TikZ: Understanding the usage of calc library
up vote
7
down vote
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For the following MWE, I need to place block (yaw) {C}
midway between (1) the middle point between (output)
and (integrator)
(2) and (sum2)
.
So, how can I correct this syntax node [block] (yaw) at ([yshift=-2cm]$(integrator)+0.5*{(output)-(integrator)}!0.5!(sum2)$) {C};
to make it work?
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at ([yshift=-2cm]$(integrator)+0.5*{(output)-(integrator)}!0.5!(sum2)$) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Additionally, is it possible to create a new node using node [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
without creating auxiliary nodes/coordinates?
tikz-pgf tikz-calc
add a comment |
up vote
7
down vote
favorite
For the following MWE, I need to place block (yaw) {C}
midway between (1) the middle point between (output)
and (integrator)
(2) and (sum2)
.
So, how can I correct this syntax node [block] (yaw) at ([yshift=-2cm]$(integrator)+0.5*{(output)-(integrator)}!0.5!(sum2)$) {C};
to make it work?
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at ([yshift=-2cm]$(integrator)+0.5*{(output)-(integrator)}!0.5!(sum2)$) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Additionally, is it possible to create a new node using node [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
without creating auxiliary nodes/coordinates?
tikz-pgf tikz-calc
Hey! Did($.25*(output)+.25*(integrator)+.5*(sum2)$)
work for you?
– Vinzza
Nov 15 at 10:26
@Vinzza It does. But, why does my approach not work?
– Diaa
Nov 15 at 10:33
Comments do not allow enough characters, so I have replied with an answer! I hope it will help you! :)
– Vinzza
Nov 15 at 13:27
Your approach does not work because you try to use{
and}
where you should use($
and$)
. Try($(0,-2cm)+(integrator)+0.5*($(output)-(integrator)$)!0.5!(sum2)$)
to have something that does not throw an error. However, from your description in words I think you wantnode [block] (yaw) at ($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
, yet this can be done withoutcalc
:node [block] (yaw) at ([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
.
– marmot
Nov 15 at 14:27
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
For the following MWE, I need to place block (yaw) {C}
midway between (1) the middle point between (output)
and (integrator)
(2) and (sum2)
.
So, how can I correct this syntax node [block] (yaw) at ([yshift=-2cm]$(integrator)+0.5*{(output)-(integrator)}!0.5!(sum2)$) {C};
to make it work?
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at ([yshift=-2cm]$(integrator)+0.5*{(output)-(integrator)}!0.5!(sum2)$) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Additionally, is it possible to create a new node using node [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
without creating auxiliary nodes/coordinates?
tikz-pgf tikz-calc
For the following MWE, I need to place block (yaw) {C}
midway between (1) the middle point between (output)
and (integrator)
(2) and (sum2)
.
So, how can I correct this syntax node [block] (yaw) at ([yshift=-2cm]$(integrator)+0.5*{(output)-(integrator)}!0.5!(sum2)$) {C};
to make it work?
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at ([yshift=-2cm]$(integrator)+0.5*{(output)-(integrator)}!0.5!(sum2)$) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Additionally, is it possible to create a new node using node [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
without creating auxiliary nodes/coordinates?
tikz-pgf tikz-calc
tikz-pgf tikz-calc
edited Nov 15 at 11:11
asked Nov 15 at 10:21
Diaa
2,51011644
2,51011644
Hey! Did($.25*(output)+.25*(integrator)+.5*(sum2)$)
work for you?
– Vinzza
Nov 15 at 10:26
@Vinzza It does. But, why does my approach not work?
– Diaa
Nov 15 at 10:33
Comments do not allow enough characters, so I have replied with an answer! I hope it will help you! :)
– Vinzza
Nov 15 at 13:27
Your approach does not work because you try to use{
and}
where you should use($
and$)
. Try($(0,-2cm)+(integrator)+0.5*($(output)-(integrator)$)!0.5!(sum2)$)
to have something that does not throw an error. However, from your description in words I think you wantnode [block] (yaw) at ($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
, yet this can be done withoutcalc
:node [block] (yaw) at ([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
.
– marmot
Nov 15 at 14:27
add a comment |
Hey! Did($.25*(output)+.25*(integrator)+.5*(sum2)$)
work for you?
– Vinzza
Nov 15 at 10:26
@Vinzza It does. But, why does my approach not work?
– Diaa
Nov 15 at 10:33
Comments do not allow enough characters, so I have replied with an answer! I hope it will help you! :)
– Vinzza
Nov 15 at 13:27
Your approach does not work because you try to use{
and}
where you should use($
and$)
. Try($(0,-2cm)+(integrator)+0.5*($(output)-(integrator)$)!0.5!(sum2)$)
to have something that does not throw an error. However, from your description in words I think you wantnode [block] (yaw) at ($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
, yet this can be done withoutcalc
:node [block] (yaw) at ([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
.
– marmot
Nov 15 at 14:27
Hey! Did
($.25*(output)+.25*(integrator)+.5*(sum2)$)
work for you?– Vinzza
Nov 15 at 10:26
Hey! Did
($.25*(output)+.25*(integrator)+.5*(sum2)$)
work for you?– Vinzza
Nov 15 at 10:26
@Vinzza It does. But, why does my approach not work?
– Diaa
Nov 15 at 10:33
@Vinzza It does. But, why does my approach not work?
– Diaa
Nov 15 at 10:33
Comments do not allow enough characters, so I have replied with an answer! I hope it will help you! :)
– Vinzza
Nov 15 at 13:27
Comments do not allow enough characters, so I have replied with an answer! I hope it will help you! :)
– Vinzza
Nov 15 at 13:27
Your approach does not work because you try to use
{
and }
where you should use ($
and $)
. Try ($(0,-2cm)+(integrator)+0.5*($(output)-(integrator)$)!0.5!(sum2)$)
to have something that does not throw an error. However, from your description in words I think you want node [block] (yaw) at ($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
, yet this can be done without calc
: node [block] (yaw) at ([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
.– marmot
Nov 15 at 14:27
Your approach does not work because you try to use
{
and }
where you should use ($
and $)
. Try ($(0,-2cm)+(integrator)+0.5*($(output)-(integrator)$)!0.5!(sum2)$)
to have something that does not throw an error. However, from your description in words I think you want node [block] (yaw) at ($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
, yet this can be done without calc
: node [block] (yaw) at ([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
.– marmot
Nov 15 at 14:27
add a comment |
5 Answers
5
active
oldest
votes
up vote
5
down vote
accepted
I don't know how complex expression can be understood by calc
but instead of trying to understand how to write such expression, I think it's easier to use an auxiliar coordinate and solve the problem:
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
coordinate (aux) at ($(integrator.east)!.5!(output)$);
node [block] (yaw) at ([yshift=-2cm]$(aux)!0.5!(sum2)$) {C};
draw (aux) |- (yaw);
draw (yaw)-|(sum2);
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Many thanks. For my question edit, can I make a new node using this syntaxnode [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
?
