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Add region constraint to Graphics
How to add fill color to a 2D Graphics object?How to use Graphics@Rotate@Show@Graphics without “Graphics is not a Graphics primitive or directive.”How to embed a filled Region in a Graphics?Integrated Solution GraphicsEdgeForm with GraphicsCombine graphics with GraphicsGridHow can I add engineering-style dimensioning to graphics?Is it possible to overlay pixel graphics with vector graphicsHow to add graphics lines in a grid?How can I get high precision circle drawing in Graphics, across ten orders of magnitude
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$begingroup$
I would like to only show the part of Black
circles within the Red
circle.
Can I do that with Graphics
, Circle
and/or some sort of Region Constraint?
step = 2 Degree;
[Alpha] = Range[2 Degree, 80 Degree, step];
x = ConstantArray[0, Length@[Alpha]];
y = Sec@[Alpha];
radius = Tan@[Alpha];
range = 1.01;
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range]
At the moment, I use ContourPlot
with RegionFunction
option. However, my plot contains a large number of these circles (The amount of the circles shown here is only a quarter for the current step
), which makes ContourPlot
approach very slow. Moreover, when I zoom in, quite often I find the circles drawn by ContourPlot
are not circular, presumably due to PlotPoints
and MaxRecursion
etc. I tried to play with these two options, but did not succeed in terms of quality (being a circle) and speed.
Thank you!
graphics
$endgroup$
add a comment |
$begingroup$
I would like to only show the part of Black
circles within the Red
circle.
Can I do that with Graphics
, Circle
and/or some sort of Region Constraint?
step = 2 Degree;
[Alpha] = Range[2 Degree, 80 Degree, step];
x = ConstantArray[0, Length@[Alpha]];
y = Sec@[Alpha];
radius = Tan@[Alpha];
range = 1.01;
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range]
At the moment, I use ContourPlot
with RegionFunction
option. However, my plot contains a large number of these circles (The amount of the circles shown here is only a quarter for the current step
), which makes ContourPlot
approach very slow. Moreover, when I zoom in, quite often I find the circles drawn by ContourPlot
are not circular, presumably due to PlotPoints
and MaxRecursion
etc. I tried to play with these two options, but did not succeed in terms of quality (being a circle) and speed.
Thank you!
graphics
$endgroup$
add a comment |
$begingroup$
I would like to only show the part of Black
circles within the Red
circle.
Can I do that with Graphics
, Circle
and/or some sort of Region Constraint?
step = 2 Degree;
[Alpha] = Range[2 Degree, 80 Degree, step];
x = ConstantArray[0, Length@[Alpha]];
y = Sec@[Alpha];
radius = Tan@[Alpha];
range = 1.01;
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range]
At the moment, I use ContourPlot
with RegionFunction
option. However, my plot contains a large number of these circles (The amount of the circles shown here is only a quarter for the current step
), which makes ContourPlot
approach very slow. Moreover, when I zoom in, quite often I find the circles drawn by ContourPlot
are not circular, presumably due to PlotPoints
and MaxRecursion
etc. I tried to play with these two options, but did not succeed in terms of quality (being a circle) and speed.
Thank you!
graphics
$endgroup$
I would like to only show the part of Black
circles within the Red
circle.
Can I do that with Graphics
, Circle
and/or some sort of Region Constraint?
step = 2 Degree;
[Alpha] = Range[2 Degree, 80 Degree, step];
x = ConstantArray[0, Length@[Alpha]];
y = Sec@[Alpha];
radius = Tan@[Alpha];
range = 1.01;
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range]
At the moment, I use ContourPlot
with RegionFunction
option. However, my plot contains a large number of these circles (The amount of the circles shown here is only a quarter for the current step
), which makes ContourPlot
approach very slow. Moreover, when I zoom in, quite often I find the circles drawn by ContourPlot
are not circular, presumably due to PlotPoints
and MaxRecursion
etc. I tried to play with these two options, but did not succeed in terms of quality (being a circle) and speed.
