用相同的编号依次写出方程式

Question

我已经写了 3 个方程式，我希望它显示如下：

equation1=....     (1)

equation2=....     (2)
equation3=....

这意味着方程 2 和方程 3 是相同的\begin{equation}

我尝试了以下代码但出现错误：

\begin{equation}
    \gamma_{T}=tan^{-1}\frac{-V_{T2}}{V_{T1}} \
\end{equation}

The acceleration components of the ballistic target can be expressed in terms of the target weight $W$ , reference area $S_{ref}$, zero lift-drag $C_{D0}$ and gravity $g$.

\begin{align*} 

    \frac{dV_{T1}}{dt}=\frac{-F_{Drag}}{m}cos\gamma_{T}=\frac{-QS_{ref}C_{D0}g}{W}cos\gamma_{T}=
    \frac{-Qg}{\beta}cos\gamma_{T}

     \\

    \frac{dV_{T2}}{dt}=\frac{F_{Drag}}{m}sin\gamma_{T}-g=\frac{QS_{ref}C_{D0}g}{W}sin\gamma_{T}-g=
    \frac{Qg}{\beta}sin\gamma_{T}-g\

\end{align*}

Answer 1

错误是由于在数学模式中留下空行而产生的。我猜你想要的是这样的：

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\gamma_{T}=\tan^{-1}\frac{-V_{T_2}}{V_{T_1}}
\end{equation}
The acceleration components of the ballistic target can be expressed in terms of the
target weight $W$, reference area $S_\mathrm{ref}$, zero lift-drag $C_{D0}$ and gravity $g$.
\begin{align} 
\frac{dV_{T_1}}{dt} &= \frac{-F_\mathrm{Drag}}{m} \cos\gamma_{T}
 = \frac{-Q S_\mathrm{ref} C_{D0} g}{W} \cos\gamma_{T}
 = \frac{-Q g}{\beta} \cos\gamma_{T}
  \\
\frac{dV_{T_2}}{dt} &= \frac{F_\mathrm{Drag}}{m} \sin\gamma_{T}-g
 = \frac{Q S_\mathrm{ref} C_{D0} g}{W} \sin\gamma_{T}-g 
 = \frac{Q g}{\beta} \sin\gamma_{T}-g \nonumber
\end{align}
\end{document}

由于这两个方程式有点相似，因此split给出中心数字的环境可能更合适：

\documentclass{article}
\usepackage{amsmath}
\begin{document}
\begin{equation}
\gamma_{T}=\tan^{-1}\frac{-V_{T_2}}{V_{T_1}}
\end{equation}
The acceleration components of the ballistic target can be expressed in terms of the
target weight $W$, reference area $S_\mathrm{ref}$, zero lift-drag $C_{D0}$ and gravity $g$.
\begin{equation} 
\begin{split}
\frac{dV_{T_1}}{dt} &= \frac{-F_\mathrm{Drag}}{m} \cos\gamma_{T}
 = \frac{-Q S_\mathrm{ref} C_{D0} g}{W} \cos\gamma_{T}
 = \frac{-Q g}{\beta} \cos\gamma_{T}
  \\
\frac{dV_{T_2}}{dt} &= \frac{F_\mathrm{Drag}}{m} \sin\gamma_{T}-g
 = \frac{Q S_\mathrm{ref} C_{D0} g}{W} \sin\gamma_{T}-g 
 = \frac{Q g}{\beta} \sin\gamma_{T}-g
\end{split}
\end{equation}
\end{document}