# Dot over A not in the right place

When I write $\dot{A}(t)$, the dot is not aligned with the top of the letter A. is there a way to correct this?

• – Werner Jun 16 '16 at 22:05
• with default fonts you get this which doesn't seem too bad the dot is offset to the right to account for the slope, perhaps you want it offset further, but such visual effects are hard to automatically control since tex sees every character as a rectangle – David Carlisle Jun 16 '16 at 22:13
• Try $\skew{6}\dot{A}(t)$ or $\skew{7}\dot{A}(t)$ (according to your taste). – GuM Jun 16 '16 at 22:24
• @DavidCarlisle I don't think this is correct. TeX know at least something about the character shape. Consider for example this comparison: $\dot{\phi}$ vs $\dot{\hbox{$\phi$}}$ – Henri Menke Jun 17 '16 at 8:50
• @HenriMenke it knows nothing about the shape. As far as tex is concerned every character is just four lengths (height, depth, width and italic correction) you are seeing the effect of italic correction (as (ab)used in math mode) but as seen with the . over the A it's quite a blunt instrument and can not really fine tune optical effects around the small dot and the exact position of the apex of the A. – David Carlisle Jun 17 '16 at 9:02

Not sure if this breaks other things, but here, I redefine \dot to check if the argument is a lone A and skew the dot only if that is the case (the skew level can be adjusted, of course, for the particular font).

\documentclass{article}
\let\svdot\dot
\def\dot#1{\dothelp#1\relax}
\def\dothelp#1#2\relax{%
\if\relax#2\relax%
\ifx A#1\skew{6}\svdot#1\else\svdot#1\fi
\else
\svdot{#1#2}%
\fi%
}
\begin{document}
$\dot{A}(t)$

$\dot A(t)$

$\dot{AG}(t)$

$\dot{G}(t)$
\end{document}


The unaltered \dot, for comparison