– Diaa
Nov 15 at 11:13
1
@Diaa Something likenode [block, yshift=-2cm] (yaw) at ($(aux)!.5!(sum2)$) {C};
works for me. Instead ofyshift
you could also usebelow=2cm
but with a different result. Thecalc
expression inpositioning
option didn't work for me.
– Ignasi
Nov 15 at 11:24
@Diaa In any case I don't see the problem in using auxiliary coordinates/nodes. What's wrong with them?
– Ignasi
Nov 15 at 11:25
Nothing wrong; I just want to teach myself how to reduce my code :)
– Diaa
Nov 15 at 11:31
add a comment |
up vote
6
down vote
Your approach does not work because you try to use {
and }
where you should use ($
and $)
. You can definitely do that without auxiliary coordinates and actually even without calc
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
bigskip
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
end{document}
I am sorry, but could you tell me where I can find more explanation on this line([yshift=-2cm] barycentric cs:output=1,integrator=1,sum2=2)
?
– Diaa
Nov 15 at 14:45
1
@Diaa Section 13.2.2 Barycentric Systems of the pgfmanual. Come on, it only has 1161 pages. (Just kidding! ;-)
– marmot
Nov 15 at 14:47
XD. If you don't mind, I have off-topic question: when sayingnode distance = 2 cm
, it measures this distance between the nodes centers. Is it possible to make this distance imply the spacing between(left node.east)
and(right node.west)
instead?
– Diaa
Nov 15 at 14:58
@Diaa I am not sure I agree with your statement. You are already loadingpositioning
, in which case the distances are measured between the node boundaries (modulo a very tiny bit of fine print). Try e.g.node [block, right=of sum2](ractuator){$frac{2}{s+2}$};
in your settings. Then you will see that the distance between the node boundaries, and not centers, is 2cm, which is the value ofnode distance
in your code.
– marmot
Nov 15 at 15:03
I tried drawing a new nodenode [block, above of = yaw, draw=none, node distance=1mm] {Yaw Rate\Sensor};
and the result is two overlapping nodes as seen here.
– Diaa
Nov 15 at 15:28
|
show 2 more comments
up vote
5
down vote
Here, to simplify the code, I'll replace (integrator)
with (A)
, (output)
with (B)
and (sum2)
with (C)
.
There is two things not right with
($ (A) + 0.5*{ (B)-(A) }!0.5!(C) $)
.
First, I don't think you can use
{}
, with the calc package, for the coordinate part. For me, it only works for with scalar. So($ {2+2}*(A) $)
will compute, but not($ 2*{(A)+(B)} $)
(or am I wrong?)The second thing is that this formula doesn't seem to correspond to the point you want.
I kind of get that you want to start from (A), "move" to the middle of [AB] and continue like that, but you mix relative (B-A) and absolute positioning (C).
One right formula would have been($ { (A) + 0.5*{(B)-(A)} }!0.5!(C) $)
.
But because tikz can't do the computation, you'll have to give the expanded formula:($ .25*(A) + .25*(B) + .5*(C)$)
.
One other way to do it is ($ (A) !.5! (B) !.5! (C) $)
. Here, we take the middle of (A) and (B), and then the middle of the result and (C).
I hope this will answer your interrogations!
You can test the three solutions here (the last one with temporary coordinate):
documentclass[tikz,margin=10pt]{standalone}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[line width=1]
draw[black!10] (0,0) grid (4,4);
node (A) at (1,1) {A};
node (B) at (3,1) {B};
node (C) at (2,3) {C};
%% 1
draw[red] ($ (A) !.5! (B) !.5! (C) $) circle (.05);
%% 2
draw[orange] ($ .25*(A) + .25*(B) + .5*(C) $) circle (.1);
%% 3
coordinate (foo) at ($ (A) !.5! (B) $);
draw[yellow] ($ (foo) !.5! (C) $) circle (.15);
end{tikzpicture}
end{document}
which gives
Thanks for the answer, but I didn't interrogate :)
– Diaa
Nov 15 at 15:29
add a comment |
up vote
5
down vote
The calc
library allows you to apply Parway Modifiers
repeatedly. Thus, the following syntax
($(integrator)!.5!!(output)!0.5!(sum2)$)
does the following:
- pgf calculates the middle of
(integrator)
and(output)
- then calculates the middle of this last calculated point and the next one
(sum2)
We can continue like this as many times as we want.
Here is for example page 144 of the manual 3.0.1a modified by adding two more points.
documentclass{article}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[every node/.style={draw,circle,inner sep=1pt}]
draw [help lines] (0,0) grid (3,2);
%first node
draw[densely dotted] (0,0) -- (3,2);
node at ($(0,0)!.3!(3,2)$) {1};
%second node
draw[densely dotted] ($(0,0)!.3!(3,2)$) -- (3,0);
node at ($(0,0)!.3!(3,2)!.7!(3,0)$){2};
%third node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)$)--(3,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$) {3};
%fourth node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$)--(0,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)!.5!(0,2)$) {4};
end{tikzpicture}
end{document}
Unfortunately, this does not simplify the writing of the code. The use of an auxiliary point as @Ignasi did is therefore more elegant.
Updated just for fun: A complete solution with the calc
library
And without using yshift=-2cm
and without intermediate point (It's really complicated and unreadable!)
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
But which places the point in the same place with the syntax indicated in the manual 3.0.1a p143, i quote:
The general meaning of
<a>!<factor>!<angle>:<b>
is “First, consider
the line from<a>
to<b>
. Then rotate this line by<angle>
around the
point<a>
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Old answer:
Nevertheless, here is a solution that includes a series of Parway Modifiers
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-([yshift=-2cm]$(integrator)!.5!(output)!0.5!(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Translated with www.DeepL.com/Translator
add a comment |
up vote
3
down vote
one way how to reduce your code:
- use tikz library
chains
placement nodes in chain and draw lines between them by macrojoin
- node "c" in feedback simple pace below of node
vdynamics
- put coordinates in image directly and not via nodes
- coordinates can contain labels, exploit this for label
$Psi$
define nodes distance only ones and than use it all all nodes positioning
documentclass{article}
usepackage{tikz,nccmath,amssymb}
usetikzlibrary{arrows,
calc, chains,
positioning,
shapes}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum size=3em,
on chain, join=by ->},
sum/.style = {draw, fill=white, circle},
}
makeatletter
tikzset{suspend join/.code={deftikz@after@path{}}} % <--- for dicountinue of jon macro
makeatother
begin{tikzpicture}[
node distance = 0.5cm and 1cm,
start chain = going right,
> = latex']
coordinate (in);
node [sum,right=of in, on chain] (sum2) {};
node [block] (ractuator) {$mfrac{2}{s+2}$};
node [block] (vdynamics) {$mfrac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block] (integrator) {$mfrac{1}{s}$};
coordinate[right=of integrator] (out) {};
node [block, suspend join,
below = of vdynamics] (yaw) {C};
%
draw[->] (in) -- (sum2);
draw[->] (integrator) -- coordinate[label=$Psi(s)$] (psi) (out);
draw[->] (psi) |- (yaw);
draw[->] (yaw) -| (sum2);
end{tikzpicture}
end{document}
off-topic: for fraction is used mfrac
(medium sized fraction) defined in the nccmath
package
Perfect! Thanks for this beautiful answer.
