How to ternary Plot3D a function












5












$begingroup$


I ploted 3D the function Sin[A/2]Sin[B/2]Sin[C/2] with A, B, C > 0 and A + B + C = Pi.



A basic approach in the post How to plot ternary density plots answers with the use of FindGeometricTransform. How can I transform the Plot3Ded function inside the equilateral triangle with FindGeometricTransform? If there is a simpler method, I would like to know it.










share|improve this question









New contributor




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







$endgroup$












  • $begingroup$
    I can make ternary density plots and ternary contour plots according to the post above. But I would like to draw 3D plots with one of the point in the equilateral triangle as the coordinates. The coordinates are essentially 2 dimensional with A + B + C = Pi
    $endgroup$
    – seiichikiri
    Apr 20 at 8:01










  • $begingroup$
    If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
    $endgroup$
    – xzczd
    Apr 20 at 8:19










  • $begingroup$
    Welcome to Mathematica.SE, seiichikiri! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    $endgroup$
    – Chris K
    Apr 21 at 14:08
















5












$begingroup$


I ploted 3D the function Sin[A/2]Sin[B/2]Sin[C/2] with A, B, C > 0 and A + B + C = Pi.



A basic approach in the post How to plot ternary density plots answers with the use of FindGeometricTransform. How can I transform the Plot3Ded function inside the equilateral triangle with FindGeometricTransform? If there is a simpler method, I would like to know it.










share|improve this question









New contributor




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







$endgroup$












  • $begingroup$
    I can make ternary density plots and ternary contour plots according to the post above. But I would like to draw 3D plots with one of the point in the equilateral triangle as the coordinates. The coordinates are essentially 2 dimensional with A + B + C = Pi
    $endgroup$
    – seiichikiri
    Apr 20 at 8:01










  • $begingroup$
    If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
    $endgroup$
    – xzczd
    Apr 20 at 8:19










  • $begingroup$
    Welcome to Mathematica.SE, seiichikiri! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    $endgroup$
    – Chris K
    Apr 21 at 14:08














5












5








5





$begingroup$


I ploted 3D the function Sin[A/2]Sin[B/2]Sin[C/2] with A, B, C > 0 and A + B + C = Pi.



A basic approach in the post How to plot ternary density plots answers with the use of FindGeometricTransform. How can I transform the Plot3Ded function inside the equilateral triangle with FindGeometricTransform? If there is a simpler method, I would like to know it.










share|improve this question









New contributor




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







$endgroup$




I ploted 3D the function Sin[A/2]Sin[B/2]Sin[C/2] with A, B, C > 0 and A + B + C = Pi.



A basic approach in the post How to plot ternary density plots answers with the use of FindGeometricTransform. How can I transform the Plot3Ded function inside the equilateral triangle with FindGeometricTransform? If there is a simpler method, I would like to know it.







plotting






share|improve this question









New contributor




seiichikiri 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




seiichikiri 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 Apr 20 at 6:42









xzczd

27.9k576258




27.9k576258






New contributor




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









asked Apr 20 at 6:31









seiichikiriseiichikiri

282




282




New contributor




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





New contributor





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






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












  • $begingroup$
    I can make ternary density plots and ternary contour plots according to the post above. But I would like to draw 3D plots with one of the point in the equilateral triangle as the coordinates. The coordinates are essentially 2 dimensional with A + B + C = Pi
    $endgroup$
    – seiichikiri
    Apr 20 at 8:01










  • $begingroup$
    If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
    $endgroup$
    – xzczd
    Apr 20 at 8:19










  • $begingroup$
    Welcome to Mathematica.SE, seiichikiri! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    $endgroup$
    – Chris K
    Apr 21 at 14:08


















  • $begingroup$
    I can make ternary density plots and ternary contour plots according to the post above. But I would like to draw 3D plots with one of the point in the equilateral triangle as the coordinates. The coordinates are essentially 2 dimensional with A + B + C = Pi
    $endgroup$
    – seiichikiri
    Apr 20 at 8:01










  • $begingroup$
    If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
    $endgroup$
    – xzczd
    Apr 20 at 8:19










  • $begingroup$
    Welcome to Mathematica.SE, seiichikiri! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
    $endgroup$
    – Chris K
    Apr 21 at 14:08
















