Problem bending arrows in xypic
Solution 1
Here are two xypic solution.
\documentclass{article}
\usepackage[all]{xy}
\begin{document}
\[\xymatrix{
A \ar `r [r]`[ddd][ddd]&\\ LongEntry \\ LongEntry \\ LongEntry
}
\qquad
\xymatrix{
A \ar `r []+<5em,0cm>`[ddd][ddd]\\ LongEntry \\ LongEntry \\ LongEntry
}\]
\end{document}
The first one is the easier to understand, but you need to add one extra column.
Here it is how it works : \ar `r [r]`[ddd][ddd]
means that the arrow begin in the r
direction, then goes to [r]
(relative coordinate) — hence the need for one extra column — and makes a quarter of turn, then goes to `[ddd] position (relative coordinates) and do another quarter of turn, and finally ends on [ddd] ("absolute" position, i.e. relative to the beginning of the arrow).
The second example avoid to use one extra column and work sensibly the same. The only difference is the use of []_<5em,0em>
instead of [r]
. But that means : the actual coordinate ([]
) plus <5em,0em>
.
Finally — for the bonus point and to be complete — the general syntax is
\ar `d1[p1]`d2[p2] ... `dn[pn] [f]
which means : start from here in direction d1 and go to [p1], then take the direction d2 and go to [p2] (relatively from p1), etc, and finally end in [f], where [f] is the position off the end node, relatively from your start node.
Usually, d2
to dn
are not necessary (and produces ugly results).
You can control the direction off the turn with ^d[p]`` or
_d[p].
You can also control the radius as in ```d/16pt[p]
.
N.B. You can also change the tip of arrows with (for example) \usepackage[all,cmtip]{xy}
For more explanations you can consult the xypic user's guide. If you want more example of diagram in xypic with bended arrows, you can look for example at here and here (if I may cite myself).
The results :
Solution 2
With a special to path
You can do this with a special to path
. I am assuming that this will only be used on nodes (which is true for tikzcd
).
I have included two implementations: One with the calc
library (called rl
), one without it with plain PGF macros (slightly faster).
The approach is the same. I extract the x and y value of the most right point of the path (the east
anchor), then I check which of both x values is greater (more to the right) and add the passed length. This will be the x value of the vertical line part.
The nodes are placed (only) on this vertical part (whereas with the rl
path operator below you could nodes along the whole path).
Code
\documentclass[tikz,convert=false]{standalone}
\usepackage{tikzcd}
\usetikzlibrary{calc}
\makeatletter
\tikzset{
RL/.style={% without calc
rounded corners={#1.1pt},
to path={% we asume that you use this path only on nodes (otherwise it will get tricky)
\pgfextra
\pgf@process{\pgfpointanchor{\tikztostart}{east}}%
\pgf@xa\pgf@x\pgf@ya\pgf@y
\pgf@process{\pgfpointanchor{\tikztotarget}{east}}%
\pgf@xb\pgf@x\pgf@yb\pgf@y
\ifdim\pgf@xb>\pgf@xa
\pgf@xa\pgf@xb
\fi
\pgfmathsetlength\pgf@xc{#1}%
\advance\pgf@xa\pgf@xc
\endpgfextra
 (+\pgf@xa,+\pgf@ya)  (+\pgf@xa,+\pgf@yb) \tikztonodes  (\tikztotarget)
}
},
RL/.default=.5cm}
\makeatother
\tikzset{
rl/.style={% with calc
rounded corners={#1.1pt},
to path={% we asume that you use this path only on nodes (otherwise it will get tricky)
let \p{@aux1}=(\tikztostart.east),
\p{@aux2}=(\tikztotarget.east),
\n{@aux@x}={#1+max(\x{@aux1},\x{@aux2})}
in
 (+\n{@aux@x},+\y{@aux1})  (+\n{@aux@x},+\y{@aux2}) \tikztonodes  (\tikztotarget)
}
},
rl/.default=.5cm,
}
\begin{document}
\begin{tikzcd}
A \arrow[RL]{ddd}\\
LongEntry \\
LongEntry \\
LongEntry
\end{tikzcd}
\begin{tikzcd}
A \arrow[rl]{ddd}\\
LongEntry \\
LongEntry \\
LongEntry
\end{tikzcd}
\end{document}
Path Operator rl
With my paths.ortho
(→ another answer) library you can do this very easily.
The path operator rl
means that both start and target of a line is connected by a combination of a horizontal, a vertical and a horizontal line. In this case (rl
instead of lr
) means that the vertical line is to the right of both nodes (or coordinates). I.e first to the r
ight, then down or up, then to the l
eft. The given distance is measured from the node/coordinate that is furthest to the right.
As the paths.ortho
already installs a rl
style that uses the rl
path operator, I’ll ad another RL
style that sets an appropriate amount for the rounded corners
radius and the rl distance
. The .1pt
was necessary so that the arrow tip is placed correctly.
This does not take other nodes in account. So if you have another very long node between start and target you will need to make manual adjustments either way.
Code
\documentclass[tikz,convert=false]{standalone}
\usepackage{tikzcd}
\usetikzlibrary{paths.ortho}
\tikzset{
RL/.style={
rounded corners={#1.1pt},
rl distance={#1},
rl
},
RL/.default=.5cm
}
\usepackage[all]{xy}
\begin{document}
\begin{tikzcd}
A \arrow[RL]{ddd}\\
LongEntry \\
LongEntry \\
LongEntry
\end{tikzcd}
\end{document}
Outputs
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user21952isagreatname
Updated on June 06, 2020Comments

