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ContourPlot — How do I color by contour curvature?
Custom contour labels in ContourPlotListContourPlot is blocking my geometryHow to plot the contour of f[x,y]==0 if always f[x,y]>=0Contour coloring and (List)ContourPlot projectionMore stream lines in a ListStreamPlotContourPlot - unequal contour spacingContourPlot color problems3D Stack of Disks with dedicated height plotsHow to color Contours in ContourPlot with custom ColorFunctionChanging the color of a specific curve in ContourPlot
$begingroup$
I'm plotting the stream lines of fluid flow past a cylinder, and I would like the colors to increase with contour curvature (i.e. increase as the velocity of the flow increases. Here's a MWE that seems to color it based on the the y-axis value:
ψ[r_, θ_] := U (r - a^2/r) Sin[θ]
r = Sqrt[x^2 + y^2];
θ = ArcSin[y/r];
stream = ContourPlot[
ψ[r, θ] /. U -> 10, a -> 1,
x, -5,5, y, -5, 5,
Contours -> 10 Table[i, i, -10, 10, 0.025]
];
cyl = Graphics[Disk[0, 0, 1]];
Show[stream, cyl]
plotting color
edited 2 hours ago
m_goldberg
87.7k872198
asked 4 hours ago
dpholmesdpholmes
19610
$endgroup$
add a comment |
$begingroup$
I'm plotting the stream lines of fluid flow past a cylinder, and I would like the colors to increase with contour curvature (i.e. increase as the velocity of the flow increases. Here's a MWE that seems to color it based on the the y-axis value:
ψ[r_, θ_] := U (r - a^2/r) Sin[θ]
r = Sqrt[x^2 + y^2];
θ = ArcSin[y/r];
stream = ContourPlot[
ψ[r, θ] /. U -> 10, a -> 1,
x, -5,5, y, -5, 5,
Contours -> 10 Table[i, i, -10, 10, 0.025]
];
cyl = Graphics[Disk[0, 0, 1]];
Show[stream, cyl]
plotting color
edited 2 hours ago
m_goldberg
87.7k872198
asked 4 hours ago
dpholmesdpholmes
19610
$endgroup$
add a comment |
$begingroup$
I'm plotting the stream lines of fluid flow past a cylinder, and I would like the colors to increase with contour curvature (i.e. increase as the velocity of the flow increases. Here's a MWE that seems to color it based on the the y-axis value:
ψ[r_, θ_] := U (r - a^2/r) Sin[θ]
r = Sqrt[x^2 + y^2];
θ = ArcSin[y/r];
stream = ContourPlot[
ψ[r, θ] /. U -> 10, a -> 1,
x, -5,5, y, -5, 5,
Contours -> 10 Table[i, i, -10, 10, 0.025]
];
cyl = Graphics[Disk[0, 0, 1]];
Show[stream, cyl]
plotting color
edited 2 hours ago
m_goldberg
87.7k872198
asked 4 hours ago
dpholmesdpholmes
19610
$endgroup$
I'm plotting the stream lines of fluid flow past a cylinder, and I would like the colors to increase with contour curvature (i.e. increase as the velocity of the flow increases. Here's a MWE that seems to color it based on the the y-axis value:
ψ[r_, θ_] := U (r - a^2/r) Sin[θ]
r = Sqrt[x^2 + y^2];
θ = ArcSin[y/r];
stream = ContourPlot[
ψ[r, θ] /. U -> 10, a -> 1,
x, -5,5, y, -5, 5,
Contours -> 10 Table[i, i, -10, 10, 0.025]
];
cyl = Graphics[Disk[0, 0, 1]];
Show[stream, cyl]
plotting color
plotting color
edited 2 hours ago
m_goldberg
87.7k872198
asked 4 hours ago
dpholmesdpholmes
19610
edited 2 hours ago
m_goldberg
87.7k872198
asked 4 hours ago
dpholmesdpholmes
19610
edited 2 hours ago
m_goldberg
87.7k872198
edited 2 hours ago
m_goldberg
87.7k872198
edited 2 hours ago
m_goldberg
87.7k872198
87.7k872198
asked 4 hours ago
dpholmesdpholmes
19610
asked 4 hours ago
dpholmesdpholmes
19610
asked 4 hours ago
dpholmesdpholmes
19610
19610
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
f = ψ[r, θ] /. U -> 10, a -> 1;
gradf = D[f, x, y, 1];
Hessf = D[f, x, y, 2];
normal = gradf[[1]]/Sqrt[gradf[[1]].gradf[[1]]];
secondfundamentalform = -PseudoInverse[gradf].Hessf // ComplexExpand // Simplify;
tangent = RotationMatrix[Pi/2].normal // Simplify;
curvaturevector = Simplify[(secondfundamentalform.tangent).tangent];
signedcurvature = curvaturevector.normal;
stream = ContourPlot[
ψ[r, θ] /. U -> 10, a -> 1, x, -5, 5, y, -5, 5,
Contours -> 10 Table[i, i, -10, 10, 0.2],
ContourShading -> None
];
curvatureplot = DensityPlot[signedcurvature, x, -5, 5, y, -5, 5,
ColorFunction -> "DarkRainbow",
PlotPoints -> 50,
PlotRange -> -1, 1 2
];
Show[
curvatureplot,
stream,
cyl
]
The white regions are peaks in the curvature distribution. You may increase PlotRange to make the white regions smaller, however, at the price of less contrast.
