How to ternary Plot3D a function
$begingroup$
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
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
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 Apr 20 at 6:42
xzczd
27.9k576258
27.9k576258
New contributor
asked Apr 20 at 6:31
seiichikiriseiichikiri
282
282
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
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
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
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
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 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?
$endgroup$
$begingroup$
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 ofMap[...]
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 addedPlotRange -> All
and the defaultBoxRatios
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
add a comment |
$begingroup$
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}]}]
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
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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 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?
$endgroup$
$begingroup$
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 ofMap[...]
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 addedPlotRange -> All
and the defaultBoxRatios
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
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 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?
$endgroup$
$begingroup$
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 ofMap[...]
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 addedPlotRange -> All
and the defaultBoxRatios
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
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 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?
$endgroup$
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]
Hmm… the result doesn't look that great on Wolfram cloud, perhaps it'll be better on Mathematica Desktop?
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 ofMap[...]
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 addedPlotRange -> All
and the defaultBoxRatios
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
add a comment |
$begingroup$
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 ofMap[...]
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 addedPlotRange -> All
and the defaultBoxRatios
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
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@xzcxd I obtained what I wanted. Thank you very much!
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– seiichikiri
Apr 22 at 0:25
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@xzcxd I obtained what I wanted. Thank you very much!
$endgroup$
– seiichikiri
Apr 22 at 0:25
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@seiichikiri Glad it help. If my answer resolves your problem, you can accept it by clicking the checkmark sign.
$endgroup$
– xzczd
2 days ago
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@seiichikiri Glad it help. If my answer resolves your problem, you can accept it by clicking the checkmark sign.
$endgroup$
– xzczd
2 days ago
add a comment |
$begingroup$
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}]}]
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
$$
A = frac13 - x - frac{y}{sqrt{3}}\
B = frac13 + x - frac{y}{sqrt{3}}\
C = frac13 + frac{2 y}{sqrt{3}}
$$
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,
$endgroup$
add a comment |
$begingroup$
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}]}]
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
$$
A = frac13 - x - frac{y}{sqrt{3}}\
B = frac13 + x - frac{y}{sqrt{3}}\
C = frac13 + frac{2 y}{sqrt{3}}
$$
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,
edited Apr 20 at 10:19
answered Apr 20 at 7:46
RomanRoman
6,03611132
6,03611132
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 Apr 20 at 7:02
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.
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seiichikiri is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
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