– Diaa
Nov 15 at 16:48
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
I don't know how complex expression can be understood by calc
but instead of trying to understand how to write such expression, I think it's easier to use an auxiliar coordinate and solve the problem:
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
coordinate (aux) at ($(integrator.east)!.5!(output)$);
node [block] (yaw) at ([yshift=-2cm]$(aux)!0.5!(sum2)$) {C};
draw (aux) |- (yaw);
draw (yaw)-|(sum2);
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Many thanks. For my question edit, can I make a new node using this syntaxnode [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
?
– Diaa
Nov 15 at 11:13
1
@Diaa Something likenode [block, yshift=-2cm] (yaw) at ($(aux)!.5!(sum2)$) {C};
works for me. Instead ofyshift
you could also usebelow=2cm
but with a different result. Thecalc
expression inpositioning
option didn't work for me.
– Ignasi
Nov 15 at 11:24
@Diaa In any case I don't see the problem in using auxiliary coordinates/nodes. What's wrong with them?
– Ignasi
Nov 15 at 11:25
Nothing wrong; I just want to teach myself how to reduce my code :)
– Diaa
Nov 15 at 11:31
add a comment |
up vote
5
down vote
accepted
I don't know how complex expression can be understood by calc
but instead of trying to understand how to write such expression, I think it's easier to use an auxiliar coordinate and solve the problem:
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
coordinate (aux) at ($(integrator.east)!.5!(output)$);
node [block] (yaw) at ([yshift=-2cm]$(aux)!0.5!(sum2)$) {C};
draw (aux) |- (yaw);
draw (yaw)-|(sum2);
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Many thanks. For my question edit, can I make a new node using this syntaxnode [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
?
– Diaa
Nov 15 at 11:13
1
@Diaa Something likenode [block, yshift=-2cm] (yaw) at ($(aux)!.5!(sum2)$) {C};
works for me. Instead ofyshift
you could also usebelow=2cm
but with a different result. Thecalc
expression inpositioning
option didn't work for me.
– Ignasi
Nov 15 at 11:24
@Diaa In any case I don't see the problem in using auxiliary coordinates/nodes. What's wrong with them?
– Ignasi
Nov 15 at 11:25
Nothing wrong; I just want to teach myself how to reduce my code :)
– Diaa
Nov 15 at 11:31
add a comment |
up vote
5
down vote
accepted
up vote
5
down vote
accepted
I don't know how complex expression can be understood by calc
but instead of trying to understand how to write such expression, I think it's easier to use an auxiliar coordinate and solve the problem:
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
coordinate (aux) at ($(integrator.east)!.5!(output)$);
node [block] (yaw) at ([yshift=-2cm]$(aux)!0.5!(sum2)$) {C};
draw (aux) |- (yaw);
draw (yaw)-|(sum2);
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
I don't know how complex expression can be understood by calc
but instead of trying to understand how to write such expression, I think it's easier to use an auxiliar coordinate and solve the problem:
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
coordinate (aux) at ($(integrator.east)!.5!(output)$);
node [block] (yaw) at ([yshift=-2cm]$(aux)!0.5!(sum2)$) {C};
draw (aux) |- (yaw);
draw (yaw)-|(sum2);
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
answered Nov 15 at 11:01
Ignasi
89.9k4162302
89.9k4162302
Many thanks. For my question edit, can I make a new node using this syntaxnode [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
?
– Diaa
Nov 15 at 11:13
1
@Diaa Something likenode [block, yshift=-2cm] (yaw) at ($(aux)!.5!(sum2)$) {C};
works for me. Instead ofyshift
you could also usebelow=2cm
but with a different result. Thecalc
expression inpositioning
option didn't work for me.
– Ignasi
Nov 15 at 11:24
@Diaa In any case I don't see the problem in using auxiliary coordinates/nodes. What's wrong with them?
– Ignasi
Nov 15 at 11:25
Nothing wrong; I just want to teach myself how to reduce my code :)
– Diaa
Nov 15 at 11:31
add a comment |
Many thanks. For my question edit, can I make a new node using this syntaxnode [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
?
– Diaa
Nov 15 at 11:13
1
@Diaa Something likenode [block, yshift=-2cm] (yaw) at ($(aux)!.5!(sum2)$) {C};
works for me. Instead ofyshift
you could also usebelow=2cm
but with a different result. Thecalc
expression inpositioning
option didn't work for me.
– Ignasi
Nov 15 at 11:24
@Diaa In any case I don't see the problem in using auxiliary coordinates/nodes. What's wrong with them?
– Ignasi
Nov 15 at 11:25
Nothing wrong; I just want to teach myself how to reduce my code :)
– Diaa
Nov 15 at 11:31
Many thanks. For my question edit, can I make a new node using this syntax
node [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
?– Diaa
Nov 15 at 11:13
Many thanks. For my question edit, can I make a new node using this syntax
node [tmp, below = 2cm of ($(output)!0.5!(integrator)$) ] (tmp1) {};
?– Diaa
Nov 15 at 11:13
1
1
@Diaa Something like
node [block, yshift=-2cm] (yaw) at ($(aux)!.5!(sum2)$) {C};
works for me. Instead of yshift
you could also use below=2cm
but with a different result. The calc
expression in positioning
option didn't work for me.– Ignasi
Nov 15 at 11:24
@Diaa Something like
node [block, yshift=-2cm] (yaw) at ($(aux)!.5!(sum2)$) {C};
works for me. Instead of yshift
you could also use below=2cm
but with a different result. The calc
expression in positioning
option didn't work for me.– Ignasi
Nov 15 at 11:24
@Diaa In any case I don't see the problem in using auxiliary coordinates/nodes. What's wrong with them?
