
이 코드에는 두 부분이 있습니다. 첫 번째는 이후의 수치와 계산입니다. (두 개의 의견으로 표시됨) 하지만 실행해 보니 반대 순서로 나타났습니다.
\subsection*{Figures}
\begin{center}
%should appear first
\begin{figure}[ht]
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{../figures/Ptime.jpg}
\caption{Pressure vs Step}
\vspace{4ex}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{../figures/temptime.jpg}
\caption{Temperature vs Step}
\vspace{4ex}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{../figures/KEtime.jpg}
\caption{Kinetic Energy vs Step}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{../figures/PEtime.jpg}
\caption{Potential Energy vs Step}
\end{subfigure}
\caption{}
\end{figure}
\end{center}
%should appear next
\textit{Calculation of Self Diffusion Coefficient}
\begin{equation*}
\begin{split}
\text{slope} & = 0.0021 \\
\text{Self Diffusion Coeff} & =\frac{1}{6}\times \text{(Slope of MSD at long time)}\\
& = \frac{1}{6} \times \text{0.0021}\\
& = \SI{3.5e-4} {\text{LJ units}}
\end{split}
\end{equation*}
\text{Considering the LJ liquid to be liquid Ar, we have $\sigma = 0.34 \textit{ nm}$, $\frac{\epsilon}{k_B}=120 \textit{ K}$. }\\
\text{$\therefore \frac{{(L^*)}^2}{t^*}=\frac{{L^2}/{\sigma^2}}{t\sqrt{\frac{\epsilon}{m\sigma^2}}}\implies \frac{L^2}{t}= (\sigma^3\sqrt{m/\epsilon}) \frac{(L^*)^2}{t^*} \implies \frac{L^2}{\epsilon}=\num{2.49e-31}\frac{(L^*)^2}{t^*}$}
\linebreak
\begin{equation*}
\begin{split}
\text{Diffusion Coefficient in SI Units} &= \num{3.5e-4}\times \SI{2.489e-31}{\meter^2 /\second}\\
&=\SI{8.7e-34}{\meter^2 /\second}
\end{split}
\end{equation*}
답변1
두 번째 eqn 위의 텍스트가 단 하나의 텍스트 단어로 단축되면 올바르게 작동합니다 considering
. 아마도 eqn에 일부 오류가 있을 수 있습니다. 그 외에는 올바르게 작동합니다.
\documentclass[]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{mathtools}
\usepackage{amssymb,booktabs}
\usepackage{caption,subcaption}
\begin{document}
\subsection*{Figures}
\begin{center}
%should appear first
\begin{figure}[ht]
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{example-image-a}
\caption{Pressure vs Step}
\vspace{4ex}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{example-image-a}
\caption{Temperature vs Step}
\vspace{4ex}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{example-image-a}
\caption{Kinetic Energy vs Step}
\end{subfigure}
\begin{subfigure}[b]{0.5\linewidth}
\centering
\includegraphics[width=0.9\linewidth]{example-image-a}
\caption{Potential Energy vs Step}
\end{subfigure}
\caption{}
\end{figure}
\end{center}
%should appear next
\textit{Calculation of Self Diffusion Coefficient}
\begin{equation}
\begin{split}
TCU & = \frac{\text{purchasing cost} + \text{setup cost} + \text{holding cost}}{time} \\
& = \frac{cy+K+h\frac{y}{2}t_0}{t_0}
\end{split}
\end{equation}
\text{Considering the LJ liquid to be liquid Ar, we have $\sigma = 0.34 \textit{ nm}$, $\frac{\epsilon}{k_B}=120 \textit{ K}$. }\\
\text{$\therefore \frac{{(L^*)}^2}{t^*}=\frac{{L^2}/{\sigma^2}}{t\sqrt{\frac{\epsilon}{m\sigma^2}}}\implies \frac{L^2}{t}= (\sigma^3\sqrt{m/\epsilon}) \frac{(L^*)^2}{t^*} \implies \frac{L^2}{\epsilon}=\num{2.49e-31}\frac{(L^*)^2}{t^*}$}
\linebreak
\begin{equation}
\begin{split}
TCU & = \frac{\text{purchasing cost} + \text{setup cost} + \text{holding cost}}{time} \\
& = \frac{cy+K+h\frac{y}{2}t_0}{t_0}
\end{split}
\end{equation}
\end{document}