answered 3 hours ago
Henrik SchumacherHenrik Schumacher
57.2k577157
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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oldest
votes
active
oldest
votes
$begingroup$
f = ψ[r, θ] /. U -> 10, a -> 1;
gradf = D[f, x, y, 1];
Hessf = D[f, x, y, 2];
normal = gradf[[1]]/Sqrt[gradf[[1]].gradf[[1]]];
secondfundamentalform = -PseudoInverse[gradf].Hessf // ComplexExpand // Simplify;
tangent = RotationMatrix[Pi/2].normal // Simplify;
curvaturevector = Simplify[(secondfundamentalform.tangent).tangent];
signedcurvature = curvaturevector.normal;
stream = ContourPlot[
ψ[r, θ] /. U -> 10, a -> 1, x, -5, 5, y, -5, 5,
Contours -> 10 Table[i, i, -10, 10, 0.2],
ContourShading -> None
];
curvatureplot = DensityPlot[signedcurvature, x, -5, 5, y, -5, 5,
ColorFunction -> "DarkRainbow",
PlotPoints -> 50,
PlotRange -> -1, 1 2
];
Show[
curvatureplot,
stream,
cyl
]
The white regions are peaks in the curvature distribution. You may increase PlotRange to make the white regions smaller, however, at the price of less contrast.
answered 3 hours ago
Henrik SchumacherHenrik Schumacher
57.2k577157
$endgroup$
add a comment |
$begingroup$
f = ψ[r, θ] /. U -> 10, a -> 1;
gradf = D[f, x, y, 1];
Hessf = D[f, x, y, 2];
normal = gradf[[1]]/Sqrt[gradf[[1]].gradf[[1]]];
secondfundamentalform = -PseudoInverse[gradf].Hessf // ComplexExpand // Simplify;
tangent = RotationMatrix[Pi/2].normal // Simplify;
curvaturevector = Simplify[(secondfundamentalform.tangent).tangent];
signedcurvature = curvaturevector.normal;
stream = ContourPlot[
ψ[r, θ] /. U -> 10, a -> 1, x, -5, 5, y, -5, 5,
Contours -> 10 Table[i, i, -10, 10, 0.2],
ContourShading -> None
];
curvatureplot = DensityPlot[signedcurvature, x, -5, 5, y, -5, 5,
ColorFunction -> "DarkRainbow",
PlotPoints -> 50,
PlotRange -> -1, 1 2
];
Show[
curvatureplot,
stream,
cyl
]
The white regions are peaks in the curvature distribution. You may increase PlotRange to make the white regions smaller, however, at the price of less contrast.
answered 3 hours ago
Henrik SchumacherHenrik Schumacher
57.2k577157
$endgroup$
add a comment |
$begingroup$
f = ψ[r, θ] /. U -> 10, a -> 1;
gradf = D[f, x, y, 1];
Hessf = D[f, x, y, 2];
normal = gradf[[1]]/Sqrt[gradf[[1]].gradf[[1]]];
secondfundamentalform = -PseudoInverse[gradf].Hessf // ComplexExpand // Simplify;
tangent = RotationMatrix[Pi/2].normal // Simplify;
curvaturevector = Simplify[(secondfundamentalform.tangent).tangent];
signedcurvature = curvaturevector.normal;
stream = ContourPlot[
ψ[r, θ] /. U -> 10, a -> 1, x, -5, 5, y, -5, 5,
Contours -> 10 Table[i, i, -10, 10, 0.2],
ContourShading -> None
];
curvatureplot = DensityPlot[signedcurvature, x, -5, 5, y, -5, 5,
ColorFunction -> "DarkRainbow",
PlotPoints -> 50,
PlotRange -> -1, 1 2
];
Show[
curvatureplot,
stream,
cyl
]
The white regions are peaks in the curvature distribution. You may increase PlotRange to make the white regions smaller, however, at the price of less contrast.
answered 3 hours ago
Henrik SchumacherHenrik Schumacher
57.2k577157
$endgroup$
f = ψ[r, θ] /. U -> 10, a -> 1;
gradf = D[f, x, y, 1];
Hessf = D[f, x, y, 2];
normal = gradf[[1]]/Sqrt[gradf[[1]].gradf[[1]]];
secondfundamentalform = -PseudoInverse[gradf].Hessf // ComplexExpand // Simplify;
tangent = RotationMatrix[Pi/2].normal // Simplify;
curvaturevector = Simplify[(secondfundamentalform.tangent).tangent];
signedcurvature = curvaturevector.normal;
stream = ContourPlot[
ψ[r, θ] /. U -> 10, a -> 1, x, -5, 5, y, -5, 5,
Contours -> 10 Table[i, i, -10, 10, 0.2],
ContourShading -> None
];
curvatureplot = DensityPlot[signedcurvature, x, -5, 5, y, -5, 5,
ColorFunction -> "DarkRainbow",
PlotPoints -> 50,
PlotRange -> -1, 1 2
];
Show[
curvatureplot,
stream,
cyl
]
The white regions are peaks in the curvature distribution. You may increase PlotRange to make the white regions smaller, however, at the price of less contrast.
answered 3 hours ago
Henrik SchumacherHenrik Schumacher
57.2k577157
edited 3 hours ago
answered 3 hours ago
Henrik SchumacherHenrik Schumacher
57.2k577157
answered 3 hours ago
Henrik SchumacherHenrik Schumacher
57.2k577157
answered 3 hours ago
Henrik SchumacherHenrik Schumacher
57.2k577157
57.2k577157
add a comment |
add a comment |
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