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请帮我将公式对齐如下

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请帮我将公式对齐如下 
For Crushing Glass,
Mini pressure applied in the cylinder (P) = 6.86x105N/m2       
Diameter of the cylinder (D) = 32 mm 
Diameter of the piston rod (d) = 17 mm               
Stroke length = 100 mm      
Area of cylinder (A) = (3.14/4*(D2)
                 = (0.785x.0322)
               A= 8.0384 x 10-4 m2
Force during forward stroke,
F = {π/4x D^2 x P}
F = (6.86 x 105 N /m2) (3.14/4*(0.0322)
F = 551.43 N 
       1kg = 9.81 N 
    Load capacity during forward stroke W= F / a
                                      W =    551.43 / 9.81
                                      W =     56.21 KG  
Force during return stroke,
F = {π/4x (D^2 – d^2) x P}
F = {π/4x (0.032^2 – 0.017^2) x 6.86x105}
 F = 395.99 N
Load capacity during return stroke W= F / a
                                      W =    395.99 / 9.81
                                      W =     40.36 KG 
For  Plastic bottles,
Mini pressure applied in the cylinder (P) = 5.39x105N/m2       
Diameter of the cylinder (D) = 32 mm 
Diameter of the piston rod (d) = 17 mm               
Stroke length = 100 mm      
Area of cylinder (A) = (3.14/4*(D2)
                 = (0.785x.0322)
               A= 8.0384 x 10-4 m2
Force during forward stroke,
F = {π/4x D^2 x P}
F = (5.39 x 105 N /m2) (3.14/4*(0.0322)
F = 433.26 N 
       1kg = 9.81 N 
    Load capacity during forward stroke W= F / a
                                      W =    433.26 / 9.81
                                      W =     44.16 KG  
Force during return stroke,
F = {π/4x (D^2 – d^2) x P}
F = {π/4x (0.032^2 – 0.017^2) x 5.39x105}
 F = 311.13 N
Load capacity during return stroke W= F / a
                                      W =    311.13/ 9.81
                                      W =     31.71 KG  

For Crushing aluminium cans,
Mini pressure applied in the cylinder (P) = 4.41x105N/m2       
Diameter of the cylinder (D) = 32 mm 
Diameter of the piston rod (d) = 17 mm               
Stroke length = 100 mm      
Area of cylinder (A) = (3.14/4*(D2)
                 = (0.785x.0322)
               A= 8.0384 x 10-4 m2
Force during forward stroke,
F = {π/4x D^2 x P}
F = (4.41 x 105 N /m2) (3.14/4*(0.0322)
F = 354.49 N 
       1kg = 9.81 N 
    Load capacity during forward stroke W= F / a
                                      W =    354.49 / 9.81
                                      W =     36.135 KG  
Force during return stroke,
F = {π/4x (D^2 – d^2) x P}
F = {π/4x (0.032^2 – 0.017^2) x 4.41x105}
 F = 254.56 N
Load capacity during return stroke W= F / a
                                      W =    254.56 / 9.81
                                      W =     25.94 KG

答案1

我认为你应该设置一个两列表格，左边是文字描述，右边是公式和数值示例。还要注意正确排版科学单位。

前几行可能如下所示：

在此处输入图片描述

\documentclass{report}
\usepackage{amsmath,longtable,booktabs,array,siunitx}
\sisetup{per-mode=symbol}
\begin{document}

\begin{longtable}{@{}l >{$}l<{$} @{}}
\toprule
\endhead
\bottomrule
\endfoot
For Crushing Glass,\\
Mini pressure applied in the cylinder & P = 6.86 \cdot 105\, \si{\newton\per\meter\squared} \\
Diameter of the cylinder &D = \SI{32}{\milli\meter}\\ 
Diameter of the piston rod & d = \SI{17}{\milli\meter}\\               
Stroke length & \SI{100}{\milli\meter}   \\    
Area of cylinder & 
\!\begin{aligned}[t] 
A &= \pi/4\cdot D^2\\
  &= 0.785\times .0322\\
  &= \SI{8.0384e-4}{\meter\squared}
\end{aligned} \\
Force during forward stroke &
\!\begin{aligned}[t]
F &= \pi /4\cdot D^2 \cdot P\\
  &= 6.86 \cdot 105\,\si{\newton\per\meter\squared} \quad(3.14/4\cdot 0.0322)\\
 &= \SI{551.43}{\newton} \quad (\SI{1}{\kilo\gram} =\SI{9.81}{\newton})
\end{aligned} \\ 
Load capacity during forward stroke & W=\dots \\
\end{longtable}
\end{document}

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