Specify the range of GridLines
$begingroup$
Graphics[Circle, Frame -> True, GridLines -> Automatic]
This puts grids across a 2D graphic (a circle here).
Is there any way to specify the range of values of GridLines? More specifically I want the part of GridLines that are inside the circle. That is the part of them that is outside the circle should be removed.
Thanks a lot!
grid-layouts grid-lines
asked 7 hours ago
DimitrisDimitris
2,3381332
$endgroup$
add a comment |
$begingroup$
Graphics[Circle, Frame -> True, GridLines -> Automatic]
This puts grids across a 2D graphic (a circle here).
Is there any way to specify the range of values of GridLines? More specifically I want the part of GridLines that are inside the circle. That is the part of them that is outside the circle should be removed.
Thanks a lot!
grid-layouts grid-lines
asked 7 hours ago
DimitrisDimitris
2,3381332
$endgroup$
add a comment |
$begingroup$
Graphics[Circle, Frame -> True, GridLines -> Automatic]
This puts grids across a 2D graphic (a circle here).
Is there any way to specify the range of values of GridLines? More specifically I want the part of GridLines that are inside the circle. That is the part of them that is outside the circle should be removed.
Thanks a lot!
grid-layouts grid-lines
asked 7 hours ago
DimitrisDimitris
2,3381332
$endgroup$
Graphics[Circle, Frame -> True, GridLines -> Automatic]
This puts grids across a 2D graphic (a circle here).
Is there any way to specify the range of values of GridLines? More specifically I want the part of GridLines that are inside the circle. That is the part of them that is outside the circle should be removed.
Thanks a lot!
grid-layouts grid-lines
grid-layouts grid-lines
asked 7 hours ago
DimitrisDimitris
2,3381332
asked 7 hours ago
DimitrisDimitris
2,3381332
asked 7 hours ago
DimitrisDimitris
2,3381332
asked 7 hours ago
DimitrisDimitris
2,3381332
asked 7 hours ago
DimitrisDimitris
2,3381332
2,3381332
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
{
White,
FilledCurve[{
{Line[{Scaled[{0,0}],Scaled[{1,0}],Scaled[{1,1}],Scaled[{0,1}],Scaled[{0,0}]}]},
{Line@CirclePoints[.5, 100]}
}],
Blue,
Circle[{0,0},.5]
},
Frame -> True,
GridLines -> Automatic
]
answered 6 hours ago
Carl WollCarl Woll
75.4k3100197
$endgroup$
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
6 hours ago
add a comment |
$begingroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = {Table[
ParametricPlot[{x, x + a}, {x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])}], {a, -1, 1, .5}],
Table[ParametricPlot[{x, -x + a}, {x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])}], {a, -1, 1, .5}]} // Flatten;
Show[{pt, Graphics[Circle]}, PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 5 hours ago
egwene sedaiegwene sedai
1,8261021
$endgroup$
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
2 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
{
White,
FilledCurve[{
{Line[{Scaled[{0,0}],Scaled[{1,0}],Scaled[{1,1}],Scaled[{0,1}],Scaled[{0,0}]}]},
{Line@CirclePoints[.5, 100]}
}],
Blue,
Circle[{0,0},.5]
},
Frame -> True,
GridLines -> Automatic
]
answered 6 hours ago
Carl WollCarl Woll
75.4k3100197
$endgroup$
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
6 hours ago
add a comment |
$begingroup$
You can use a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
{
White,
FilledCurve[{
{Line[{Scaled[{0,0}],Scaled[{1,0}],Scaled[{1,1}],Scaled[{0,1}],Scaled[{0,0}]}]},
{Line@CirclePoints[.5, 100]}
}],
Blue,
Circle[{0,0},.5]
},
Frame -> True,
GridLines -> Automatic
]
answered 6 hours ago
Carl WollCarl Woll
75.4k3100197
$endgroup$
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
6 hours ago
add a comment |
$begingroup$
You can use a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
{
White,
FilledCurve[{
