can't blend gradient colors with a stream
$begingroup$
The following function generates a plot of the 3d function indicated in the example.
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> {Texture[
StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, Frame -> None, ImageSize -> Large,
StreamStyle -> Black]]}]
However, when I choose a different ColorFunction parameter the texture (that only consists of arrows) disappears. Any idea how to correct this? I tried to make the background transparent, combine two 3D plots etc without success. Also, I have no idea why this is happening.
Here is the 3D plot without the gradient field.
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> {Texture[
StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, Frame -> None, ImageSize -> Large,
StreamStyle -> Black]]}, ColorFunction -> "Rainbow"]
plotting style textures
New contributor
$endgroup$
add a comment |
$begingroup$
The following function generates a plot of the 3d function indicated in the example.
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> {Texture[
StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, Frame -> None, ImageSize -> Large,
StreamStyle -> Black]]}]
However, when I choose a different ColorFunction parameter the texture (that only consists of arrows) disappears. Any idea how to correct this? I tried to make the background transparent, combine two 3D plots etc without success. Also, I have no idea why this is happening.
Here is the 3D plot without the gradient field.
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> {Texture[
StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, Frame -> None, ImageSize -> Large,
StreamStyle -> Black]]}, ColorFunction -> "Rainbow"]
plotting style textures
New contributor
$endgroup$
add a comment |
$begingroup$
The following function generates a plot of the 3d function indicated in the example.
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> {Texture[
StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, Frame -> None, ImageSize -> Large,
StreamStyle -> Black]]}]
However, when I choose a different ColorFunction parameter the texture (that only consists of arrows) disappears. Any idea how to correct this? I tried to make the background transparent, combine two 3D plots etc without success. Also, I have no idea why this is happening.
Here is the 3D plot without the gradient field.
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> {Texture[
StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, Frame -> None, ImageSize -> Large,
StreamStyle -> Black]]}, ColorFunction -> "Rainbow"]
plotting style textures
New contributor
$endgroup$
The following function generates a plot of the 3d function indicated in the example.
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> {Texture[
StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, Frame -> None, ImageSize -> Large,
StreamStyle -> Black]]}]
However, when I choose a different ColorFunction parameter the texture (that only consists of arrows) disappears. Any idea how to correct this? I tried to make the background transparent, combine two 3D plots etc without success. Also, I have no idea why this is happening.
Here is the 3D plot without the gradient field.
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> {Texture[
StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, Frame -> None, ImageSize -> Large,
StreamStyle -> Black]]}, ColorFunction -> "Rainbow"]
plotting style textures
plotting style textures
New contributor
New contributor
edited 2 hours ago
Mr.Wizard♦
231k294741041
231k294741041
New contributor
asked 2 hours ago
user17164user17164
1212
1212
New contributor
New contributor
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You can use StreamDensityPlot
(which accepts the ColorFunction
option) to produce the texture:
sdp = StreamDensityPlot[Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, StreamStyle -> Black,
ColorFunction -> (ColorData["Rainbow"][(#^2 + #2^2) Exp[1 - #^2 - #2^2]] &),
ColorFunctionScaling -> False, Frame -> False, Axes -> False, PlotRangePadding -> None];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> Texture[sdp], Lighting -> "Neutral"]
$endgroup$
$begingroup$
Slightly shorter:sdp = StreamDensityPlot[ Evaluate[{-D[#, {{x, y}}], #} &[(x^2 + y^2) Exp[ 1 - x^2 - y^2]]], {x, -3, 3}, {y, -3, 3}, StreamStyle -> Black, ColorFunction -> "Rainbow", Frame -> False, Axes -> False, PlotRangePadding -> None];
$endgroup$
– Michael E2
57 mins ago
add a comment |
$begingroup$
The color is not quite right but the idea seems to work. Edit: much closer now.
dp = DensityPlot[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
ColorFunction -> "Rainbow", PlotPoints -> 100];
sp = StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3, 3}, {y, -3, 3},
Frame -> None, ImageSize -> Large, StreamStyle -> Black];
tex = Show[dp, sp, Frame -> None, PlotRangePadding -> 0, ImageSize -> 500];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3}, Mesh -> None,
ImageSize -> Large, PlotPoints -> 35
, PlotStyle -> {Texture[Lighter[tex, 0.15]]}
, Lighting -> "Neutral"
]
$endgroup$
add a comment |
$begingroup$
PlotStyle -> Texture[...]
relies on VertexTextureCoordinates
to map the texture to polygon vertices.
