带有方程式的表格中的附加空格单元格

带有方程式的表格中的附加空格单元格

我尝试按照一些建议为带有方程式的单元格添加垂直空间,但无法实现。

这是我的 MWE:

\documentclass[12pt,a4paper]{article}
    \usepackage{array}
\usepackage{makecell}
\renewcommand\theadfont{\normalsize\bfseries}
\usepackage[math]{cellspace}
\setlength\cellspacetoplimit{4pt}
\setlength\cellspacebottomlimit{4pt}

\usepackage{color,colortbl,hhline}
\definecolor{Gray}{gray}{0.9}
\usepackage{caption}
\usepackage{amsmath}

\begin{document}    
\begin{table}[htpb]
    \centering
    \caption{Solución de la ecuación }
        \begin{tabular}{|p{1cm}|p{12cm}|}
        \hline
        \rowcolor{Gray} \thead{Caso} & \thead{Solución} \\
        \hline
        $\zeta = 1$ & $u(t) = [u(0) + (\dot u(0) + u(0) \omega_n )t]e^{-\omega_n t}$\\\hline
        $\zeta > 1$ & $u(t) = \dfrac{e^{-\omega_n\zeta}}{2\omega_n\sqrt{\zeta^2-1}}\left\lbrace\left[\omega_n(\zeta+\sqrt{\zeta^2-1})u(0)+\dot u(0)\right]e^{\omega_n\sqrt{\zeta^2-1}}-\big[\omega_n(\zeta - \sqrt{\zeta^2-1})u(0)+\dot u(0)\big]e^{-\omega_n\sqrt{\zeta^2-1}}\right\rbrace$ \\\hline
        $\zeta < 1$ & $u(t) = e^{-\omega_n\zeta t}\left[u(0)\cos{\omega_Dt} + \left(\dfrac{\dot u(0) + \omega_n\zeta u(0)}{\omega_D}\right)\right]$ \\\hline
        \end{tabular}
    \end{table}
\end{document}

更新

也许我无法实现解决方案,因为某些包之间存在冲突。这是我的序言,Typeset Engine 是pdfLaTeX

\documentclass[12pt,a4paper]{article}

%\usepackage{polyglossia}
%\setdefaultlanguage[babelshorthands]{spanish}
%\usepackage[Ligatures=TeX]{fontspec}
\usepackage[spanish,es-noshorthands]{babel}
\usepackage[T1]{fontenc}%http://tex.stackexchange.com/questions/44694/fontenc-vs-inputenc
\usepackage[utf8]{inputenc}
\usepackage{mathptmx}%http://tex.stackexchange.com/questions/669/how-do-i-make-my-document-use-the-times-font-both-for-the-text-and-the-math
%\fontspec{Times New Roman}
%\usepackage{fontspec}
%\setmainfont{Times New Roman}
%\usepackage[T1]{fontenc}
%\usepackage{ifthen}
%\usepackage{fontspec}
%\setmainfont{Arial}

\usepackage[table]{xcolor}

%\usepackage[htt]{hyphenat}
%
%\usepackage{pdflscape}

%\usepackage[title]{appendix}

\usepackage[style=apa]{biblatex}
\addbibresource{bibltarea6.bib}

%\usepackage{array,multirow}
\usepackage{makecell}
\renewcommand\theadfont{\normalsize\bfseries}
\renewcommand\theadgape{}
\usepackage[math]{cellspace}
\setlength\cellspacetoplimit{4pt}
\setlength\cellspacebottomlimit{4pt}

\usepackage{listings}
\lstset{
basicstyle = \ttfamily,
keywordstyle= \color{blue}\textbf,
commentstyle= \color{gray},
stringstyle= \color{green!70!black},
stringstyle= \color{red},
columns= fullflexible,
numbers= left,
numberstyle= \scriptsize\sffamily\color{gray},
xleftmargin= 0.07\textwidth,
xrightmargin= 0.05\textwidth,
showstringspaces = false,
}

%\usepackage{inconsolata}

%\newcommand{\Cpp}{C\texttt{++}}
%\renewcommand{\lstlistingname}{Código}
%\renewcommand{\lstlistlistingname}{Índice de códigos}

