File:Chaotic Bunimovich stadium.png

Page contents not supported in other languages.
This is a file from the Wikimedia Commons
From Wikipedia, the free encyclopedia

Chaotic_Bunimovich_stadium.png(758 × 379 pixels, file size: 7 KB, MIME type: image/png)

Summary

Description
English: billiards in a Bunimovich stadium, initial deviation is an angle of one degree

Mathematica source code

In[403]:= NN[v_]:=Sqrt[v[[1]]^2+v[[2]]^2]; Ang[v0_,va_,vb_]:=(va-v0).(vb-v0)/NN[va-v0]/NN[vb-v0]; 1st trajectory p0={0,0}; q0=\[Pi]/9; In[334]:= NSolve[(p0[[1]]+t Cos[q0]-1)^2+(p0[[2]]+t Sin[q0])^2==1,t] Out[334]= {{t\[Rule]0.},{t\[Rule]1.87939}} In[335]:= t0=1.8793852415718169`; p1=p0+t0{Cos[q0],Sin[q0]}; q1=-\[Pi]+(ArcCos[p1[[1]]-1]+q0); NSolve[p1[[2]]+t Sin[q1]\[Equal]-1,t] Out[338]= {{t\[Rule]1.89693}} In[180]:= t1=1.896927737347811; p2=p1+t1{Cos[q1],Sin[q1]}; q2=2\[Pi]-q1; NSolve[p2[[2]]+t Sin[q2]\[Equal]1,t] Out[183]= {{t\[Rule]2.3094}} In[202]:= t2=2.3094010767585043; p3=p2+t2{Cos[q2],Sin[q2]}; q3=2\[Pi]-q2; NSolve[(p3[[1]]+t Cos[q3]+1)^2+(p3[[2]]+t Sin[q3])^2==1,t] Out[205]= {{t\[Rule]0.200212},{t\[Rule]2.19472}} In[405]:= t3=2.194718395858327; p4=p3+t3{Cos[q3],Sin[q3]}; Solve[Ang[p4,p3,{-1,0}]\[Equal]Ang[p4,({Cos[t],Sin[t]}+p4),{-1,0}],t] From In[405]:= \!\(\* RowBox[{\(Power::"infy"\), \(\(:\)\(\ \)\), "\<\"Infinite expression \ \\!\\(1\\/0\\^2\\) encountered. \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", \ ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Power::infy\\\"]\\)\"\>"}]\) From In[405]:= \!\(\* RowBox[{\(Solve::"ifun"\), \(\(:\)\(\ \)\), "\<\"Inverse functions are \ being used by \\!\\(Solve\\), so some solutions may not be found; use Reduce \ for complete solution information. \ \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", ButtonStyle->\\\"RefGuideLinkText\ \\\", ButtonFrame->None, ButtonData:>\\\"Solve::ifun\\\"]\\)\"\>"}]\) Out[407]= {{t\[Rule]1.0472},{t\[Rule]1.19548}} In[328]:= q4=1.1954752520981573; NSolve[p4[[2]]+t Sin[q4]\[Equal]1,t] Out[329]= {{t\[Rule]2.04289}} In[440]:= t4=2.0428873267106815`; p5=p4+t4{Cos[q4],Sin[q4]}; q5=2\[Pi]-q4; 2 nd trajectory In[384]:= P0={0,0}; Q0=\[Pi]/9+\[Pi]/180; In[386]:= NSolve[(P0[[1]]+t Cos[Q0]-1)^2+(P0[[2]]+t Sin[Q0])^2==1,t] Out[386]= {{t\[Rule]0.},{t\[Rule]1.86716}} In[387]:= T0=1.8671608529944035`; P1=P0+T0{Cos[Q0],Sin[Q0]}; Q1=-\[Pi]+(ArcCos[P1[[1]]-1]+Q0); NSolve[P1[[2]]+t Sin[Q1]\[Equal]-1,t] Out[390]= {{t\[Rule]1.87331}} In[391]:= T1=1.8733090735550966`; P2=P1+T1{Cos[Q1],Sin[Q1]}; Q2=2\[Pi]-Q1; NSolve[P2[[2]]+t Sin[Q2]\[Equal]1,t] Out[394]= {{t\[Rule]2.24465}} In[395]:= T2=2.2446524752687225`; P3=P2+T2{Cos[Q2],Sin[Q2]}; Q3=2\[Pi]-Q2; NSolve[(P3[[1]]+t Cos[Q3]+1)^2+(P3[[2]]+t Sin[Q3])^2==1,t] Out[398]= {{t\[Rule]0.341712},{t\[Rule]2.23354}} In[419]:= T3=2.233539454680641`; P4=P3+T3{Cos[Q3],Sin[Q3]}; Solve[Ang[P4,P3,{-1,0}]\[Equal]Ang[P4,({Cos[t],Sin[t]}+P4),{-1,0}],t] From In[419]:= \!\(\* RowBox[{\(Power::"infy"\), \(\(:\)\(\ \)\), "\<\"Infinite expression \ \\!\\(1\\/0\\^2\\) encountered. \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", \ ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Power::infy\\\"]\\)\"\>"}]\) From In[419]:= \!\(\* RowBox[{\(Solve::"ifun"\), \(\(:\)\(\ \)\), "\<\"Inverse functions are \ being used by \\!\\(Solve\\), so some solutions may not be found; use Reduce \ for complete solution information. \ \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", ButtonStyle->\\\"RefGuideLinkText\ \\\", ButtonFrame->None, ButtonData:>\\\"Solve::ifun\\\"]\\)\"\>"}]\) Out[421]= {{t\[Rule]1.09956},{t\[Rule]1.76035}} In[423]:= Q4=1.786499618850784`; NSolve[(P4[[1]]+t Cos[Q4]+1)^2+(P4[[2]]+t Sin[Q4])^2==1,t] Out[424]= \!\({{t \[Rule] \(-2.961831812996791`*^-16\)}, {t \[Rule] 1.874216860919306`}}\) In[428]:= T4=1.874216860919306`; P5=P4+T4{Cos[Q4],Sin[Q4]}; Solve[Ang[P5,P4,{-1,0}]\[Equal]Ang[P5,({Cos[t],Sin[t]}+P5),{-1,0}],t] From In[428]:= \!\(\* RowBox[{\(Power::"infy"\), \(\(:\)\(\ \)\), "\<\"Infinite expression \ \\!\\(1\\/0\\^2\\) encountered. \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", \ ButtonStyle->\\\"RefGuideLinkText\\\", ButtonFrame->None, \ ButtonData:>\\\"Power::infy\\\"]\\)\"\>"}]\) From In[428]:= \!\(\* RowBox[{\(Solve::"ifun"\), \(\(:\)\(\ \)\), "\<\"Inverse functions are \ being used by \\!\\(Solve\\), so some solutions may not be found; use Reduce \ for complete solution information. \ \\!\\(\\*ButtonBox[\\\"More\[Ellipsis]\\\", ButtonStyle->\\\"RefGuideLinkText\ \\\", ButtonFrame->None, ButtonData:>\\\"Solve::ifun\\\"]\\)\"\>"}]\) Out[430]= {{t\[Rule]-1.35509},{t\[Rule]-0.642004}} In[432]:= Q5=-0.6420035368814776`; Illustration In[451]:= Show[Graphics[{ Thickness[.003], Line[{{-1,-1},{1,-1}}], Line[{{-1,1},{1,1}}], Circle[{1,0},1,{-\[Pi]/2,\[Pi]/2}], Circle[{-1,0},1,{\[Pi]/2,3\[Pi]/2}], RGBColor[254/256,194/256,0], Thickness[.0051],PointSize[.03], Line[{p0,p0+t0{Cos[q0],Sin[q0]}}], Line[{p1,p1+t1{Cos[q1],Sin[q1]}}], Line[{p2,p2+t2{Cos[q2],Sin[q2]}}], Line[{p3,p3+t3{Cos[q3],Sin[q3]}}], Line[{p4,p4+t4{Cos[q4],Sin[q4]}}], Line[{p5,p5+1.9{Cos[q5],Sin[q5]}}], Point[p5+1.9{Cos[q5],Sin[q5]}], RGBColor[188/256,30/256,71/256], Line[{P0,P0+T0{Cos[Q0],Sin[Q0]}}], Line[{P1,P1+T1{Cos[Q1],Sin[Q1]}}], Line[{P2,P2+T2{Cos[Q2],Sin[Q2]}}], Line[{P3,P3+T3{Cos[Q3],Sin[Q3]}}], Line[{P4,P4+T4{Cos[Q4],Sin[Q4]}}], Line[{P5,P5+1.9{Cos[Q5],Sin[Q5]}}], Point[P5+1.9{Cos[Q5],Sin[Q5]}] }],AspectRatio\[Rule]Automatic]
Source Own work
Author Jakob.scholbach

