Putting a 2D region plot under a 3D plot The Next CEO of Stack OverflowPutting ticks with values on a parametric plotPlot above desired regionIntersecting RegionPlotsRegion plot with predicatePlot region from given coordinatesPutting arrows in 3d plotFilling the region under a lineHow Plot this RegionPlot a function under constraintregion plot with region function
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Putting a 2D region plot under a 3D plot
The Next CEO of Stack OverflowPutting ticks with values on a parametric plotPlot above desired regionIntersecting RegionPlotsRegion plot with predicatePlot region from given coordinatesPutting arrows in 3d plotFilling the region under a lineHow Plot this RegionPlot a function under constraintregion plot with region function
$begingroup$
I am trying to plot this 3D function over a hexagonal region:
a1 = Sqrt[3] 1, 0;
a2 = Sqrt[3] 1/2, Sqrt[3]/2;
k = kx, ky;
S = 1 + Exp[I k. a2] + Exp[I k.(a2 - a1)];
EE = Abs[S]
R = 4 Pi/(3 Sqrt[3]);
ep = Plot3D[EE, -EE, kx, ky [Element] RegularPolygon[R, 6], Axes -> False, Boxed -> False, AspectRatio -> 2]
This works fine, but I would also like to draw the region under the 3D graph, something like:
bz = Graphics[RegularPolygon[R, 6]];
or
bz = RegionPlot[RegularPolygon[R, 6]];
However, using
Show[ep, bz]
doesn't work. I have found a few similar questions but they mostly seem to be about contours, I didn't know how to extend this for something as simple as a regular polygon.
plotting regions
edited 12 mins ago
J. M. is slightly pensive♦
98.7k10310467
asked 5 hours ago
AshAsh
185
$endgroup$
add a comment |
$begingroup$
I am trying to plot this 3D function over a hexagonal region:
a1 = Sqrt[3] 1, 0;
a2 = Sqrt[3] 1/2, Sqrt[3]/2;
k = kx, ky;
S = 1 + Exp[I k. a2] + Exp[I k.(a2 - a1)];
EE = Abs[S]
R = 4 Pi/(3 Sqrt[3]);
ep = Plot3D[EE, -EE, kx, ky [Element] RegularPolygon[R, 6], Axes -> False, Boxed -> False, AspectRatio -> 2]
This works fine, but I would also like to draw the region under the 3D graph, something like:
bz = Graphics[RegularPolygon[R, 6]];
or
bz = RegionPlot[RegularPolygon[R, 6]];
However, using
Show[ep, bz]
doesn't work. I have found a few similar questions but they mostly seem to be about contours, I didn't know how to extend this for something as simple as a regular polygon.
plotting regions
edited 12 mins ago
J. M. is slightly pensive♦
98.7k10310467
asked 5 hours ago
AshAsh
185
$endgroup$
$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
$endgroup$
– Alrubaie
4 hours ago
add a comment |
$begingroup$
I am trying to plot this 3D function over a hexagonal region:
a1 = Sqrt[3] 1, 0;
a2 = Sqrt[3] 1/2, Sqrt[3]/2;
k = kx, ky;
S = 1 + Exp[I k. a2] + Exp[I k.(a2 - a1)];
EE = Abs[S]
R = 4 Pi/(3 Sqrt[3]);
ep = Plot3D[EE, -EE, kx, ky [Element] RegularPolygon[R, 6], Axes -> False, Boxed -> False, AspectRatio -> 2]
This works fine, but I would also like to draw the region under the 3D graph, something like:
bz = Graphics[RegularPolygon[R, 6]];
or
bz = RegionPlot[RegularPolygon[R, 6]];
However, using
Show[ep, bz]
doesn't work. I have found a few similar questions but they mostly seem to be about contours, I didn't know how to extend this for something as simple as a regular polygon.
plotting regions
edited 12 mins ago
J. M. is slightly pensive♦
98.7k10310467
asked 5 hours ago
AshAsh
185
$endgroup$
I am trying to plot this 3D function over a hexagonal region:
a1 = Sqrt[3] 1, 0;
a2 = Sqrt[3] 1/2, Sqrt[3]/2;
k = kx, ky;
S = 1 + Exp[I k. a2] + Exp[I k.(a2 - a1)];
EE = Abs[S]
R = 4 Pi/(3 Sqrt[3]);
ep = Plot3D[EE, -EE, kx, ky [Element] RegularPolygon[R, 6], Axes -> False, Boxed -> False, AspectRatio -> 2]
This works fine, but I would also like to draw the region under the 3D graph, something like:
bz = Graphics[RegularPolygon[R, 6]];
or
bz = RegionPlot[RegularPolygon[R, 6]];
However, using
Show[ep, bz]
doesn't work. I have found a few similar questions but they mostly seem to be about contours, I didn't know how to extend this for something as simple as a regular polygon.
