Coordinate position not precise












3















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
    14 hours ago
















3















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
    14 hours ago














3












3








3








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 14 hours ago









Thevesh ThevaThevesh Theva

524114




524114








  • 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
    14 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
    14 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
14 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
14 hours ago










2 Answers
2






active

oldest

votes


















5














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
    13 hours ago








  • 3





    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
    12 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
    11 hours ago



















8














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

























    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "85"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f481586%2fcoordinate-position-not-precise%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5














    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
      13 hours ago








    • 3





      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
      12 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
      11 hours ago
















    5














    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
      13 hours ago








    • 3





      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
      12 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
      11 hours ago














    5












    5








    5







    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 13 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
      13 hours ago








    • 3





      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
      12 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
      11 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
      13 hours ago








    • 3





      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
      12 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
      11 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
    13 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
    13 hours ago






    3




    3





    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
    12 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
    12 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
    11 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
    11 hours ago











    8














    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






























      8














      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




























        8












        8








        8







        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 13 hours ago

























        answered 14 hours ago









        ZarkoZarko

        128k868167




        128k868167






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to TeX - LaTeX Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f481586%2fcoordinate-position-not-precise%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

            He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

            Slayer Innehåll Historia | Stil, komposition och lyrik | Bandets betydelse och framgångar | Sidoprojekt och samarbeten | Kontroverser | Medlemmar | Utmärkelser och nomineringar | Turnéer och festivaler | Diskografi | Referenser | Externa länkar | Navigeringsmenywww.slayer.net”Metal Massacre vol. 1””Metal Massacre vol. 3””Metal Massacre Volume III””Show No Mercy””Haunting the Chapel””Live Undead””Hell Awaits””Reign in Blood””Reign in Blood””Gold & Platinum – Reign in Blood””Golden Gods Awards Winners”originalet”Kerrang! Hall Of Fame””Slayer Looks Back On 37-Year Career In New Video Series: Part Two””South of Heaven””Gold & Platinum – South of Heaven””Seasons in the Abyss””Gold & Platinum - Seasons in the Abyss””Divine Intervention””Divine Intervention - Release group by Slayer””Gold & Platinum - Divine Intervention””Live Intrusion””Undisputed Attitude””Abolish Government/Superficial Love””Release “Slatanic Slaughter: A Tribute to Slayer” by Various Artists””Diabolus in Musica””Soundtrack to the Apocalypse””God Hates Us All””Systematic - Relationships””War at the Warfield””Gold & Platinum - War at the Warfield””Soundtrack to the Apocalypse””Gold & Platinum - Still Reigning””Metallica, Slayer, Iron Mauden Among Winners At Metal Hammer Awards””Eternal Pyre””Eternal Pyre - Slayer release group””Eternal Pyre””Metal Storm Awards 2006””Kerrang! Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029