JimboBimbo

Mathematica solution, with larger tip

Oct 10th, 2012
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  1. angle = 35; lst = {{0.6, 0},{0.35,.05},{-.1,.15}, {-.4,.375},{-.15,.1},{-.15,-.1},{-.4,-.375},{-.1,-.15},{.35,-.05},{.6,0}};
  2. tlst = Function[x, {Extract[x, 1] + .6, Extract[x, 2]}] /@Function[x, RotationMatrix[angle*Pi/180].x] /@Function[x, {Extract[x, 1] - .6, Extract[x, 2]}] /@lst;
  3. WriteString["stdout","\\pgfarrowsdeclare{anglelatex'}{anglelatex'}\n{\n \\newdimen\\len\n \\len=\\pgfgetarrowoptions{anglelatex'}\n \\pgfarrowsleftextend{-0.4\\len}\n pgfarrowsrightextend{0.6\\len}\n}\n{\n \\newdimen\\len\n \\len=\\pgfgetarrowoptions{anglelatex'}\n \\pgfpathmoveto{\n \\pgfqpoint{", ToString[Extract[Extract[tlst, 1], 1]] "\\len}{", ToString[Extract[Extract[tlst, 1], 2]] "\\len}}\n \\pgfpathcurveto\n {\\pgfqpoint{", ToString[Extract[Extract[tlst, 2], 1]] "\\len}{", ToString[Extract[Extract[tlst, 2], 2]] "\\len}}\n {\\pgfqpoint{", ToString[Extract[Extract[tlst, 3], 1]] "\\len}{", ToString[Extract[Extract[tlst, 3], 2]] "\\len}}\n {\\pgfqpoint{", ToString[Extract[Extract[tlst, 4], 1]] "\\len}{", ToString[Extract[Extract[tlst, 4], 2]] "\\len}}\n \\pgfpathcurveto\n {\\pgfqpoint{", ToString[Extract[Extract[tlst, 5], 1]] "\\len}{", ToString[Extract[Extract[tlst, 5], 2]] "\\len}}\n {\\pgfqpoint{", ToString[Extract[Extract[tlst, 6], 1]] "\\len}{", ToString[Extract[Extract[tlst, 6], 2]] "\\len}}\n {\\pgfqpoint{", ToString[Extract[Extract[tlst, 7], 1]] "\\len}{", ToString[Extract[Extract[tlst, 7], 2]] "\\len}}\n \\pgfpathcurveto\n {\\pgfqpoint{", ToString[Extract[Extract[tlst, 8], 1]] "\\len}{", ToString[Extract[Extract[tlst, 8], 2]] "\\len}}\n {\\pgfqpoint{", ToString[Extract[Extract[tlst, 9], 1]] "\\len}{", ToString[Extract[Extract[tlst, 9], 2]] "\\len}}\n {\\pgfqpoint{", ToString[Extract[Extract[tlst, 10], 1]] "\\len}{", ToString[Extract[Extract[tlst, 10], 2]] "\\len}}\n pgfusepathqfill\n} \\pgfsetarrowoptions{anglelatex'}{8pt}\n\\pgfkeys{/tiplen/.default=8pt, /tiplen/.code={\\pgfsetarrowoptions{anglelatex'}{#1}}}"]
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