$begingroup$
I can make ternary density plots and ternary contour plots according to the post above. But I would like to draw 3D plots with one of the point in the equilateral triangle as the coordinates. The coordinates are essentially 2 dimensional with A + B + C = Pi
$endgroup$
– seiichikiri
Apr 20 at 8:01




$begingroup$
I can make ternary density plots and ternary contour plots according to the post above. But I would like to draw 3D plots with one of the point in the equilateral triangle as the coordinates. The coordinates are essentially 2 dimensional with A + B + C = Pi
$endgroup$
– seiichikiri
Apr 20 at 8:01












$begingroup$
If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
$endgroup$
– xzczd
Apr 20 at 8:19




$begingroup$
If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
$endgroup$
– xzczd
Apr 20 at 8:19












$begingroup$
Welcome to Mathematica.SE, seiichikiri! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
$endgroup$
– Chris K
Apr 21 at 14:08




$begingroup$
Welcome to Mathematica.SE, seiichikiri! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour and check the faqs! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign!
$endgroup$
– Chris K
Apr 21 at 14:08










3 Answers
3






active

oldest

votes


















5












$begingroup$

It's not hard to transform the Graphics3D generated by Plot3D if you understand its structure. We already have numbers of posts about this issue so I'd like not to talk about it in this answer, you may check e.g. this post for more info. Here comes the code, notice I've made use of the new-in-v12 feature of Callout to create ticks, which is more troublesome compared to the transforming part in my opinion:



old = Pi First@Triangle    
begin = {##, 0} & @@@ (π AnglePath[{0, 120 °, 120 °}])
direction = Normalize /@ Differences@begin;
p3 = Plot3D[Sin[a/2] Sin[b/2] Sin[(Pi - a - b)/2], {a, b} ∈ Triangle@old];

{error, tr} = FindGeometricTransform[Most /@ Most@begin, old];

newp3 = p3 /.
GraphicsComplex[pts_, rest__] :>GraphicsComplex[SubsetMap[tr, #, {1, 2}] & /@ pts, rest];

ticks = ListPointPlot3D@Flatten@With[{n = 5},
Table[Callout[begin[[i]] + direction[[i]] j Pi/n, j Pi/n], {i, 3}, {j, 0, n}]];

Show[newp3, ticks, Axes -> False, Boxed -> False, PlotRange -> All]


enter image description here



Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?






share|improve this answer











$endgroup$













  • $begingroup$
    My intension is to compare LHS and RHS of the inequalities. In this case, 6r/R = 24 Sin[A/2]Sin[B/2]Sin[C/2] <= CyclicSum (Sin[A]+Sin[B])/Cot[C/2], where r is the incenter and circumcenter of a triangle. I deleted 3rd line and 7th line. I replaced 6th line by newp3 = p3/. GraphicsComplex[pts_,rest_] -> GraphicsComplex[Map[tr, #,{1,2}]&/@pts,rest]. Is this change allowable? My version of Mathematica 8 does not include SubsetMap. I could plot similar figures as yours. But the base of the plot is isosceles, not equilateral. I found this by rotation.
    $endgroup$
    – seiichikiri
    Apr 21 at 3:01










  • $begingroup$
    @seiichikiri No, your usage of Map[...] is incorrect. It should be e.g. newp3 = p3 /. GraphicsComplex[pts_, rest__] :> GraphicsComplex[Join[tr[{#, #2}], {#3}] & @@@ pts, rest]. The shape of triangle looks incorrect because I've added PlotRange -> All and the default BoxRatios is {1, 1, 0.4} in this case.
    $endgroup$
    – xzczd
    Apr 21 at 13:29










  • $begingroup$
    @xzcxd I obtained what I wanted. Thank you very much!
    $endgroup$
    – seiichikiri
    Apr 22 at 0:25










  • $begingroup$
    @seiichikiri Glad it help. If my answer resolves your problem, you can accept it by clicking the checkmark sign.
    $endgroup$
    – xzczd
    2 days ago



















5












$begingroup$

Without using any transformations, you have



$$
A = frac13 - x - frac{y}{sqrt{3}}\
B = frac13 + x - frac{y}{sqrt{3}}\
C = frac13 + frac{2 y}{sqrt{3}}
$$



In this form, they span the ranges $[0,1]$ over an equilateral triangle with unit edges, and satisfy $A+B+C=1$. In what follows I'll use $a$, $b$, $c$ instead of the capital letters because it's not a good idea to use capital letters for your own definitions in Mathematica.



