# Table centering

Could someone please help me to center the equations in this table? This is the first table I'm making in LaTeX, and I can't find the right way to do it.

\begin{center}
\begin{tabular}{ |p{2cm}||p{5cm}|p{5cm}|  }
\hline
\multicolumn{3}{|c|}{Erdseilström} \\
\hline
&Einfachleitung &Doppelleitung\10pt] \hline 1 Erdseil & I_N = -\frac{3Z_{NX}}{Z_{NN}} I_0 & I_N= -\frac{6Z_{NX}}{Z_{NN}} I_0\\[20pt] 2 Erdseile & I_{N_1} = I_{N_2}= -\frac{3Z_{NX}}{Z_{NN}+Z_{N_1N_2}} I_0 & I_{N_1}=I_{N_2}= -\frac{3I_{0}(Z_{NX}+Z_{NY})}{Z_{NN}+Z_{N_1N_2}} \\[20pt] \hline \end{tabular} \end{center}  • p columns aren't centered – user31729 Mar 6, 2018 at 10:17 • Since line-breaking inside cells wouldn't seem to be desirable, don't use a column type that allows automatic line-breaking. – Mico Mar 6, 2018 at 10:21 ## 4 Answers similar to Mico solution. small differences is use amsmath for math writing and makecell for vertical line spacing: \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage{booktabs, makecell} \setcellgapes{3pt} \begin{document} {\noindent \makegapedcells \begin{tabular*}{\linewidth}{@{\extracolsep{\fill}} l >{}c<{} >{}c<{}} \toprule & \multicolumn{2}{c}{Erdseilström} \\ \cmidrule(lr){2-3} & \text{Einfachleitung} & \text{Doppelleitung} \\ \midrule 1 Erdseil & I_N = -\dfrac{3Z_{NX}}{Z_{NN}} I_0 & I_N = -\dfrac{6Z_{NX}}{Z_{NN}} I_0 \\ 2 Erdseile & I_{N_{1,2}}= -\dfrac{3Z_{NX}}{Z_{NN}+Z_{N_1N_2}} I_0 & I_{N_{1,2}}= -\dfrac{3I_{0}(Z_{NX}+Z_{NY})}{Z_{NN}+Z_{N_1N_2}} \\ \bottomrule \end{tabular*} } \end{document}  (red lines indicate text border). I'd like to suggest that you use an array environment, since most of the cells employ math mode. Note that I don't use any vertical lines and fewer, but well-spaced, horizontal lines. \documentclass{article} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{array,booktabs} \newcolumntype{L}{>{}l<{}} % left-aligned text mode in an "array" \newcolumntype{C}{>{\displaystyle}c} \begin{document} \[ \begin{array}{@{} LCC @{}} \toprule & \multicolumn{2}{c@{}}{Erdseilström} \\ \cmidrule(l){2-3} &Einfachleitung &Doppelleitung\\ \midrule 1 Erdseil & I_N = -\frac{3Z_{NX}}{Z_{NN}} I_0 & I_N= -\frac{6Z_{NX}}{Z_{NN}} I_0\\ \addlinespace 2 Erdseile & I_{N_1} = I_{N_2}= -\frac{3Z_{NX}}{Z_{NN}+Z_{N_1N_2}} I_0 & I_{N_1}=I_{N_2}= -\frac{3I_{0}(Z_{NX}+Z_{NY})}{Z_{NN}+Z_{N_1N_2}} \\ \bottomrule \end{array}
\end{document}


Addendum to address the OP's follow-up question. If the table must span the full width of the textblock, I'd suggest using a tabular* environment, with its full width set to \textwidth. (The @{\extracolsep{\fill}} bit instructs LaTeX to insert as much intercolumn whitespace as is needed to fill up the target width. The array and tabular environments aren't set up to take a target width as one of their main parameters.)