user21952isagreatname almost 3 years
I'm having an annoying problem bending an arrow with
xypic
. I read the part 2.8 of the manual (http://wwwbcf.usc.edu/~cautis/tools/xypic.pdf) about bending arrows, but could not understand it. Here's the best I could do (UPDATE: the first version is what I had before; the output mysteriously looks different on different runs of LaTeX. The second version is based on a suggestion from a comment below. It's still ugly, but perhaps that's the best I can do?):\documentclass{article} \usepackage[all]{xy} \begin{document} $$\xymatrix{ A \ar `/40pt[d] `[ddd] [ddd] \\ LongEntry \\ LongEntry \\ LongEntry } \quad \xymatrix{ A \ar `/40pt[ddd] `/10pt[ddd] [ddd] \\ LongEntry \\ LongEntry \\ LongEntry }$$ \end{document}
The output looks as follows:
How do I make the ending of the arrow sane? For bonus points, how does one bend arrows in general?
(I'm more interested in an
xypic
answer, since this is part of a large diagram, but atikzcd
answer would be nice too if it's particularly clear. The misadventure with this diagram is encouraging me to switch).
Qrrbrbirlbel almost 10 yearsCould you add a few properties of this arrow so that one can recreate it most accurate in TikZ? A minimal working example (MWE) would be highly appreciated, too.

Qrrbrbirlbel almost 10 years
\ar `/40pt[ddd] `/15pt[ddd] [ddd]
seems to do a pretty good job. 
user21952isagreatname almost 10 years@Qrrbrbirlbel: Thank you! With your suggestion, I get almost what I want, but the arrow points in the wrong direction. I will post a MWE in a second.


user21952isagreatname almost 10 yearsThank you for your answer! Do you know if this would be hard to implement in pure tikzcd? As nice as your library is, I'm not sure I want to drag around an extra file with my paper...

Qrrbrbirlbel almost 10 years@user21952isagreatname I have added both a PGF and a
calc
solution. Both of them can easily be adjusted to work for all other directions, too. If you also want to avoid overlapping with nodes between start and target, more work is involved. Will this be needed? 
user21952isagreatname almost 10 yearsThank you! I appreciate all the different recipes; they will be very useful if I need to make a sufficiently weird arrow in tikz.