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Coordinate position not precise
Rotate a node but not its content: the case of the ellipse decorationHow to define the default vertical distance between nodes?Calculate the intersection between a path enclosed by a `scope` and another pathUsing current path position in coordinate calculationNumerical 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 themtikz: not so precise graphic
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
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:
Why is this occurring?
tikz-pgf coordinates
add a comment |
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:
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
Mar 26 at 20:32
add a comment |
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:
Why is this occurring?
tikz-pgf coordinates
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:
Why is this occurring?
tikz-pgf coordinates
tikz-pgf coordinates
asked Mar 26 at 20:26
Thevesh ThevaThevesh Theva
5261 silver badge14 bronze badges
5261 silver badge14 bronze badges
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
Mar 26 at 20:32
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
Mar 26 at 20:32
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
Mar 26 at 20:32
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
Mar 26 at 20:32
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.
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
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
Mar 26 at 20:48
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
Mar 26 at 22:03
@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
Mar 26 at 22:41
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:
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)
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.
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
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
Mar 26 at 20:48
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
Mar 26 at 22:03
@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
Mar 26 at 22:41
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.
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
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
Mar 26 at 20:48
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
Mar 26 at 22:03
@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
Mar 26 at 22:41
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.
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
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
answered Mar 26 at 20:41
marmotmarmot
134k6 gold badges190 silver badges352 bronze badges
134k6 gold badges190 silver badges352 bronze badges
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
Mar 26 at 20:48
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
Mar 26 at 22:03
@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
Mar 26 at 22:41
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
Mar 26 at 20:48
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
Mar 26 at 22:03
@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
Mar 26 at 22:41
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
Mar 26 at 20:48
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
Mar 26 at 20:48
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
Mar 26 at 22:03
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
Mar 26 at 22:03
@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
Mar 26 at 22:41
@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
Mar 26 at 22:41
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:
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)
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:
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)
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:
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)
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:
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)
edited Mar 26 at 20:39
answered Mar 26 at 20:32
ZarkoZarko
141k8 gold badges78 silver badges190 bronze badges
141k8 gold badges78 silver badges190 bronze badges
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
Mar 26 at 20:32