Dynamic filling of a region of a polar plot
4
$begingroup$
I would like to shade area of region as a function of angle using PolarPlot.
Here is my attempt.
With[
{pts =
Cases[PolarPlot[1 + 2 Sin[θ], {θ, 0, 2 π}], _Line, {0, Infinity}][[1, 1]]},
Manipulate[
Show[
ListLinePlot[{{{0, 0}, pts[[n]]}, pts[[1 ;; n]]},
Filling -> {2 -> {Axis, LightBlue}, 1 -> {Axis, LightBlue}},
PlotRange -> {{-2, 2}, {-0.5, 3.2}}, AspectRatio -> 1,
PlotStyle -> Directive[AbsoluteThickness@3, Magenta, Magenta],
ImageSize -> 500, AxesStyle -> Directive[Black, 18],
PlotLabel -> Style["r=1+2 sin(θ)", Black, 20]],
PolarPlot[1 + 2 Sin[θ], {θ, 0, 2.2 π},
AspectRatio -> 1, PlotStyle -> {Black, AbsoluteThickness@3}]],
{n, 1, Length @ pts, 1}]]
Two thing I would like to achieve:
- I don't want to see the yellow highlited region.
- When inner loop is shaded twice, I would like to make it darker to emphasize that it is the 2nd time.
Any suggestion..
plotting filling
asked 4 hours ago
Okkes DulgerciOkkes Dulgerci
5,4691919
$endgroup$
add a comment |
4
$begingroup$
I would like to shade area of region as a function of angle using PolarPlot.
Here is my attempt.
With[
{pts =
Cases[PolarPlot[1 + 2 Sin[θ], {θ, 0, 2 π}], _Line, {0, Infinity}][[1, 1]]},
Manipulate[
Show[
ListLinePlot[{{{0, 0}, pts[[n]]}, pts[[1 ;; n]]},
Filling -> {2 -> {Axis, LightBlue}, 1 -> {Axis, LightBlue}},
PlotRange -> {{-2, 2}, {-0.5, 3.2}}, AspectRatio -> 1,
PlotStyle -> Directive[AbsoluteThickness@3, Magenta, Magenta],
ImageSize -> 500, AxesStyle -> Directive[Black, 18],
PlotLabel -> Style["r=1+2 sin(θ)", Black, 20]],
PolarPlot[1 + 2 Sin[θ], {θ, 0, 2.2 π},
AspectRatio -> 1, PlotStyle -> {Black, AbsoluteThickness@3}]],
{n, 1, Length @ pts, 1}]]
Two thing I would like to achieve:
- I don't want to see the yellow highlited region.
- When inner loop is shaded twice, I would like to make it darker to emphasize that it is the 2nd time.
Any suggestion..
plotting filling
asked 4 hours ago
Okkes DulgerciOkkes Dulgerci
5,4691919
$endgroup$
add a comment |
4
4
4
1
$begingroup$
I would like to shade area of region as a function of angle using PolarPlot.
Here is my attempt.
With[
{pts =
Cases[PolarPlot[1 + 2 Sin[θ], {θ, 0, 2 π}], _Line, {0, Infinity}][[1, 1]]},
Manipulate[
Show[
ListLinePlot[{{{0, 0}, pts[[n]]}, pts[[1 ;; n]]},
Filling -> {2 -> {Axis, LightBlue}, 1 -> {Axis, LightBlue}},
PlotRange -> {{-2, 2}, {-0.5, 3.2}}, AspectRatio -> 1,
PlotStyle -> Directive[AbsoluteThickness@3, Magenta, Magenta],
ImageSize -> 500, AxesStyle -> Directive[Black, 18],
PlotLabel -> Style["r=1+2 sin(θ)", Black, 20]],
PolarPlot[1 + 2 Sin[θ], {θ, 0, 2.2 π},
AspectRatio -> 1, PlotStyle -> {Black, AbsoluteThickness@3}]],
{n, 1, Length @ pts, 1}]]
Two thing I would like to achieve:
- I don't want to see the yellow highlited region.
- When inner loop is shaded twice, I would like to make it darker to emphasize that it is the 2nd time.
Any suggestion..
plotting filling
asked 4 hours ago
Okkes DulgerciOkkes Dulgerci
5,4691919
$endgroup$
I would like to shade area of region as a function of angle using PolarPlot.
Here is my attempt.
