1

Two of these equations do not fit the column of the two-column layout.

enter image description here

They are coded with \align such that they all align at the = following the scheme

\begin{align}
...
\dot\omega &= ...
\dot f &=
...
\end{align}

Now I can of course brute force make a line break as in

\begin{align}
...
\dot\omega &= ... \notag \\
&- ...
\dot f &= ... \notag \\
&+ ...
...
\end{align}

which yields

enter image description here

However I do not find that ideal. It would look much better if the second line of each multi-line equation was rather aligned to the right.

How can I do that?

  • 1
    Please, provide code for shoved equation in form of small, complete and compilable document that we not need to retype equations. – Zarko Aug 11 '20 at 15:55
  • 4
    Take a look at multlined from mathtools. That will allow you to divide just part of a long equation. You will also need to use \biggl (or the appropriate size) instead of \left, etc. (This is the best I can do without knowing what document class and other code you are using.) – barbara beeton Aug 11 '20 at 15:56
1

Other than pushing the second line of a two-line equation to the far right, you appear to consider only one other alignment possibility, viz., to place the first item on the second line smack dab below the = symbol in the first line; see equation (A.5') in the following screenshot.

I'd like to point out that there are at least two additional alignment possibilities: (a) pushing the second line to the right by \quad, and (b) pushing the second line to the right by \qquad. These two possibilities are illustrated by equations (A.5'') and (A.5''') in the following screenshot. The final equation, labelled (A.5''''), is generated with the help of a multlined environment, a possibility already mentioned in a comment provided by @barabarabeeton.

enter image description here

Speaking for myself, I can't see anything wrong with the fixed indentation amounts of \quad and \qquad shown in equations (A.5'') and (A.5'''). There's nothing wrong with equation (A.5'''') either. However, achieving that look involves considerably more setting-up overhead than either (A.5'') or (A.5''') do. For sure, the look of (A.5') isn't as good as that of either of the following three equations.

\documentclass{article}
%% trying to replicate the look of the OP's screenshots:
\usepackage[letterpaper,twocolumn,margin=0.667in]{geometry}
\usepackage{mathtools}
\usepackage{newtxtext,newtxmath}
\renewcommand\theequation{A.\arabic{equation}}

\begin{document}
\begin{align}
\addtocounter{equation}{3}
\dot{\Omega} 
&= \sqrt{\frac{p}{Gm}}\frac{\sin(\omega+f)}{ 1+e\cos f}\frac{1}{\sin\imath}\mathcal{W}\\
%% no linebreak
\dot{\omega} 
&= \sqrt{\frac{p}{Gm}}\frac{1}{e}\biggl(-\cos f\mathcal{R} + \frac{2+e\cos f}{1+e\cos f}\sin f\mathcal{S} -e\cot\imath \frac{\sin(\omega+f)}{1+e\cos(f)} \mathcal{W}\biggr) \label{a5}\\
%% 1 linebreak, no indentation
\dot{\omega} 
&= \sqrt{\frac{p}{Gm}}\frac{1}{e}\biggl(-\cos f\mathcal{R} + \frac{2+e\cos f}{1+e\cos f}\sin f\mathcal{S} \tag{\ref{a5}$'$} \\
&-e\cot\imath \frac{\sin(\omega+f)}{1+e\cos(f)} \mathcal{W}\biggr) \notag\\
%% \quad indentation
\dot{\omega} 
&= \sqrt{\frac{p}{Gm}}\frac{1}{e}\biggl(-\cos f\mathcal{R} + \frac{2+e\cos f}{1+e\cos f}\sin f\mathcal{S} \tag{\ref{a5}$''$} \\
&\quad-e\cot\imath \frac{\sin(\omega+f)}{1+e\cos(f)} \mathcal{W}\biggr) \notag \\
%% \qquad indentation
\dot{\omega} 
&= \sqrt{\frac{p}{Gm}}\frac{1}{e}\biggl(-\cos f\mathcal{R} + \frac{2+e\cos f}{1+e\cos f}\sin f\mathcal{S} \tag{\ref{a5}$'''$} \\
&\qquad-e\cot\imath \frac{\sin(\omega+f)}{1+e\cos(f)} \mathcal{W}\biggr) \notag \\
%% multlined solution
\dot{\omega} 
&= \begin{multlined}[t]
\sqrt{\frac{p}{Gm}}\frac{1}{e}\biggl(-\cos f\mathcal{R} + \frac{2+e\cos f}{1+e\cos f}\sin f\mathcal{S} \tag{\ref{a5}$''''$} \\
-e\cot\imath \frac{\sin(\omega+f)}{1+e\cos(f)} \mathcal{W}\biggr)
\end{multlined}
\end{align}

\end{document}
  • Great list of solutions. Thank you so much @Micro! I will try out each of them and see what looks best to me. Cheers! – Britzel Aug 11 '20 at 19:52
  • 1
    It is indeed, sorry for that! – Britzel Aug 11 '20 at 20:02

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.