How to ternary Plot3D a function Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How to plot ternary density plots?Where can I get detailed information on how the Plot command works?How to plot a barycentric lineHow to add a common color legend to a grid of density plots?How to plot ternary density plots?Extending a ternary plot to a tetrahedron (using ListPlot)Mysterious spikes in Plot3DHow to plot the following print?How to plot a region defined by corner pointsDifferent Mesh color in multiple Plot3DMake Plot3D only the real values of a functionGraphic representation of a triangle using ArrayPlot

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How to ternary Plot3D a function



Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?How to plot ternary density plots?Where can I get detailed information on how the Plot command works?How to plot a barycentric lineHow to add a common color legend to a grid of density plots?How to plot ternary density plots?Extending a ternary plot to a tetrahedron (using ListPlot)Mysterious spikes in Plot3DHow to plot the following print?How to plot a region defined by corner pointsDifferent Mesh color in multiple Plot3DMake Plot3D only the real values of a functionGraphic representation of a triangle using ArrayPlot










4












$begingroup$


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
    18 hours ago










  • $begingroup$
    If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
    $endgroup$
    – xzczd
    18 hours ago















4












$begingroup$


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
    18 hours ago










  • $begingroup$
    If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
    $endgroup$
    – xzczd
    18 hours ago













4












4








4





$begingroup$


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 19 hours ago









xzczd

27.8k575258




27.8k575258






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 19 hours ago









seiichikiriseiichikiri

211




211




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
    18 hours ago










  • $begingroup$
    If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
    $endgroup$
    – xzczd
    18 hours ago
















  • $begingroup$
    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
    18 hours ago










  • $begingroup$
    If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
    $endgroup$
    – xzczd
    18 hours ago















$begingroup$
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
18 hours ago




$begingroup$
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
18 hours ago












$begingroup$
If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
$endgroup$
– xzczd
18 hours ago




$begingroup$
If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
$endgroup$
– xzczd
18 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

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 answer, 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, 0, Pi, b, 0, Pi,
RegionFunction -> Function[a, b, Pi - a - b > 0]];

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











$endgroup$




















    3












    $begingroup$

    Without using any transformations, you have



    $$
    A = frac13 - x - fracysqrt3\
    B = frac13 + x - fracysqrt3\
    C = frac13 + frac2 ysqrt3
    $$



    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









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        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 answer, 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, 0, Pi, b, 0, Pi,
        RegionFunction -> Function[a, b, Pi - a - b > 0]];

        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











        $endgroup$

















          4












          $begingroup$

          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 answer, 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, 0, Pi, b, 0, Pi,
          RegionFunction -> Function[a, b, Pi - a - b > 0]];

          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











          $endgroup$















            4












            4








            4





            $begingroup$

            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 answer, 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, 0, Pi, b, 0, Pi,
            RegionFunction -> Function[a, b, Pi - a - b > 0]];

            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 answer, 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, 0, Pi, b, 0, Pi,
            RegionFunction -> Function[a, b, Pi - a - b > 0]];

            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 14 hours ago

























            answered 18 hours ago









            xzczdxzczd

            27.8k575258




            27.8k575258





















                3












                $begingroup$

                Without using any transformations, you have



                $$
                A = frac13 - x - fracysqrt3\
                B = frac13 + x - fracysqrt3\
                C = frac13 + frac2 ysqrt3
                $$



                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











                $endgroup$

















                  3












                  $begingroup$

                  Without using any transformations, you have



                  $$
                  A = frac13 - x - fracysqrt3\
                  B = frac13 + x - fracysqrt3\
                  C = frac13 + frac2 ysqrt3
                  $$



                  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











                  $endgroup$















                    3












                    3








                    3





                    $begingroup$

                    Without using any transformations, you have



                    $$
                    A = frac13 - x - fracysqrt3\
                    B = frac13 + x - fracysqrt3\
                    C = frac13 + frac2 ysqrt3
                    $$



                    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



                    $$
                    A = frac13 - x - fracysqrt3\
                    B = frac13 + x - fracysqrt3\
                    C = frac13 + frac2 ysqrt3
                    $$



                    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 16 hours ago

























                    answered 18 hours ago









                    RomanRoman

                    5,80111131




                    5,80111131





















                        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 19 hours ago









                            David G. StorkDavid G. Stork

                            24.9k22155




                            24.9k22155




















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