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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
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
asked 1 hour ago
Thevesh ThevaThevesh Theva
519114
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
asked 1 hour ago
Thevesh ThevaThevesh Theva
519114
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 whenA1is a coordinate.
– marmot
1 hour ago
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
asked 1 hour ago
Thevesh ThevaThevesh Theva
519114
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 1 hour ago
Thevesh ThevaThevesh Theva
519114
asked 1 hour ago
Thevesh ThevaThevesh Theva
519114
asked 1 hour ago
Thevesh ThevaThevesh Theva
519114
asked 1 hour ago
Thevesh ThevaThevesh Theva
519114
asked 1 hour ago
Thevesh ThevaThevesh Theva
519114
519114
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 whenA1is a coordinate.
– marmot
1 hour 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 whenA1is a coordinate.
– marmot
1 hour ago
1
1
Node's have finite size and you can simply say
draw[thick, blue, name path = ACLt] (0,0|-H)-- (H) --(A1); which will work fine when A1 is a coordinate.– marmot
1 hour ago
Node's have finite size and you can simply say
draw[thick, blue, name path = ACLt] (0,0|-H)-- (H) --(A1); which will work fine when A1 is a coordinate.– marmot
1 hour ago
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
answered 1 hour ago
marmotmarmot
112k5141267
1
Upvoted both answers; accepted yours because your comment was sufficient for me to solve my problem before I saw either answer (I hadn't realised that nodes had a finite size even when I left the label empty). Thanks!
– Thevesh Theva
1 hour ago
I think that Zarko do not need your comment to see the difference betweennodeandcoordinate. 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
18 mins 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:
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)
answered 1 hour ago
ZarkoZarko
128k868167
add a comment |
Your Answer
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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
answered 1 hour ago
marmotmarmot
112k5141267
1
Upvoted both answers; accepted yours because your comment was sufficient for me to solve my problem before I saw either answer (I hadn't realised that nodes had a finite size even when I left the label empty). Thanks!
– Thevesh Theva
1 hour ago
I think that Zarko do not need your comment to see the difference betweennodeandcoordinate. 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
18 mins 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.
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 1 hour ago
marmotmarmot
112k5141267
1
Upvoted both answers; accepted yours because your comment was sufficient for me to solve my problem before I saw either answer (I hadn't realised that nodes had a finite size even when I left the label empty). Thanks!
– Thevesh Theva
1 hour ago
I think that Zarko do not need your comment to see the difference betweennodeandcoordinate. 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
18 mins 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.
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 1 hour ago
marmotmarmot
112k5141267
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 1 hour ago
marmotmarmot
112k5141267
answered 1 hour ago
marmotmarmot
112k5141267
answered 1 hour ago
marmotmarmot
112k5141267
answered 1 hour ago
marmotmarmot
112k5141267
112k5141267
1
Upvoted both answers; accepted yours because your comment was sufficient for me to solve my problem before I saw either answer (I hadn't realised that nodes had a finite size even when I left the label empty). Thanks!
– Thevesh Theva
1 hour ago
I think that Zarko do not need your comment to see the difference betweennodeandcoordinate. 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
18 mins 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
1 hour ago
I think that Zarko do not need your comment to see the difference betweennodeandcoordinate. 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
18 mins ago
1
1
Upvoted both answers; accepted yours because your comment was sufficient for me to solve my problem before I saw either answer (I hadn't realised that nodes had a finite size even when I left the label empty). Thanks!
– Thevesh Theva
1 hour ago
Upvoted both answers; accepted yours because your comment was sufficient for me to solve my problem before I saw either answer (I hadn't realised that nodes had a finite size even when I left the label empty). Thanks!
– Thevesh Theva
1 hour ago
I think that Zarko do not need your comment to see the difference between
node and coordinate. TeX.SX is not a competition site "who will answer first" and we are not here to claim paternity of "greate" ideas, but just to help, IMHO.– Kpym
18 mins ago
I think that Zarko do not need your comment to see the difference between
node and coordinate. TeX.SX is not a competition site "who will answer first" and we are not here to claim paternity of "greate" ideas, but just to help, IMHO.– Kpym
18 mins 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:
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)
answered 1 hour ago
ZarkoZarko
128k868167
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)
answered 1 hour ago
ZarkoZarko
128k868167
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)
answered 1 hour ago
ZarkoZarko
128k868167
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)
answered 1 hour ago
ZarkoZarko
128k868167
edited 1 hour ago
answered 1 hour ago
ZarkoZarko
128k868167
answered 1 hour ago
ZarkoZarko
128k868167
answered 1 hour ago
ZarkoZarko
128k868167
128k868167
add a comment |
add a comment |
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Required, but never shown
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 whenA1is a coordinate.– marmot
1 hour ago