%\usepackage{tikz}
%\usepgflibrary{arrows.meta}
%\tikzset{%
%  newTip/.tip={Bar[sep=-1.45pt, width=10pt] Circle[sep=-1.25pt,width=2.5pt,length=2.5pt]}
%  }
%\usetikzlibrary{patterns}
%
%\usepackage{tikz}
%\usepgflibrary{arrows.meta,shapes.arrows}
%\tikzset{%
%  newTip/.tip={Bar[sep=-1.45pt, width=10pt] Circle[sep=-1.25pt,width=2.5pt,length=2.5pt]}
%  }
%\usetikzlibrary{patterns}
%\usetikzlibrary{calc}

%\usepackage{caption}
%\usepackage{subcaption}

\usepackage{amsmath}
\usepackage{mathtools}
%\setcounter{MaxMatrixCols}{25}

\usepackage{siunitx}
\sisetup{per-mode = symbol}

%\usepackage{pgfplots}

%\newcounter{myrow}

%\usepackage{longtable}
%\usepackage{threeparttable}

%\usepackage[inline]{enumitem}
%\usepackage{hyperref}

\usepackage{titlesec}
\titleformat*{\section}{\bfseries\normalsize}
\titleformat*{\subsection}{\bfseries\normalsize}
\titleformat*{\subsubsection}{\bfseries\normalsize}

\usepackage[ampersand]{easylist}

%\usepackage{layout}

\usepackage{color,colortbl,hhline}
\definecolor{Gray}{gray}{0.9}

%\usepackage[a4paper,left=3cm, top=3cm, right=2.5cm, bottom=2.5cm]{geometry}

\setlength{\hoffset}{.46cm}
\setlength{\oddsidemargin}{13.08pt}
\setlength{\evensidemargin}{0cm}
\setlength{\marginparwidth}{0cm}
\setlength{\topmargin}{-7mm}
\setlength{\marginparsep}{0cm}
\setlength{\headheight}{8mm}
\setlength{\headsep}{4.6mm}
\setlength{\textheight}{24cm}
\setlength{\footskip}{8mm}
\setlength{\textwidth}{15.5cm}

\usepackage{fancyhdr}
\pagestyle{fancy}
 \lhead{\footnotesize{Pontificia Universidad Católica del Perú}\\ \footnotesize{Escuela de Posgrado - Ingeniería Civil}}%Significa que est
 \chead{}%Vac
 \rhead{\footnotesize{Dinámica de Estructuras\\ Tarea 1}}
 \lfoot{\footnotesize{}}%Pie izquierdo
 \cfoot{\thepage}
 \rfoot{\footnotesize{Fernández Yopla, Isai}}

 \renewcommand{\headrulewidth}{0.4pt}%Falta investigar
 \renewcommand{\footrulewidth}{0.4pt}

\usepackage[pass]{geometry}%pass tells output driver which paper size use.

答案1

第二行/列中的方程太长了。您需要将其分成两行。例如,使用aligned包中的环境mathtools。为了获得更多的垂直空间,您应该使用S中的列说明符cellspace。例如对于第二列S{p{12cm}}::

编辑:

  • 在某些情况下,当siunitx使用包时,包S中的说明符cellspace必须替换为C,即C{p(12cm)}
  • 在下面的 mwe 中我仅考虑与表相关的包
  • 离题:由于您也使用该geometry包,因此您可以使用它来定义页面布局,并删除序言中的所有其他低级页面布局定义

\documentclass[12pt,a4paper]{article}
\usepackage[T1]{fontenc}%http://tex.stackexchange.com/questions/44694/fontenc-vs-inputenc
\usepackage[utf8]{inputenc}
\usepackage[spanish,es-noshorthands]{babel}
\usepackage{mathptmx}%http://tex.stackexchange.com/questions/669/how-do-i-make-my-document-use-the-times-font-both-for-the-text-and-the-math

\usepackage{makecell,hhline}        % <---
\renewcommand\theadfont{\normalsize\bfseries}
\renewcommand\theadgape{}
\usepackage{cellspace}              % <---
\setlength\cellspacetoplimit{4pt}
\setlength\cellspacebottomlimit{4pt}

\usepackage{mathtools}              % it load "amsmath" too
\usepackage{siunitx}                % <--- used it, the "S" from the "cellspace# package
                                    %  had to be replaced with "C"
\usepackage[table]{xcolor}          % <---
\definecolor{Gray}{gray}{0.9}
\usepackage{geometry}               % for simple determination of page layout.
                                    % see its documentation!

