Calculating list of areas between the curves in an intersection region
4
$begingroup$
I have 2 tables;
table 1:
ctr = {(Exp[Pi*0.5] + 1)/2, 0};
radius = (Exp[Pi*0.5] - 1)/2;
pp = PolarPlot[
Evaluate@Table[Exp[(t + 2*Pi*i/120)*0.5], {i, 1, 120}], {t, -Pi,
Pi}, RegionFunction -> (Norm[{#, #2} - ctr] <= radius &)]
Total[ArcLength /@ Cases[pp, _Line, All]]
and table 2:
ctr = {(Exp[Pi*0.5] - 1)/2, 0};
radius = (Exp[Pi*0.5] - 1)/2;
pp2 = ParametricPlot[
Evaluate[Table[{t, t (Tan[2 j Pi / 120])}, {j, 1, 60}]], {t, -5, 5},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &),
PlotRange -> All]
Total[ArcLength /@ Cases[pp2, _Line, All]]
I want to calculate each area between the curves and show them in histogram so I can evaluate maximum and minimum gaps between the tables. I illustrated what I want as below:
can you show me how?
regions intersection
asked Feb 23 at 12:09
Alper91Alper91
1356
$endgroup$
add a comment |
4
$begingroup$
I have 2 tables;
table 1:
ctr = {(Exp[Pi*0.5] + 1)/2, 0};
radius = (Exp[Pi*0.5] - 1)/2;
pp = PolarPlot[
Evaluate@Table[Exp[(t + 2*Pi*i/120)*0.5], {i, 1, 120}], {t, -Pi,
Pi}, RegionFunction -> (Norm[{#, #2} - ctr] <= radius &)]
Total[ArcLength /@ Cases[pp, _Line, All]]
and table 2:
ctr = {(Exp[Pi*0.5] - 1)/2, 0};
radius = (Exp[Pi*0.5] - 1)/2;
pp2 = ParametricPlot[
Evaluate[Table[{t, t (Tan[2 j Pi / 120])}, {j, 1, 60}]], {t, -5, 5},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &),
PlotRange -> All]
Total[ArcLength /@ Cases[pp2, _Line, All]]
I want to calculate each area between the curves and show them in histogram so I can evaluate maximum and minimum gaps between the tables. I illustrated what I want as below:
can you show me how?
regions intersection
asked Feb 23 at 12:09
Alper91Alper91
1356
$endgroup$
add a comment |
4
4
4
$begingroup$
I have 2 tables;
table 1:
ctr = {(Exp[Pi*0.5] + 1)/2, 0};
radius = (Exp[Pi*0.5] - 1)/2;
pp = PolarPlot[
Evaluate@Table[Exp[(t + 2*Pi*i/120)*0.5], {i, 1, 120}], {t, -Pi,
Pi}, RegionFunction -> (Norm[{#, #2} - ctr] <= radius &)]
Total[ArcLength /@ Cases[pp, _Line, All]]
and table 2:
ctr = {(Exp[Pi*0.5] - 1)/2, 0};
radius = (Exp[Pi*0.5] - 1)/2;
pp2 = ParametricPlot[
Evaluate[Table[{t, t (Tan[2 j Pi / 120])}, {j, 1, 60}]], {t, -5, 5},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &),
PlotRange -> All]
Total[ArcLength /@ Cases[pp2, _Line, All]]
I want to calculate each area between the curves and show them in histogram so I can evaluate maximum and minimum gaps between the tables. I illustrated what I want as below:
can you show me how?
