\documentclass[a4paper, 12pt, fleqn]{book}
\usepackage[utf8,ansinew]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{siunitx}
\usepackage[fleqn,centertags]{mathtools}
\begin{document}
This is ordinary text and below 3 blocks of equations. The first one without using subequations.
\begin{alignat}{2}
&a &&=b ~,\label{1a}\\[1ex]
&alpha &&=beta~,\label{1b} \\[1ex]
&c &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923}{kN/m^2} \label{1c} \\[1ex]
&cc &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923.08}{kN/m^2} \label{1d}
\end{alignat}%
The second block uses the subequation environment.
{\allowdisplaybreaks [4]
\begin{subequations}\label{1}
%\noeqref{1a,1b,1c,1d}
\begin{alignat}{2}
&a &&=b ~,\label{1a}\\[1ex]
&alpha &&=beta~,\label{1b} \\[1ex]
&c &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923}{kN/m^2} \label{1c} \\[1ex]
&cc &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923.08}{kN/m^2} \label{1d}
\end{alignat}
\end{subequations}}%
The alignment on the left side is fine in both cases. However, in the second case there is too much horizontal space left of the "=" sign. When the equation (5d) is removed, or if I reduce the length of (5d) by deleting one or more digits again, the alignment is fine, as can be seen below.
{\allowdisplaybreaks [4]
\begin{subequations}\label{1}
%\noeqref{1a,1b,1c,1d}
\begin{alignat}{2}
&a &&=b ~,\label{1a}\\[1ex]
&alpha &&=beta~,\label{1b} \\[1ex]
&c &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923}{kN/m^2} \label{1c} \\[1ex]
&cc &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923}{kN/m^2} \label{1d}
\end{alignat}
\end{subequations}}
\end{document}
이것이 내가 얻는 것입니다:
답변1
당신은 운이 좋지 않습니다. 보시다시피 두 번째 디스플레이의 태그가 잘못 정렬되어 있습니다. amsmath이 경우에는 더 나은 결과를 얻을 수 있으며 적어도 정렬이 너무 넓다는 경고를 표시해야 합니다.
를 사용하고 있으므로 fleqn긴 방정식이 실제로는 조금 더 짧은 것처럼 가정하여 문제를 해결할 수 있습니다.
\documentclass[a4paper, 12pt, fleqn]{book}
\usepackage[T1]{fontenc}
\usepackage{siunitx}
\usepackage[fleqn,centertags]{mathtools}
\usepackage{amsmath}
\begin{document}
This is ordinary text and below 3 blocks of equations. The first one without using subequations.
\begin{alignat}{2}
&a &&=b ~,\label{1a}\\[1ex]
&alpha &&=beta~,\label{1b} \\[1ex]
&c &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923}{kN/m^2} \label{1c} \\[1ex]
&cc &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923.08}{kN/m^2} \label{1d}
\end{alignat}
The second block uses the subequation environment.
\begin{subequations}\label{1}
\begin{alignat}{2}
&a &&=b ~,\label{x1a}\\[1ex]
&alpha &&=beta~,\label{x1b} \\[1ex]
&c &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923}{kN/m^2} \label{x1c} \\[1ex]
&cc &&=\frac{\num{72000}\cdot\num{50000}\,(1 - \num{0.30})}{\num{50000}\,(1 - \num{0.30}) -
2\cdot\num{72000}\cdot\num{0.25}^2} = \SI{96923.08}{kN/m^2}
\hspace{-2em}
\label{x1d}
\end{alignat}
\end{subequations}
\end{document}
\allowdisplaybreaks문서의 최종 버전에는 사용하지 마세요 . 문서를 작성할 때 서문에서 사용할 수 있지만 마지막에는 표시되는 페이지 나누기를 신중하게 선택해야 합니다.로드하는 것은 의미가 없습니다
ansinew. 통화 를 삭제하세요inputenc.왜 왼쪽 정렬이 왼쪽입니까?