\begin{document}
    \begin{table}[htpb]
    \centering
\caption{Solución de la ecuación }
    \begin{tabular}{|p{1cm}|C{p{12cm}}|}
        \hline
        \rowcolor{Gray}
        \thead{Caso} & \thead{Solución}             \\
        \hline
        $\zeta = 1$  & $u(t) = \bigl[u(0) + \bigl(\dot{u}(0) + u(0) \omega_n\bigr) t\bigr] e^{-\omega_n t}$     \\ \hline
        $\zeta > 1$ & $\begin{multlined}
            u(t) = \dfrac{e^{-\omega_n\zeta}}{2\omega_n\sqrt{\zeta^2-1}}
            \biggl\{\Bigl[\omega_n\Bigl(\zeta+\sqrt{\zeta^2-1}\Bigr)u(0)+\dot{u}(0)\Bigr] e^{\omega_n\sqrt{\zeta^2-1}} -  \\
            \Bigl[\omega_n\Bigl(\zeta - \sqrt{\zeta^2-1}\Bigr)u(0)+\dot{u}(0)\Bigr] e^{-\omega_n\sqrt{\zeta^2-1}}
            \biggr\}
                        \end{multlined}$                \\  \hline
        $\zeta < 1$ & $u(t) = e^{-\omega_n\zeta t}
        \biggl[u(0)\cos{\omega_Dt} + 
            \biggl(\dfrac{\dot{u}(0) + \omega_n\zeta u(0)}{\omega_D}\biggr)
        \biggr]$                                        \\  \hline
        \end{tabular}
    \end{table}
\end{document}

在此处输入图片描述

答案2

正如@Zarko 已经指出的那样他的回答,主要问题是其中一个主要方程太长,无法放在一行中。使用aligned环境引入适当选择的换行符。

一旦您对长方程式进行了换行,就不需要使用列类型;相反,只需对两列p使用基本列类型。l

为了使表格材料更加整洁,我会u(t) =从单元格中删除重复的部分,然后将这些信息粘贴到表格的标题中。

我还想建议您让表格看起来更加开放,主要是通过删除所有垂直和水平黑线——它们根本就不需要。我还会加载包booktabs并使用其\addlinespace宏来进一步增加行之间的空白。

此外,我会用明确的尺寸说明替换所有指令\left\right

在此处输入图片描述

\documentclass[12pt,a4paper]{article}
\usepackage[spanish]{babel} % new
\usepackage{array}
\usepackage{booktabs} % for "\addlinespace" macro
\usepackage[table]{xcolor}
\definecolor{Gray}{gray}{0.9}
\usepackage[skip=0.333\baselineskip]{caption}
\usepackage{amsmath} % for 'aligned' environment

\begin{document}    
\begin{table}[htpb]
\setlength\extrarowheight{3pt}
\centering
\caption{Solución de la ecuación por $u(t)$}
\begin{tabular}{ll}
\rowcolor{Gray} \textbf{Caso} & \textbf{Solución } \\
\addlinespace
$\zeta = 1$ & 
$\displaystyle
  \bigl[u(0) + (\dot u(0) + u(0) \omega_n )t\bigr]
         e^{-\omega_n t}
$\\ \addlinespace
$\zeta > 1$ & 
$\begin{aligned}[t]
\frac{e^{-\omega_n\zeta}}{2\omega_n\sqrt{\zeta^2-1}}\Bigl\lbrace
  &\bigr[\omega_n(\zeta+\sqrt{\zeta^2-1})u(0)+\dot u(0)\bigr] 
    e^{\omega_n\sqrt{\zeta^2-1}}\\
-&\bigl[\omega_n(\zeta - \sqrt{\zeta^2-1})u(0)+\dot u(0)\bigr]
    e^{-\omega_n\sqrt{\zeta^2-1}}\,
\Bigr\rbrace
\end{aligned}$ \\ \addlinespace
$\zeta < 1$ & 
$\displaystyle 
e^{-\omega_n\zeta t}\bigl[u(0)\cos{\omega_Dt} 
+ \bigl(\dot u(0) + \omega_n\zeta u(0)\bigr)\big/\omega_D
\bigr]$ \\
\end{tabular}
\end{table}
\end{document}

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