如何将所有方程式对齐到左侧而不是右侧？

Question

我试图将所有方程式左对齐。但它却自动右对齐。我试过了，flaligned但没用。请帮忙。

\begin{equation} 
\begin{aligned}
\dot{u} = -g\sin(\theta) + rv - qw \\ 
\dot{v} = g\sin(\phi)\cos(\theta) -ru + pw \\  
\dot{w} = \frac{1}{m}(-F_{z}) + g\cos(\phi)\cos(\theta) + qu - pv \\  
\dot{p} = \frac{1}{I_{xx}}(L + (I_{yy}-I_{zz})qr) \\  
\dot{q} = \frac{1}{I_{yy}}(M + (I_{zz}-I_{xx})pr) \\  
\dot{r} = \frac{1}{I_{zz}}(N + (I_{xx}-I_{yy})pq) \\  
\dot{\phi} = p + (q \sin\phi + r \cos\phi) \tan\theta \\  \dot{\theta} = q \cos\phi - r \sin\phi \\ 
\dot{\psi} = (q \sin\phi + r \cos\phi) \sec\theta \\  
\dot{x}^{E} = cos{\theta}cos{\psi}u^{b} + (-cos{\phi}sin{\psi}+sin{\phi}sin{\theta}cos{\psi})v^{b} + (sin{\phi}sin{\psi}+cos{\phi}sin{\theta}cos{\psi})w^{b} \\  
\dot{y}^{E} = cos{\theta}sin{\psi}u^{b} + (cos{\phi}cos{\psi}+sin{\phi}sin{\theta}sin{\psi})v^{b} + (-sin{\phi}cos{\psi}+cos{\phi}sin{\theta}sin{\psi})w^{b} \\  \dot{h}^{E} = -1*(-sin{\theta}u^{b} + sin{\phi}cos{\theta}v^{b} + cos{\phi}cos{\theta}w^{b}) ,
\end{aligned}
\end{equation}

Answer 1

这个怎么样？

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation} 
\begin{aligned}
\dot{u} &= -g\sin(\theta) + rv - qw \\ 
\dot{v} &= g\sin(\phi)\cos(\theta) -ru + pw \\  
\dot{w} &= \frac{1}{m}(-F_{z}) + g\cos(\phi)\cos(\theta) + qu - pv \\  
\dot{p} &= \frac{1}{I_{xx}}(L + (I_{yy}-I_{zz})qr) \\  
\dot{q} &= \frac{1}{I_{yy}}(M + (I_{zz}-I_{xx})pr) \\  
\dot{r} &= \frac{1}{I_{zz}}(N + (I_{xx}-I_{yy})pq) \\  
\dot{\phi} &= p + (q \sin\phi + r \cos\phi) \tan\theta \\  \dot{\theta} &= q \cos\phi - r \sin\phi \\ 
\dot{\psi} &= (q \sin\phi + r \cos\phi) \sec\theta \\  
\dot{x}^{E} &= \cos{\theta}\cos{\psi}u^{b} + (-\cos{\phi}\sin{\psi}+\sin{\phi}\sin{\theta}\cos{\psi})v^{b} + (\sin{\phi}\sin{\psi}+\cos{\phi}\sin{\theta}\cos{\psi})w^{b} \\  
\dot{y}^{E} &= \cos{\theta}\sin{\psi}u^{b} + (\cos{\phi}\cos{\psi}+\sin{\phi}\sin{\theta}\sin{\psi})v^{b} + (-\sin{\phi}\cos{\psi}+\cos{\phi}\sin{\theta}\sin{\psi})w^{b} \\  \dot{h}^{E} &= -1*(-\sin{\theta}u^{b} + \sin{\phi}\cos{\theta}v^{b} + \cos{\phi}\cos{\theta}w^{b}) ,
\end{aligned}
\end{equation}
\end{document}

Answer 2

• 你的意思不太清楚左对齐。
• 您错误地使用了aligned没有锚点的环境来对齐方程式，例如在符号处=（如其他答案中所建议的那样）。
• 如果你喜欢让所有方程式左对齐（在左侧\@mathmargin，默认为 30pt），那么你可以使用fleqn并修改“聚集环境”，如建议的那样@egreg 回答：

\documentclass[fleqn]{article} % <---
\usepackage{geometry}
\usepackage{amsmath}
\makeatletter
% amsmath.sty, line 1253: taken from https://tex.stackexchange.com/questions/272930/
\renewenvironment{gathered}[1][c]{%
    \RIfM@\else
        \nonmatherr@{\begin{gathered}}%
    \fi
    \null\,%
    \if #1t\vtop \else \if#1b\vbox \else \vcenter \fi\fi \bgroup
        \Let@ \chardef\dspbrk@context\@ne \restore@math@cr
        \spread@equation
        \ialign\bgroup
            %%% In the original there is just \hfil
            \if@fleqn\else\hfil\fi\strut@$\m@th\displaystyle##$\hfil
            \crcr
}{%
  \endaligned
}
\makeatother

\begin{document}
    \begin{equation}
\begin{gathered}
\dot{u} = -g\sin\theta + rv - qw \\
\dot{v} =  g\sin\phi\cos\theta -ru + pw \\
\dot{w} = \frac{1}{m}(-F_{z}) + g\cos\phi\cos\theta + qu - pv \\
\dot{p} = \frac{1}{I_{xx}}(L + (I_{yy}-I_{zz})qr) \\
\dot{q} = \frac{1}{I_{yy}}(M + (I_{zz}-I_{xx})pr) \\
\dot{r} = \frac{1}{I_{zz}}(N + (I_{xx}-I_{yy})pq) \\
\dot\phi = p + (q \sin\phi + r \cos\phi) \tan\theta \\
\dot\theta = q \cos\phi - r \sin\phi \\
\dot\psi = (q \sin\phi + r \cos\phi) \sec\theta \\
\dot{x}^{E} = \cos\theta\cos\psi u^{b} + 
                (-\cos\phi\sin\psi+\sin\phi\sin\theta\cos\psi)v^{b} + 
                (\sin\phi\sin\psi+\cos\phi\sin\theta\cos\psi)w^{b} \\
\dot{y}^{E} = \cos\theta\sin\psi u^{b} + 
                (\cos\phi\cos\psi+\sin\phi\sin\theta\sin\psi)v^{b} + 
                (-\sin\phi\cos\psi+\cos\phi\sin\theta\sin\psi)w^{b} \\  
\dot{h}^{E} = -1{\cdot}(-\sin\theta u^{b} + 
                 \sin\phi\cos\theta v^{b} + \cos\phi\cos\theta w^{b}) ,
\end{gathered}
    \end{equation}
\end{document}