Re: question concerning coordinate transformations
Alain Matthes <alain.matthes <at> mac.com>
2009-10-03 18:39:31 GMT
Le 3 oct. 2009 à 13:09, koslowj a écrit :
>
> If somebody knows a better way of realizing these diagrams, please
> let me
> know.
Hi, (Sorry for my bad english)
I don't know if my way is better ...
Firstly, I think it's better to avoid the relevant
coordinates explicitly.
For example :
\begin{tikzpicture}[ baseline=(current bounding box.east),scale=1.75]
\node (00) at (0,0) {$A$};
\node (20) at (2,0) {$B$};
\node (01) at (0,1) {$C$};
\node (21) at (2,1) {$D$};
\draw[->] (00) to node[auto]{$f$} (01);
\draw[->] (01) to node[auto]{$g$} (21);
% (m1)
\path[] (00) to node(m1){} (20);
\draw[->] (00) to node[auto]{$h$} (20);
\draw[->,auto] (20) to node[swap]{$k$} (21);
\draw[->,out=-60,in=-120] (00) to node[label=below:$l$](m2){} (20);
\draw[-implies,double distance=1.2pt]
(m1) to node[auto,swap]{$\tau$} (m2);
\end{tikzpicture}
If I use a transformation, you are right : there is problem
with the nodes and the cm transformation. Perhaps the transformation
is applied to the intersection point of the curved
arrows and the node. So, I used shorten >=10pt,shorten <=10pt
and the anchor "center" but there is another problem (see the
end of the mail)
\begin{tikzpicture}[ baseline=(current bounding box.east),%
scale=1.75,cm={cos{30},-.5,cos{30},.5,(0,0)}]
\draw[help lines] (0,0) grid (3,2);
\node (00) at (0,0) {$A$};
\node (20) at (2,0) {$B$};
\node (01) at (0,1) {$C$};
\node (21) at (2,1) {$D$};
\node (a) at ($(1,-sqrt(2)$) {$a$};
\draw[lightgray,->](00)--(a);
\draw[lightgray,->](20)--(a);
\draw[->] (00) to node[auto]{$f$} (01);
\draw[->] (01) to node[auto]{$g$} (21);
\path[] (00) to node(m1){} (20);
\draw[->] (00) to node[auto]{$h$} (20);
\draw[->,auto] (20) to node[swap]{$k$} (21);
\draw[->,out=-60,in=-120,shorten >=10pt,shorten <=10pt]%
(00.center) to node[label=below:$l$](m2){} (20.center);
\draw[-implies,double distance=1.2pt]
(m1) to node[auto,swap]{$\tau$} (m2);
\end{tikzpicture}
I don't know exactly how work x= ... and y=... with polar coordinates
and
options out =... and in = ....
With cm, it's possible to use
\draw[help lines] (0,0) grid (3,2);
Perhaps you want exactly your first drawing and perhaps
you want the tau curved arrow vertically. It's possible to
calculate the intersection of tau with l (pgf 2 cvs)
Another way is to place the first part of the drawing in a scope
but it's not automatically
\begin{tikzpicture}[ baseline=(current bounding box.east),scale=1.75]
\begin{scope}[cm={cos{30},-.5,cos{30},.5,(0,0)}]
\draw[help lines] (0,0) grid (3,2);
\node (00) at (0,0) {$A$};
\node (20) at (2,0) {$B$};
\node (01) at (0,1) {$C$};
\node (21) at (2,1) {$D$};
\draw[->] (00) to node[auto]{$f$} (01);
\draw[->] (01) to node[auto]{$g$} (21);
\path[] (00) to node(m1){} (20);
\draw[->] (00) to node[auto]{$h$} (20);
\draw[->,auto] (20) to node[swap]{$k$} (21);
\end{scope}
\draw[reset cm,->,out=-80,in=-150,shorten >=8pt,shorten <=8pt]%
(00.center) to node[pos=.60](m2){} (20.center);
\draw[-implies,double distance=1.2pt]
(m1) to node[auto,swap]{$\tau$} (m2) node[]{$a$};
\end{tikzpicture}
but now there is another problem
\draw[reset cm,->,out=-80,in=-150,red]%
(00.center) to node[pos=.60](m2){} (20.center);
instead of
\draw[reset cm,->,out=-80,in=-150,shorten >=8pt,shorten <=8pt]%
(00.center) to node[pos=.60](m2){} (20.center);
gives a different arrow (perhaps it's normal)
Conclusion: I prefer to avoid coordinates and to draw automatically
the diagram but I seems there are some difficulties to use cm.
Best Regards
Alain Matthes
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> > -- Jürgen > Hi, postactions and the markings decoration are your freinds here. See the docs for details: \documentclass{article} \usepackage{tikz} \usetikzlibrary{decorations} \usetikzlibrary{decorations.markings} %\usepackage{textcomp} \begin{document} \pgfdeclaredecoration{single line}{initial}{ \state{initial}[width=\pgfdecoratedpathlength-1sp]{\pgfpathmoveto{\pgfpointorigin}} \state{final}{\pgfpathlineto{\pgfpointorigin}} } \tikzset{ raise line/.style={
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