regions intersection
asked Feb 23 at 12:09
Alper91Alper91
1356
$endgroup$
I have 2 tables;
table 1:
ctr = {(Exp[Pi*0.5] + 1)/2, 0};
radius = (Exp[Pi*0.5] - 1)/2;
pp = PolarPlot[
Evaluate@Table[Exp[(t + 2*Pi*i/120)*0.5], {i, 1, 120}], {t, -Pi,
Pi}, RegionFunction -> (Norm[{#, #2} - ctr] <= radius &)]
Total[ArcLength /@ Cases[pp, _Line, All]]
and table 2:
ctr = {(Exp[Pi*0.5] - 1)/2, 0};
radius = (Exp[Pi*0.5] - 1)/2;
pp2 = ParametricPlot[
Evaluate[Table[{t, t (Tan[2 j Pi / 120])}, {j, 1, 60}]], {t, -5, 5},
RegionFunction -> (Norm[{#, #2} - ctr] <= radius &),
PlotRange -> All]
Total[ArcLength /@ Cases[pp2, _Line, All]]
I want to calculate each area between the curves and show them in histogram so I can evaluate maximum and minimum gaps between the tables. I illustrated what I want as below:
can you show me how?
regions intersection
regions intersection
asked Feb 23 at 12:09
Alper91Alper91
1356
asked Feb 23 at 12:09
Alper91Alper91
1356
asked Feb 23 at 12:09
Alper91Alper91
1356
asked Feb 23 at 12:09
Alper91Alper91
1356
asked Feb 23 at 12:09
Alper91Alper91
1356
1356
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
6
$begingroup$
You could define a ParametricRegion , examplary between curves #10 and #11
reg10 = ParametricRegion[{{Cos[t], Sin[t]} Exp[(t + 2*Pi*ii/120)/2], (Sqrt[#.#] < radius) &[{Cos[t],Sin[t]} Exp[(t + 2*Pi*ii/120)/2] - ctr]}
, {{t, -Pi, Pi}, {ii, 10, 11}}]
DiscretizeRegion[reg10]
Area[DiscretizeRegion[reg10]]
(*0.0592725*)
addendum
Knowing the plot pp a very fast straightforward solution is possible:
First get the Line-objects from the plot
linien = Cases[pp, _Line , Infinity];
Two neighboring lines form a polygon from which the area can be calculated
areas=Table[ Area@Polygon[Join[linien[[i]][[1]], Reverse[linien[[i + 1]][[1]]]]], {i,1, Length[linien] - 1}]
(*{0.0243947, 0.0281509, 0.0318023, 0.0353948, 0.039143, 0.0430014,
0.0469289, 0.050997, 0.0551564, 0.0594015, 0.0639209, 0.0684968,
0.0731468, 0.0780328, 0.0830698, 0.0882845, 0.0936695, 0.0992263,
0.104955, 0.111007, 0.117283, 0.123424, 0.129694, 0.136315, 0.143126,
0.150108, 0.157259, 0.164582, 0.172953, 0.1806, 0.187462, 0.195366,
0.203387, 0.21151, 0.22005, 0.228322, 0.23627, 0.24457, 0.252866,
0.261078, 0.269163, 0.27711, 0.284845, 0.292302, 0.299405, 0.306072,
0.312203, 0.317686, 0.322383, 0.326143, 0.329235, 0.330515, 0.329747,
0.327478, 0.322887, 0.315273, 0.304139, 0.288177, 0.265912, 0.234501,
0.187173}*)
That's it! Hope it helps.
answered Feb 23 at 14:21
Ulrich NeumannUlrich Neumann
9,613617
$endgroup$
$begingroup$
Thank you very much! However, how could I store each of the ares into a table so I can show the histogram could you help me with that as well?
$endgroup$
– Alper91
Feb 23 at 15:30
$begingroup$
Put it in a Table[...,{i,1,119}]
$endgroup$
– Ulrich Neumann
Feb 23 at 17:57
$begingroup$
@ Alper91 Look at the edit of my answer, I added a solution which only needs to knowpp
$endgroup$
– Ulrich Neumann
Feb 23 at 19:44
$begingroup$
Thanks. It worked easily.
