Is it possible to rotate the Isolines on a Surface Using `MeshFunction`?












5












$begingroup$


This came up in a different context but some expertise in 3D surfaces or the graphic options would be appreciated. I'm trying to extrapolate the curves from any given surface and things seem to be going quite smoothly. All the curves can be grabbed in one more step as a GraphicsComplex. Perfect for more processing. However, now I'm trying to rotate the isolines to get even more control. This is possible in other software but I'm not sure how it was achieved. I assume there is some way to use the MeshFunction to rotate the Mesh through at least 45 degrees but all my searching hasn't brought up anything helpful. A less practical approach might be to find the intersecting curve of a regularly spaced vertical planes.



Plot3D[Cos[(x y)/2], {x, 0, 4}, {y, 0, 8},
BoxRatios->{4,8,1},
Boxed->False,
Axes->False,
ImageSize->Large,
Mesh->{3,8},
PlotStyle->Directive[Lighting->"Neutral",FaceForm[White,Specularity[0.2,10]]]]


enter image description here



enter image description here










share|improve this question









$endgroup$

















    5












    $begingroup$


    This came up in a different context but some expertise in 3D surfaces or the graphic options would be appreciated. I'm trying to extrapolate the curves from any given surface and things seem to be going quite smoothly. All the curves can be grabbed in one more step as a GraphicsComplex. Perfect for more processing. However, now I'm trying to rotate the isolines to get even more control. This is possible in other software but I'm not sure how it was achieved. I assume there is some way to use the MeshFunction to rotate the Mesh through at least 45 degrees but all my searching hasn't brought up anything helpful. A less practical approach might be to find the intersecting curve of a regularly spaced vertical planes.



    Plot3D[Cos[(x y)/2], {x, 0, 4}, {y, 0, 8},
    BoxRatios->{4,8,1},
    Boxed->False,
    Axes->False,
    ImageSize->Large,
    Mesh->{3,8},
    PlotStyle->Directive[Lighting->"Neutral",FaceForm[White,Specularity[0.2,10]]]]


    enter image description here



    enter image description here










    share|improve this question









    $endgroup$















      5












      5








      5





      $begingroup$


      This came up in a different context but some expertise in 3D surfaces or the graphic options would be appreciated. I'm trying to extrapolate the curves from any given surface and things seem to be going quite smoothly. All the curves can be grabbed in one more step as a GraphicsComplex. Perfect for more processing. However, now I'm trying to rotate the isolines to get even more control. This is possible in other software but I'm not sure how it was achieved. I assume there is some way to use the MeshFunction to rotate the Mesh through at least 45 degrees but all my searching hasn't brought up anything helpful. A less practical approach might be to find the intersecting curve of a regularly spaced vertical planes.



      Plot3D[Cos[(x y)/2], {x, 0, 4}, {y, 0, 8},
      BoxRatios->{4,8,1},
      Boxed->False,
      Axes->False,
      ImageSize->Large,
      Mesh->{3,8},
      PlotStyle->Directive[Lighting->"Neutral",FaceForm[White,Specularity[0.2,10]]]]


      enter image description here



      enter image description here










      share|improve this question









      $endgroup$




      This came up in a different context but some expertise in 3D surfaces or the graphic options would be appreciated. I'm trying to extrapolate the curves from any given surface and things seem to be going quite smoothly. All the curves can be grabbed in one more step as a GraphicsComplex. Perfect for more processing. However, now I'm trying to rotate the isolines to get even more control. This is possible in other software but I'm not sure how it was achieved. I assume there is some way to use the MeshFunction to rotate the Mesh through at least 45 degrees but all my searching hasn't brought up anything helpful. A less practical approach might be to find the intersecting curve of a regularly spaced vertical planes.



      Plot3D[Cos[(x y)/2], {x, 0, 4}, {y, 0, 8},
      BoxRatios->{4,8,1},
      Boxed->False,
      Axes->False,
      ImageSize->Large,
      Mesh->{3,8},
      PlotStyle->Directive[Lighting->"Neutral",FaceForm[White,Specularity[0.2,10]]]]


      enter image description here



      enter image description here







      plotting graphics






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 2 hours ago









      BBirdsellBBirdsell

      430313




      430313






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Plot3D[Cos[(x y)/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> {4, 8, 1}, 
          Boxed -> False, Axes -> False, ImageSize -> Large,
          MeshFunctions -> {# + #2 &, # - #2 &},
          Mesh -> {3, 8},
          PlotStyle -> Directive[Lighting -> "Neutral", FaceForm[White, Specularity[0.2, 10]]]]


          enter image description here






          share|improve this answer









          $endgroup$





















            0












            $begingroup$

            Since we have the identity



            RotationMatrix[θ] == {AngleVector[-θ], AngleVector[π/2 - θ]}


            one can use this to construct a mesh that is arbitrarily oriented; e.g.



