Coordinate position not precise
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
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
1
Node's have finite size and you can simply saydraw[thick, blue, name path = ACLt] (0,0|-H)-- (H) --(A1);
which will work fine whenA1
is a coordinate.
– marmot
14 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
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
tikz-pgf coordinates
asked 14 hours ago
Thevesh ThevaThevesh Theva
524114
524114
1
Node's have finite size and you can simply saydraw[thick, blue, name path = ACLt] (0,0|-H)-- (H) --(A1);
which will work fine whenA1
is a coordinate.
– marmot
14 hours ago
add a comment |
1
Node's have finite size and you can simply saydraw[thick, blue, name path = ACLt] (0,0|-H)-- (H) --(A1);
which will work fine whenA1
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
add a comment |
2 Answers
2
active
oldest
votes
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}
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 betweennode
andcoordinate
. 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
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)
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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}
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 betweennode
andcoordinate
. 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
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}
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 betweennode
andcoordinate
. 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
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}
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}
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 betweennode
andcoordinate
. 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
add a comment |
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 betweennode
andcoordinate
. 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
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)
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)
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)
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)
edited 13 hours ago
answered 14 hours ago
ZarkoZarko
128k868167
128k868167
add a comment |
add a comment |
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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 whenA1
is a coordinate.– marmot
14 hours ago