– Ignasi
Nov 15 at 11:25
@Diaa In any case I don't see the problem in using auxiliary coordinates/nodes. What's wrong with them?
– Ignasi
Nov 15 at 11:25
Nothing wrong; I just want to teach myself how to reduce my code :)
– Diaa
Nov 15 at 11:31
Nothing wrong; I just want to teach myself how to reduce my code :)
– Diaa
Nov 15 at 11:31
add a comment |
up vote
6
down vote
Your approach does not work because you try to use {
and }
where you should use ($
and $)
. You can definitely do that without auxiliary coordinates and actually even without calc
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
bigskip
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
end{document}
I am sorry, but could you tell me where I can find more explanation on this line([yshift=-2cm] barycentric cs:output=1,integrator=1,sum2=2)
?
– Diaa
Nov 15 at 14:45
1
@Diaa Section 13.2.2 Barycentric Systems of the pgfmanual. Come on, it only has 1161 pages. (Just kidding! ;-)
– marmot
Nov 15 at 14:47
XD. If you don't mind, I have off-topic question: when sayingnode distance = 2 cm
, it measures this distance between the nodes centers. Is it possible to make this distance imply the spacing between(left node.east)
and(right node.west)
instead?
– Diaa
Nov 15 at 14:58
@Diaa I am not sure I agree with your statement. You are already loadingpositioning
, in which case the distances are measured between the node boundaries (modulo a very tiny bit of fine print). Try e.g.node [block, right=of sum2](ractuator){$frac{2}{s+2}$};
in your settings. Then you will see that the distance between the node boundaries, and not centers, is 2cm, which is the value ofnode distance
in your code.
– marmot
Nov 15 at 15:03
I tried drawing a new nodenode [block, above of = yaw, draw=none, node distance=1mm] {Yaw Rate\Sensor};
and the result is two overlapping nodes as seen here.
– Diaa
Nov 15 at 15:28
|
show 2 more comments
up vote
6
down vote
Your approach does not work because you try to use {
and }
where you should use ($
and $)
. You can definitely do that without auxiliary coordinates and actually even without calc
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
bigskip
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
end{document}
I am sorry, but could you tell me where I can find more explanation on this line([yshift=-2cm] barycentric cs:output=1,integrator=1,sum2=2)
?
– Diaa
Nov 15 at 14:45
1
@Diaa Section 13.2.2 Barycentric Systems of the pgfmanual. Come on, it only has 1161 pages. (Just kidding! ;-)
– marmot
Nov 15 at 14:47
XD. If you don't mind, I have off-topic question: when sayingnode distance = 2 cm
, it measures this distance between the nodes centers. Is it possible to make this distance imply the spacing between(left node.east)
and(right node.west)
instead?
– Diaa
Nov 15 at 14:58
@Diaa I am not sure I agree with your statement. You are already loadingpositioning
, in which case the distances are measured between the node boundaries (modulo a very tiny bit of fine print). Try e.g.node [block, right=of sum2](ractuator){$frac{2}{s+2}$};
in your settings. Then you will see that the distance between the node boundaries, and not centers, is 2cm, which is the value ofnode distance
in your code.
– marmot
Nov 15 at 15:03
I tried drawing a new nodenode [block, above of = yaw, draw=none, node distance=1mm] {Yaw Rate\Sensor};
and the result is two overlapping nodes as seen here.
– Diaa
Nov 15 at 15:28
|
show 2 more comments
up vote
6
down vote
up vote
6
down vote
Your approach does not work because you try to use {
and }
where you should use ($
and $)
. You can definitely do that without auxiliary coordinates and actually even without calc
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
bigskip
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
end{document}
Your approach does not work because you try to use {
and }
where you should use ($
and $)
. You can definitely do that without auxiliary coordinates and actually even without calc
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
bigskip
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
node [block] (yaw) at
([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
%
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output)
coordinate[midway] (aux);
draw (aux) |- (yaw) -| (sum2);
end{tikzpicture}
end{document}
answered Nov 15 at 14:34
marmot
76.2k486160
76.2k486160
I am sorry, but could you tell me where I can find more explanation on this line([yshift=-2cm] barycentric cs:output=1,integrator=1,sum2=2)
?
– Diaa
Nov 15 at 14:45
1
@Diaa Section 13.2.2 Barycentric Systems of the pgfmanual. Come on, it only has 1161 pages. (Just kidding! ;-)
– marmot
Nov 15 at 14:47
XD. If you don't mind, I have off-topic question: when sayingnode distance = 2 cm
, it measures this distance between the nodes centers. Is it possible to make this distance imply the spacing between(left node.east)
and(right node.west)
instead?
– Diaa
Nov 15 at 14:58
@Diaa I am not sure I agree with your statement. You are already loadingpositioning
, in which case the distances are measured between the node boundaries (modulo a very tiny bit of fine print). Try e.g.node [block, right=of sum2](ractuator){$frac{2}{s+2}$};
in your settings. Then you will see that the distance between the node boundaries, and not centers, is 2cm, which is the value ofnode distance
in your code.
– marmot
Nov 15 at 15:03
I tried drawing a new nodenode [block, above of = yaw, draw=none, node distance=1mm] {Yaw Rate\Sensor};
and the result is two overlapping nodes as seen here.
– Diaa
Nov 15 at 15:28
|
show 2 more comments
I am sorry, but could you tell me where I can find more explanation on this line([yshift=-2cm] barycentric cs:output=1,integrator=1,sum2=2)
?
– Diaa
Nov 15 at 14:45
1
@Diaa Section 13.2.2 Barycentric Systems of the pgfmanual. Come on, it only has 1161 pages. (Just kidding! ;-)
– marmot
Nov 15 at 14:47
XD. If you don't mind, I have off-topic question: when sayingnode distance = 2 cm
, it measures this distance between the nodes centers. Is it possible to make this distance imply the spacing between(left node.east)
and(right node.west)
instead?
– Diaa
Nov 15 at 14:58
@Diaa I am not sure I agree with your statement. You are already loadingpositioning
, in which case the distances are measured between the node boundaries (modulo a very tiny bit of fine print). Try e.g.node [block, right=of sum2](ractuator){$frac{2}{s+2}$};
in your settings. Then you will see that the distance between the node boundaries, and not centers, is 2cm, which is the value ofnode distance
in your code.
– marmot
Nov 15 at 15:03
I tried drawing a new nodenode [block, above of = yaw, draw=none, node distance=1mm] {Yaw Rate\Sensor};
and the result is two overlapping nodes as seen here.