            With[{θ = π/4}, 
            Plot3D[Cos[x y/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> Automatic,
            MeshFunctions -> {AngleVector[-θ].{#, #2} &,
            AngleVector[π/2 - θ].{#, #2} &}]]


            and you can change the value of θ for other orientations.






            share|improve this answer









            $endgroup$













            • $begingroup$
              (If anyone is kind enough to edit my post to include the resulting image, please do so.)
              $endgroup$
              – J. M. is computer-less
              21 mins ago











            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "387"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192157%2fis-it-possible-to-rotate-the-isolines-on-a-surface-using-meshfunction%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Plot3D[Cos[(x y)/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> {4, 8, 1}, 
            Boxed -> False, Axes -> False, ImageSize -> Large,
            MeshFunctions -> {# + #2 &, # - #2 &},
            Mesh -> {3, 8},
            PlotStyle -> Directive[Lighting -> "Neutral", FaceForm[White, Specularity[0.2, 10]]]]


            enter image description here






            share|improve this answer









            $endgroup$


















              3












              $begingroup$

              Plot3D[Cos[(x y)/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> {4, 8, 1}, 
              Boxed -> False, Axes -> False, ImageSize -> Large,
              MeshFunctions -> {# + #2 &, # - #2 &},
              Mesh -> {3, 8},
              PlotStyle -> Directive[Lighting -> "Neutral", FaceForm[White, Specularity[0.2, 10]]]]


              enter image description here






              share|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Plot3D[Cos[(x y)/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> {4, 8, 1}, 
                Boxed -> False, Axes -> False, ImageSize -> Large,
                MeshFunctions -> {# + #2 &, # - #2 &},
                Mesh -> {3, 8},
                PlotStyle -> Directive[Lighting -> "Neutral", FaceForm[White, Specularity[0.2, 10]]]]


                enter image description here






                share|improve this answer









                $endgroup$



                Plot3D[Cos[(x y)/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> {4, 8, 1}, 
                Boxed -> False, Axes -> False, ImageSize -> Large,
                MeshFunctions -> {# + #2 &, # - #2 &},
                Mesh -> {3, 8},
                PlotStyle -> Directive[Lighting -> "Neutral", FaceForm[White, Specularity[0.2, 10]]]]


                enter image description here







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 2 hours ago









                kglrkglr

                185k10202421




                185k10202421























                    0












                    $begingroup$

                    Since we have the identity



                    RotationMatrix[θ] == {AngleVector[-θ], AngleVector[π/2 - θ]}


                    one can use this to construct a mesh that is arbitrarily oriented; e.g.



                    With[{θ = π/4}, 
                    Plot3D[Cos[x y/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> Automatic,
                    MeshFunctions -> {AngleVector[-θ].{#, #2} &,
                    AngleVector[π/2 - θ].{#, #2} &}]]


                    and you can change the value of θ for other orientations.






                    share|improve this answer









                    $endgroup$













                    • $begingroup$
                      (If anyone is kind enough to edit my post to include the resulting image, please do so.)
                      $endgroup$
                      – J. M. is computer-less
                      21 mins ago
















                    0












                    $begingroup$

                    Since we have the identity



                    RotationMatrix[θ] == {AngleVector[-θ], AngleVector[π/2 - θ]}


                    one can use this to construct a mesh that is arbitrarily oriented; e.g.



                    With[{θ = π/4}, 
                    Plot3D[Cos[x y/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> Automatic,
                    MeshFunctions -> {AngleVector[-θ].{#, #2} &,
                    AngleVector[π/2 - θ].{#, #2} &}]]


                    and you can change the value of θ for other orientations.






                    share|improve this answer









                    $endgroup$













                    • $begingroup$
                      (If anyone is kind enough to edit my post to include the resulting image, please do so.)
                      $endgroup$
                      – J. M. is computer-less
                      21 mins ago














                    0












                    0








                    0





                    $begingroup$

                    Since we have the identity



                    RotationMatrix[θ] == {AngleVector[-θ], AngleVector[π/2 - θ]}


                    one can use this to construct a mesh that is arbitrarily oriented; e.g.



                    With[{θ = π/4}, 
                    Plot3D[Cos[x y/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> Automatic,
                    MeshFunctions -> {AngleVector[-θ].{#, #2} &,
                    AngleVector[π/2 - θ].{#, #2} &}]]


                    and you can change the value of θ for other orientations.






                    share|improve this answer









                    $endgroup$



                    Since we have the identity



                    RotationMatrix[θ] == {AngleVector[-θ], AngleVector[π/2 - θ]}


                    one can use this to construct a mesh that is arbitrarily oriented; e.g.



                    With[{θ = π/4}, 
                    Plot3D[Cos[x y/2], {x, 0, 4}, {y, 0, 8}, BoxRatios -> Automatic,
                    MeshFunctions -> {AngleVector[-θ].{#, #2} &,
                    AngleVector[π/2 - θ].{#, #2} &}]]


                    and you can change the value of θ for other orientations.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 22 mins ago









                    J. M. is computer-lessJ. M. is computer-less

                    96.8k10303462




                    96.8k10303462












                    • $begingroup$
                      (If anyone is kind enough to edit my post to include the resulting image, please do so.)
                      $endgroup$
                      – J. M. is computer-less
                      21 mins ago


















                    • $begingroup$
                      (If anyone is kind enough to edit my post to include the resulting image, please do so.)
                      $endgroup$
                      – J. M. is computer-less
                      21 mins ago
















                    $begingroup$
                    (If anyone is kind enough to edit my post to include the resulting image, please do so.)
                    $endgroup$
                    – J. M. is computer-less
                    21 mins ago




                    $begingroup$
                    (If anyone is kind enough to edit my post to include the resulting image, please do so.)
                    $endgroup$
                    – J. M. is computer-less
                    21 mins ago


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematica Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f192157%2fis-it-possible-to-rotate-the-isolines-on-a-surface-using-meshfunction%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Accessing regular linux commands in Huawei's Dopra Linux

                    Can't connect RFCOMM socket: Host is down

                    Kernel panic - not syncing: Fatal Exception in Interrupt