I have two (maybe simple) questions concerning the following LaTeX document
\documentclass{scrartcl}
\usepackage[ansinew]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{tikz}
\usepackage{pgfplots}
\usetikzlibrary{intersections, calc}
\makeatletter
\DeclareMathSizes{\@xipt}{\@xipt}{6}{5}
\makeatother
\begin{document}
This is a nice picture with $\underline{U}_\mathrm{N}$, $U_\mathrm{L}$, $\varphi_\mathrm{U}$ and $\varphi_\mathrm{L}.$
\begin{tikzpicture}
\def\angle{acos(0.95)}
\draw[->, name path=ul] (0,0) -- (5,0) coordinate (A) node[near end, anchor=north] {$U_\mathrm{L}$};
\draw[->, name path=ux] (A) -- +({90-\angle}:3) coordinate (B) node[midway, anchor=west] {$\mathrm{j} \, X_\mathrm{N} \, \underline{I}$};
\draw[->, name path=un] (0,0) -- (B) node[near end, anchor=south east] {$\underline{U}_\mathrm{N}$};
\draw[->, name path=i] (0,0) -- +({-\angle}:1.5) coordinate (C) node[near end, anchor=north east] {$\underline{I}$};
\path (0,0) let \p1 = ($(B)$) in +(15:{veclen(\x1,\y1)}) coordinate (D);
\draw[gray, dashed, name path=circ] let \p1 = ($(B)$) in (D) arc (15:50:{veclen(\x1,\y1)});
\draw[->, gray, name path=ii] (C) -- +({-\angle}:0.5) coordinate (CC) node[anchor=north east] {$\underline{I}'$};
\draw[->, gray, name path=uull] (0,0) -- (4,0) coordinate (AA) node[near end, anchor=south] {$U'_\mathrm{L}$};
\path[name path=uuxx] (AA) -- +({90-\angle}:8);
\draw[->, gray, name intersections={of=uuxx and circ}] (AA) -- (intersection-1) coordinate (BB);
\draw[->, gray, name path=uunn] (0,0) -- (BB) node[near end, anchor=south east] {$\underline{U}'_\mathrm{N}$};
\draw[->] (1,0) let \p1 = ($(B)$) in arc (0:{atan(\y1/\x1)}:1) node[midway, anchor=west] {\scriptsize{$\varphi_\mathrm{U}$}};
\draw[->] (1,0) let \p1 = ($(C)$) in arc (0:{atan(\y1/\x1)}:1) node[near end, anchor=west] {\scriptsize{$-\varphi_\mathrm{L}$}};
\end{tikzpicture}
\end{document}}
which yields something like the following:
- Why is the prime symbol of
$U'_\mathrm{L}$
in the tikzpicture lower than it is for the$\underline{U}'_\mathrm{N}$
? (The latter prime positioning seems correct to me...) Obviously, this must have something to do with the\underline
command - but what can I do to solve the problem of this prime symbol misplacement? Raising the prime symbol by writing$U^{'}_\mathrm{L}$
lifts it too high - and is most probably not the optimal way of solving this problem anyway. - I used
\DeclareMathSizes{\@xipt}{\@xipt}{6}{5}
to reduce the size of subscripts and subsubscripts in math mode. However, this seems not to work when I use\scriptsize{}
in the math environment as one can see from the angle label: The U and the L in the angle labels are appreciably bigger than they are in the text line above the diagram. I guess that this is because of my "static" setting in\DeclareMathSizes{\@xipt}{\@xipt}{6}{5}
where I set the sizes to the fixed values of 6 and 5? If so, how can I reduce the size of subscripts and subsubscripts in math mode in a more flexible way? Writing\DeclareMathSizes{\@xipt}{\@xipt}{0.8*\@xipt}{0.7*\@xipt}
didn't work...
$\smash{\underline{U}}'_\mathrm{L}$}
. This makes TeX ignore the extra height generated by the\underline
, but the prime marks are then both lower.\usepackage{lmodern}
, for instance); just do the math:\DeclareMathSizes{\@xipt}{\@xipt}{8.8}{7.7}
(but these values are too big).