How to ternary Plot3D a function

Question

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.

---

An example of ternary 3D plot of a geometric inequality

Let ABC be a triangle and its radius of circumcircle R = 1.

In ternary 3D plot, based on A, B, C > 0 and A + B + C = Pi, the value of a function is plotted 3-dimensionally on the equilateral domain.

Denote:

a, b, c = the sidelengths of ABC respectively.

s = (a + b + c)/2.

r = the radii of incenter.

ha, hb, hc = the altitudes of ABC respectively.

The bottom (yellow), middle (magenta) and top (cyan) surfaces show the values of left-hand-side, middle and right-hand-side functions of the following inequality by Mr. George Apostolopouls.

Ga19Mar30: 6rs/R ≤ Sin[A](hb + hc) + Sin[B](hc + ha) + Sin[C](ha + ha) ≤ 3s.

Answer 1

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]

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

Answer 2

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}]}]

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,

Answer 3

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