Thank you!
graphics
graphics
asked Jul 14 at 3:36
Bemtevi77Bemtevi77
1457 bronze badges
1457 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can use the three-argument form of Circle
:
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
x, y, radius],
PlotRange -> -range, range, -range, range]
Alternatively, use RegionIntersection
with Disk[]
to get the needed portions of black circles:
circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];
circles2 = RegionIntersection[Disk[], #] & /@ N[circles];
Graphics[Red, Circle[0, 0, 1], Black, circles2,
PlotRange -> -range, range, -range, range]
same picture
Update: An alternative way to hide unwanted portions of circles
using FilledCurve
:
filledCurve = FilledCurve[Line[Append[#, First @ #]& @
CirclePoints[range Sqrt @2, 4]],
Line[Append[#, First @ #]& @ CirclePoints[200]]];
Graphics[Red, Circle[0, 0, 1], Black, circles,
EdgeForm[None], White, filledCurve,
PlotRange -> -range, range, -range, range]
same picture as above
$endgroup$
$begingroup$
I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems theArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]]
can be shortened toArcTan[#3]
. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
$endgroup$
– Bemtevi77
Jul 14 at 6:33
$begingroup$
@Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
$endgroup$
– kglr
Jul 14 at 6:42
add a comment |
$begingroup$
g = Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range];
Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
PlotStyle -> White]]
$endgroup$
$begingroup$
I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
$endgroup$
– Bemtevi77
Jul 14 at 6:43
add a comment |
Your Answer
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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 the three-argument form of Circle
:
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
x, y, radius],
PlotRange -> -range, range, -range, range]
Alternatively, use RegionIntersection
with Disk[]
to get the needed portions of black circles:
circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];
circles2 = RegionIntersection[Disk[], #] & /@ N[circles];
Graphics[Red, Circle[0, 0, 1], Black, circles2,
PlotRange -> -range, range, -range, range]
same picture
Update: An alternative way to hide unwanted portions of circles
using FilledCurve
:
filledCurve = FilledCurve[Line[Append[#, First @ #]& @
CirclePoints[range Sqrt @2, 4]],
Line[Append[#, First @ #]& @ CirclePoints[200]]];
Graphics[Red, Circle[0, 0, 1], Black, circles,
EdgeForm[None], White, filledCurve,
PlotRange -> -range, range, -range, range]
same picture as above
$endgroup$
$begingroup$
I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems theArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]]
can be shortened toArcTan[#3]
. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
$endgroup$
– Bemtevi77
Jul 14 at 6:33
$begingroup$
@Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
$endgroup$
– kglr
Jul 14 at 6:42
add a comment |
$begingroup$
You can use the three-argument form of Circle
:
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
x, y, radius],
PlotRange -> -range, range, -range, range]
Alternatively, use RegionIntersection
with Disk[]
to get the needed portions of black circles:
circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];
circles2 = RegionIntersection[Disk[], #] & /@ N[circles];
Graphics[Red, Circle[0, 0, 1], Black, circles2,
PlotRange -> -range, range, -range, range]
same picture
Update: An alternative way to hide unwanted portions of circles
using FilledCurve
:
filledCurve = FilledCurve[Line[Append[#, First @ #]& @
CirclePoints[range Sqrt @2, 4]],
Line[Append[#, First @ #]& @ CirclePoints[200]]];
Graphics[Red, Circle[0, 0, 1], Black, circles,
EdgeForm[None], White, filledCurve,
PlotRange -> -range, range, -range, range]
same picture as above
$endgroup$
$begingroup$
I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems theArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]]
can be shortened toArcTan[#3]
. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
$endgroup$
– Bemtevi77
Jul 14 at 6:33
$begingroup$
@Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
$endgroup$
– kglr
Jul 14 at 6:42
add a comment |
$begingroup$
You can use the three-argument form of Circle
:
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
x, y, radius],
PlotRange -> -range, range, -range, range]
Alternatively, use RegionIntersection
with Disk[]
to get the needed portions of black circles:
circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];
circles2 = RegionIntersection[Disk[], #] & /@ N[circles];
Graphics[Red, Circle[0, 0, 1], Black, circles2,
PlotRange -> -range, range, -range, range]
same picture
Update: An alternative way to hide unwanted portions of circles
using FilledCurve
:
filledCurve = FilledCurve[Line[Append[#, First @ #]& @
CirclePoints[range Sqrt @2, 4]],
