6

I'm displaying some figures with different bounding box. The figures seems to be bottom aligned, but I would like to have them vertically aligned. I though subcaption automatically take care of that, but it seems that does not.

Here's a piece of my code:

\documentclass[final,5p,times]{elsarticle}
\usepackage{subcaption}

\begin{document}

\begin{figure*}[htpb]
\subcaptionbox{Curl-free potential singularities}{\includegraphics[width=.3\textwidth]{figs/Case1/Singularities/rotational_pot_sing}} \hfill
\subcaptionbox{Divergence-free potential singularities}{\includegraphics[width=.3\textwidth]{figs/Case1/Singularities/divergence_pot_sing}} \hfill
\subcaptionbox{Vector field potentials singularities}{\includegraphics[width=.3\textwidth]{figs/Case1/Singularities/vector_field_sing}}
\label{fig:case1_sing}
\end{figure*}

\end{document}

And that is the result:

enter image description here

How could I get the desired result?

Thank you.

9

You can capture the tallest image in a box, which you can then use to measure the height by which you want/have to raise the other not-so-tall images:

enter image description here

\documentclass[final,5p,times]{elsarticle}
\usepackage{subcaption}

\begin{document}

\begin{figure*}[htpb]
  \setbox9=\hbox{\includegraphics[width=.3\linewidth]{example-image-1x1}}% Capture tallest image in box 9
  \subcaptionbox{Curl-free potential singularities}
    {\raisebox{\dimexpr\ht9-\height}{\includegraphics[width=.3\linewidth]{example-image-a}}} \hfill
  \subcaptionbox{Divergence-free potential singularities}
    {\raisebox{\dimexpr\ht9-\height}{\includegraphics[width=.3\linewidth]{example-image-b}}} \hfill
  \subcaptionbox{Vector field potentials singularities}{\includegraphics[width=.3\linewidth]
    {example-image-1x1}}
\end{figure*}

\end{document}

\ht9 represents the height of box 9, while \height is the height of <stuff> in \raisebox{<height>}{<stuff>}.

As reference see References for \dimexpr and \numexpr.

  • 1
    Not as clean as I would have like but it gets the job done ... – Amxx Apr 23 '15 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.