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2

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL] (0,0) plot [domain=0.7:9] (x,0.75*x + 0.8) node[label=right:{AC$_L$ = S$_L$}](A1){};

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] let p1 = (H) in (0,y1) node[left, red]{$w_u$} -- (x1,y1) --(A1);

end{tikzpicture}   



end{document}

My problem is as follows:

• I have attempted to define a coordinate (A1) at the end of the first line (name path ACL).


• I then use this coordinate (A1) as the last coordinate in the second line (name path ACLt).

It seems that the coordinate is not truly at the end of the line, since the dotted line is extending slightly beyond the thick blue line:

Why is this occurring?

tikz-pgf coordinates

share|improve this question

asked 2 hours ago

Thevesh ThevaThevesh Theva

519114

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Node's have finite size and you can simply say draw[thick, blue, name path = ACLt] (0,0|-H)-- (H) --(A1); which will work fine when A1 is a coordinate.
  
  – marmot
  2 hours ago
  

  
  
  
  

add a comment | 

2

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL] (0,0) plot [domain=0.7:9] (x,0.75*x + 0.8) node[label=right:{AC$_L$ = S$_L$}](A1){};

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] let p1 = (H) in (0,y1) node[left, red]{$w_u$} -- (x1,y1) --(A1);

end{tikzpicture}   



end{document}

My problem is as follows:

• I have attempted to define a coordinate (A1) at the end of the first line (name path ACL).


• I then use this coordinate (A1) as the last coordinate in the second line (name path ACLt).

It seems that the coordinate is not truly at the end of the line, since the dotted line is extending slightly beyond the thick blue line:

Why is this occurring?

tikz-pgf coordinates

share|improve this question

asked 2 hours ago

Thevesh ThevaThevesh Theva

519114

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Node's have finite size and you can simply say draw[thick, blue, name path = ACLt] (0,0|-H)-- (H) --(A1); which will work fine when A1 is a coordinate.
  
  – marmot
  2 hours ago
  

  
  
  
  

add a comment | 

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL] (0,0) plot [domain=0.7:9] (x,0.75*x + 0.8) node[label=right:{AC$_L$ = S$_L$}](A1){};

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] let p1 = (H) in (0,y1) node[left, red]{$w_u$} -- (x1,y1) --(A1);

end{tikzpicture}   



end{document}

My problem is as follows:

• I have attempted to define a coordinate (A1) at the end of the first line (name path ACL).


• I then use this coordinate (A1) as the last coordinate in the second line (name path ACLt).

It seems that the coordinate is not truly at the end of the line, since the dotted line is extending slightly beyond the thick blue line:

Why is this occurring?

tikz-pgf coordinates

share|improve this question

asked 2 hours ago

Thevesh ThevaThevesh Theva

519114

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL] (0,0) plot [domain=0.7:9] (x,0.75*x + 0.8) node[label=right:{AC$_L$ = S$_L$}](A1){};

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] let p1 = (H) in (0,y1) node[left, red]{$w_u$} -- (x1,y1) --(A1);

end{tikzpicture}   



end{document}

share|improve this question

asked 2 hours ago

Thevesh ThevaThevesh Theva

519114

share|improve this question

asked 2 hours ago

Thevesh ThevaThevesh Theva

519114

asked 2 hours ago

Thevesh ThevaThevesh Theva

519114

asked 2 hours ago

Thevesh ThevaThevesh Theva

519114

2 Answers
2

active

oldest

votes

3

Well, Zarko has turned my comment into "his" answer, but here it is again: use coordinate instead of node because the latter has a finite size (by default). And my answer and comment come with an arguably simpler construction of the path.

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL] (0,0) plot [domain=0.7:9] (x,0.75*x + 0.8) 

coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] (0,0|-H) node[left, red]{$w_u$} -- (H) --(A1);

end{tikzpicture}   



end{document}

share|improve this answer

answered 1 hour ago

marmotmarmot

112k5141267

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Upvoted both answers; accepted yours because your comment was sufficient for me to solve my problem before I saw either answer (I hadn't realised that nodes had a finite size even when I left the label empty). Thanks!
  
  – Thevesh Theva
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think that Zarko do not need your comment to see the difference between node and coordinate. TeX.SX is not a competition site "who will answer first" and we are not here to claim paternity of "greate" ideas, but just to help, IMHO.
  
  – Kpym
  30 mins ago
  

  
  
  
  

add a comment | 

5

calculation of coordinate is sufficient accurate, problem is that for it you use node, replace node with coordinate and result will become as you wish:

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL]

    plot [domain=0.7:9] (x,0.75*x + 0.8) coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] let p1 = (H) in (0,y1) node[left, red]{$w_u$} -- (x1,y1) --(A1);

end{tikzpicture}



end{document}

difference between node and coordinate arise since node has some width regardless if it is empty. it is determined by default value of inner sep. if you set it to zero: inner sep=0pt the result will be the same.

off-topic: in your diagram you not use intersections library, so you can remove line names from code (as i do in aboveo mwe)

share|improve this answer

edited 1 hour ago

answered 2 hours ago

ZarkoZarko

128k868167

add a comment | 

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3

Well, Zarko has turned my comment into "his" answer, but here it is again: use coordinate instead of node because the latter has a finite size (by default). And my answer and comment come with an arguably simpler construction of the path.