With[
{pts =
Cases[PolarPlot[1 + 2 Sin[θ], {θ, 0, 2 π}], _Line, {0, Infinity}][[1, 1]]},
Manipulate[
Show[
ListLinePlot[{{{0, 0}, pts[[n]]}, pts[[1 ;; n]]},
Filling -> {2 -> {Axis, LightBlue}, 1 -> {Axis, LightBlue}},
PlotRange -> {{-2, 2}, {-0.5, 3.2}}, AspectRatio -> 1,
PlotStyle -> Directive[AbsoluteThickness@3, Magenta, Magenta],
ImageSize -> 500, AxesStyle -> Directive[Black, 18],
PlotLabel -> Style["r=1+2 sin(θ)", Black, 20]],
PolarPlot[1 + 2 Sin[θ], {θ, 0, 2.2 π},
AspectRatio -> 1, PlotStyle -> {Black, AbsoluteThickness@3}]],
{n, 1, Length @ pts, 1}]]
Two thing I would like to achieve:
- I don't want to see the yellow highlited region.
- When inner loop is shaded twice, I would like to make it darker to emphasize that it is the 2nd time.
Any suggestion..
plotting filling
plotting filling
asked 4 hours ago
Okkes DulgerciOkkes Dulgerci
5,4691919
asked 4 hours ago
Okkes DulgerciOkkes Dulgerci
5,4691919
edited 2 hours ago
Okkes Dulgerci
asked 4 hours ago
Okkes DulgerciOkkes Dulgerci
5,4691919
asked 4 hours ago
Okkes DulgerciOkkes Dulgerci
5,4691919
asked 4 hours ago
Okkes DulgerciOkkes Dulgerci
5,4691919
5,4691919
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
2
$begingroup$
This is what you need:
Manipulate[ParametricPlot[
r (1 + 2 Sin[θ]) {Cos[θ], Sin[θ]},
{θ, 0, thmax},
{r, 0, 1},
PlotRange -> {{-2.25, 2.25}, {-0.5, 3.5}},
PerformanceGoal -> "Quality"
], {thmax, 0.01, 2 Pi}]
edited 1 hour ago
m_goldberg
88.9k873200
answered 1 hour ago
C. E.C. E.
51.2k3101207
$endgroup$
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
2
$begingroup$
This is what you need:
Manipulate[ParametricPlot[
r (1 + 2 Sin[θ]) {Cos[θ], Sin[θ]},
{θ, 0, thmax},
{r, 0, 1},
PlotRange -> {{-2.25, 2.25}, {-0.5, 3.5}},
PerformanceGoal -> "Quality"
], {thmax, 0.01, 2 Pi}]
edited 1 hour ago
m_goldberg
88.9k873200
answered 1 hour ago
C. E.C. E.
51.2k3101207
$endgroup$
add a comment |
2
$begingroup$
This is what you need:
Manipulate[ParametricPlot[
r (1 + 2 Sin[θ]) {Cos[θ], Sin[θ]},
{θ, 0, thmax},
{r, 0, 1},
PlotRange -> {{-2.25, 2.25}, {-0.5, 3.5}},
PerformanceGoal -> "Quality"
], {thmax, 0.01, 2 Pi}]
edited 1 hour ago
m_goldberg
88.9k873200
answered 1 hour ago
C. E.C. E.
51.2k3101207
$endgroup$
add a comment |
2
2
2
$begingroup$
This is what you need:
Manipulate[ParametricPlot[
r (1 + 2 Sin[θ]) {Cos[θ], Sin[θ]},
{θ, 0, thmax},
{r, 0, 1},
PlotRange -> {{-2.25, 2.25}, {-0.5, 3.5}},
PerformanceGoal -> "Quality"
], {thmax, 0.01, 2 Pi}]
edited 1 hour ago
m_goldberg
88.9k873200
answered 1 hour ago
C. E.C. E.
51.2k3101207
$endgroup$
This is what you need:
Manipulate[ParametricPlot[
r (1 + 2 Sin[θ]) {Cos[θ], Sin[θ]},
{θ, 0, thmax},
{r, 0, 1},
PlotRange -> {{-2.25, 2.25}, {-0.5, 3.5}},
PerformanceGoal -> "Quality"
], {thmax, 0.01, 2 Pi}]
edited 1 hour ago
m_goldberg
88.9k873200
answered 1 hour ago
C. E.C. E.
51.2k3101207
edited 1 hour ago
m_goldberg
88.9k873200
edited 1 hour ago
m_goldberg
88.9k873200
edited 1 hour ago
m_goldberg
88.9k873200
88.9k873200
answered 1 hour ago
C. E.C. E.
51.2k3101207
answered 1 hour ago
C. E.C. E.
51.2k3101207
answered 1 hour ago
C. E.C. E.
51.2k3101207
51.2k3101207
add a comment |
add a comment |
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