– Diaa
Nov 15 at 15:28
I am sorry, but could you tell me where I can find more explanation on this line
([yshift=-2cm] barycentric cs:output=1,integrator=1,sum2=2)
?– Diaa
Nov 15 at 14:45
I am sorry, but could you tell me where I can find more explanation on this line
([yshift=-2cm] barycentric cs:output=1,integrator=1,sum2=2)
?– Diaa
Nov 15 at 14:45
1
1
@Diaa Section 13.2.2 Barycentric Systems of the pgfmanual. Come on, it only has 1161 pages. (Just kidding! ;-)
– marmot
Nov 15 at 14:47
@Diaa Section 13.2.2 Barycentric Systems of the pgfmanual. Come on, it only has 1161 pages. (Just kidding! ;-)
– marmot
Nov 15 at 14:47
XD. If you don't mind, I have off-topic question: when saying
node distance = 2 cm
, it measures this distance between the nodes centers. Is it possible to make this distance imply the spacing between (left node.east)
and (right node.west)
instead?– Diaa
Nov 15 at 14:58
XD. If you don't mind, I have off-topic question: when saying
node distance = 2 cm
, it measures this distance between the nodes centers. Is it possible to make this distance imply the spacing between (left node.east)
and (right node.west)
instead?– Diaa
Nov 15 at 14:58
@Diaa I am not sure I agree with your statement. You are already loading
positioning
, in which case the distances are measured between the node boundaries (modulo a very tiny bit of fine print). Try e.g. node [block, right=of sum2](ractuator){$frac{2}{s+2}$};
in your settings. Then you will see that the distance between the node boundaries, and not centers, is 2cm, which is the value of node distance
in your code.– marmot
Nov 15 at 15:03
@Diaa I am not sure I agree with your statement. You are already loading
positioning
, in which case the distances are measured between the node boundaries (modulo a very tiny bit of fine print). Try e.g. node [block, right=of sum2](ractuator){$frac{2}{s+2}$};
in your settings. Then you will see that the distance between the node boundaries, and not centers, is 2cm, which is the value of node distance
in your code.– marmot
Nov 15 at 15:03
I tried drawing a new node
node [block, above of = yaw, draw=none, node distance=1mm] {Yaw Rate\Sensor};
and the result is two overlapping nodes as seen here.– Diaa
Nov 15 at 15:28
I tried drawing a new node
node [block, above of = yaw, draw=none, node distance=1mm] {Yaw Rate\Sensor};
and the result is two overlapping nodes as seen here.– Diaa
Nov 15 at 15:28
|
show 2 more comments
up vote
5
down vote
Here, to simplify the code, I'll replace (integrator)
with (A)
, (output)
with (B)
and (sum2)
with (C)
.
There is two things not right with
($ (A) + 0.5*{ (B)-(A) }!0.5!(C) $)
.
First, I don't think you can use
{}
, with the calc package, for the coordinate part. For me, it only works for with scalar. So($ {2+2}*(A) $)
will compute, but not($ 2*{(A)+(B)} $)
(or am I wrong?)The second thing is that this formula doesn't seem to correspond to the point you want.
I kind of get that you want to start from (A), "move" to the middle of [AB] and continue like that, but you mix relative (B-A) and absolute positioning (C).
One right formula would have been($ { (A) + 0.5*{(B)-(A)} }!0.5!(C) $)
.
But because tikz can't do the computation, you'll have to give the expanded formula:($ .25*(A) + .25*(B) + .5*(C)$)
.
One other way to do it is ($ (A) !.5! (B) !.5! (C) $)
. Here, we take the middle of (A) and (B), and then the middle of the result and (C).
I hope this will answer your interrogations!
You can test the three solutions here (the last one with temporary coordinate):
documentclass[tikz,margin=10pt]{standalone}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[line width=1]
draw[black!10] (0,0) grid (4,4);
node (A) at (1,1) {A};
node (B) at (3,1) {B};
node (C) at (2,3) {C};
%% 1
draw[red] ($ (A) !.5! (B) !.5! (C) $) circle (.05);
%% 2
draw[orange] ($ .25*(A) + .25*(B) + .5*(C) $) circle (.1);
%% 3
coordinate (foo) at ($ (A) !.5! (B) $);
draw[yellow] ($ (foo) !.5! (C) $) circle (.15);
end{tikzpicture}
end{document}
which gives
Thanks for the answer, but I didn't interrogate :)
– Diaa
Nov 15 at 15:29
add a comment |
up vote
5
down vote
Here, to simplify the code, I'll replace (integrator)
with (A)
, (output)
with (B)
and (sum2)
with (C)
.
There is two things not right with
($ (A) + 0.5*{ (B)-(A) }!0.5!(C) $)
.
First, I don't think you can use
{}
, with the calc package, for the coordinate part. For me, it only works for with scalar. So($ {2+2}*(A) $)
will compute, but not($ 2*{(A)+(B)} $)
(or am I wrong?)The second thing is that this formula doesn't seem to correspond to the point you want.
I kind of get that you want to start from (A), "move" to the middle of [AB] and continue like that, but you mix relative (B-A) and absolute positioning (C).
One right formula would have been($ { (A) + 0.5*{(B)-(A)} }!0.5!(C) $)
.
But because tikz can't do the computation, you'll have to give the expanded formula:($ .25*(A) + .25*(B) + .5*(C)$)
.
One other way to do it is ($ (A) !.5! (B) !.5! (C) $)
. Here, we take the middle of (A) and (B), and then the middle of the result and (C).
I hope this will answer your interrogations!
You can test the three solutions here (the last one with temporary coordinate):
documentclass[tikz,margin=10pt]{standalone}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[line width=1]
draw[black!10] (0,0) grid (4,4);
node (A) at (1,1) {A};
node (B) at (3,1) {B};
node (C) at (2,3) {C};
%% 1
draw[red] ($ (A) !.5! (B) !.5! (C) $) circle (.05);
%% 2
draw[orange] ($ .25*(A) + .25*(B) + .5*(C) $) circle (.1);
%% 3
coordinate (foo) at ($ (A) !.5! (B) $);
draw[yellow] ($ (foo) !.5! (C) $) circle (.15);
end{tikzpicture}
end{document}
which gives
Thanks for the answer, but I didn't interrogate :)
– Diaa
Nov 15 at 15:29
add a comment |
up vote
5
down vote
up vote
5
down vote
Here, to simplify the code, I'll replace (integrator)
with (A)
, (output)
with (B)
and (sum2)
with (C)
.
There is two things not right with
($ (A) + 0.5*{ (B)-(A) }!0.5!(C) $)
.
First, I don't think you can use
{}
, with the calc package, for the coordinate part. For me, it only works for with scalar. So($ {2+2}*(A) $)
will compute, but not($ 2*{(A)+(B)} $)
(or am I wrong?)The second thing is that this formula doesn't seem to correspond to the point you want.
