为什么我会收到“额外的对齐制表符已更改为 \cr”的错误？

Question

我正在尝试在两点上对齐方程，我的代码如下：

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
    \begin{cases}
    \omega_W \geq \omega_{W0} \hspace{2mm} &\Leftrightarrow \hspace{2mm} S_x \rightarrow \infty & \longrightarrow \text{Wheel is spinning} \\
    \omega_W > \omega_{W0} \hspace{2mm} &\Leftrightarrow \hspace{2mm} S_x > 0 & \longrightarrow \text{Tire driving force} (F_x >0) \\
    \omega_W = \omega_{W0} \hspace{2mm} &\Leftrightarrow \hspace{2mm} S_x =0 & \longrightarrow \text{Free-rolling tire} (F_x=0) \\
    \omega_W \leq \omega_{W0} \hspace{2mm} &\Leftrightarrow \hspace{2mm} S_x < 0 & \longrightarrow \text{Tire braking force} (F_x <0) \\
    \omega_W =0 \hspace{2mm} &\Leftrightarrow \hspace{2mm} S_x= -1 & \longrightarrow \text{Wheel lock-up} \\
    \end{cases}
\end{equation}
\end{document}

问题是我想要在“Leftrightarrow”和“longrightarrow”中进行对齐。但是，我得到的输出如下（当然还有“额外的对齐标签已更改为 \cr”的错误）：

知道为什么它不起作用吗？或者我该如何修复它？我写了所有内容，\\并在&我想要对齐的点中添加了。

Answer 1

cases&每行仅接受一个。您可以使用 轻松模拟它array。r>{{}}l诀窍在于获得关系符号周围的正确间距。我删除了长箭头，因为我觉得它们是多余的。

\documentclass{article}
\usepackage{amsmath,array}

\begin{document}

\begin{equation}
\left\{
  \renewcommand{\arraystretch}{1.2}% like cases
  \setlength{\arraycolsep}{0pt}%
  \begin{array}{ r>{{}}l @{\quad} c @{\quad} r>{{}}l @{\quad} l }
  \omega_W &\geq \omega_{W0} &\Leftrightarrow & S_x &\to \infty & \text{Wheel is spinning} \\
  \omega_W &>    \omega_{W0} &\Leftrightarrow & S_x &>   0      & \text{Tire driving force ($F_x>0$)} \\
  \omega_W &=    \omega_{W0} &\Leftrightarrow & S_x &=   0      & \text{Free-rolling tire ($F_x=0$)} \\
  \omega_W &\leq \omega_{W0} &\Leftrightarrow & S_x &<   0      & \text{Tire braking force ($F_x<0$)} \\
  \omega_W &=    0           &\Leftrightarrow & S_x &=  -1      & \text{Wheel lock-up}
  \end{array}
\right.
\end{equation}

\end{document}

Answer 2

empheq带有环境的简单代码alignedat：

\documentclass{article}

\usepackage{empheq}

\begin{document}

\begin{empheq}[left =\empheqlbrace\,]{equation}
    \begin{alignedat}{3}
        & \omega_W \geq \omega_{W0} & \: &\Leftrightarrow\: S_x \rightarrow \infty &\: & \longrightarrow \text{Wheel is spinning} \\
        & \omega_W > \omega_{W0} & &\Leftrightarrow \: S_x > 0 & & \longrightarrow \text{Tire driving force} (F_x >0) \\
        & \omega_W = \omega_{W0} & &\Leftrightarrow \: S_x =0 & & \longrightarrow \text{Free-rolling tire} (F_x=0) \\
        & \omega_W \leq \omega_{W0} & &\Leftrightarrow \: S_x < 0 & & \longrightarrow \text{Tire braking force} (F_x <0) \\
        & \omega_W =0 & &\Leftrightarrow \: S_x= -1 & & \longrightarrow \text{Wheel lock-up} \\
    \end{alignedat}
\end{empheq}

\end{document} 

Answer 3

&在环境中每行只能使用一个cases：

\documentclass{article}

\usepackage{amsmath,eqparbox}

\begin{document}

\begin{equation}
  \begin{cases}
    \omega_W \geq \omega_{W0} &\Leftrightarrow \quad S_x \rightarrow \infty \longrightarrow \text{Wheel is spinning} \\
    \omega_W > \omega_{W0}    &\Leftrightarrow \quad S_x >  0               \longrightarrow \text{Tire driving force ($F_x > 0$)} \\
    \omega_W = \omega_{W0}    &\Leftrightarrow \quad S_x =  0               \longrightarrow \text{Free-rolling tire ($F_x = 0$)} \\
    \omega_W \leq \omega_{W0} &\Leftrightarrow \quad S_x <  0               \longrightarrow \text{Tire braking force ($F_x < 0$)} \\
    \omega_W = 0              &\Leftrightarrow \quad S_x = -1               \longrightarrow \text{Wheel lock-up}
  \end{cases}
\end{equation}

\newcommand{\Scond}[1]{\eqmakebox[Scond][l]{$#1$}}

\begin{equation}
  \begin{cases}
    \omega_W \geq \omega_{W0} &\Leftrightarrow \quad \Scond{S_x \rightarrow \infty} \longrightarrow \text{Wheel is spinning} \\
    \omega_W > \omega_{W0}    &\Leftrightarrow \quad \Scond{S_x >  0}               \longrightarrow \text{Tire driving force ($F_x > 0$)} \\
    \omega_W = \omega_{W0}    &\Leftrightarrow \quad \Scond{S_x =  0}               \longrightarrow \text{Free-rolling tire ($F_x = 0$)} \\
    \omega_W \leq \omega_{W0} &\Leftrightarrow \quad \Scond{S_x <  0}               \longrightarrow \text{Tire braking force ($F_x < 0$)} \\
    \omega_W = 0              &\Leftrightarrow \quad \Scond{S_x = -1}               \longrightarrow \text{Wheel lock-up}
  \end{cases}
\end{equation}

\end{document}

上面提供了两个选项，具体取决于您的对齐要求。

Answer 4

案例只能处理一个&。

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation*}
\left\{
    \begin{array}{lll@{}}
    \omega_W \geq \omega_{W0} &\Leftrightarrow S_x \rightarrow \infty &\longrightarrow \text{Wheel is spinning} \\
    \omega_W > \omega_{W0} &\Leftrightarrow S_x > 0  &\longrightarrow \text{Tire driving force} (F_x >0) \\
    \omega_W = \omega_{W0} &\Leftrightarrow S_x =0 &\longrightarrow \text{Free-rolling tire} (F_x=0) \\
    \omega_W \leq \omega_{W0} &\Leftrightarrow S_x < 0 &\longrightarrow \text{Tire braking force} (F_x <0) \\
    \omega_W = 0 &\Leftrightarrow S_x= -1 &\longrightarrow \text{Wheel lock-up} \\
    \end{array}
\right.
\end{equation*}
\end{document}