can't blend gradient colors with a stream












4












$begingroup$


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"]









share|improve this question









New contributor




user17164 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    4












    $begingroup$


    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"]









    share|improve this question









    New contributor




    user17164 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4





      $begingroup$


      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"]









      share|improve this question









      New contributor




      user17164 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      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






      share|improve this question









      New contributor




      user17164 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|improve this question









      New contributor




      user17164 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|improve this question




      share|improve this question








      edited 2 hours ago









      Mr.Wizard

      231k294741041




      231k294741041






      New contributor




      user17164 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 2 hours ago









      user17164user17164

      1212




      1212




      New contributor




      user17164 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      user17164 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes


















          7












          $begingroup$

          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"]


          enter image description here






          share|improve this answer











          $endgroup$













          • $begingroup$
            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



















          6












          $begingroup$

          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"
          ]


          enter image description here






          share|improve this answer











          $endgroup$





















            2












            $begingroup$

            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}]


            enter image description here



            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...






            share|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              7












              $begingroup$

              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"]


              enter image description here






              share|improve this answer











              $endgroup$













              • $begingroup$
                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
















              7












              $begingroup$

              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"]


              enter image description here






              share|improve this answer











              $endgroup$













              • $begingroup$
                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














              7












              7








              7





              $begingroup$

              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"]


              enter image description here






              share|improve this answer











              $endgroup$



              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"]


              enter image description here







              share|improve this answer














              share|improve this answer



              share|improve this answer








              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


















              • $begingroup$
                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











              6












              $begingroup$

              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"
              ]


              enter image description here






              share|improve this answer











              $endgroup$


















                6












                $begingroup$

                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"
                ]


                enter image description here






                share|improve this answer











                $endgroup$
















                  6












                  6








                  6





                  $begingroup$

                  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"
                  ]


                  enter image description here






                  share|improve this answer











                  $endgroup$



                  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"
                  ]


                  enter image description here







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 2 hours ago

























                  answered 2 hours ago









                  Mr.WizardMr.Wizard

                  231k294741041




                  231k294741041























                      2












                      $begingroup$

                      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}]


                      enter image description here



                      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...






                      share|improve this answer









                      $endgroup$


















                        2












                        $begingroup$

                        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}]


                        enter image description here



                        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...






                        share|improve this answer









                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          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}]


                          enter image description here



                          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...






                          share|improve this answer









                          $endgroup$



                          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}]


                          enter image description here



                          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...







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 2 hours ago









                          Brett ChampionBrett Champion

                          17.2k251114




                          17.2k251114






















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