Putting a 2D region plot under a 3D plot












3












$begingroup$


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.










share|improve this question











$endgroup$












  • $begingroup$
    do you want you Polygon flat under the region of 3D !?
    $endgroup$
    – Alrubaie
    Mar 28 at 20:28










  • $begingroup$
    Try This p = Graphics3D[Polygon[{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}]]
    $endgroup$
    – Alrubaie
    Mar 28 at 20:30
















3












$begingroup$


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.










share|improve this question











$endgroup$












  • $begingroup$
    do you want you Polygon flat under the region of 3D !?
    $endgroup$
    – Alrubaie
    Mar 28 at 20:28










  • $begingroup$
    Try This p = Graphics3D[Polygon[{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}]]
    $endgroup$
    – Alrubaie
    Mar 28 at 20:30














3












3








3





$begingroup$


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.










share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Mar 29 at 1:16









J. M. is slightly pensive

98.8k10311467




98.8k10311467










asked Mar 28 at 20:02









AshAsh

235




235












  • $begingroup$
    do you want you Polygon flat under the region of 3D !?
    $endgroup$
    – Alrubaie
    Mar 28 at 20:28










  • $begingroup$
    Try This p = Graphics3D[Polygon[{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}]]
    $endgroup$
    – Alrubaie
    Mar 28 at 20:30


















  • $begingroup$
    do you want you Polygon flat under the region of 3D !?
    $endgroup$
    – Alrubaie
    Mar 28 at 20:28










  • $begingroup$
    Try This p = Graphics3D[Polygon[{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}]]
    $endgroup$
    – Alrubaie
    Mar 28 at 20:30
















$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
Mar 28 at 20:28




$begingroup$
do you want you Polygon flat under the region of 3D !?
$endgroup$
– Alrubaie
Mar 28 at 20:28












$begingroup$
Try This p = Graphics3D[Polygon[{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}]]
$endgroup$
– Alrubaie
Mar 28 at 20:30




$begingroup$
Try This p = Graphics3D[Polygon[{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}]]
$endgroup$
– Alrubaie
Mar 28 at 20:30










1 Answer
1






active

oldest

votes


















5












$begingroup$

Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. {x_, y_} :> {x, y, -3}]];
Show[ep,region]


Mathematica graphics






share|improve this answer









$endgroup$













  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    Mar 28 at 20:41










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    Mar 28 at 20:42












  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    Mar 28 at 21:49










  • $begingroup$
    @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
    Mar 28 at 22:18












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. {x_, y_} :> {x, y, -3}]];
Show[ep,region]


Mathematica graphics






share|improve this answer









$endgroup$













  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    Mar 28 at 20:41










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    Mar 28 at 20:42












  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    Mar 28 at 21:49










  • $begingroup$
    @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
    Mar 28 at 22:18
















5












$begingroup$

Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. {x_, y_} :> {x, y, -3}]];
Show[ep,region]


Mathematica graphics






share|improve this answer









$endgroup$













  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    Mar 28 at 20:41










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    Mar 28 at 20:42












  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    Mar 28 at 21:49










  • $begingroup$
    @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
    Mar 28 at 22:18














5












5








5





$begingroup$

Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. {x_, y_} :> {x, y, -3}]];
Show[ep,region]


Mathematica graphics






share|improve this answer









$endgroup$



Try this:



region = Graphics3D[Polygon[CirclePoints[R, 6] /. {x_, y_} :> {x, y, -3}]];
Show[ep,region]


Mathematica graphics







share|improve this answer












share|improve this answer



share|improve this answer










answered Mar 28 at 20:35









MarcoBMarcoB

38.2k556114




38.2k556114












  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    Mar 28 at 20:41










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    Mar 28 at 20:42












  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    Mar 28 at 21:49










  • $begingroup$
    @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
    Mar 28 at 22:18


















  • $begingroup$
    Works perfectly, thank you!
    $endgroup$
    – Ash
    Mar 28 at 20:41










  • $begingroup$
    @Ash You are very welcome! Thank you for the accept as well!
    $endgroup$
    – MarcoB
    Mar 28 at 20:42












  • $begingroup$
    That's nice! What AspectRatio did you use?
    $endgroup$
    – mjw
    Mar 28 at 21:49










  • $begingroup$
    @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
    Mar 28 at 22:18
















$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
Mar 28 at 20:41




$begingroup$
Works perfectly, thank you!
$endgroup$
– Ash
Mar 28 at 20:41












$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
Mar 28 at 20:42






$begingroup$
@Ash You are very welcome! Thank you for the accept as well!
$endgroup$
– MarcoB
Mar 28 at 20:42














$begingroup$
That's nice! What AspectRatio did you use?
$endgroup$
– mjw
Mar 28 at 21:49




$begingroup$
That's nice! What AspectRatio did you use?
$endgroup$
– mjw
Mar 28 at 21:49












$begingroup$
@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
Mar 28 at 22:18




$begingroup$
@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
Mar 28 at 22:18


















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