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 1 hour 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 1 hour 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.
      Check out our Code of Conduct.






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






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          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











          $endgroup$





















            3












            $begingroup$

            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









            $endgroup$





















              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









              $endgroup$













                Your Answer





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                3 Answers
                3






                active

                oldest

                votes








                3 Answers
                3






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                4












                $begingroup$

                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











                $endgroup$


















                  4












                  $begingroup$

                  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











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    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 1 hour ago

























                    answered 1 hour ago









                    Mr.WizardMr.Wizard

                    231k294741041




                    231k294741041























                        3












                        $begingroup$

                        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









                        $endgroup$


















                          3












                          $begingroup$

                          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









                          $endgroup$
















                            3












                            3








                            3





                            $begingroup$

                            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










                            answered 47 mins ago









                            kglrkglr

                            178k9198409




                            178k9198409























                                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 1 hour ago









                                    Brett ChampionBrett Champion

                                    17.2k251114




                                    17.2k251114






















                                        user17164 is a new contributor. Be nice, and check out our Code of Conduct.










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