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Coordinate position not precise


With TikZ, How do I use a labeled coordinate that's inside a node?Rotate a node but not its content: the case of the ellipse decorationTikZ: text along path as nodeHow to define the default vertical distance between nodes?TikZ scaling graphic and adjust node position and keep font sizeNumerical conditional within tikz keys?TikZ/ERD: node (=Entity) label on the insideTikZ: Drawing an arc from an intersection to an intersectionDrawing rectilinear curves in Tikz, aka an Etch-a-Sketch drawingLine up nested tikz enviroments or how to get rid of them













2















documentclassstandalone
usepackagetikz
usetikzlibrarycalc,intersections
usepackageamsmath

begindocument

begintikzpicture[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);
endtikzpicture

enddocument


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
    4 hours ago















2















documentclassstandalone
usepackagetikz
usetikzlibrarycalc,intersections
usepackageamsmath

begindocument

begintikzpicture[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);
endtikzpicture

enddocument


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
    4 hours ago













2












2








2








documentclassstandalone
usepackagetikz
usetikzlibrarycalc,intersections
usepackageamsmath

begindocument

begintikzpicture[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);
endtikzpicture

enddocument


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














documentclassstandalone
usepackagetikz
usetikzlibrarycalc,intersections
usepackageamsmath

begindocument

begintikzpicture[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);
endtikzpicture

enddocument


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 4 hours 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
    4 hours 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
    4 hours 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
4 hours 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
4 hours ago










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.



documentclassstandalone
usepackagetikz
usetikzlibrarycalc,intersections
usepackageamsmath

begindocument

begintikzpicture[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);
endtikzpicture

enddocument


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
    4 hours ago







  • 1





    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
    3 hours ago











  • @Kpym This is IMHO not the question. The question is whether or not one should acknowledge an earlier post saying the same thing. If you acknowledge this, it does not mean you didn't know that. It is just fair play.

    – marmot
    2 hours ago


















6














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



documentclassstandalone
usepackagetikz
usetikzlibrarycalc,intersections
usepackageamsmath

begindocument

begintikzpicture[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);
endtikzpicture

enddocument


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









    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.



    documentclassstandalone
    usepackagetikz
    usetikzlibrarycalc,intersections
    usepackageamsmath

    begindocument

    begintikzpicture[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);
    endtikzpicture

    enddocument


    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
      4 hours ago







    • 1





      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
      3 hours ago











    • @Kpym This is IMHO not the question. The question is whether or not one should acknowledge an earlier post saying the same thing. If you acknowledge this, it does not mean you didn't know that. It is just fair play.

      – marmot
      2 hours ago















    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.



    documentclassstandalone
    usepackagetikz
    usetikzlibrarycalc,intersections
    usepackageamsmath

    begindocument

    begintikzpicture[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);
    endtikzpicture

    enddocument


    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
      4 hours ago







    • 1





      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
      3 hours ago











    • @Kpym This is IMHO not the question. The question is whether or not one should acknowledge an earlier post saying the same thing. If you acknowledge this, it does not mean you didn't know that. It is just fair play.

      – marmot
      2 hours ago













    3












    3








    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.



    documentclassstandalone
    usepackagetikz
    usetikzlibrarycalc,intersections
    usepackageamsmath

    begindocument

    begintikzpicture[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);
    endtikzpicture

    enddocument


    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.



    documentclassstandalone
    usepackagetikz
    usetikzlibrarycalc,intersections
    usepackageamsmath

    begindocument

    begintikzpicture[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);
    endtikzpicture

    enddocument


    enter image description here







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 4 hours 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
      4 hours ago







    • 1





      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
      3 hours ago











    • @Kpym This is IMHO not the question. The question is whether or not one should acknowledge an earlier post saying the same thing. If you acknowledge this, it does not mean you didn't know that. It is just fair play.

      – marmot
      2 hours 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
      4 hours ago







    • 1





      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
      3 hours ago











    • @Kpym This is IMHO not the question. The question is whether or not one should acknowledge an earlier post saying the same thing. If you acknowledge this, it does not mean you didn't know that. It is just fair play.

      – marmot
      2 hours 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
    4 hours 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
    4 hours ago





    1




    1





    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
    3 hours 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
    3 hours ago













    @Kpym This is IMHO not the question. The question is whether or not one should acknowledge an earlier post saying the same thing. If you acknowledge this, it does not mean you didn't know that. It is just fair play.

    – marmot
    2 hours ago





    @Kpym This is IMHO not the question. The question is whether or not one should acknowledge an earlier post saying the same thing. If you acknowledge this, it does not mean you didn't know that. It is just fair play.

    – marmot
    2 hours ago











    6














    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



    documentclassstandalone
    usepackagetikz
    usetikzlibrarycalc,intersections
    usepackageamsmath

    begindocument

    begintikzpicture[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);
    endtikzpicture

    enddocument


    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





























      6














      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



      documentclassstandalone
      usepackagetikz
      usetikzlibrarycalc,intersections
      usepackageamsmath

      begindocument

      begintikzpicture[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);
      endtikzpicture

      enddocument


      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



























        6












        6








        6







        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



        documentclassstandalone
        usepackagetikz
        usetikzlibrarycalc,intersections
        usepackageamsmath

        begindocument

        begintikzpicture[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);
        endtikzpicture

        enddocument


        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



        documentclassstandalone
        usepackagetikz
        usetikzlibrarycalc,intersections
        usepackageamsmath

        begindocument

        begintikzpicture[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);
        endtikzpicture

        enddocument


        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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        edited 4 hours ago

























        answered 4 hours ago









        ZarkoZarko

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