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










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.
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$endgroup$











  • $begingroup$
    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
    2 days 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
    2 days ago










  • $begingroup$
    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
    yesterday















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
    2 days 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
    2 days ago










  • $begingroup$
    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
    yesterday













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 2 days ago









xzczd

27.8k576258




27.8k576258






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 2 days ago









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
    2 days 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
    2 days ago










  • $begingroup$
    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
    yesterday
















  • $begingroup$
    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
    2 days 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
    2 days ago










  • $begingroup$
    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
    yesterday















$begingroup$
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
2 days 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
2 days 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
2 days 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
2 days ago












$begingroup$
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
yesterday




$begingroup$
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
yesterday










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
    2 days ago










  • $begingroup$
    @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
    yesterday










  • $begingroup$
    @xzcxd I obtained what I wanted. Thank you very much!
    $endgroup$
    – seiichikiri
    yesterday










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


















5












$begingroup$

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









      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
        2 days ago










      • $begingroup$
        @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
        yesterday










      • $begingroup$
        @xzcxd I obtained what I wanted. Thank you very much!
        $endgroup$
        – seiichikiri
        yesterday










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















      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
        2 days ago










      • $begingroup$
        @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
        yesterday










      • $begingroup$
        @xzcxd I obtained what I wanted. Thank you very much!
        $endgroup$
        – seiichikiri
        yesterday










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













      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 yesterday

























      answered 2 days ago









      xzczdxzczd

      27.8k576258




      27.8k576258











      • $begingroup$
        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
        2 days ago










      • $begingroup$
        @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
        yesterday










      • $begingroup$
        @xzcxd I obtained what I wanted. Thank you very much!
        $endgroup$
        – seiichikiri
        yesterday










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
















      • $begingroup$
        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
        2 days ago










      • $begingroup$
        @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
        yesterday










      • $begingroup$
        @xzcxd I obtained what I wanted. Thank you very much!
        $endgroup$
        – seiichikiri
        yesterday










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















      $begingroup$
      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
      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
      2 days ago












      $begingroup$
      @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
      yesterday




      $begingroup$
      @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
      yesterday












      $begingroup$
      @xzcxd I obtained what I wanted. Thank you very much!
      $endgroup$
      – seiichikiri
      yesterday




      $begingroup$
      @xzcxd I obtained what I wanted. Thank you very much!
      $endgroup$
      – seiichikiri
      yesterday












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




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











      5












      $begingroup$

      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











      $endgroup$

















        5












        $begingroup$

        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











        $endgroup$















          5












          5








          5





          $begingroup$

          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 2 days ago

























          answered 2 days ago









          RomanRoman

          5,98111132




          5,98111132





















              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 2 days ago









                  David G. StorkDavid G. Stork

                  24.9k22155




                  24.9k22155




















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