{Line[{Scaled[{0,0}],Scaled[{1,0}],Scaled[{1,1}],Scaled[{0,1}],Scaled[{0,0}]}]},
{Line@CirclePoints[.5, 100]}
}],
Blue,
Circle[{0,0},.5]
},
Frame -> True,
GridLines -> Automatic
]
answered 6 hours ago
Carl WollCarl Woll
75.4k3100197
$endgroup$
You can use a FilledCurve to create a graphics primitive with a hole in it. For example:
Graphics[
{
White,
FilledCurve[{
{Line[{Scaled[{0,0}],Scaled[{1,0}],Scaled[{1,1}],Scaled[{0,1}],Scaled[{0,0}]}]},
{Line@CirclePoints[.5, 100]}
}],
Blue,
Circle[{0,0},.5]
},
Frame -> True,
GridLines -> Automatic
]
answered 6 hours ago
Carl WollCarl Woll
75.4k3100197
answered 6 hours ago
Carl WollCarl Woll
75.4k3100197
answered 6 hours ago
Carl WollCarl Woll
75.4k3100197
answered 6 hours ago
Carl WollCarl Woll
75.4k3100197
75.4k3100197
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
6 hours ago
add a comment |
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
6 hours ago
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
6 hours ago
$begingroup$
This is perfect! Thank you very much. Is is possible also to modify the orientation of these grid lines (for instance by π/4 rad)?
$endgroup$
– Dimitris
6 hours ago
add a comment |
$begingroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = {Table[
ParametricPlot[{x, x + a}, {x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])}], {a, -1, 1, .5}],
Table[ParametricPlot[{x, -x + a}, {x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])}], {a, -1, 1, .5}]} // Flatten;
Show[{pt, Graphics[Circle]}, PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 5 hours ago
egwene sedaiegwene sedai
1,8261021
$endgroup$
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
2 hours ago
add a comment |
$begingroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = {Table[
ParametricPlot[{x, x + a}, {x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])}], {a, -1, 1, .5}],
Table[ParametricPlot[{x, -x + a}, {x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])}], {a, -1, 1, .5}]} // Flatten;
Show[{pt, Graphics[Circle]}, PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 5 hours ago
egwene sedaiegwene sedai
1,8261021
$endgroup$
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
2 hours ago
add a comment |
$begingroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = {Table[
ParametricPlot[{x, x + a}, {x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])}], {a, -1, 1, .5}],
Table[ParametricPlot[{x, -x + a}, {x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])}], {a, -1, 1, .5}]} // Flatten;
Show[{pt, Graphics[Circle]}, PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 5 hours ago
egwene sedaiegwene sedai
1,8261021
$endgroup$
You may also use parametric plot which can give more flexibility (like the pi/4 rotation you want):
pt = {Table[
ParametricPlot[{x, x + a}, {x, 1/2 (-a - Sqrt[2 - a^2]),
1/2 (-a + Sqrt[2 - a^2])}], {a, -1, 1, .5}],
Table[ParametricPlot[{x, -x + a}, {x, 1/2 (a - Sqrt[2 - a^2]),
1/2 (a + Sqrt[2 - a^2])}], {a, -1, 1, .5}]} // Flatten;
Show[{pt, Graphics[Circle]}, PlotRange -> All, Frame -> True]
where the x range for the gridlines are from
Solve[x + a == Sqrt[1 - x^2], x]
Solve[-x + a == Sqrt[1 - x^2], x]
answered 5 hours ago
egwene sedaiegwene sedai
1,8261021
answered 5 hours ago
egwene sedaiegwene sedai
1,8261021
answered 5 hours ago
egwene sedaiegwene sedai
1,8261021
answered 5 hours ago
egwene sedaiegwene sedai
1,8261021
1,8261021
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
2 hours ago
add a comment |
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
2 hours ago
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
2 hours ago
$begingroup$
Thanks! How can we modify the code in order to have the center of circle (and the grids translated) to (0.5,0.5)?
$endgroup$
– Dimitris
2 hours ago
add a comment |
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