TRIUMF poster
Author:
Carla Barquest
Last Updated:
10 месяцев назад
License:
Creative Commons CC BY 4.0
Аннотация:
Scientific Poster Vertical-LaTex by TRIUMF
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass[portrait]{a0poster}
% Note: if package conflicts occur, try importing them before triumfposter
\usepackage{triumfposter}
% for example text in columns
\usepackage{lipsum}
\usepackage{float}
\usepackage{multicol}
\columnsep=60pt % spacing between columns
\columnseprule=0pt % vertical line between columns
\usepackage{amsfonts, amsmath, amsthm, amssymb} % For math fonts, symbols and environments
\title{An Example Title for a Poster in VeryHuge Font}
% \veryHuge \\ And I Can Also Have a Subtitle -- Smaller Font Possible \\}
% Single Affiliation
% \author{T.~Author1, T.~Author2, and T.~Author3\\TRIUMF, Vancouver, BC, Canada}
% Multiple Affiliations
\author[1]{T.~Author1}
\author[2]{V.~Author2}
\author[1]{T.~Author3}
\author[1,2]{T.V.~Author4}
\affil[1]{TRIUMF, Vancouver, BC, Canada}
\affil[2]{University of Victoria, Victoria, BC, Canada}
\date{\vspace{-5ex}}
\begin{document}
\maketitle
\thispagestyle{empty}
\darkheader
\begin{multicols}{2}
\large
\sectiona{Abstract}
\lipsum[7]
\sectionb{Section 1}
\lipsum[1]
\sectionb{Mathematical Section}
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\begin{equation}
E = mc^{2}
\label{eqn:Einstein}
\end{equation}
Curabitur mi sem, pulvinar quis aliquam rutrum. (1) edf (2)
, $\Omega=[-1,1]^3$, maecenas leo est, ornare at. $z=-1$ edf $z=1$ sed interdum felis dapibus sem. $x$ set $y$ ytruem.
Turpis $j$ amet accumsan enim $y$-lacina;
ref $k$-viverra nec porttitor $x$-lacina.
Vestibulum ac diam a odio tempus congue. Vivamus id enim nisi:
\begin{eqnarray}
\cos\bar{\phi}_k Q_{j,k+1,t} + Q_{j,k+1,x}+\frac{\sin^2\bar{\phi}_k}{T\cos\bar{\phi}_k} Q_{j,k+1} &=&\nonumber\\
-\cos\phi_k Q_{j,k,t} + Q_{j,k,x}-\frac{\sin^2\phi_k}{T\cos\phi_k} Q_{j,k}\label{edgek}
\end{eqnarray}
and
\begin{eqnarray}
\cos\bar{\phi}_j Q_{j+1,k,t} + Q_{j+1,k,y}+\frac{\sin^2\bar{\phi}_j}{T\cos\bar{\phi}_j} Q_{j+1,k}&=&\nonumber \\
-\cos\phi_j Q_{j,k,t} + Q_{j,k,y}-\frac{\sin^2\phi_j}{T\cos\phi_j} Q_{j,k}.\label{edgej}
\end{eqnarray}
Nulla sed arcu arcu. Duis et ante gravida orci venenatis tincidunt. Fusce vitae lacinia metus. Pellentesque habitant morbi. $\mathbf{A}\underline{\xi}=\underline{\beta}$ Vim $\underline{\xi}$ enum nidi $3(P+2)^{2}$ lacina. Id feugain $\mathbf{A}$ nun quis; magno.
\sectionb{Section 2}
\lipsum[2]
\sectionb{Image Section}
\begin{figure}[H]
\centering
\begin{tikzpicture}[scale=0.1]
\boomerang
\end{tikzpicture}
\caption{Figure in two column}
\end{figure}
\sectionb{Section 3}
\lipsum[3]
\sectionb{Conclusion}
\large
\lipsum[5-6]
\sectiona{Acknowledgement}
\small
\lipsum[4]
\end{multicols}
\end{document}