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{array,booktabs}
\newcolumntype{C}{>{$\displaystyle}c<{$}}
\begin{document}

\begingroup  % localize scope of the next instruction
\setlength\tabcolsep{0pt}
\noindent
\begin{tabular*}{\textwidth}{l @{\extracolsep{\fill}} CC }
\toprule & \multicolumn{2}{c@{}}{Erdseilström} \\
\cmidrule(l){2-3}
&$Einfachleitung$ &$Doppelleitung$\\
\midrule
1 Erdseil  & I_N = -\frac{3Z_{NX}}{Z_{NN}} I_0
& I_N = -\frac{6Z_{NX}}{Z_{NN}} I_0\\
2 Erdseile & I_{N_{1,2}}= -\frac{3Z_{NX}}{Z_{NN}+Z_{N_1N_2}} I_0
& I_{N_{1,2}}= -\frac{3I_{0}(Z_{NX}+Z_{NY})}{Z_{NN}+Z_{N_1N_2}} \\
\bottomrule
\end{tabular*}
\endgroup

\end{document}

• Same idea with a little arraystretch for lisibility. Mar 6, 2018 at 10:21
• Tnx @Mico, I just changed equations a bit (they are shorter), so how can I strech the table from margin to margin? Here how it looks now:[ \begin{array}{@{} LCC @{}} \toprule & \multicolumn{2}{c@{}}{$Erdseilström$} \\ \cmidrule(l){2-3} &$Einfachleitung$ &$Doppelleitung$\\ \midrule 1 Erdseil & I_N = -\frac{3Z_{NX}}{Z_{NN}} I_0 & I_N= -\frac{6Z_{NX}}{Z_{NN}} I_0\\ \addlinespace 2 Erdseile & I_{N_{1,2}}= -\frac{3Z_{NX}}{Z_{NN}+Z_{N_1N_2}} I_0 & I_{N_{1,2}}= -\frac{3I_{0}(Z_{NX}+Z_{NY})}{Z_{NN}+Z_{N_1N_2}} \\ \bottomrule \end{array} ] Mar 6, 2018 at 10:26
• @elsol, please edit your question and put changes there that will be visible to all Mar 6, 2018 at 10:38

Mico's layout is more professional and is to be preferred, but if you want to keep yours:

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{array}
\newcolumntype{C}[1]{>{\centering\arraybackslash$}p{#1}<{$}}
\begin{document}

\begin{center}
\setlength\extrarowheight{10pt}
\begin{tabular}{ |p{2cm}||C{5cm}|C{5cm}|  }
\hline
\multicolumn{3}{|c|}{Erdseilström} \\[10pt]
\hline
&\multicolumn{1}{c|}{Einfachleitung} &\multicolumn{1}{c|}{Doppelleitung}\\[10pt]
\hline
1 Erdseil   & I_N = -\frac{3Z_{NX}}{Z_{NN}} I_0  & I_N= -\frac{6Z_{NX}}{Z_{NN}} I_0\\[20pt]
2 Erdseile &  I_{N_1} = I_{N_2}= -\frac{3Z_{NX}}{Z_{NN}+Z_{N_1N_2}} I_0   & I_{N_1}=I_{N_2}=  -\frac{3I_{0}(Z_{NX}+Z_{NY})}{Z_{NN}+Z_{N_1N_2}} \\[20pt]
\hline
\end{tabular}
\end{center}

\end{document}


Here is another simple possibility, with the array package. I added some improvements as to the vertical spacing, with the cellspace package, which ensures a minimal padding at the top and bottom of cells in columns with specifier preffixed withe letter S (or C if you load siunitx). Also, I used the medium-sized fractions from nccmath, and hhline for the intersection of double lines:

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{array}
\usepackage{mathtools}
\usepackage{nccmath} % provides medium sized formulae and acontrol of intertext spacing via an optional argument

\usepackage{cellspace, hhline}
\setlength{\cellspacetoplimit}{6pt}
\setlength{\cellspacebottomlimit}{6pt}

\begin{document}

\begin{center}
\begin{tabular}{ |p{2cm}||S{>{\centering\arraybackslash} p{5cm}}|>{\centering\arraybackslash}p{5cm}| }
\hline
\multicolumn{3}{|Sc|}{Erdseilström} \\
\hline
&Einfachleitung &Doppelleitung\\
\hhline{-||--}
1 Erdseil & $I_N = -\mfrac{3Z_{NX}}{Z_{NN}} I_0$ & $I_N= -\mfrac{6Z_{NX}}{Z_{NN}} I_0$\\
2 Erdseile & $I_{N_1} = I_{N_2}= -\mfrac{3Z_{NX}}{Z_{NN}+Z_{N_1N_2}} I_0$ & $I_{N_1}=I_{N_2}= -\mfrac{3I_{0}(Z_{NX}+Z_{NY})}{Z_{NN}+Z_{N_1N_2}}$ \\
\hline
\end{tabular}
\end{center}

\end{document}