I kind of get that you want to start from (A), "move" to the middle of [AB] and continue like that, but you mix relative (B-A) and absolute positioning (C).
One right formula would have been($ { (A) + 0.5*{(B)-(A)} }!0.5!(C) $)
.
But because tikz can't do the computation, you'll have to give the expanded formula:($ .25*(A) + .25*(B) + .5*(C)$)
.
One other way to do it is ($ (A) !.5! (B) !.5! (C) $)
. Here, we take the middle of (A) and (B), and then the middle of the result and (C).
I hope this will answer your interrogations!
You can test the three solutions here (the last one with temporary coordinate):
documentclass[tikz,margin=10pt]{standalone}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[line width=1]
draw[black!10] (0,0) grid (4,4);
node (A) at (1,1) {A};
node (B) at (3,1) {B};
node (C) at (2,3) {C};
%% 1
draw[red] ($ (A) !.5! (B) !.5! (C) $) circle (.05);
%% 2
draw[orange] ($ .25*(A) + .25*(B) + .5*(C) $) circle (.1);
%% 3
coordinate (foo) at ($ (A) !.5! (B) $);
draw[yellow] ($ (foo) !.5! (C) $) circle (.15);
end{tikzpicture}
end{document}
which gives
Here, to simplify the code, I'll replace (integrator)
with (A)
, (output)
with (B)
and (sum2)
with (C)
.
There is two things not right with
($ (A) + 0.5*{ (B)-(A) }!0.5!(C) $)
.
First, I don't think you can use
{}
, with the calc package, for the coordinate part. For me, it only works for with scalar. So($ {2+2}*(A) $)
will compute, but not($ 2*{(A)+(B)} $)
(or am I wrong?)The second thing is that this formula doesn't seem to correspond to the point you want.
I kind of get that you want to start from (A), "move" to the middle of [AB] and continue like that, but you mix relative (B-A) and absolute positioning (C).
One right formula would have been($ { (A) + 0.5*{(B)-(A)} }!0.5!(C) $)
.
But because tikz can't do the computation, you'll have to give the expanded formula:($ .25*(A) + .25*(B) + .5*(C)$)
.
One other way to do it is ($ (A) !.5! (B) !.5! (C) $)
. Here, we take the middle of (A) and (B), and then the middle of the result and (C).
I hope this will answer your interrogations!
You can test the three solutions here (the last one with temporary coordinate):
documentclass[tikz,margin=10pt]{standalone}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[line width=1]
draw[black!10] (0,0) grid (4,4);
node (A) at (1,1) {A};
node (B) at (3,1) {B};
node (C) at (2,3) {C};
%% 1
draw[red] ($ (A) !.5! (B) !.5! (C) $) circle (.05);
%% 2
draw[orange] ($ .25*(A) + .25*(B) + .5*(C) $) circle (.1);
%% 3
coordinate (foo) at ($ (A) !.5! (B) $);
draw[yellow] ($ (foo) !.5! (C) $) circle (.15);
end{tikzpicture}
end{document}
which gives
answered Nov 15 at 13:12
Vinzza
1808
1808
Thanks for the answer, but I didn't interrogate :)
– Diaa
Nov 15 at 15:29
add a comment |
Thanks for the answer, but I didn't interrogate :)
– Diaa
Nov 15 at 15:29
Thanks for the answer, but I didn't interrogate :)
– Diaa
Nov 15 at 15:29
Thanks for the answer, but I didn't interrogate :)
– Diaa
Nov 15 at 15:29
add a comment |
up vote
5
down vote
The calc
library allows you to apply Parway Modifiers
repeatedly. Thus, the following syntax
($(integrator)!.5!!(output)!0.5!(sum2)$)
does the following:
- pgf calculates the middle of
(integrator)
and(output)
- then calculates the middle of this last calculated point and the next one
(sum2)
We can continue like this as many times as we want.
Here is for example page 144 of the manual 3.0.1a modified by adding two more points.
documentclass{article}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[every node/.style={draw,circle,inner sep=1pt}]
draw [help lines] (0,0) grid (3,2);
%first node
draw[densely dotted] (0,0) -- (3,2);
node at ($(0,0)!.3!(3,2)$) {1};
%second node
draw[densely dotted] ($(0,0)!.3!(3,2)$) -- (3,0);
node at ($(0,0)!.3!(3,2)!.7!(3,0)$){2};
%third node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)$)--(3,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$) {3};
%fourth node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$)--(0,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)!.5!(0,2)$) {4};
end{tikzpicture}
end{document}
Unfortunately, this does not simplify the writing of the code. The use of an auxiliary point as @Ignasi did is therefore more elegant.
Updated just for fun: A complete solution with the calc
library
And without using yshift=-2cm
and without intermediate point (It's really complicated and unreadable!)
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
But which places the point in the same place with the syntax indicated in the manual 3.0.1a p143, i quote:
The general meaning of
<a>!<factor>!<angle>:<b>
is “First, consider
the line from<a>
to<b>
. Then rotate this line by<angle>
around the
point<a>
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Old answer:
Nevertheless, here is a solution that includes a series of Parway Modifiers
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-([yshift=-2cm]$(integrator)!.5!(output)!0.5!(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Translated with www.DeepL.com/Translator
add a comment |
up vote
5
down vote
The calc
library allows you to apply Parway Modifiers
repeatedly. Thus, the following syntax
($(integrator)!.5!!(output)!0.5!(sum2)$)
does the following:
- pgf calculates the middle of
(integrator)
and(output)
- then calculates the middle of this last calculated point and the next one
(sum2)
We can continue like this as many times as we want.
Here is for example page 144 of the manual 3.0.1a modified by adding two more points.
documentclass{article}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[every node/.style={draw,circle,inner sep=1pt}]
draw [help lines] (0,0) grid (3,2);
%first node
draw[densely dotted] (0,0) -- (3,2);
node at ($(0,0)!.3!(3,2)$) {1};
%second node
draw[densely dotted] ($(0,0)!.3!(3,2)$) -- (3,0);
node at ($(0,0)!.3!(3,2)!.7!(3,0)$){2};
%third node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)$)--(3,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$) {3};
%fourth node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$)--(0,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)!.5!(0,2)$) {4};
end{tikzpicture}
end{document}
Unfortunately, this does not simplify the writing of the code. The use of an auxiliary point as @Ignasi did is therefore more elegant.
Updated just for fun: A complete solution with the calc
library
And without using yshift=-2cm
and without intermediate point (It's really complicated and unreadable!)