ColorFunction -> colorfunction
relies on VertexColors
to associate colors with the polygon vertices.
Only one of them actually gets to style the polygon. In my case, it seems to be the texture:
Graphics3D[{Texture[RandomImage[1, 100]],
Polygon[{{0, 0, 0}, {1, 0, 0}, {1, 1, 0}},
VertexColors -> {Red, Green, Blue},
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}}]},
Lighting -> "Neutral", BoxRatios -> {1, 1, 1}]
It sounds like the color function is winning in your case. It wouldn't surprise me if that was dependent on things like OS, software version, phase of the moon, etc...
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use StreamDensityPlot
(which accepts the ColorFunction
option) to produce the texture:
sdp = StreamDensityPlot[Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, StreamStyle -> Black,
ColorFunction -> (ColorData["Rainbow"][(#^2 + #2^2) Exp[1 - #^2 - #2^2]] &),
ColorFunctionScaling -> False, Frame -> False, Axes -> False, PlotRangePadding -> None];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> Texture[sdp], Lighting -> "Neutral"]
$endgroup$
$begingroup$
Slightly shorter:sdp = StreamDensityPlot[ Evaluate[{-D[#, {{x, y}}], #} &[(x^2 + y^2) Exp[ 1 - x^2 - y^2]]], {x, -3, 3}, {y, -3, 3}, StreamStyle -> Black, ColorFunction -> "Rainbow", Frame -> False, Axes -> False, PlotRangePadding -> None];
$endgroup$
– Michael E2
57 mins ago
add a comment |
$begingroup$
You can use StreamDensityPlot
(which accepts the ColorFunction
option) to produce the texture:
sdp = StreamDensityPlot[Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, StreamStyle -> Black,
ColorFunction -> (ColorData["Rainbow"][(#^2 + #2^2) Exp[1 - #^2 - #2^2]] &),
ColorFunctionScaling -> False, Frame -> False, Axes -> False, PlotRangePadding -> None];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> Texture[sdp], Lighting -> "Neutral"]
$endgroup$
$begingroup$
Slightly shorter:sdp = StreamDensityPlot[ Evaluate[{-D[#, {{x, y}}], #} &[(x^2 + y^2) Exp[ 1 - x^2 - y^2]]], {x, -3, 3}, {y, -3, 3}, StreamStyle -> Black, ColorFunction -> "Rainbow", Frame -> False, Axes -> False, PlotRangePadding -> None];
$endgroup$
– Michael E2
57 mins ago
add a comment |
$begingroup$
You can use StreamDensityPlot
(which accepts the ColorFunction
option) to produce the texture:
sdp = StreamDensityPlot[Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, StreamStyle -> Black,
ColorFunction -> (ColorData["Rainbow"][(#^2 + #2^2) Exp[1 - #^2 - #2^2]] &),
ColorFunctionScaling -> False, Frame -> False, Axes -> False, PlotRangePadding -> None];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> Texture[sdp], Lighting -> "Neutral"]
$endgroup$
You can use StreamDensityPlot
(which accepts the ColorFunction
option) to produce the texture:
sdp = StreamDensityPlot[Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3,
3}, {y, -3, 3}, StreamStyle -> Black,
ColorFunction -> (ColorData["Rainbow"][(#^2 + #2^2) Exp[1 - #^2 - #2^2]] &),
ColorFunctionScaling -> False, Frame -> False, Axes -> False, PlotRangePadding -> None];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
Mesh -> None, ImageSize -> Large, PlotPoints -> 35,
PlotStyle -> Texture[sdp], Lighting -> "Neutral"]
edited 1 hour ago
