Asymptote: 3d graph over a disc
5
Is there a straightforward way to draw a 3D graph over a disc domain? Say
z=x^2-y^2 for x^2+y^2<1.
[I just started to use asymptote; this page explained me how to do it for a rectangular domain. I hope it is an easy question.]
graphs asymptote
asked Apr 1 at 3:36
Anton PetruninAnton Petrunin
542313
add a comment |
5
Is there a straightforward way to draw a 3D graph over a disc domain? Say
z=x^2-y^2 for x^2+y^2<1.
[I just started to use asymptote; this page explained me how to do it for a rectangular domain. I hope it is an easy question.]
graphs asymptote
asked Apr 1 at 3:36
Anton PetruninAnton Petrunin
542313
add a comment |
5
5
5
Is there a straightforward way to draw a 3D graph over a disc domain? Say
z=x^2-y^2 for x^2+y^2<1.
[I just started to use asymptote; this page explained me how to do it for a rectangular domain. I hope it is an easy question.]
graphs asymptote
asked Apr 1 at 3:36
Anton PetruninAnton Petrunin
542313
Is there a straightforward way to draw a 3D graph over a disc domain? Say
z=x^2-y^2 for x^2+y^2<1.
[I just started to use asymptote; this page explained me how to do it for a rectangular domain. I hope it is an easy question.]
graphs asymptote
graphs asymptote
asked Apr 1 at 3:36
Anton PetruninAnton Petrunin
542313
asked Apr 1 at 3:36
Anton PetruninAnton Petrunin
542313
asked Apr 1 at 3:36
Anton PetruninAnton Petrunin
542313
asked Apr 1 at 3:36
Anton PetruninAnton Petrunin
542313
asked Apr 1 at 3:36
Anton PetruninAnton Petrunin
542313
542313
add a comment |
add a comment |
1 Answer
1
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5
One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi).
documentclass[variwidth,border=3.14mm]{standalone}
usepackage{asypictureB}
begin{document}
begin{asypicture}{name=discgraph}
usepackage("mathrsfs");
import graph3;
import solids;
import interpolate;
settings.outformat="pdf";
size(500);
defaultpen(0.5mm);
pen darkgreen=rgb(0,138/255,122/255);
draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);
//function: call the radial coordinate r=t.x and the angle phi=t.y
triple f(pair t) {
return ((t.x)*cos(t.y), (t.x)*sin(t.y),
((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
}
surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
usplinetype=new splinetype {notaknot,notaknot,monotonic},
vsplinetype=Spline);
pen p=rgb(0,0,.7);
draw(s,lightolive+white);
end{asypicture}
end{document}
answered Apr 1 at 3:58
marmotmarmot
114k5145276
Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
– Anton Petrunin
Apr 1 at 4:31
@marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
– Marian G.
2 days ago
2
A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
– O.G.
2 days 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
5
One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi).
documentclass[variwidth,border=3.14mm]{standalone}
usepackage{asypictureB}
begin{document}
begin{asypicture}{name=discgraph}
usepackage("mathrsfs");
import graph3;
import solids;
import interpolate;
settings.outformat="pdf";
size(500);
defaultpen(0.5mm);
pen darkgreen=rgb(0,138/255,122/255);
draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);
//function: call the radial coordinate r=t.x and the angle phi=t.y
triple f(pair t) {
return ((t.x)*cos(t.y), (t.x)*sin(t.y),
((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
}
surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
usplinetype=new splinetype {notaknot,notaknot,monotonic},
vsplinetype=Spline);
pen p=rgb(0,0,.7);
draw(s,lightolive+white);
end{asypicture}
end{document}
answered Apr 1 at 3:58
marmotmarmot
114k5145276
Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
– Anton Petrunin
Apr 1 at 4:31
@marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
– Marian G.
2 days ago
2
A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
– O.G.
