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
$endgroup$
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
$endgroup$
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
$endgroup$
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
edited Jan 14 at 19:58
Mr.Wizard♦
231k294761044
231k294761044
asked Jan 14 at 19:54
user17164user17164
1262
1262
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 |
$begingroup$
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^2 + y^2) Exp[1 - x^2 - y^2]}], {x, -3, 3}, {y, -3, 3},
StreamStyle -> Black,
ColorFunction -> "Rainbow",
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[Lighter@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
Jan 14 at 21:55
$begingroup$
@MichaelE2, I tried that version; but the colors do not match the colors in Plot3D.
$endgroup$
– kglr
Jan 14 at 23:01
$begingroup$
Odd, they match your code above, for me. I switched between the two images and saw no (perceptible) difference.
$endgroup$
– Michael E2
Jan 14 at 23:02
$begingroup$
@MichaelE2, maybe version/os difference (i am using v 11.3 windows 10/64bit).
$endgroup$
– kglr
Jan 14 at 23:04
$begingroup$
@MichaelE2,ColorFunction -> "Rainbow"
does work if the first argument ofStreamDensityPlot
has the form ${{v_x, v_y}, s }$.
$endgroup$
– kglr
Jan 15 at 1:39
|
show 2 more comments
$begingroup$
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 |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Jan 14 at 20:25
answered Jan 14 at 20:12
Mr.Wizard♦Mr.Wizard
231k294761044
231k294761044
add a comment |
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^2 + y^2) Exp[1 - x^2 - y^2]}], {x, -3, 3}, {y, -3, 3},
StreamStyle -> Black,
ColorFunction -> "Rainbow",
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[Lighter@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
Jan 14 at 21:55
$begingroup$
@MichaelE2, I tried that version; but the colors do not match the colors in Plot3D.
$endgroup$
– kglr
Jan 14 at 23:01
$begingroup$
Odd, they match your code above, for me. I switched between the two images and saw no (perceptible) difference.
$endgroup$
– Michael E2
Jan 14 at 23:02
$begingroup$
@MichaelE2, maybe version/os difference (i am using v 11.3 windows 10/64bit).
$endgroup$
– kglr
Jan 14 at 23:04
$begingroup$
@MichaelE2,ColorFunction -> "Rainbow"
does work if the first argument ofStreamDensityPlot
has the form ${{v_x, v_y}, s }$.
$endgroup$
– kglr
Jan 15 at 1:39
|
show 2 more comments
$begingroup$
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^2 + y^2) Exp[1 - x^2 - y^2]}], {x, -3, 3}, {y, -3, 3},
StreamStyle -> Black,
ColorFunction -> "Rainbow",
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[Lighter@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
Jan 14 at 21:55
$begingroup$
@MichaelE2, I tried that version; but the colors do not match the colors in Plot3D.
$endgroup$
– kglr
Jan 14 at 23:01
$begingroup$
Odd, they match your code above, for me. I switched between the two images and saw no (perceptible) difference.
$endgroup$
– Michael E2
Jan 14 at 23:02
$begingroup$
@MichaelE2, maybe version/os difference (i am using v 11.3 windows 10/64bit).
$endgroup$
– kglr
Jan 14 at 23:04
$begingroup$
@MichaelE2,ColorFunction -> "Rainbow"
does work if the first argument ofStreamDensityPlot
has the form ${{v_x, v_y}, s }$.
$endgroup$
– kglr
Jan 15 at 1:39
|
show 2 more comments
$begingroup$
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^2 + y^2) Exp[1 - x^2 - y^2]}], {x, -3, 3}, {y, -3, 3},
StreamStyle -> Black,
ColorFunction -> "Rainbow",
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[Lighter@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^2 + y^2) Exp[1 - x^2 - y^2]}], {x, -3, 3}, {y, -3, 3},
StreamStyle -> Black,
ColorFunction -> "Rainbow",
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[Lighter@sdp], Lighting -> "Neutral"]
edited Jan 15 at 1:36
answered Jan 14 at 20:56
kglrkglr
181k10200413
181k10200413
$begingroup$
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
Jan 14 at 21:55
$begingroup$
@MichaelE2, I tried that version; but the colors do not match the colors in Plot3D.
$endgroup$
– kglr
Jan 14 at 23:01
$begingroup$
Odd, they match your code above, for me. I switched between the two images and saw no (perceptible) difference.
$endgroup$
– Michael E2
Jan 14 at 23:02
$begingroup$
@MichaelE2, maybe version/os difference (i am using v 11.3 windows 10/64bit).
$endgroup$
– kglr
Jan 14 at 23:04
$begingroup$
@MichaelE2,ColorFunction -> "Rainbow"
does work if the first argument ofStreamDensityPlot
has the form ${{v_x, v_y}, s }$.
$endgroup$
– kglr
Jan 15 at 1:39
|
show 2 more comments
$begingroup$
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
Jan 14 at 21:55
$begingroup$
@MichaelE2, I tried that version; but the colors do not match the colors in Plot3D.
$endgroup$
– kglr
Jan 14 at 23:01
$begingroup$
Odd, they match your code above, for me. I switched between the two images and saw no (perceptible) difference.
$endgroup$
– Michael E2
Jan 14 at 23:02
$begingroup$
@MichaelE2, maybe version/os difference (i am using v 11.3 windows 10/64bit).
$endgroup$
– kglr
Jan 14 at 23:04
$begingroup$
@MichaelE2,ColorFunction -> "Rainbow"
does work if the first argument ofStreamDensityPlot
has the form ${{v_x, v_y}, s }$.
$endgroup$
– kglr
Jan 15 at 1:39
$begingroup$
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
Jan 14 at 21:55
$begingroup$
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
Jan 14 at 21:55
$begingroup$
@MichaelE2, I tried that version; but the colors do not match the colors in Plot3D.
$endgroup$
– kglr
Jan 14 at 23:01
$begingroup$
@MichaelE2, I tried that version; but the colors do not match the colors in Plot3D.
$endgroup$
– kglr
Jan 14 at 23:01
$begingroup$
Odd, they match your code above, for me. I switched between the two images and saw no (perceptible) difference.
$endgroup$
– Michael E2
Jan 14 at 23:02
$begingroup$
Odd, they match your code above, for me. I switched between the two images and saw no (perceptible) difference.
$endgroup$
– Michael E2
Jan 14 at 23:02
$begingroup$
@MichaelE2, maybe version/os difference (i am using v 11.3 windows 10/64bit).
$endgroup$
– kglr
Jan 14 at 23:04
$begingroup$
@MichaelE2, maybe version/os difference (i am using v 11.3 windows 10/64bit).
$endgroup$
– kglr
Jan 14 at 23:04
$begingroup$
@MichaelE2,
ColorFunction -> "Rainbow"
does work if the first argument of StreamDensityPlot
has the form ${{v_x, v_y}, s }$.$endgroup$
– kglr
Jan 15 at 1:39
$begingroup$
@MichaelE2,
ColorFunction -> "Rainbow"
does work if the first argument of StreamDensityPlot
has the form ${{v_x, v_y}, s }$.$endgroup$
– kglr
Jan 15 at 1:39
|
show 2 more comments
$begingroup$
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 Jan 14 at 20:13
Brett ChampionBrett Champion
17.3k252114
17.3k252114
add a comment |
add a comment |
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