Plotting your function, you need to multiply these with $pi$ to get your desired range.



Here's a very simplistic way of plotting that does not generate any tick marks. It is mostly for getting a quick overview. If you want proper tick marks you need to follow some of the other recommendations, for example on question 39733. Also, MeshFunctions can give interesting meshes when combined with the effective coordinates $a$, $b$, $c$.



f[a_, b_, c_] = Sin[π*a/2] Sin[π*b/2] Sin[π*c/2];
DensityPlot[f[1/3-x-y/Sqrt[3], 1/3+x-y/Sqrt[3], 1/3+2y/Sqrt[3]],
{x, -0.6, 0.6}, {y, -0.4, 0.7},
RegionFunction -> Function[{x, y}, 0<=1/3-x-y/Sqrt[3]<=1 && 0<=1/3+x-y/Sqrt[3]<=1 && 0<=1/3+2y/Sqrt[3]<=1],
AspectRatio -> Automatic,
Epilog -> {Text["A", {-1/2, -1/(2 Sqrt[3])}, {Sqrt[3]/2, 1/2}],
Text["B", {1/2, -1/(2 Sqrt[3])}, {-Sqrt[3]/2, 1/2}],
Text["C", {0, 1/Sqrt[3]}, {0, -1}]}]


enter image description here



Here is what happens if we set the function $f(a,b,c)$ to either $a$, $b$, or $c$: you can see the behavior of these coordinates,



enter image description here






share|improve this answer











$endgroup$





















    2












    $begingroup$

    DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
    {a, 0, 2}, {b, 0, 2}, {c, 0, 2}]


    enter image description here






    share|improve this answer









    $endgroup$














      Your Answer








      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
      });


      }
      });






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










      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathematica.stackexchange.com%2fquestions%2f195633%2fhow-to-ternary-plot3d-a-function%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      It's not hard to transform the Graphics3D generated by Plot3D if you understand its structure. We already have numbers of posts about this issue so I'd like not to talk about it in this answer, you may check e.g. this post for more info. Here comes the code, notice I've made use of the new-in-v12 feature of Callout to create ticks, which is more troublesome compared to the transforming part in my opinion:



      old = Pi First@Triangle    
      begin = {##, 0} & @@@ (π AnglePath[{0, 120 °, 120 °}])
      direction = Normalize /@ Differences@begin;
      p3 = Plot3D[Sin[a/2] Sin[b/2] Sin[(Pi - a - b)/2], {a, b} ∈ Triangle@old];

      {error, tr} = FindGeometricTransform[Most /@ Most@begin, old];

      newp3 = p3 /.
      GraphicsComplex[pts_, rest__] :>GraphicsComplex[SubsetMap[tr, #, {1, 2}] & /@ pts, rest];

      ticks = ListPointPlot3D@Flatten@With[{n = 5},
      Table[Callout[begin[[i]] + direction[[i]] j Pi/n, j Pi/n], {i, 3}, {j, 0, n}]];

      Show[newp3, ticks, Axes -> False, Boxed -> False, PlotRange -> All]


      enter image description here



      Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?






      share|improve this answer











      $endgroup$













      • $begingroup$
        My intension is to compare LHS and RHS of the inequalities. In this case, 6r/R = 24 Sin[A/2]Sin[B/2]Sin[C/2] <= CyclicSum (Sin[A]+Sin[B])/Cot[C/2], where r is the incenter and circumcenter of a triangle. I deleted 3rd line and 7th line. I replaced 6th line by newp3 = p3/. GraphicsComplex[pts_,rest_] -> GraphicsComplex[Map[tr, #,{1,2}]&/@pts,rest]. Is this change allowable? My version of Mathematica 8 does not include SubsetMap. I could plot similar figures as yours. But the base of the plot is isosceles, not equilateral. I found this by rotation.
        $endgroup$
        – seiichikiri
        Apr 21 at 3:01










      • $begingroup$
        @seiichikiri No, your usage of Map[...] is incorrect. It should be e.g. newp3 = p3 /. GraphicsComplex[pts_, rest__] :> GraphicsComplex[Join[tr[{#, #2}], {#3}] & @@@ pts, rest]. The shape of triangle looks incorrect because I've added PlotRange -> All and the default BoxRatios is {1, 1, 0.4} in this case.
        $endgroup$
        – xzczd
        Apr 21 at 13:29