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL] (0,0) plot [domain=0.7:9] (x,0.75*x + 0.8) 

coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] (0,0|-H) node[left, red]{$w_u$} -- (H) --(A1);

end{tikzpicture}   



end{document}

share|improve this answer

answered 1 hour ago

marmotmarmot

112k5141267

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Upvoted both answers; accepted yours because your comment was sufficient for me to solve my problem before I saw either answer (I hadn't realised that nodes had a finite size even when I left the label empty). Thanks!
  
  – Thevesh Theva
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think that Zarko do not need your comment to see the difference between node and coordinate. TeX.SX is not a competition site "who will answer first" and we are not here to claim paternity of "greate" ideas, but just to help, IMHO.
  
  – Kpym
  30 mins ago
  

  
  
  
  

add a comment | 

3

Well, Zarko has turned my comment into "his" answer, but here it is again: use coordinate instead of node because the latter has a finite size (by default). And my answer and comment come with an arguably simpler construction of the path.

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL] (0,0) plot [domain=0.7:9] (x,0.75*x + 0.8) 

coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] (0,0|-H) node[left, red]{$w_u$} -- (H) --(A1);

end{tikzpicture}   



end{document}

share|improve this answer

answered 1 hour ago

marmotmarmot

112k5141267

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Upvoted both answers; accepted yours because your comment was sufficient for me to solve my problem before I saw either answer (I hadn't realised that nodes had a finite size even when I left the label empty). Thanks!
  
  – Thevesh Theva
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I think that Zarko do not need your comment to see the difference between node and coordinate. TeX.SX is not a competition site "who will answer first" and we are not here to claim paternity of "greate" ideas, but just to help, IMHO.
  
  – Kpym
  30 mins ago
  

  
  
  
  

add a comment | 

Well, Zarko has turned my comment into "his" answer, but here it is again: use coordinate instead of node because the latter has a finite size (by default). And my answer and comment come with an arguably simpler construction of the path.

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL] (0,0) plot [domain=0.7:9] (x,0.75*x + 0.8) 

coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] (0,0|-H) node[left, red]{$w_u$} -- (H) --(A1);

end{tikzpicture}   



end{document}

share|improve this answer

answered 1 hour ago

marmotmarmot

112k5141267

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL] (0,0) plot [domain=0.7:9] (x,0.75*x + 0.8) 

coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] (0,0|-H) node[left, red]{$w_u$} -- (H) --(A1);

end{tikzpicture}   



end{document}

share|improve this answer

answered 1 hour ago

marmotmarmot

112k5141267

answered 1 hour ago

marmotmarmot

112k5141267

answered 1 hour ago

marmotmarmot

112k5141267

5

calculation of coordinate is sufficient accurate, problem is that for it you use node, replace node with coordinate and result will become as you wish:

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL]

    plot [domain=0.7:9] (x,0.75*x + 0.8) coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] let p1 = (H) in (0,y1) node[left, red]{$w_u$} -- (x1,y1) --(A1);

end{tikzpicture}



end{document}

difference between node and coordinate arise since node has some width regardless if it is empty. it is determined by default value of inner sep. if you set it to zero: inner sep=0pt the result will be the same.

off-topic: in your diagram you not use intersections library, so you can remove line names from code (as i do in aboveo mwe)

share|improve this answer

edited 1 hour ago

answered 2 hours ago

ZarkoZarko

128k868167

add a comment | 

5

calculation of coordinate is sufficient accurate, problem is that for it you use node, replace node with coordinate and result will become as you wish:

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL]

    plot [domain=0.7:9] (x,0.75*x + 0.8) coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] let p1 = (H) in (0,y1) node[left, red]{$w_u$} -- (x1,y1) --(A1);

end{tikzpicture}



end{document}

difference between node and coordinate arise since node has some width regardless if it is empty. it is determined by default value of inner sep. if you set it to zero: inner sep=0pt the result will be the same.

off-topic: in your diagram you not use intersections library, so you can remove line names from code (as i do in aboveo mwe)

share|improve this answer

edited 1 hour ago

answered 2 hours ago

ZarkoZarko

128k868167

add a comment | 

calculation of coordinate is sufficient accurate, problem is that for it you use node, replace node with coordinate and result will become as you wish:

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL]

    plot [domain=0.7:9] (x,0.75*x + 0.8) coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] let p1 = (H) in (0,y1) node[left, red]{$w_u$} -- (x1,y1) --(A1);

end{tikzpicture}



end{document}

difference between node and coordinate arise since node has some width regardless if it is empty. it is determined by default value of inner sep. if you set it to zero: inner sep=0pt the result will be the same.

off-topic: in your diagram you not use intersections library, so you can remove line names from code (as i do in aboveo mwe)

share|improve this answer

edited 1 hour ago

answered 2 hours ago

ZarkoZarko

128k868167

documentclass{standalone}

usepackage{tikz}

usetikzlibrary{calc,intersections}

usepackage{amsmath}



begin{document}



begin{tikzpicture}[scale=0.5]

draw[<->] (0,12) node[above]{$w$} |- (12,0) node[right]{$q_L$};



draw[dotted, name path = ACL]

    plot [domain=0.7:9] (x,0.75*x + 0.8) coordinate[label=right:{AC$_L$ = S$_L$}] (A1);

coordinate (H) at ($({5.6/0.75},5.6) - ({0.8/0.75},0)$);

draw[thick, blue, name path = ACLt] let p1 = (H) in (0,y1) node[left, red]{$w_u$} -- (x1,y1) --(A1);

end{tikzpicture}



end{document}

share|improve this answer

edited 1 hour ago

answered 2 hours ago

ZarkoZarko

128k868167

answered 2 hours ago

ZarkoZarko

128k868167

answered 2 hours ago

ZarkoZarko

128k868167