$endgroup$
– Alper91
Feb 24 at 7:29
add a comment |
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1 Answer
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1 Answer
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6
$begingroup$
You could define a ParametricRegion , examplary between curves #10 and #11
reg10 = ParametricRegion[{{Cos[t], Sin[t]} Exp[(t + 2*Pi*ii/120)/2], (Sqrt[#.#] < radius) &[{Cos[t],Sin[t]} Exp[(t + 2*Pi*ii/120)/2] - ctr]}
, {{t, -Pi, Pi}, {ii, 10, 11}}]
DiscretizeRegion[reg10]
Area[DiscretizeRegion[reg10]]
(*0.0592725*)
addendum
Knowing the plot pp a very fast straightforward solution is possible:
First get the Line-objects from the plot
linien = Cases[pp, _Line , Infinity];
Two neighboring lines form a polygon from which the area can be calculated
areas=Table[ Area@Polygon[Join[linien[[i]][[1]], Reverse[linien[[i + 1]][[1]]]]], {i,1, Length[linien] - 1}]
(*{0.0243947, 0.0281509, 0.0318023, 0.0353948, 0.039143, 0.0430014,
0.0469289, 0.050997, 0.0551564, 0.0594015, 0.0639209, 0.0684968,
0.0731468, 0.0780328, 0.0830698, 0.0882845, 0.0936695, 0.0992263,
0.104955, 0.111007, 0.117283, 0.123424, 0.129694, 0.136315, 0.143126,
0.150108, 0.157259, 0.164582, 0.172953, 0.1806, 0.187462, 0.195366,
0.203387, 0.21151, 0.22005, 0.228322, 0.23627, 0.24457, 0.252866,
0.261078, 0.269163, 0.27711, 0.284845, 0.292302, 0.299405, 0.306072,
0.312203, 0.317686, 0.322383, 0.326143, 0.329235, 0.330515, 0.329747,
0.327478, 0.322887, 0.315273, 0.304139, 0.288177, 0.265912, 0.234501,
0.187173}*)
That's it! Hope it helps.
answered Feb 23 at 14:21
Ulrich NeumannUlrich Neumann
9,613617
$endgroup$
$begingroup$
Thank you very much! However, how could I store each of the ares into a table so I can show the histogram could you help me with that as well?
$endgroup$
– Alper91
Feb 23 at 15:30
$begingroup$
Put it in a Table[...,{i,1,119}]
$endgroup$
– Ulrich Neumann
Feb 23 at 17:57
$begingroup$
@ Alper91 Look at the edit of my answer, I added a solution which only needs to knowpp
$endgroup$
– Ulrich Neumann
Feb 23 at 19:44
$begingroup$
Thanks. It worked easily.
$endgroup$
– Alper91
Feb 24 at 7:29
add a comment |
6
$begingroup$
You could define a ParametricRegion , examplary between curves #10 and #11
reg10 = ParametricRegion[{{Cos[t], Sin[t]} Exp[(t + 2*Pi*ii/120)/2], (Sqrt[#.#] < radius) &[{Cos[t],Sin[t]} Exp[(t + 2*Pi*ii/120)/2] - ctr]}
, {{t, -Pi, Pi}, {ii, 10, 11}}]
DiscretizeRegion[reg10]
Area[DiscretizeRegion[reg10]]
(*0.0592725*)
addendum
Knowing the plot pp a very fast straightforward solution is possible:
First get the Line-objects from the plot
linien = Cases[pp, _Line , Infinity];
Two neighboring lines form a polygon from which the area can be calculated
areas=Table[ Area@Polygon[Join[linien[[i]][[1]], Reverse[linien[[i + 1]][[1]]]]], {i,1, Length[linien] - 1}]
(*{0.0243947, 0.0281509, 0.0318023, 0.0353948, 0.039143, 0.0430014,
0.0469289, 0.050997, 0.0551564, 0.0594015, 0.0639209, 0.0684968,
0.0731468, 0.0780328, 0.0830698, 0.0882845, 0.0936695, 0.0992263,
0.104955, 0.111007, 0.117283, 0.123424, 0.129694, 0.136315, 0.143126,
0.150108, 0.157259, 0.164582, 0.172953, 0.1806, 0.187462, 0.195366,
0.203387, 0.21151, 0.22005, 0.228322, 0.23627, 0.24457, 0.252866,
0.261078, 0.269163, 0.27711, 0.284845, 0.292302, 0.299405, 0.306072,
0.312203, 0.317686, 0.322383, 0.326143, 0.329235, 0.330515, 0.329747,
0.327478, 0.322887, 0.315273, 0.304139, 0.288177, 0.265912, 0.234501,
0.187173}*)
That's it! Hope it helps.
