic, then you could use an Arrow with a custom Arrowheads directive. For example, here is a Polygon object:

poly = Polygon[
 7.361093855790543, 0.2777667989114292, 8.37955832193947, 4.837661706721057,
 8.064449639390588, 8.016729782194783, 3.518023168010062, 8.74578450698846, 
 0.8343834304870779, 2.2170380309534554, 4.807772222882807, 0.09815872747936005, 
 7.361093855790543, 0.2777667989114292
];

Graphics[RegionBoundary @ poly]

enter image description here

Show 10 evenly spaced points:

Graphics[
 Arrowheads @ Append[
 ConstantArray[
 .1, Automatic, Graphics[PointSize[Large], Point[0,0]],
 10
 ],
 0
 ],
 Arrow @@ RegionBoundary[poly]
]

enter image description here

answered Aug 30 at 20:52

Carl Woll

56k272146

answered Aug 30 at 20:52

Carl Woll

56k272146

answered Aug 30 at 20:52

Carl Woll

56k272146

answered Aug 30 at 20:52

Carl Woll

56k272146

add a commentÂ |Â

up vote
5
down vote

You can use BSplineFunction and ParametricPlot with ArcLength as option value for the option MeshFunctions:

SeedRandom[77]
pts = RandomReal[-1, 1, 7, 2];
coords = pts[[FindShortestTour[pts][[2]]]];
n = 10;

bsF = BSplineFunction[coords, SplineDegree -> 1];
ParametricPlot[bsF[t], t, 0, 1, Frame -> True, Axes -> False, AspectRatio -> Automatic, 
 MeshFunctions -> ArcLength, 
 Mesh -> n, 
 MeshStyle -> Directive[Red, PointSize[Large]], 
 Epilog -> Red, PointSize[Large] , Point@bsF[0]]

enter image description here

edited Aug 30 at 21:17

answered Aug 30 at 21:12

kglr

159k8183382

add a commentÂ |Â

up vote
5
down vote

You can use BSplineFunction and ParametricPlot with ArcLength as option value for the option MeshFunctions:

SeedRandom[77]
pts = RandomReal[-1, 1, 7, 2];
coords = pts[[FindShortestTour[pts][[2]]]];
n = 10;

bsF = BSplineFunction[coords, SplineDegree -> 1];
ParametricPlot[bsF[t], t, 0, 1, Frame -> True, Axes -> False, AspectRatio -> Automatic, 
 MeshFunctions -> ArcLength, 
 Mesh -> n, 
 MeshStyle -> Directive[Red, PointSize[Large]], 
 Epilog -> Red, PointSize[Large] , Point@bsF[0]]

enter image description here

edited Aug 30 at 21:17

answered Aug 30 at 21:12

kglr

159k8183382

add a commentÂ |Â

up vote
5
down vote

You can use BSplineFunction and ParametricPlot with ArcLength as option value for the option MeshFunctions:

SeedRandom[77]
pts = RandomReal[-1, 1, 7, 2];
coords = pts[[FindShortestTour[pts][[2]]]];
n = 10;

bsF = BSplineFunction[coords, SplineDegree -> 1];
ParametricPlot[bsF[t], t, 0, 1, Frame -> True, Axes -> False, AspectRatio -> Automatic, 
 MeshFunctions -> ArcLength, 
 Mesh -> n, 
 MeshStyle -> Directive[Red, PointSize[Large]], 
 Epilog -> Red, PointSize[Large] , Point@bsF[0]]

enter image description here

edited Aug 30 at 21:17

answered Aug 30 at 21:12

kglr

159k8183382

You can use BSplineFunction and ParametricPlot with ArcLength as option value for the option MeshFunctions:

SeedRandom[77]
pts = RandomReal[-1, 1, 7, 2];
coords = pts[[FindShortestTour[pts][[2]]]];
n = 10;

bsF = BSplineFunction[coords, SplineDegree -> 1];
ParametricPlot[bsF[t], t, 0, 1, Frame -> True, Axes -> False, AspectRatio -> Automatic, 
 MeshFunctions -> ArcLength, 
 Mesh -> n, 
 MeshStyle -> Directive[Red, PointSize[Large]], 
 Epilog -> Red, PointSize[Large] , Point@bsF[0]]

enter image description here

edited Aug 30 at 21:17

answered Aug 30 at 21:12

kglr

159k8183382

edited Aug 30 at 21:17

answered Aug 30 at 21:12

kglr

159k8183382

answered Aug 30 at 21:12

kglr

159k8183382

answered Aug 30 at 21:12

kglr

159k8183382

add a commentÂ |Â

up vote
4
down vote

We could also use mesh functions, as Vitaliy showed here:

plot = ListLinePlot[
 coords,
 Mesh -> 20,
 MeshFunctions -> "ArcLength"
 ];
pts = Cases[Normal[plot], Point[pt_] :> pt, Infinity];
pts = Prepend[pts, First[coords]];