      • $begingroup$
        @xzcxd I obtained what I wanted. Thank you very much!
        $endgroup$
        – seiichikiri
        Apr 22 at 0:25










      • $begingroup$
        @seiichikiri Glad it help. If my answer resolves your problem, you can accept it by clicking the checkmark sign.
        $endgroup$
        – xzczd
        2 days ago
















      5












      $begingroup$

      It's not hard to transform the Graphics3D generated by Plot3D if you understand its structure. We already have numbers of posts about this issue so I'd like not to talk about it in this answer, you may check e.g. this post for more info. Here comes the code, notice I've made use of the new-in-v12 feature of Callout to create ticks, which is more troublesome compared to the transforming part in my opinion:



      old = Pi First@Triangle    
      begin = {##, 0} & @@@ (π AnglePath[{0, 120 °, 120 °}])
      direction = Normalize /@ Differences@begin;
      p3 = Plot3D[Sin[a/2] Sin[b/2] Sin[(Pi - a - b)/2], {a, b} ∈ Triangle@old];

      {error, tr} = FindGeometricTransform[Most /@ Most@begin, old];

      newp3 = p3 /.
      GraphicsComplex[pts_, rest__] :>GraphicsComplex[SubsetMap[tr, #, {1, 2}] & /@ pts, rest];

      ticks = ListPointPlot3D@Flatten@With[{n = 5},
      Table[Callout[begin[[i]] + direction[[i]] j Pi/n, j Pi/n], {i, 3}, {j, 0, n}]];

      Show[newp3, ticks, Axes -> False, Boxed -> False, PlotRange -> All]


      enter image description here



      Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?






      share|improve this answer











      $endgroup$













      • $begingroup$
        My intension is to compare LHS and RHS of the inequalities. In this case, 6r/R = 24 Sin[A/2]Sin[B/2]Sin[C/2] <= CyclicSum (Sin[A]+Sin[B])/Cot[C/2], where r is the incenter and circumcenter of a triangle. I deleted 3rd line and 7th line. I replaced 6th line by newp3 = p3/. GraphicsComplex[pts_,rest_] -> GraphicsComplex[Map[tr, #,{1,2}]&/@pts,rest]. Is this change allowable? My version of Mathematica 8 does not include SubsetMap. I could plot similar figures as yours. But the base of the plot is isosceles, not equilateral. I found this by rotation.
        $endgroup$
        – seiichikiri
        Apr 21 at 3:01










      • $begingroup$
        @seiichikiri No, your usage of Map[...] is incorrect. It should be e.g. newp3 = p3 /. GraphicsComplex[pts_, rest__] :> GraphicsComplex[Join[tr[{#, #2}], {#3}] & @@@ pts, rest]. The shape of triangle looks incorrect because I've added PlotRange -> All and the default BoxRatios is {1, 1, 0.4} in this case.
        $endgroup$
        – xzczd
        Apr 21 at 13:29










      • $begingroup$
        @xzcxd I obtained what I wanted. Thank you very much!
        $endgroup$
        – seiichikiri
        Apr 22 at 0:25










      • $begingroup$
        @seiichikiri Glad it help. If my answer resolves your problem, you can accept it by clicking the checkmark sign.
        $endgroup$
        – xzczd
        2 days ago














      5












      5








      5





      $begingroup$

      It's not hard to transform the Graphics3D generated by Plot3D if you understand its structure. We already have numbers of posts about this issue so I'd like not to talk about it in this answer, you may check e.g. this post for more info. Here comes the code, notice I've made use of the new-in-v12 feature of Callout to create ticks, which is more troublesome compared to the transforming part in my opinion:



      old = Pi First@Triangle    
      begin = {##, 0} & @@@ (π AnglePath[{0, 120 °, 120 °}])
      direction = Normalize /@ Differences@begin;
      p3 = Plot3D[Sin[a/2] Sin[b/2] Sin[(Pi - a - b)/2], {a, b} ∈ Triangle@old];