answered 1 hour ago
kglrkglr
178k9198409
178k9198409
$begingroup$
Slightly shorter:sdp = StreamDensityPlot[ Evaluate[{-D[#, {{x, y}}], #} &[(x^2 + y^2) Exp[ 1 - x^2 - y^2]]], {x, -3, 3}, {y, -3, 3}, StreamStyle -> Black, ColorFunction -> "Rainbow", Frame -> False, Axes -> False, PlotRangePadding -> None];
$endgroup$
– Michael E2
57 mins ago
add a comment |
$begingroup$
Slightly shorter:sdp = StreamDensityPlot[ Evaluate[{-D[#, {{x, y}}], #} &[(x^2 + y^2) Exp[ 1 - x^2 - y^2]]], {x, -3, 3}, {y, -3, 3}, StreamStyle -> Black, ColorFunction -> "Rainbow", Frame -> False, Axes -> False, PlotRangePadding -> None];
$endgroup$
– Michael E2
57 mins ago
$begingroup$
Slightly shorter:
sdp = StreamDensityPlot[ Evaluate[{-D[#, {{x, y}}], #} &[(x^2 + y^2) Exp[ 1 - x^2 - y^2]]], {x, -3, 3}, {y, -3, 3}, StreamStyle -> Black, ColorFunction -> "Rainbow", Frame -> False, Axes -> False, PlotRangePadding -> None];
$endgroup$
– Michael E2
57 mins ago
$begingroup$
Slightly shorter:
sdp = StreamDensityPlot[ Evaluate[{-D[#, {{x, y}}], #} &[(x^2 + y^2) Exp[ 1 - x^2 - y^2]]], {x, -3, 3}, {y, -3, 3}, StreamStyle -> Black, ColorFunction -> "Rainbow", Frame -> False, Axes -> False, PlotRangePadding -> None];
$endgroup$
– Michael E2
57 mins ago
add a comment |
$begingroup$
The color is not quite right but the idea seems to work. Edit: much closer now.
dp = DensityPlot[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
ColorFunction -> "Rainbow", PlotPoints -> 100];
sp = StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3, 3}, {y, -3, 3},
Frame -> None, ImageSize -> Large, StreamStyle -> Black];
tex = Show[dp, sp, Frame -> None, PlotRangePadding -> 0, ImageSize -> 500];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3}, Mesh -> None,
ImageSize -> Large, PlotPoints -> 35
, PlotStyle -> {Texture[Lighter[tex, 0.15]]}
, Lighting -> "Neutral"
]
$endgroup$
add a comment |
$begingroup$
The color is not quite right but the idea seems to work. Edit: much closer now.
dp = DensityPlot[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
ColorFunction -> "Rainbow", PlotPoints -> 100];
sp = StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3, 3}, {y, -3, 3},
Frame -> None, ImageSize -> Large, StreamStyle -> Black];
tex = Show[dp, sp, Frame -> None, PlotRangePadding -> 0, ImageSize -> 500];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3}, Mesh -> None,
ImageSize -> Large, PlotPoints -> 35
, PlotStyle -> {Texture[Lighter[tex, 0.15]]}
, Lighting -> "Neutral"
]
$endgroup$
add a comment |
$begingroup$
The color is not quite right but the idea seems to work. Edit: much closer now.
dp = DensityPlot[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
ColorFunction -> "Rainbow", PlotPoints -> 100];
sp = StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3, 3}, {y, -3, 3},
Frame -> None, ImageSize -> Large, StreamStyle -> Black];
tex = Show[dp, sp, Frame -> None, PlotRangePadding -> 0, ImageSize -> 500];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3}, Mesh -> None,
ImageSize -> Large, PlotPoints -> 35
, PlotStyle -> {Texture[Lighter[tex, 0.15]]}
, Lighting -> "Neutral"
]
$endgroup$
The color is not quite right but the idea seems to work. Edit: much closer now.
