# Making an anti-clockwise circled arrow around a + sign in a math environment

\makeatletter
\newlength\@SizeOfCirc%
\newcommand{\CricArrowRight}[1]{%
\setlength{\@SizeOfCirc}{\maxof{\widthof{#1}}{\heightof{#1}}}%
\tikz [x=1.0ex,y=1.0ex,line width=.15ex, draw=black]%
\draw [->,anchor=center]%
node (0,0) {#1}%
(0,1.2\@SizeOfCirc) arc (-90:180:1.2\@SizeOfCirc);%
}%
\makeatother


This being in the premeable and the code below used in the document,

$$\CricArrowRight{+}\sum M_{z})_{o}&=0$$


what I have produced is this,

but what I need to produce is the following:

The arc starts at (0, 1.2\@SizeOfCirc), which is above the + sign. As you want it to start below the +, simply change the arc to start at (0, -1.2\@SizeOfCirc). This gives the following result:

To position the symbol such that the + is aligned correctly, you'll have to change the baseline option of \tikz to baseline=-\the\dimexpr\fontdimen22\textfont2\relax, as described in this answer. The result then looks like this:

The complete code:

\documentclass{article}

\usepackage{amsmath}
\usepackage{tikz,calc}

\makeatletter
\newlength\@SizeOfCirc%
\newcommand{\CricArrowRight}[1]{%
\setlength{\@SizeOfCirc}{\maxof{\widthof{#1}}{\heightof{#1}}}%
\tikz [x=1.0ex,y=1.0ex,line width=.15ex, draw=black, baseline=-\the\dimexpr\fontdimen22\textfont2\relax]%
\draw [->,anchor=center]%
node (0,0) {#1}%
(0,-1.2\@SizeOfCirc) arc (-90:180:1.2\@SizeOfCirc);
}%
\makeatother

\begin{document}
$$\CricArrowRight{+} \sum M_{z})_{o}=0$$
\end{document}


Done without tikz, by overlaying the amsmath symbol \circlearrowleft atop the argument. The circled arrow is scaled and rotated.

\documentclass{article}
\usepackage{stackengine,amssymb,graphicx}
\newcommand\CircArrowLeft[1]{\stackengine{-.3ex}{#1}{\CAL}{O}{c}{F}{F}{L}}
\newcommand\CAL{\scalebox{2}{\rotatebox[origin=center]{90}{$\circlearrowleft$}}}
\stackMath
\begin{document}
$$\CircArrowLeft{+}\sum M_{z})_{o}=0$$
\end{document}


Similar to hbaderts' but with a bit of extras (e.g., \DOTSB for correct autopositioning of \dots, usage of pgfmath macros to calculate the width, etc.)

\documentclass{scrartcl}
\usepackage{mathtools,amssymb,lmodern,tikz}
\usetikzlibrary{arrows,bending}

\newcommand*\carr[1]
{\DOTSB\mathbin{\mkern2mu\tikz[x=1ex, y=1ex, line width=.1ex, line cap=round, draw=black, baseline=-\the\fontdimen22\textfont2] {
\pgfmathsetmacro\circlesize{max(width("$#1$"),height("$#1$"))*.8}
\draw[->, >=stealth', anchor=center] node (0,0) {$#1$}
++(-100:\circlesize pt) arc (-100:195:\circlesize pt);}}}

\begin{document}
$\Bigl( A \carr{+} \sum M_z \Bigr)_{0} = 0$
\end{document}

• Nice. I think *.8 might look better, such that it does not rival the size of the symmation symbol. Jul 28, 2015 at 11:57
• The angles of the arc is another thing to consider, I chose what I chose because it looked good with + but may be if one uses another symbol another values are better (the default -90:180 might be the most general). @daleif That might be true (I edited the answer), in any case, it's quite easy to change the *.8 by whatever one prefers. Jul 28, 2015 at 12:04
• @user82745 The way of saying thanks here is upvoting answers; and accepting the one that works better for you. In any case, you should change your username to something more telling than user12345. Jul 28, 2015 at 19:46

Another TikZ solution.

\documentclass[varwidth,border=7pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,bending}
\def\CricArrowRight#1{\tikz[baseline=(A.base)]
\draw[-stealth,line width=.035em]
(0,0) node[circle, inner sep=0cm](A){$#1$}
let \p1=(A.center),\p2=(A.west), \n1={\x1-\x2} in
(-90:\n1) arc(-90:190:\n1);}
\begin{document}
$\CricArrowRight{+}\sum M_{z})_{o}=0$\\[1em]
\CricArrowRight{+} \CricArrowRight{-} \CricArrowRight{\#}
\end{document}


\usepackage{tikz}
\usetikzlibrary{calc,bending}
\def\CricArrowNCW#1{\tikz[baseline=(A.base)]
\draw[-stealth,line width=.035em]
(0,0) node[circle, inner sep=.05cm](A){$#1$}
let \p1=(A.center),\p2=(A.west), \n1={\x1-\x2} in
(90:\n1) arc(90:-190:\n1);}

\def\CricArrowACW#1{\tikz[baseline=(A.base)]
\draw[-stealth,line width=.035em]
(0,0) node[circle, inner sep=.05cm](A){$#1$}
let \p1=(A.center),\p2=(A.west), \n1={\x1-\x2} in
(-90:\n1) arc(-90:180:\n1);}

\begin{document}
\CricArrowNCW{+} %normal clock wise
\CricArrowACW{-} %  anti clock wise
\end{document}


• Hello old friend! Could you please point out in what way this answer differs from the one given by @Kpym below. Just adding a counter clockwise part by inverting the arc is not really enough to justify a new answer. Feb 9 at 0:15