      {error, tr} = FindGeometricTransform[Most /@ Most@begin, old];

      newp3 = p3 /.
      GraphicsComplex[pts_, rest__] :>GraphicsComplex[SubsetMap[tr, #, {1, 2}] & /@ pts, rest];

      ticks = ListPointPlot3D@Flatten@With[{n = 5},
      Table[Callout[begin[[i]] + direction[[i]] j Pi/n, j Pi/n], {i, 3}, {j, 0, n}]];

      Show[newp3, ticks, Axes -> False, Boxed -> False, PlotRange -> All]


      enter image description here



      Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?






      share|improve this answer











      $endgroup$



      It's not hard to transform the Graphics3D generated by Plot3D if you understand its structure. We already have numbers of posts about this issue so I'd like not to talk about it in this answer, you may check e.g. this post for more info. Here comes the code, notice I've made use of the new-in-v12 feature of Callout to create ticks, which is more troublesome compared to the transforming part in my opinion:



      old = Pi First@Triangle    
      begin = {##, 0} & @@@ (π AnglePath[{0, 120 °, 120 °}])
      direction = Normalize /@ Differences@begin;
      p3 = Plot3D[Sin[a/2] Sin[b/2] Sin[(Pi - a - b)/2], {a, b} ∈ Triangle@old];

      {error, tr} = FindGeometricTransform[Most /@ Most@begin, old];

      newp3 = p3 /.
      GraphicsComplex[pts_, rest__] :>GraphicsComplex[SubsetMap[tr, #, {1, 2}] & /@ pts, rest];

      ticks = ListPointPlot3D@Flatten@With[{n = 5},
      Table[Callout[begin[[i]] + direction[[i]] j Pi/n, j Pi/n], {i, 3}, {j, 0, n}]];

      Show[newp3, ticks, Axes -> False, Boxed -> False, PlotRange -> All]


      enter image description here



      Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Apr 21 at 13:36

























      answered Apr 20 at 8:15









      xzczdxzczd

      27.9k576258




      27.9k576258












      • $begingroup$
        My intension is to compare LHS and RHS of the inequalities. In this case, 6r/R = 24 Sin[A/2]Sin[B/2]Sin[C/2] <= CyclicSum (Sin[A]+Sin[B])/Cot[C/2], where r is the incenter and circumcenter of a triangle. I deleted 3rd line and 7th line. I replaced 6th line by newp3 = p3/. GraphicsComplex[pts_,rest_] -> GraphicsComplex[Map[tr, #,{1,2}]&/@pts,rest]. Is this change allowable? My version of Mathematica 8 does not include SubsetMap. I could plot similar figures as yours. But the base of the plot is isosceles, not equilateral. I found this by rotation.
        $endgroup$
        – seiichikiri
        Apr 21 at 3:01










      • $begingroup$
        @seiichikiri No, your usage of Map[...] is incorrect. It should be e.g. newp3 = p3 /. GraphicsComplex[pts_, rest__] :> GraphicsComplex[Join[tr[{#, #2}], {#3}] & @@@ pts, rest]. The shape of triangle looks incorrect because I've added PlotRange -> All and the default BoxRatios is {1, 1, 0.4} in this case.
        $endgroup$
        – xzczd
        Apr 21 at 13:29










      • $begingroup$
        @xzcxd I obtained what I wanted. Thank you very much!
        $endgroup$
        – seiichikiri
        Apr 22 at 0:25










      • $begingroup$
        @seiichikiri Glad it help. If my answer resolves your problem, you can accept it by clicking the checkmark sign.
        $endgroup$
        – xzczd
        2 days ago


















      • $begingroup$
        My intension is to compare LHS and RHS of the inequalities. In this case, 6r/R = 24 Sin[A/2]Sin[B/2]Sin[C/2] <= CyclicSum (Sin[A]+Sin[B])/Cot[C/2], where r is the incenter and circumcenter of a triangle. I deleted 3rd line and 7th line. I replaced 6th line by newp3 = p3/. GraphicsComplex[pts_,rest_] -> GraphicsComplex[Map[tr, #,{1,2}]&/@pts,rest]. Is this change allowable? My version of Mathematica 8 does not include SubsetMap. I could plot similar figures as yours. But the base of the plot is isosceles, not equilateral. I found this by rotation.
        $endgroup$
        – seiichikiri
        Apr 21 at 3:01