2 days ago
add a comment |
5
One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi).
documentclass[variwidth,border=3.14mm]{standalone}
usepackage{asypictureB}
begin{document}
begin{asypicture}{name=discgraph}
usepackage("mathrsfs");
import graph3;
import solids;
import interpolate;
settings.outformat="pdf";
size(500);
defaultpen(0.5mm);
pen darkgreen=rgb(0,138/255,122/255);
draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);
//function: call the radial coordinate r=t.x and the angle phi=t.y
triple f(pair t) {
return ((t.x)*cos(t.y), (t.x)*sin(t.y),
((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
}
surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
usplinetype=new splinetype {notaknot,notaknot,monotonic},
vsplinetype=Spline);
pen p=rgb(0,0,.7);
draw(s,lightolive+white);
end{asypicture}
end{document}
answered Apr 1 at 3:58
marmotmarmot
114k5145276
Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
– Anton Petrunin
Apr 1 at 4:31
@marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
– Marian G.
2 days ago
2
A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
– O.G.
2 days ago
add a comment |
5
5
5
One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi).
documentclass[variwidth,border=3.14mm]{standalone}
usepackage{asypictureB}
begin{document}
begin{asypicture}{name=discgraph}
usepackage("mathrsfs");
import graph3;
import solids;
import interpolate;
settings.outformat="pdf";
size(500);
defaultpen(0.5mm);
pen darkgreen=rgb(0,138/255,122/255);
draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);
//function: call the radial coordinate r=t.x and the angle phi=t.y
triple f(pair t) {
return ((t.x)*cos(t.y), (t.x)*sin(t.y),
((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
}
surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
usplinetype=new splinetype {notaknot,notaknot,monotonic},
vsplinetype=Spline);
pen p=rgb(0,0,.7);
draw(s,lightolive+white);
end{asypicture}
end{document}
answered Apr 1 at 3:58
marmotmarmot
114k5145276
One way to make sure that x^2+y^2<1 is to use polar coordinates. Then x=r cos(phi) and y=r sin(phi).
documentclass[variwidth,border=3.14mm]{standalone}
usepackage{asypictureB}
begin{document}
begin{asypicture}{name=discgraph}
usepackage("mathrsfs");
import graph3;
import solids;
import interpolate;
settings.outformat="pdf";
size(500);
defaultpen(0.5mm);
pen darkgreen=rgb(0,138/255,122/255);
draw(Label("$x$",1),(0,0,0)--(1.2,0,0),darkgreen,Arrow3);
draw(Label("$y$",1),(0,0,0)--(0,1.2,0),darkgreen,Arrow3);
draw(Label("$f(x,y)$",1),(0,0,0)--(0,0,0.6),darkgreen,Arrow3);
//function: call the radial coordinate r=t.x and the angle phi=t.y
triple f(pair t) {
return ((t.x)*cos(t.y), (t.x)*sin(t.y),
((t.x)*cos(t.y))^2-((t.x)*sin(t.y))^2);
}
surface s=surface(f,(0,1),(0.49,2.5*pi),32,16,
usplinetype=new splinetype {notaknot,notaknot,monotonic},
vsplinetype=Spline);
pen p=rgb(0,0,.7);
draw(s,lightolive+white);
end{asypicture}
end{document}
answered Apr 1 at 3:58
marmotmarmot
114k5145276
answered Apr 1 at 3:58
marmotmarmot
114k5145276
answered Apr 1 at 3:58
marmotmarmot
114k5145276
answered Apr 1 at 3:58
marmotmarmot
114k5145276
114k5145276
Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
– Anton Petrunin
Apr 1 at 4:31
@marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
– Marian G.
2 days ago
2
A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
– O.G.
2 days ago
add a comment |
Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
– Anton Petrunin
Apr 1 at 4:31
@marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
– Marian G.
2 days ago
2
A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
– O.G.
2 days ago
Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
– Anton Petrunin
Apr 1 at 4:31
Thank you, but is there a direct way to make a condition x^2+y^2<1 for the arguments?
– Anton Petrunin
Apr 1 at 4:31
@marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
– Marian G.
2 days ago
@marmot: The x-axis near origin should be hidden from the given point of view. Is there any way to improve this issue? E.g., by setting some samples-option?
– Marian G.
2 days ago
2
2
A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
– O.G.
2 days ago
A line has a thickness, a surface not. It is why you see the x-axis near origin. You can observe the same behavior with a simple square surface and the x-axis. Perhaps it is possible to avoid its by creating two z translated surfaces, but you have to manage the boundary...
– O.G.
2 days ago
add a comment |
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