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
$begingroup$
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 Plot3D
ed function inside the equilateral triangle with FindGeometricTransform
? If there is a simpler method, I would like to know it.
plotting
New contributor
$endgroup$
add a comment |
$begingroup$
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 Plot3D
ed function inside the equilateral triangle with FindGeometricTransform
? If there is a simpler method, I would like to know it.
plotting
New contributor
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 Plot3D
ed function inside the equilateral triangle with FindGeometricTransform
? If there is a simpler method, I would like to know it.
plotting
New contributor
$endgroup$
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 Plot3D
ed function inside the equilateral triangle with FindGeometricTransform
? If there is a simpler method, I would like to know it.
plotting
plotting
New contributor
New contributor
edited 19 hours ago
xzczd
27.8k575258
27.8k575258
New contributor
asked 19 hours ago
seiichikiriseiichikiri
211
211
New contributor
New contributor
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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]
Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?
$endgroup$
add a comment |
$begingroup$
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]]
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,
$endgroup$
add a comment |
$begingroup$
DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
a, 0, 2, b, 0, 2, c, 0, 2]
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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]
Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?
$endgroup$
add a comment |
$begingroup$
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]
Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?
$endgroup$
add a comment |
$begingroup$
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]
Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?
$endgroup$
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]
Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?
edited 14 hours ago
answered 18 hours ago
xzczdxzczd
27.8k575258
27.8k575258
add a comment |
add a comment |
$begingroup$
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]]
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,
$endgroup$
add a comment |
$begingroup$
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]]
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,
$endgroup$
add a comment |
$begingroup$
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]]
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,
$endgroup$
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]]
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,
edited 16 hours ago
answered 18 hours ago
RomanRoman
5,80111131
5,80111131
add a comment |
add a comment |
$begingroup$
DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
a, 0, 2, b, 0, 2, c, 0, 2]
$endgroup$
add a comment |
$begingroup$
DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
a, 0, 2, b, 0, 2, c, 0, 2]
$endgroup$
add a comment |
$begingroup$
DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
a, 0, 2, b, 0, 2, c, 0, 2]
$endgroup$
DensityPlot3D[Sin[a/2] Sin[b/2] Sin[c/2],
a, 0, 2, b, 0, 2, c, 0, 2]
answered 19 hours ago
David G. StorkDavid G. Stork
24.9k22155
24.9k22155
add a comment |
add a comment |
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.
seiichikiri is a new contributor. Be nice, and check out our Code of Conduct.
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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
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– seiichikiri
18 hours ago
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If you want to clarify the question further, you can click the edit button in the left-bottom corner of your question .
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– xzczd
18 hours ago