      • $begingroup$
        @seiichikiri No, your usage of Map[...] is incorrect. It should be e.g. newp3 = p3 /. GraphicsComplex[pts_, rest__] :> GraphicsComplex[Join[tr[{#, #2}], {#3}] & @@@ pts, rest]. The shape of triangle looks incorrect because I've added PlotRange -> All and the default BoxRatios is {1, 1, 0.4} in this case.
        $endgroup$
        – xzczd
        Apr 21 at 13:29










      • $begingroup$
        @xzcxd I obtained what I wanted. Thank you very much!
        $endgroup$
        – seiichikiri
        Apr 22 at 0:25










      • $begingroup$
        @seiichikiri Glad it help. If my answer resolves your problem, you can accept it by clicking the checkmark sign.
        $endgroup$
        – xzczd
        2 days ago
















      $begingroup$
      My intension is to compare LHS and RHS of the inequalities. In this case, 6r/R = 24 Sin[A/2]Sin[B/2]Sin[C/2] <= CyclicSum (Sin[A]+Sin[B])/Cot[C/2], where r is the incenter and circumcenter of a triangle. I deleted 3rd line and 7th line. I replaced 6th line by newp3 = p3/. GraphicsComplex[pts_,rest_] -> GraphicsComplex[Map[tr, #,{1,2}]&/@pts,rest]. Is this change allowable? My version of Mathematica 8 does not include SubsetMap. I could plot similar figures as yours. But the base of the plot is isosceles, not equilateral. I found this by rotation.
      $endgroup$
      – seiichikiri
      Apr 21 at 3:01




      $begingroup$
      My intension is to compare LHS and RHS of the inequalities. In this case, 6r/R = 24 Sin[A/2]Sin[B/2]Sin[C/2] <= CyclicSum (Sin[A]+Sin[B])/Cot[C/2], where r is the incenter and circumcenter of a triangle. I deleted 3rd line and 7th line. I replaced 6th line by newp3 = p3/. GraphicsComplex[pts_,rest_] -> GraphicsComplex[Map[tr, #,{1,2}]&/@pts,rest]. Is this change allowable? My version of Mathematica 8 does not include SubsetMap. I could plot similar figures as yours. But the base of the plot is isosceles, not equilateral. I found this by rotation.
      $endgroup$
      – seiichikiri
      Apr 21 at 3:01












      $begingroup$
      @seiichikiri No, your usage of Map[...] is incorrect. It should be e.g. newp3 = p3 /. GraphicsComplex[pts_, rest__] :> GraphicsComplex[Join[tr[{#, #2}], {#3}] & @@@ pts, rest]. The shape of triangle looks incorrect because I've added PlotRange -> All and the default BoxRatios is {1, 1, 0.4} in this case.
      $endgroup$
      – xzczd
      Apr 21 at 13:29




      $begingroup$
      @seiichikiri No, your usage of Map[...] is incorrect. It should be e.g. newp3 = p3 /. GraphicsComplex[pts_, rest__] :> GraphicsComplex[Join[tr[{#, #2}], {#3}] & @@@ pts, rest]. The shape of triangle looks incorrect because I've added PlotRange -> All and the default BoxRatios is {1, 1, 0.4} in this case.
      $endgroup$
      – xzczd
      Apr 21 at 13:29












      $begingroup$
      @xzcxd I obtained what I wanted. Thank you very much!
      $endgroup$
      – seiichikiri
      Apr 22 at 0:25




      $begingroup$
      @xzcxd I obtained what I wanted. Thank you very much!
      $endgroup$
      – seiichikiri
      Apr 22 at 0:25












      $begingroup$
      @seiichikiri Glad it help. If my answer resolves your problem, you can accept it by clicking the checkmark sign.
      $endgroup$
      – xzczd
      2 days ago




      $begingroup$
      @seiichikiri Glad it help. If my answer resolves your problem, you can accept it by clicking the checkmark sign.
      $endgroup$
      – xzczd
      2 days ago











      5












      $begingroup$

      Without using any transformations, you have



      $$
      A = frac13 - x - frac{y}{sqrt{3}}\
      B = frac13 + x - frac{y}{sqrt{3}}\
      C = frac13 + frac{2 y}{sqrt{3}}
      $$



      In this form, they span the ranges $[0,1]$ over an equilateral triangle with unit edges, and satisfy $A+B+C=1$. In what follows I'll use $a$, $b$, $c$ instead of the capital letters because it's not a good idea to use capital letters for your own definitions in Mathematica.