Line[Append[#, First @ #]& @ CirclePoints[200]]];
Graphics[Red, Circle[0, 0, 1], Black, circles,
EdgeForm[None], White, filledCurve,
PlotRange -> -range, range, -range, range]
same picture as above
$endgroup$
You can use the three-argument form of Circle
:
Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3, π + ArcTan[#3], 2 π - ArcTan[#3]] &,
x, y, radius],
PlotRange -> -range, range, -range, range]
Alternatively, use RegionIntersection
with Disk[]
to get the needed portions of black circles:
circles = MapThread[Circle[#1, #2, #3] &, x, y, radius];
circles2 = RegionIntersection[Disk[], #] & /@ N[circles];
Graphics[Red, Circle[0, 0, 1], Black, circles2,
PlotRange -> -range, range, -range, range]
same picture
Update: An alternative way to hide unwanted portions of circles
using FilledCurve
:
filledCurve = FilledCurve[Line[Append[#, First @ #]& @
CirclePoints[range Sqrt @2, 4]],
Line[Append[#, First @ #]& @ CirclePoints[200]]];
Graphics[Red, Circle[0, 0, 1], Black, circles,
EdgeForm[None], White, filledCurve,
PlotRange -> -range, range, -range, range]
same picture as above
edited Jul 14 at 6:59
answered Jul 14 at 5:07
kglrkglr
207k10 gold badges236 silver badges468 bronze badges
207k10 gold badges236 silver badges468 bronze badges
$begingroup$
I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems theArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]]
can be shortened toArcTan[#3]
. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
$endgroup$
– Bemtevi77
Jul 14 at 6:33
$begingroup$
@Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
$endgroup$
– kglr
Jul 14 at 6:42
add a comment |
$begingroup$
I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems theArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]]
can be shortened toArcTan[#3]
. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.
$endgroup$
– Bemtevi77
Jul 14 at 6:33
$begingroup$
@Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
$endgroup$
– kglr
Jul 14 at 6:42
$begingroup$
I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems the
ArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]]
can be shortened to ArcTan[#3]
. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.$endgroup$
– Bemtevi77
Jul 14 at 6:33
$begingroup$
I used the 1st method of yours, which is the fastest among all the provided answers at the moment. But it seems the
ArcTan[#3 Sqrt[1/(1 + #3^2)] Sqrt[1 + #3^2]]
can be shortened to ArcTan[#3]
. Also, the center of the circles lies along the y axis, which is a special case. I wonder what this method could be if the center of the circle is arbitrary. Could you please elaborate if possible? Thanks. I realized that for the "arbitrary center" case, it requires the actual information of the circle, and therefore, it maybe changes from case to case.$endgroup$
– Bemtevi77
Jul 14 at 6:33
$begingroup$
@Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
$endgroup$
– kglr
Jul 14 at 6:42
$begingroup$
@Bemtevi77, updated with the simpler form. Re arbitrary centers, I think the second and third methods should work as is. I think the first method should also work but I am not sure.
$endgroup$
– kglr
Jul 14 at 6:42
add a comment |
$begingroup$
g = Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range];
Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
PlotStyle -> White]]
$endgroup$
$begingroup$
I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
$endgroup$
– Bemtevi77
Jul 14 at 6:43
add a comment |
$begingroup$
g = Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range];
Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
PlotStyle -> White]]
$endgroup$
$begingroup$
I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
$endgroup$
– Bemtevi77
Jul 14 at 6:43
add a comment |
$begingroup$
g = Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range];
Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
PlotStyle -> White]]
$endgroup$
g = Graphics[Red, Circle[0, 0, 1], Black,
MapThread[Circle[#1, #2, #3] &, x, y, radius],
PlotRange -> -range, range, -range, range];
Show[g, RegionPlot[x^2 + y^2 > 1, x, -1.2, 1.2, y, -1.2, 1.2,
PlotStyle -> White]]
answered Jul 14 at 4:57
MelaGoMelaGo
2,1161 gold badge1 silver badge7 bronze badges
2,1161 gold badge1 silver badge7 bronze badges
$begingroup$
I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
$endgroup$
– Bemtevi77
Jul 14 at 6:43
add a comment |
$begingroup$
I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
$endgroup$
– Bemtevi77
Jul 14 at 6:43
$begingroup$
I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
$endgroup$
– Bemtevi77
Jul 14 at 6:43
$begingroup$
I realized that this method can be easily applied to "circles with arbitrary center and radius" as asked in my comment to kglr's answer. It may not be the fastest, but is more versatile, in my opinion. Thank you very much!
$endgroup$
– Bemtevi77
Jul 14 at 6:43
add a comment |
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