question concerning coordinate transformations

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question concerning coordinate transformations

by koslowj :: Rate this Message:

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Hi,

I'm still in the process of converting old xy-pic diagrams into TikZ/PGF and ran into the following problem. Some diagrams are based on equilateral triangles being pasted together. Either I can compute the relevant coordinates explicitly:

\begin{tikzpicture}
[baseline=(current bounding box.east),scale=1.75]
\node (00) at (0,0) {$A$};
\node (20) at (2*cos{30},-1) {$B$};
\node (01) at (cos{30},.5) {$C$};
\node (21) at (3*cos{30},-.5){$D$};
\draw[->]
(00) -- node[auto]{$f$}
(01);
\draw[->]
(01) -- node[auto]{$g$}
(21);
\draw[->]
(00) -- node[auto]{$h$}
(20);
\draw[->]
(20) -- node[auto,swap]{$k$}
(21);
\draw[->,out=-60,in=210]
(00) to node[auto,swap,pos=.72]{$l$}
(20);
\pgfsetlinewidth{2 pt}\pgfsetinnerlinewidth{1.2 pt}
\draw[-implies]
(cos{30}+.15,-.7) -- node[auto,swap]{$\tau$}
(cos{30}+.15,-1);
\end{tikzpicture}

which is somewhat messy, or I could change to an appropriate coordinate system. Initially I tried to set the new x- and y-vectors separately, but found out that theses specifications influenced each other in unexpected ways. Actually, after setting the x-vector, the y-vector has to be specified in the already modified coordinate system. Then I tried the cm-method of specifying a matrix, which worked fine for positioning the vertices, but produced problems with the out= and in= angles for curved arrows. In particular, the target of the arrow "l" seems to be the south-east corner of "B", rather than the center of "B". Could this be a bug? Here are the examples with attempted coordinate transformations, first the cm-variant:

\documentclass[a4paper,10pt]{article}
\usepackage{tikz}
\usetikzlibrary{arrows,calc}

\begin{document}
\centering

\[\begin{tikzpicture}
[baseline=(current bounding box.east),
cm={cos{30},-.5,cos{30},.5,(0,0)},scale=1.75]
\node (00) at (0,0) {$A$};
\node (20) at (2,0) {$B$};
\node (01) at (0,1) {$C$};
\node (21) at (2,1) {$D$};
\draw[->]
(00) -- node[auto]{$f$}
(01);
\draw[->]
(01) -- node[auto]{$g$}
(21);
\draw[->]
(00) -- node[auto]{$h$}
(20);
\draw[->]
(20) -- node[auto,swap]{$k$}
(21);
\draw[->,out=-30,in=-105]
(00) to node[auto,swap,pos=.77]{$l$}
(20);
\pgfsetlinewidth{2 pt}\pgfsetinnerlinewidth{1.2 pt}
\draw[-implies]
(1.25,-.1) -- node[auto,swap]{$\tau$}
(1.55,-.4);
\end{tikzpicture}
\qquad
\begin{tikzpicture}
[baseline=(current bounding box.east),
x={(-30:1)},y={(45:1.41)},scale=1.75] % both variants work
% y={(cos{30},.5)},x={(2*cos{30},-1)},scale=1.75]
\node (00) at (0,0) {$A$};
\node (20) at (2,0) {$B$};
\node (01) at (0,1) {$C$};
\node (21) at (2,1) {$D$};
\draw[->]
(00) -- node[auto]{$f$}
(01);
\draw[->]
(01) -- node[auto]{$g$}
(21);
\draw[->]
(00) -- node[auto]{$h$}
(20);
\draw[->]
(20) -- node[auto,swap]{$k$}
(21);
\draw[->,out=-60,in=-150]
(00) to node[auto,swap,pos=.72]{$l$}
(20);
\pgfsetlinewidth{2 pt}\pgfsetinnerlinewidth{1.2 pt}
\draw[-implies]
(1.25,-.1) -- node[auto,swap]{$\tau$}
(1.55,-.4);
\end{tikzpicture}
\]

\end{document}

If somebody knows a better way of realizing these diagrams, please let me know.

-- Jürgen

Re: question concerning coordinate transformations

by Alain Matthes-2 :: Rate this Message:

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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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