plotting regions
plotting regions
edited 12 mins ago
J. M. is slightly pensive♦
98.7k10310467
asked 5 hours ago
AshAsh
185
edited 12 mins ago
J. M. is slightly pensive♦
98.7k10310467
asked 5 hours ago
AshAsh
185
edited 12 mins ago
J. M. is slightly pensive♦
98.7k10310467
edited 12 mins ago
J. M. is slightly pensive♦
98.7k10310467
edited 12 mins ago
J. M. is slightly pensive♦
98.7k10310467
98.7k10310467
asked 5 hours ago
AshAsh
185
asked 5 hours ago
AshAsh
185
asked 5 hours ago
AshAsh
185
185
$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
$endgroup$
– Alrubaie
4 hours ago
add a comment |
$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
$endgroup$
– Alrubaie
4 hours ago
$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
$endgroup$
– Alrubaie
4 hours ago
$begingroup$
Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
$endgroup$
– Alrubaie
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Try this:
region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]
answered 4 hours ago
MarcoBMarcoB
38k556114
$endgroup$
$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
4 hours ago
$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
4 hours ago
$begingroup$
That's nice! WhatAspectRatiodid you use?
$endgroup$
– mjw
3 hours ago
$begingroup$
@mjw Thank you! The aspect ratio is inherited fromep, which was in the OP's original code; they had set it to $2$.
$endgroup$
– MarcoB
3 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try this:
region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]
answered 4 hours ago
MarcoBMarcoB
38k556114
$endgroup$
$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
4 hours ago
$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
4 hours ago
$begingroup$
That's nice! WhatAspectRatiodid you use?
$endgroup$
– mjw
3 hours ago
$begingroup$
@mjw Thank you! The aspect ratio is inherited fromep, which was in the OP's original code; they had set it to $2$.
$endgroup$
– MarcoB
3 hours ago
add a comment |
$begingroup$
Try this:
region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]
answered 4 hours ago
MarcoBMarcoB
38k556114
$endgroup$
$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
4 hours ago
$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
4 hours ago
$begingroup$
That's nice! WhatAspectRatiodid you use?
$endgroup$
– mjw
3 hours ago
$begingroup$
@mjw Thank you! The aspect ratio is inherited fromep, which was in the OP's original code; they had set it to $2$.
$endgroup$
– MarcoB
3 hours ago
add a comment |
$begingroup$
Try this:
region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]
answered 4 hours ago
MarcoBMarcoB
38k556114
$endgroup$
Try this:
region = Graphics3D[Polygon[CirclePoints[R, 6] /. x_, y_ :> x, y, -3]];
Show[ep,region]
answered 4 hours ago
MarcoBMarcoB
38k556114
answered 4 hours ago
MarcoBMarcoB
38k556114
answered 4 hours ago
MarcoBMarcoB
38k556114
answered 4 hours ago
MarcoBMarcoB
38k556114
38k556114
$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
4 hours ago
$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
4 hours ago
$begingroup$
That's nice! WhatAspectRatiodid you use?
$endgroup$
– mjw
3 hours ago
$begingroup$
@mjw Thank you! The aspect ratio is inherited fromep, which was in the OP's original code; they had set it to $2$.
$endgroup$
– MarcoB
3 hours ago
add a comment |
$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
4 hours ago
$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
4 hours ago
$begingroup$
That's nice! WhatAspectRatiodid you use?
$endgroup$
– mjw
3 hours ago
$begingroup$
@mjw Thank you! The aspect ratio is inherited fromep, which was in the OP's original code; they had set it to $2$.
$endgroup$
– MarcoB
3 hours ago
$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
4 hours ago
$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
4 hours ago
$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
4 hours ago
$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
4 hours ago
$begingroup$
That's nice! What
AspectRatio did you use?$endgroup$
– mjw
3 hours ago
$begingroup$
That's nice! What
AspectRatio did you use?$endgroup$
– mjw
3 hours ago
$begingroup$
@mjw Thank you! The aspect ratio is inherited from
ep, which was in the OP's original code; they had set it to $2$.$endgroup$
– MarcoB
3 hours ago
$begingroup$
@mjw Thank you! The aspect ratio is inherited from
ep, which was in the OP's original code; they had set it to $2$.$endgroup$
– MarcoB
3 hours ago
add a comment |
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$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
Try This p = Graphics3D[Polygon[1, 0, 0, 0, 1, 0, 0, 0, 1]]
$endgroup$
– Alrubaie
4 hours ago