dp = DensityPlot[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3},
ColorFunction -> "Rainbow", PlotPoints -> 100];
sp = StreamPlot[
Evaluate[-D[(x^2 + y^2) Exp[1 - x^2 - y^2], {{x, y}}]], {x, -3, 3}, {y, -3, 3},
Frame -> None, ImageSize -> Large, StreamStyle -> Black];
tex = Show[dp, sp, Frame -> None, PlotRangePadding -> 0, ImageSize -> 500];
Plot3D[(x^2 + y^2) Exp[1 - x^2 - y^2], {x, -3, 3}, {y, -3, 3}, Mesh -> None,
ImageSize -> Large, PlotPoints -> 35
, PlotStyle -> {Texture[Lighter[tex, 0.15]]}
, Lighting -> "Neutral"
]
edited 2 hours ago
answered 2 hours ago
Mr.Wizard♦Mr.Wizard
231k294741041
231k294741041
add a comment |
add a comment |
$begingroup$
PlotStyle -> Texture[...]
relies on VertexTextureCoordinates
to map the texture to polygon vertices.
ColorFunction -> colorfunction
relies on VertexColors
to associate colors with the polygon vertices.
Only one of them actually gets to style the polygon. In my case, it seems to be the texture:
Graphics3D[{Texture[RandomImage[1, 100]],
Polygon[{{0, 0, 0}, {1, 0, 0}, {1, 1, 0}},
VertexColors -> {Red, Green, Blue},
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}}]},
Lighting -> "Neutral", BoxRatios -> {1, 1, 1}]
It sounds like the color function is winning in your case. It wouldn't surprise me if that was dependent on things like OS, software version, phase of the moon, etc...
$endgroup$
add a comment |
$begingroup$
PlotStyle -> Texture[...]
relies on VertexTextureCoordinates
to map the texture to polygon vertices.
ColorFunction -> colorfunction
relies on VertexColors
to associate colors with the polygon vertices.
Only one of them actually gets to style the polygon. In my case, it seems to be the texture:
Graphics3D[{Texture[RandomImage[1, 100]],
Polygon[{{0, 0, 0}, {1, 0, 0}, {1, 1, 0}},
VertexColors -> {Red, Green, Blue},
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}}]},
Lighting -> "Neutral", BoxRatios -> {1, 1, 1}]
It sounds like the color function is winning in your case. It wouldn't surprise me if that was dependent on things like OS, software version, phase of the moon, etc...
$endgroup$
add a comment |
$begingroup$
PlotStyle -> Texture[...]
relies on VertexTextureCoordinates
to map the texture to polygon vertices.
ColorFunction -> colorfunction
relies on VertexColors
to associate colors with the polygon vertices.
Only one of them actually gets to style the polygon. In my case, it seems to be the texture:
Graphics3D[{Texture[RandomImage[1, 100]],
Polygon[{{0, 0, 0}, {1, 0, 0}, {1, 1, 0}},
VertexColors -> {Red, Green, Blue},
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}}]},
Lighting -> "Neutral", BoxRatios -> {1, 1, 1}]
It sounds like the color function is winning in your case. It wouldn't surprise me if that was dependent on things like OS, software version, phase of the moon, etc...
$endgroup$
PlotStyle -> Texture[...]
relies on VertexTextureCoordinates
to map the texture to polygon vertices.
ColorFunction -> colorfunction
relies on VertexColors
to associate colors with the polygon vertices.
Only one of them actually gets to style the polygon. In my case, it seems to be the texture:
Graphics3D[{Texture[RandomImage[1, 100]],
Polygon[{{0, 0, 0}, {1, 0, 0}, {1, 1, 0}},
VertexColors -> {Red, Green, Blue},
VertexTextureCoordinates -> {{0, 0}, {1, 0}, {1, 1}}]},
Lighting -> "Neutral", BoxRatios -> {1, 1, 1}]
It sounds like the color function is winning in your case. It wouldn't surprise me if that was dependent on things like OS, software version, phase of the moon, etc...
answered 2 hours ago
Brett ChampionBrett Champion
17.2k251114
17.2k251114
add a comment |
add a comment |
user17164 is a new contributor. Be nice, and check out our Code of Conduct.
user17164 is a new contributor. Be nice, and check out our Code of Conduct.
user17164 is a new contributor. Be nice, and check out our Code of Conduct.
user17164 is a new contributor. Be nice, and check out our Code of Conduct.
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