Graphics[
 Line[coords],
 PointSize[Large], Red, Point[pts]
 ]

Mathematica graphics

I reused the polygon in Carl's answer.

edited Aug 30 at 22:47

answered Aug 30 at 22:29

C. E.

47.6k390192

add a commentÂ |Â

up vote
4
down vote

We could also use mesh functions, as Vitaliy showed here:

plot = ListLinePlot[
 coords,
 Mesh -> 20,
 MeshFunctions -> "ArcLength"
 ];
pts = Cases[Normal[plot], Point[pt_] :> pt, Infinity];
pts = Prepend[pts, First[coords]];

Graphics[
 Line[coords],
 PointSize[Large], Red, Point[pts]
 ]

Mathematica graphics

I reused the polygon in Carl's answer.

edited Aug 30 at 22:47

answered Aug 30 at 22:29

C. E.

47.6k390192

add a commentÂ |Â

up vote
4
down vote

We could also use mesh functions, as Vitaliy showed here:

plot = ListLinePlot[
 coords,
 Mesh -> 20,
 MeshFunctions -> "ArcLength"
 ];
pts = Cases[Normal[plot], Point[pt_] :> pt, Infinity];
pts = Prepend[pts, First[coords]];

Graphics[
 Line[coords],
 PointSize[Large], Red, Point[pts]
 ]

Mathematica graphics

I reused the polygon in Carl's answer.

edited Aug 30 at 22:47

answered Aug 30 at 22:29

C. E.

47.6k390192

We could also use mesh functions, as Vitaliy showed here:

plot = ListLinePlot[
 coords,
 Mesh -> 20,
 MeshFunctions -> "ArcLength"
 ];
pts = Cases[Normal[plot], Point[pt_] :> pt, Infinity];
pts = Prepend[pts, First[coords]];

Graphics[
 Line[coords],
 PointSize[Large], Red, Point[pts]
 ]

Mathematica graphics

I reused the polygon in Carl's answer.

edited Aug 30 at 22:47

answered Aug 30 at 22:29

C. E.

47.6k390192

edited Aug 30 at 22:47

answered Aug 30 at 22:29

C. E.

47.6k390192

answered Aug 30 at 22:29

C. E.

47.6k390192

answered Aug 30 at 22:29

C. E.

47.6k390192

add a commentÂ |Â

up vote
0
down vote

I think that the broad scheme of doing it would be as follows:

Load the Polytopes package:

<< Polytopes`

Use the vertices command (e.g. assuming that you are working with an octagon):

a=Vertices[Octagon]

Set up a function which interpolates between two points using a parameter t.

Use the Map command to calculate the interpolated co-ordinates for any given value of t on successive values of your vertices.

answered Aug 30 at 20:45

GerardF123

764

add a commentÂ |Â

up vote
0
down vote

I think that the broad scheme of doing it would be as follows:

Load the Polytopes package:

<< Polytopes`

Use the vertices command (e.g. assuming that you are working with an octagon):

a=Vertices[Octagon]

Set up a function which interpolates between two points using a parameter t.

Use the Map command to calculate the interpolated co-ordinates for any given value of t on successive values of your vertices.

answered Aug 30 at 20:45

GerardF123

764

add a commentÂ |Â

up vote
0
down vote

I think that the broad scheme of doing it would be as follows:

Load the Polytopes package:

<< Polytopes`

Use the vertices command (e.g. assuming that you are working with an octagon):

a=Vertices[Octagon]

Set up a function which interpolates between two points using a parameter t.

Use the Map command to calculate the interpolated co-ordinates for any given value of t on successive values of your vertices.

answered Aug 30 at 20:45

GerardF123

764

I think that the broad scheme of doing it would be as follows:

Load the Polytopes package:

<< Polytopes`

Use the vertices command (e.g. assuming that you are working with an octagon):

a=Vertices[Octagon]

Set up a function which interpolates between two points using a parameter t.

Use the Map command to calculate the interpolated co-ordinates for any given value of t on successive values of your vertices.

answered Aug 30 at 20:45

GerardF123

764

answered Aug 30 at 20:45

GerardF123

764

answered Aug 30 at 20:45

GerardF123

764

answered Aug 30 at 20:45

GerardF123

764

add a commentÂ |Â

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