Licensing

I, the copyright holder of this work, hereby publish it under the following licenses:
w:en:Creative Commons
attribution share alike
This file is licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.
You are free:
  • to share – to copy, distribute and transmit the work
  • to remix – to adapt the work
Under the following conditions:
  • attribution – You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.
  • share alike – If you remix, transform, or build upon the material, you must distribute your contributions under the same or compatible license as the original.
GNU head Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, and no Back-Cover Texts. A copy of the license is included in the section entitled GNU Free Documentation License.
You may select the license of your choice.

Captions

Add a one-line explanation of what this file represents

Items portrayed in this file

depicts

image/png

17fb36e6aaecbcb32c4ad6d7ac31dd5f0a0276f7

6,690 byte

379 pixel

758 pixel

File history

Click on a date/time to view the file as it appeared at that time.

Date/TimeThumbnailDimensionsUserComment
current14:24, 13 February 2011Thumbnail for version as of 14:24, 13 February 2011758 × 379 (7 KB)Jakob.scholbach{{Information |Description ={{en|1=billiards in a Bunimovich stadium, initial deviation is an angle of one degree Mathematica source code <nowiki> In[403]:= NN[v_]:=Sqrt[v1^2+v2^2]; Ang[v0_,va_,vb_]:=(va-v0).(vb-v0)/NN[va-v0]/NN[vb-v0]; 1st t
No pages on the English Wikipedia use this file (pages on other projects are not listed).

Global file usage

The following other wikis use this file: