Coordinate position not precise Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraWith 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
Married in secret, can marital status in passport be changed at a later date?
Co-worker works way more than he should
First instead of 1 when referencing
Approximating integral with small parameter
What ability score does a Hexblade's Pact Weapon use for attack and damage when wielded by another character?
How would this chord from "Rocket Man" be analyzed?
Is it acceptable to use working hours to read general interest books?
Are there moral objections to a life motivated purely by money? How to sway a person from this lifestyle?
Can you stand up from being prone using Skirmisher outside of your turn?
Reattaching fallen shelf to wall?
Was Dennis Ritchie being too modest in this quote about C and Pascal?
How do I check if a string is entirely made of the same substring?
iOS App Store: Unable to download and update apps due to Terms and Conditions loop, even after agreeing
What does a straight horizontal line above a few notes, after a changed tempo mean?
Why didn't the Space Shuttle bounce back into space as many times as possible so as to lose a lot of kinetic energy up there?
"My boss was furious with me and I have been fired" vs. "My boss was furious with me and I was fired"
How to pronounce the unstressed е in прише́дшие?
Are these square matrices always diagonalisable?
What *exactly* is electrical current, voltage, and resistance?
How to keep bees out of canned beverages?
Multiple options vs single option UI
What's the difference between using dependency injection with a container and using a service locator?
A faster way to compute the largest prime factor
Retract an already submitted recommendation letter (written for an undergrad student)
Coordinate position not precise
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraWith 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
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
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
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 |
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
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
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
121k6158296
121k6158296
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
131k869170
131k869170
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
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
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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