answered Feb 23 at 14:21
Ulrich NeumannUlrich Neumann
9,613617
$endgroup$
$begingroup$
Thank you very much! However, how could I store each of the ares into a table so I can show the histogram could you help me with that as well?
$endgroup$
– Alper91
Feb 23 at 15:30
$begingroup$
Put it in a Table[...,{i,1,119}]
$endgroup$
– Ulrich Neumann
Feb 23 at 17:57
$begingroup$
@ Alper91 Look at the edit of my answer, I added a solution which only needs to knowpp
$endgroup$
– Ulrich Neumann
Feb 23 at 19:44
$begingroup$
Thanks. It worked easily.
$endgroup$
– Alper91
Feb 24 at 7:29
add a comment |
6
6
6
$begingroup$
You could define a ParametricRegion , examplary between curves #10 and #11
reg10 = ParametricRegion[{{Cos[t], Sin[t]} Exp[(t + 2*Pi*ii/120)/2], (Sqrt[#.#] < radius) &[{Cos[t],Sin[t]} Exp[(t + 2*Pi*ii/120)/2] - ctr]}
, {{t, -Pi, Pi}, {ii, 10, 11}}]
DiscretizeRegion[reg10]
Area[DiscretizeRegion[reg10]]
(*0.0592725*)
addendum
Knowing the plot pp a very fast straightforward solution is possible:
First get the Line-objects from the plot
linien = Cases[pp, _Line , Infinity];
Two neighboring lines form a polygon from which the area can be calculated
areas=Table[ Area@Polygon[Join[linien[[i]][[1]], Reverse[linien[[i + 1]][[1]]]]], {i,1, Length[linien] - 1}]
(*{0.0243947, 0.0281509, 0.0318023, 0.0353948, 0.039143, 0.0430014,
0.0469289, 0.050997, 0.0551564, 0.0594015, 0.0639209, 0.0684968,
0.0731468, 0.0780328, 0.0830698, 0.0882845, 0.0936695, 0.0992263,
0.104955, 0.111007, 0.117283, 0.123424, 0.129694, 0.136315, 0.143126,
0.150108, 0.157259, 0.164582, 0.172953, 0.1806, 0.187462, 0.195366,
0.203387, 0.21151, 0.22005, 0.228322, 0.23627, 0.24457, 0.252866,
0.261078, 0.269163, 0.27711, 0.284845, 0.292302, 0.299405, 0.306072,
0.312203, 0.317686, 0.322383, 0.326143, 0.329235, 0.330515, 0.329747,
0.327478, 0.322887, 0.315273, 0.304139, 0.288177, 0.265912, 0.234501,
0.187173}*)
That's it! Hope it helps.