      Plotting your function, you need to multiply these with $pi$ to get your desired range.



      Here's a very simplistic way of plotting that does not generate any tick marks. It is mostly for getting a quick overview. If you want proper tick marks you need to follow some of the other recommendations, for example on question 39733. Also, MeshFunctions can give interesting meshes when combined with the effective coordinates $a$, $b$, $c$.



      f[a_, b_, c_] = Sin[π*a/2] Sin[π*b/2] Sin[π*c/2];
      DensityPlot[f[1/3-x-y/Sqrt[3], 1/3+x-y/Sqrt[3], 1/3+2y/Sqrt[3]],
      {x, -0.6, 0.6}, {y, -0.4, 0.7},
      RegionFunction -> Function[{x, y}, 0<=1/3-x-y/Sqrt[3]<=1 && 0<=1/3+x-y/Sqrt[3]<=1 && 0<=1/3+2y/Sqrt[3]<=1],
      AspectRatio -> Automatic,
      Epilog -> {Text["A", {-1/2, -1/(2 Sqrt[3])}, {Sqrt[3]/2, 1/2}],
      Text["B", {1/2, -1/(2 Sqrt[3])}, {-Sqrt[3]/2, 1/2}],
      Text["C", {0, 1/Sqrt[3]}, {0, -1}]}]


      enter image description here



      Here is what happens if we set the function $f(a,b,c)$ to either $a$, $b$, or $c$: you can see the behavior of these coordinates,



      enter image description here






      share|improve this answer











      $endgroup$


















        5












        $begingroup$

        Without using any transformations, you have



        $$
        A = frac13 - x - frac{y}{sqrt{3}}\
        B = frac13 + x - frac{y}{sqrt{3}}\
        C = frac13 + frac{2 y}{sqrt{3}}
        $$



        In this form, they span the ranges $[0,1]$ over an equilateral triangle with unit edges, and satisfy $A+B+C=1$. In what follows I'll use $a$, $b$, $c$ instead of the capital letters because it's not a good idea to use capital letters for your own definitions in Mathematica.



        Plotting your function, you need to multiply these with $pi$ to get your desired range.



        Here's a very simplistic way of plotting that does not generate any tick marks. It is mostly for getting a quick overview. If you want proper tick marks you need to follow some of the other recommendations, for example on question 39733. Also, MeshFunctions can give interesting meshes when combined with the effective coordinates $a$, $b$, $c$.



        f[a_, b_, c_] = Sin[π*a/2] Sin[π*b/2] Sin[π*c/2];
        DensityPlot[f[1/3-x-y/Sqrt[3], 1/3+x-y/Sqrt[3], 1/3+2y/Sqrt[3]],
        {x, -0.6, 0.6}, {y, -0.4, 0.7},
        RegionFunction -> Function[{x, y}, 0<=1/3-x-y/Sqrt[3]<=1 && 0<=1/3+x-y/Sqrt[3]<=1 && 0<=1/3+2y/Sqrt[3]<=1],
        AspectRatio -> Automatic,
        Epilog -> {Text["A", {-1/2, -1/(2 Sqrt[3])}, {Sqrt[3]/2, 1/2}],
        Text["B", {1/2, -1/(2 Sqrt[3])}, {-Sqrt[3]/2, 1/2}],
        Text["C", {0, 1/Sqrt[3]}, {0, -1}]}]


        enter image description here



        Here is what happens if we set the function $f(a,b,c)$ to either $a$, $b$, or $c$: you can see the behavior of these coordinates,



        enter image description here






        share|improve this answer











        $endgroup$
















          5












          5








          5





          $begingroup$

          Without using any transformations, you have



          $$
          A = frac13 - x - frac{y}{sqrt{3}}\
          B = frac13 + x - frac{y}{sqrt{3}}\
          C = frac13 + frac{2 y}{sqrt{3}}
          $$



          In this form, they span the ranges $[0,1]$ over an equilateral triangle with unit edges, and satisfy $A+B+C=1$. In what follows I'll use $a$, $b$, $c$ instead of the capital letters because it's not a good idea to use capital letters for your own definitions in Mathematica.



          Plotting your function, you need to multiply these with $pi$ to get your desired range.



