Coordinate position not precise












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:



enter image description here



Why is this occurring?










share|improve this question


















  • 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
    1 hour ago
















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:



enter image description here



Why is this occurring?










share|improve this question


















  • 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
    1 hour ago














2












2








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:



enter image description here



Why is this occurring?










share|improve this question














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:



enter image description here



Why is this occurring?







tikz-pgf coordinates






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 1 hour ago









Thevesh ThevaThevesh Theva

519114




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
    1 hour ago














  • 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
    1 hour ago








1




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
1 hour ago





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
1 hour ago










2 Answers
2






active

oldest

votes


















2














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}


enter image description here






share|improve this answer



















  • 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
    8 mins ago



















4














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:



enter image description here



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

























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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    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}


    enter image description here






    share|improve this answer



















    • 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
      8 mins ago
















    2














    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}


    enter image description here






    share|improve this answer



















    • 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
      8 mins ago














    2












    2








    2







    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}


    enter image description here






    share|improve this answer













    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}


    enter image description here







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 1 hour ago









    marmotmarmot

    112k5141267




    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
      8 mins ago














    • 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
      8 mins ago








    1




    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







    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
    8 mins 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
    8 mins ago











    4














    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:



    enter image description here



    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






























      4














      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:



      enter image description here



      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




























        4












        4








        4







        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:



        enter image description here



        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















        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:



        enter image description here



        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














        share|improve this answer



        share|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        ZarkoZarko

        128k868167




        128k868167






























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