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
But which places the point in the same place with the syntax indicated in the manual 3.0.1a p143, i quote:
The general meaning of
<a>!<factor>!<angle>:<b>
is “First, consider
the line from<a>
to<b>
. Then rotate this line by<angle>
around the
point<a>
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Old answer:
Nevertheless, here is a solution that includes a series of Parway Modifiers
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-([yshift=-2cm]$(integrator)!.5!(output)!0.5!(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Translated with www.DeepL.com/Translator
add a comment |
up vote
5
down vote
up vote
5
down vote
The calc
library allows you to apply Parway Modifiers
repeatedly. Thus, the following syntax
($(integrator)!.5!!(output)!0.5!(sum2)$)
does the following:
- pgf calculates the middle of
(integrator)
and(output)
- then calculates the middle of this last calculated point and the next one
(sum2)
We can continue like this as many times as we want.
Here is for example page 144 of the manual 3.0.1a modified by adding two more points.
documentclass{article}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[every node/.style={draw,circle,inner sep=1pt}]
draw [help lines] (0,0) grid (3,2);
%first node
draw[densely dotted] (0,0) -- (3,2);
node at ($(0,0)!.3!(3,2)$) {1};
%second node
draw[densely dotted] ($(0,0)!.3!(3,2)$) -- (3,0);
node at ($(0,0)!.3!(3,2)!.7!(3,0)$){2};
%third node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)$)--(3,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$) {3};
%fourth node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$)--(0,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)!.5!(0,2)$) {4};
end{tikzpicture}
end{document}
Unfortunately, this does not simplify the writing of the code. The use of an auxiliary point as @Ignasi did is therefore more elegant.
Updated just for fun: A complete solution with the calc
library
And without using yshift=-2cm
and without intermediate point (It's really complicated and unreadable!)
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
But which places the point in the same place with the syntax indicated in the manual 3.0.1a p143, i quote:
The general meaning of
<a>!<factor>!<angle>:<b>
is “First, consider
the line from<a>
to<b>
. Then rotate this line by<angle>
around the
point<a>
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Old answer:
Nevertheless, here is a solution that includes a series of Parway Modifiers
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-([yshift=-2cm]$(integrator)!.5!(output)!0.5!(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Translated with www.DeepL.com/Translator
The calc
library allows you to apply Parway Modifiers
repeatedly. Thus, the following syntax
($(integrator)!.5!!(output)!0.5!(sum2)$)
does the following:
- pgf calculates the middle of
(integrator)
and(output)
- then calculates the middle of this last calculated point and the next one
(sum2)
We can continue like this as many times as we want.
Here is for example page 144 of the manual 3.0.1a modified by adding two more points.
documentclass{article}
usepackage{tikz}
usetikzlibrary{calc}
begin{document}
begin{tikzpicture}[every node/.style={draw,circle,inner sep=1pt}]
draw [help lines] (0,0) grid (3,2);
%first node
draw[densely dotted] (0,0) -- (3,2);
node at ($(0,0)!.3!(3,2)$) {1};
%second node
draw[densely dotted] ($(0,0)!.3!(3,2)$) -- (3,0);
node at ($(0,0)!.3!(3,2)!.7!(3,0)$){2};
%third node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)$)--(3,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$) {3};
%fourth node
draw[densely dotted] ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)$)--(0,2);
nodeat ($(0,0)!.3!(3,2)!.7!(3,0)!.6!(3,2)!.5!(0,2)$) {4};
end{tikzpicture}
end{document}
Unfortunately, this does not simplify the writing of the code. The use of an auxiliary point as @Ignasi did is therefore more elegant.
Updated just for fun: A complete solution with the calc
library
And without using yshift=-2cm
and without intermediate point (It's really complicated and unreadable!)
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
But which places the point in the same place with the syntax indicated in the manual 3.0.1a p143, i quote:
The general meaning of
<a>!<factor>!<angle>:<b>
is “First, consider
the line from<a>
to<b>
. Then rotate this line by<angle>
around the
point<a>
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-($(integrator)!.5!(output)!0.5!(sum2)!2cm!90:(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Old answer:
Nevertheless, here is a solution that includes a series of Parway Modifiers
.
documentclass{article}
usepackage{tikz,mathtools,amssymb}
usetikzlibrary{shapes,arrows,positioning,calc}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum height=3em, minimum width=3em},
tmp/.style = {coordinate},
sum/.style= {draw, fill=white, circle, node distance=1cm},
input/.style = {coordinate},
output/.style= {coordinate},
pinstyle/.style = {pin edge={to-,thin,black}
}
}
begin{tikzpicture}[auto, node distance=2cm,>=latex',align=center]
node [sum] (sum2) {};
node [block, right = 1cm of sum2](ractuator){$frac{2}{s+2}$};
node [block, right = 1cm of ractuator,] (vdynamics) {$frac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block, right = 1cm of vdynamics,] (integrator) {$frac{1}{s}$};
node [output, right = 1.5cm of integrator] (output) {};
draw (sum2)|-([yshift=-2cm]$(integrator)!.5!(output)!0.5!(sum2)$)node[block]{C}-|($(integrator)!.5!(output)$);
draw [->] (ractuator) -- (vdynamics);
draw [->] (vdynamics) -- (integrator);
draw [->] (integrator) -- node[name=heading]{$Psi(s)$} (output);
end{tikzpicture}
end{document}
Translated with www.DeepL.com/Translator
edited Nov 15 at 22:19
answered Nov 15 at 15:08
AndréC
5,7921937
5,7921937
add a comment |
add a comment |
up vote
3
down vote
one way how to reduce your code:
- use tikz library
chains
placement nodes in chain and draw lines between them by macrojoin
- node "c" in feedback simple pace below of node
vdynamics
- put coordinates in image directly and not via nodes
- coordinates can contain labels, exploit this for label
$Psi$
define nodes distance only ones and than use it all all nodes positioning
documentclass{article}
usepackage{tikz,nccmath,amssymb}
usetikzlibrary{arrows,
calc, chains,
positioning,
shapes}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum size=3em,
on chain, join=by ->},
sum/.style = {draw, fill=white, circle},
}
makeatletter
tikzset{suspend join/.code={deftikz@after@path{}}} % <--- for dicountinue of jon macro
makeatother
begin{tikzpicture}[
node distance = 0.5cm and 1cm,
start chain = going right,
> = latex']
coordinate (in);
node [sum,right=of in, on chain] (sum2) {};
node [block] (ractuator) {$mfrac{2}{s+2}$};
node [block] (vdynamics) {$mfrac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block] (integrator) {$mfrac{1}{s}$};
coordinate[right=of integrator] (out) {};
node [block, suspend join,
below = of vdynamics] (yaw) {C};
%
draw[->] (in) -- (sum2);
draw[->] (integrator) -- coordinate[label=$Psi(s)$] (psi) (out);
draw[->] (psi) |- (yaw);
draw[->] (yaw) -| (sum2);
end{tikzpicture}
end{document}
off-topic: for fraction is used mfrac
(medium sized fraction) defined in the nccmath
package
Perfect! Thanks for this beautiful answer.
