禁忌环境中的垂直文本

Question

我尝试垂直对齐第一列的文本。因此我使用了rotatebox此解决方案，但我并不完全满意，因为文本在顶部对齐。

问题

有什么可行的解决方案可以使第一列的文本居中对齐？

\documentclass[a4paper,pagesize ,landscape, fontsize=6pt]{scrartcl}
\usepackage[left=1.5cm,right=1.5cm, top=1.5cm, bottom=1.5cm]{geometry}
\usepackage{multicol}
\usepackage{amsmath, amsfonts, amssymb}
\usepackage{tikz}
\usepackage{array,multirow}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{tabu}

\begin{document}

\begin{multicols*}{3}

    \tabulinesep=1.0mm
    \begin{tabu} to \columnwidth{> {$\textstyle} X[1,l] < {$} > {$\textstyle} X[30,l] < {$}}
        \multirow{2}{*}{\rotatebox[origin=c]{90}{polar}}    & \mathbf{r}(\rho, \varphi, z) = \rho\mathbf{e}_\rho(\varphi) + z\mathbf{e}_z \\\cline{2-2}
                                                            & \mathbf{\dot{r}}(\rho, \varphi, z) = \dot{\rho}\mathbf{e}_\rho + \rho\dot{\varphi}\mathbf{e}_\varphi + \dot{z}\mathbf{e}_z \\\cline{2-2}
                                                            & \mathbf{\ddot{r}}(\rho, \varphi, z) = (\ddot{\rho} - \rho\dot{\varphi}^2)\mathbf{e}_\rho + (\rho\ddot{\varphi} + 2\dot{r}\dot{\varphi})\mathbf{e}_\varphi + \ddot{z}\mathbf{e}_z \\\hline
        \multirow{2}{*}{\rotatebox[origin=c]{90}{spherical}}& \mathbf{r}(r, \theta, \varphi) = r\mathbf{e}_r(\theta, \varphi) \\\cline{2-2}
                                                            & \mathbf{\dot{r}}(r, \theta, \varphi) = \dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_\theta + r\sin\theta\dot{\varphi}\mathbf{e}_\varphi \\\cline{2-2}
                                                            & \begin{aligned}
                                                                \mathbf{\ddot{r}}(r, \theta, \varphi)   &= (\ddot{r} - r\dot{\theta}^2\cos^2\varphi - r\dot{\varphi}^2)\mathbf{e}_r  \\
                                                                                                &+ (2\dot{r}\dot{\theta}\cos\varphi + r\ddot{\theta}\cos\varphi - 2r\dot{\theta}\dot{\varphi}\sin\varphi)\mathbf{e}_\theta \\
                                                                                                &+ (2\dot{r}\dot{\varphi} + r\dot{\theta}^2\sin\varphi\cos\varphi + r\ddot{\varphi})\mathbf{e}_\varphi
                                                                 \end{aligned} \\\hline
        \end{tabu}

\end{multicols*}

\end{document}

Answer 1

使用2行来组合单元格\multirow是不够的。这太窄了，以至于旋转后的框文本与顶部边界对齐。

3我建议对第一个旋转文本和5第二个文本使用行。

\documentclass[a4paper,pagesize ,landscape, fontsize=6pt]{scrartcl}
\usepackage[left=1.5cm,right=1.5cm, top=1.5cm, bottom=1.5cm]{geometry}
\usepackage{multicol}
\usepackage{amsmath, amsfonts, amssymb}
\usepackage{tikz}
\usepackage{array,multirow}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage{tabu}

\begin{document}

\begin{multicols*}{3}

    \tabulinesep=1.0mm
    \begin{tabu} to \columnwidth{> {$\textstyle} X[1,l] < {$} > {$\textstyle} X[30,l] < {$}}
        \multirow{3}{*}{\rotatebox[origin=c]{90}{polar}}    & \mathbf{r}(\rho, \varphi, z) = \rho\mathbf{e}_\rho(\varphi) + z\mathbf{e}_z \\\cline{2-2}
                                                            & \mathbf{\dot{r}}(\rho, \varphi, z) = \dot{\rho}\mathbf{e}_\rho + \rho\dot{\varphi}\mathbf{e}_\varphi + \dot{z}\mathbf{e}_z \\\cline{2-2}
                                                            & \mathbf{\ddot{r}}(\rho, \varphi, z) = (\ddot{\rho} - \rho\dot{\varphi}^2)\mathbf{e}_\rho + (\rho\ddot{\varphi} + 2\dot{r}\dot{\varphi})\mathbf{e}_\varphi + \ddot{z}\mathbf{e}_z \\\hline
        \multirow{5}{*}{\rotatebox[origin=c]{90}{spherical}}& \mathbf{r}(r, \theta, \varphi) = r\mathbf{e}_r(\theta, \varphi) \\\cline{2-2}
                                                            & \mathbf{\dot{r}}(r, \theta, \varphi) = \dot{r}\mathbf{e}_r + r\dot{\theta}\mathbf{e}_\theta + r\sin\theta\dot{\varphi}\mathbf{e}_\varphi \\\cline{2-2}
                                                            & \begin{aligned}
                                                                \mathbf{\ddot{r}}(r, \theta, \varphi)   &= (\ddot{r} - r\dot{\theta}^2\cos^2\varphi - r\dot{\varphi}^2)\mathbf{e}_r  \\
                                                                                                &+ (2\dot{r}\dot{\theta}\cos\varphi + r\ddot{\theta}\cos\varphi - 2r\dot{\theta}\dot{\varphi}\sin\varphi)\mathbf{e}_\theta \\
                                                                                                &+ (2\dot{r}\dot{\varphi} + r\dot{\theta}^2\sin\varphi\cos\varphi + r\ddot{\varphi})\mathbf{e}_\varphi
                                                                 \end{aligned} \\\hline
        \end{tabu}

\end{multicols*}
\end{document}

我认为\cline{2-2}应该删除（为了视觉吸引力）。我在屏幕截图中已经这样做了，但上面的代码中没有这样做：