answered Feb 23 at 14:21
Ulrich NeumannUlrich Neumann
9,613617
$endgroup$
You could define a ParametricRegion , examplary between curves #10 and #11
reg10 = ParametricRegion[{{Cos[t], Sin[t]} Exp[(t + 2*Pi*ii/120)/2], (Sqrt[#.#] < radius) &[{Cos[t],Sin[t]} Exp[(t + 2*Pi*ii/120)/2] - ctr]}
, {{t, -Pi, Pi}, {ii, 10, 11}}]
DiscretizeRegion[reg10]
Area[DiscretizeRegion[reg10]]
(*0.0592725*)
addendum
Knowing the plot pp a very fast straightforward solution is possible:
First get the Line-objects from the plot
linien = Cases[pp, _Line , Infinity];
Two neighboring lines form a polygon from which the area can be calculated
areas=Table[ Area@Polygon[Join[linien[[i]][[1]], Reverse[linien[[i + 1]][[1]]]]], {i,1, Length[linien] - 1}]
(*{0.0243947, 0.0281509, 0.0318023, 0.0353948, 0.039143, 0.0430014,
0.0469289, 0.050997, 0.0551564, 0.0594015, 0.0639209, 0.0684968,
0.0731468, 0.0780328, 0.0830698, 0.0882845, 0.0936695, 0.0992263,
0.104955, 0.111007, 0.117283, 0.123424, 0.129694, 0.136315, 0.143126,
0.150108, 0.157259, 0.164582, 0.172953, 0.1806, 0.187462, 0.195366,
0.203387, 0.21151, 0.22005, 0.228322, 0.23627, 0.24457, 0.252866,
0.261078, 0.269163, 0.27711, 0.284845, 0.292302, 0.299405, 0.306072,
0.312203, 0.317686, 0.322383, 0.326143, 0.329235, 0.330515, 0.329747,
0.327478, 0.322887, 0.315273, 0.304139, 0.288177, 0.265912, 0.234501,
0.187173}*)
That's it! Hope it helps.
answered Feb 23 at 14:21
Ulrich NeumannUlrich Neumann
9,613617
edited Feb 23 at 19:42
answered Feb 23 at 14:21
Ulrich NeumannUlrich Neumann
9,613617
answered Feb 23 at 14:21
Ulrich NeumannUlrich Neumann
9,613617
answered Feb 23 at 14:21
Ulrich NeumannUlrich Neumann
9,613617
9,613617
$begingroup$
Thank you very much! However, how could I store each of the ares into a table so I can show the histogram could you help me with that as well?
$endgroup$
– Alper91
Feb 23 at 15:30
$begingroup$
Put it in a Table[...,{i,1,119}]
$endgroup$
– Ulrich Neumann
Feb 23 at 17:57
$begingroup$
@ Alper91 Look at the edit of my answer, I added a solution which only needs to knowpp
$endgroup$
– Ulrich Neumann
Feb 23 at 19:44
$begingroup$
Thanks. It worked easily.
$endgroup$
– Alper91
Feb 24 at 7:29
add a comment |
$begingroup$
Thank you very much! However, how could I store each of the ares into a table so I can show the histogram could you help me with that as well?
$endgroup$
– Alper91
Feb 23 at 15:30
$begingroup$
Put it in a Table[...,{i,1,119}]
$endgroup$
– Ulrich Neumann
Feb 23 at 17:57
$begingroup$
@ Alper91 Look at the edit of my answer, I added a solution which only needs to knowpp
$endgroup$
– Ulrich Neumann
Feb 23 at 19:44
$begingroup$
Thanks. It worked easily.
$endgroup$
– Alper91
Feb 24 at 7:29
$begingroup$
Thank you very much! However, how could I store each of the ares into a table so I can show the histogram could you help me with that as well?
$endgroup$
– Alper91
Feb 23 at 15:30
$begingroup$
Thank you very much! However, how could I store each of the ares into a table so I can show the histogram could you help me with that as well?
$endgroup$
– Alper91
Feb 23 at 15:30
$begingroup$
Put it in a Table[...,{i,1,119}]
$endgroup$
– Ulrich Neumann
Feb 23 at 17:57
$begingroup$
Put it in a Table[...,{i,1,119}]
$endgroup$
– Ulrich Neumann
Feb 23 at 17:57
$begingroup$
@ Alper91 Look at the edit of my answer, I added a solution which only needs to know
pp$endgroup$
– Ulrich Neumann
Feb 23 at 19:44
$begingroup$
@ Alper91 Look at the edit of my answer, I added a solution which only needs to know
pp$endgroup$
– Ulrich Neumann
Feb 23 at 19:44
$begingroup$
Thanks. It worked easily.
$endgroup$
– Alper91
Feb 24 at 7:29
$begingroup$
Thanks. It worked easily.
$endgroup$
– Alper91
Feb 24 at 7:29
add a comment |
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