          Here's a very simplistic way of plotting that does not generate any tick marks. It is mostly for getting a quick overview. If you want proper tick marks you need to follow some of the other recommendations, for example on question 39733. Also, MeshFunctions can give interesting meshes when combined with the effective coordinates $a$, $b$, $c$.



          f[a_, b_, c_] = Sin[π*a/2] Sin[π*b/2] Sin[π*c/2];
          DensityPlot[f[1/3-x-y/Sqrt[3], 1/3+x-y/Sqrt[3], 1/3+2y/Sqrt[3]],
          {x, -0.6, 0.6}, {y, -0.4, 0.7},
          RegionFunction -> Function[{x, y}, 0<=1/3-x-y/Sqrt[3]<=1 && 0<=1/3+x-y/Sqrt[3]<=1 && 0<=1/3+2y/Sqrt[3]<=1],
          AspectRatio -> Automatic,
          Epilog -> {Text["A", {-1/2, -1/(2 Sqrt[3])}, {Sqrt[3]/2, 1/2}],
          Text["B", {1/2, -1/(2 Sqrt[3])}, {-Sqrt[3]/2, 1/2}],
          Text["C", {0, 1/Sqrt[3]}, {0, -1}]}]


          enter image description here



          Here is what happens if we set the function $f(a,b,c)$ to either $a$, $b$, or $c$: you can see the behavior of these coordinates,



          enter image description here






          share|improve this answer











          $endgroup$



          Without using any transformations, you have



          $$
          A = frac13 - x - frac{y}{sqrt{3}}\
          B = frac13 + x - frac{y}{sqrt{3}}\
          C = frac13 + frac{2 y}{sqrt{3}}
          $$



          In this form, they span the ranges $[0,1]$ over an equilateral triangle with unit edges, and satisfy $A+B+C=1$. In what follows I'll use $a$, $b$, $c$ instead of the capital letters because it's not a good idea to use capital letters for your own definitions in Mathematica.



          Plotting your function, you need to multiply these with $pi$ to get your desired range.



          Here's a very simplistic way of plotting that does not generate any tick marks. It is mostly for getting a quick overview. If you want proper tick marks you need to follow some of the other recommendations, for example on question 39733. Also, MeshFunctions can give interesting meshes when combined with the effective coordinates $a$, $b$, $c$.



          f[a_, b_, c_] = Sin[π*a/2] Sin[π*b/2] Sin[π*c/2];
          DensityPlot[f[1/3-x-y/Sqrt[3], 1/3+x-y/Sqrt[3], 1/3+2y/Sqrt[3]],
          {x, -0.6, 0.6}, {y, -0.4, 0.7},
          RegionFunction -> Function[{x, y}, 0<=1/3-x-y/Sqrt[3]<=1 && 0<=1/3+x-y/Sqrt[3]<=1 && 0<=1/3+2y/Sqrt[3]<=1],
          AspectRatio -> Automatic,
          Epilog -> {Text["A", {-1/2, -1/(2 Sqrt[3])}, {Sqrt[3]/2, 1/2}],
          Text["B", {1/2, -1/(2 Sqrt[3])}, {-Sqrt[3]/2, 1/2}],
          Text["C", {0, 1/Sqrt[3]}, {0, -1}]}]


          enter image description here



          Here is what happens if we set the function $f(a,b,c)$ to either $a$, $b$, or $c$: you can see the behavior of these coordinates,



          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Apr 20 at 10:19

























          answered Apr 20 at 7:46









          RomanRoman

          6,03611132




          6,03611132























              2












              $begingroup$

              DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
              {a, 0, 2}, {b, 0, 2}, {c, 0, 2}]


              enter image description here






              share|improve this answer









              $endgroup$


















                2












                $begingroup$

                DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
                {a, 0, 2}, {b, 0, 2}, {c, 0, 2}]


                enter image description here






                share|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
                  {a, 0, 2}, {b, 0, 2}, {c, 0, 2}]


                  enter image description here






                  share|improve this answer









                  $endgroup$



                  DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
                  {a, 0, 2}, {b, 0, 2}, {c, 0, 2}]


                  enter image description here







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Apr 20 at 7:02









                  David G. StorkDavid G. Stork

                  24.9k22155




                  24.9k22155






















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










                      draft saved

                      draft discarded


















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













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












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
















                      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%2f195633%2fhow-to-ternary-plot3d-a-function%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

                      Plaza Victoria

                      Puebla de Zaragoza

                      Musa