– Diaa
Nov 15 at 16:48
add a comment |
up vote
3
down vote
one way how to reduce your code:
- use tikz library
chains
placement nodes in chain and draw lines between them by macrojoin
- node "c" in feedback simple pace below of node
vdynamics
- put coordinates in image directly and not via nodes
- coordinates can contain labels, exploit this for label
$Psi$
define nodes distance only ones and than use it all all nodes positioning
documentclass{article}
usepackage{tikz,nccmath,amssymb}
usetikzlibrary{arrows,
calc, chains,
positioning,
shapes}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum size=3em,
on chain, join=by ->},
sum/.style = {draw, fill=white, circle},
}
makeatletter
tikzset{suspend join/.code={deftikz@after@path{}}} % <--- for dicountinue of jon macro
makeatother
begin{tikzpicture}[
node distance = 0.5cm and 1cm,
start chain = going right,
> = latex']
coordinate (in);
node [sum,right=of in, on chain] (sum2) {};
node [block] (ractuator) {$mfrac{2}{s+2}$};
node [block] (vdynamics) {$mfrac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block] (integrator) {$mfrac{1}{s}$};
coordinate[right=of integrator] (out) {};
node [block, suspend join,
below = of vdynamics] (yaw) {C};
%
draw[->] (in) -- (sum2);
draw[->] (integrator) -- coordinate[label=$Psi(s)$] (psi) (out);
draw[->] (psi) |- (yaw);
draw[->] (yaw) -| (sum2);
end{tikzpicture}
end{document}
off-topic: for fraction is used mfrac
(medium sized fraction) defined in the nccmath
package
Perfect! Thanks for this beautiful answer.
– Diaa
Nov 15 at 16:48
add a comment |
up vote
3
down vote
up vote
3
down vote
one way how to reduce your code:
- use tikz library
chains
placement nodes in chain and draw lines between them by macrojoin
- node "c" in feedback simple pace below of node
vdynamics
- put coordinates in image directly and not via nodes
- coordinates can contain labels, exploit this for label
$Psi$
define nodes distance only ones and than use it all all nodes positioning
documentclass{article}
usepackage{tikz,nccmath,amssymb}
usetikzlibrary{arrows,
calc, chains,
positioning,
shapes}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum size=3em,
on chain, join=by ->},
sum/.style = {draw, fill=white, circle},
}
makeatletter
tikzset{suspend join/.code={deftikz@after@path{}}} % <--- for dicountinue of jon macro
makeatother
begin{tikzpicture}[
node distance = 0.5cm and 1cm,
start chain = going right,
> = latex']
coordinate (in);
node [sum,right=of in, on chain] (sum2) {};
node [block] (ractuator) {$mfrac{2}{s+2}$};
node [block] (vdynamics) {$mfrac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block] (integrator) {$mfrac{1}{s}$};
coordinate[right=of integrator] (out) {};
node [block, suspend join,
below = of vdynamics] (yaw) {C};
%
draw[->] (in) -- (sum2);
draw[->] (integrator) -- coordinate[label=$Psi(s)$] (psi) (out);
draw[->] (psi) |- (yaw);
draw[->] (yaw) -| (sum2);
end{tikzpicture}
end{document}
off-topic: for fraction is used mfrac
(medium sized fraction) defined in the nccmath
package
one way how to reduce your code:
- use tikz library
chains
placement nodes in chain and draw lines between them by macrojoin
- node "c" in feedback simple pace below of node
vdynamics
- put coordinates in image directly and not via nodes
- coordinates can contain labels, exploit this for label
$Psi$
define nodes distance only ones and than use it all all nodes positioning
documentclass{article}
usepackage{tikz,nccmath,amssymb}
usetikzlibrary{arrows,
calc, chains,
positioning,
shapes}
begin{document}
tikzset{
block/.style = {draw, fill=white, rectangle, minimum size=3em,
on chain, join=by ->},
sum/.style = {draw, fill=white, circle},
}
makeatletter
tikzset{suspend join/.code={deftikz@after@path{}}} % <--- for dicountinue of jon macro
makeatother
begin{tikzpicture}[
node distance = 0.5cm and 1cm,
start chain = going right,
> = latex']
coordinate (in);
node [sum,right=of in, on chain] (sum2) {};
node [block] (ractuator) {$mfrac{2}{s+2}$};
node [block] (vdynamics) {$mfrac{-0.125(s+0.437)}{(s+1.29)(s+0.193)}$};
node [block] (integrator) {$mfrac{1}{s}$};
coordinate[right=of integrator] (out) {};
node [block, suspend join,
below = of vdynamics] (yaw) {C};
%
draw[->] (in) -- (sum2);
draw[->] (integrator) -- coordinate[label=$Psi(s)$] (psi) (out);
draw[->] (psi) |- (yaw);
draw[->] (yaw) -| (sum2);
end{tikzpicture}
end{document}
off-topic: for fraction is used mfrac
(medium sized fraction) defined in the nccmath
package
edited Nov 15 at 20:14
answered Nov 15 at 16:42
Zarko
116k865154
116k865154
Perfect! Thanks for this beautiful answer.
– Diaa
Nov 15 at 16:48
add a comment |
Perfect! Thanks for this beautiful answer.
– Diaa
Nov 15 at 16:48
Perfect! Thanks for this beautiful answer.
– Diaa
Nov 15 at 16:48
Perfect! Thanks for this beautiful answer.
– Diaa
Nov 15 at 16:48
add a comment |
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Hey! Did
($.25*(output)+.25*(integrator)+.5*(sum2)$)
work for you?– Vinzza
Nov 15 at 10:26
@Vinzza It does. But, why does my approach not work?
– Diaa
Nov 15 at 10:33
Comments do not allow enough characters, so I have replied with an answer! I hope it will help you! :)
– Vinzza
Nov 15 at 13:27
Your approach does not work because you try to use
{
and}
where you should use($
and$)
. Try($(0,-2cm)+(integrator)+0.5*($(output)-(integrator)$)!0.5!(sum2)$)
to have something that does not throw an error. However, from your description in words I think you wantnode [block] (yaw) at ($(0,-2cm)+($(output)!0.5!(integrator)$)!0.5!(sum2)$) {C};
, yet this can be done withoutcalc
:node [block] (yaw) at ([yshift=-2cm]barycentric cs:output=1,integrator=1,sum2=2) {C};
.– marmot
Nov 15 at 14:27