## 5 Truncation Error and Accuracy Analysis
\begin{table}
\begin{tabular}{c c c c c c} \hline  & \(\Delta\mathbf{B(\phi)}\) & \(\Delta\mathbf{u(\phi)}\) & \(\Delta\mathbf{\psi(\phi)}\) & \(\Delta\mathbf{\vartheta(\phi)}\) & \(\Delta\mathbf{\varphi(\phi)}\) \\ \hline \(e^{10}\) & 2.61934 \(\times\)\(10^{-9}\) & 9.31323 \(\times\)\(10^{-10}\) & 9.74978 \(\times\)\(10^{-10}\) & 9.02219 \(\times\)\(10^{-10}\) & 1.01863 \(\times\)\(10^{-9}\) \\ \(e^{8}\) & 4.53001 \(\times\)\(10^{-7}\) & 1.63243 \(\times\)\(10^{-7}\) & 1.42056 \(\times\)\(10^{-7}\) & 1.35537 \(\times\)\(10^{-7}\) & 1.56637 \(\times\)\(10^{-7}\) \\ \(e^{6}\) & 8.10159 \(\times\)\(10^{-5}\) & 3.01542 \(\times\)\(10^{-5}\) & 2.25394 \(\times\)\(10^{-5}\) & 2.19159 \(\times\)\(10^{-5}\) & 2.53659 \(\times\)\(10^{-5}\) \\ \(e^{4}\) & 0.0151037 & 0.00590163 & 0.0035465 & 0.00367417 & 0.00392772 \\ \(n^{5}\) & 1.89175 \(\times\)\(10^{-10}\) & 1.16415 \(\times\)\(10^{-10}\) & 1.164263 \(\times\)\(10^{-10}\) & 1.16415 \(\times\)\(10^{-10}\) & 1.74623 \(\times\)\(10^{-10}\) \\ \(n^{4}\) & 4.52274 \(\times\)\(10^{-8}\) & 6.40284 \(\times\)\(10^{-10}\) & 7.01402 \(\times\)\(10^{-9}\) & 6.72298 \(\times\)\(10^{-9}\) & 5.52973 \(\times\)\(10^{-9}\) \\ \(n^{3}\) & 1.15434 \(\times\)\(10^{-5}\) & 4.0984 \(\times\)\(10^{-7}\) & 2.18944 \(\times\)\(10^{-6}\) & 2.0791 \(\times\)\(10^{-6}\) & 2.29341 \(\times\)\(10^{-6}\) \\ \(n^{2}\) & 0.00455778 & 0.000325912 & 0.00109875 & 0.000766065 & 0.00107432 \\ \hline \end{tabular}
\end{table} 
Table 12: Truncation errors of common latitude expanded to different orders with geocentric latitude as the variable (units: arc seconds). \[ \begin{array}{rl}\theta(\phi)&=\phi+\left(\frac{1}{6}e^{2}+\frac{7}{90}e^{4}+ \frac{281}{5040}e^{6}\right)\sin 2\phi+\left(\frac{1}{180}e^{4}+\frac{1}{252}e^{6} \right)\sin 4\phi-\frac{131}{25360}e^{6}\sin 6\phi\\ &=\phi+\left(\frac{2}{3}n-\frac{4}{45}n^{2}+\frac{62}{105}n^{3}\right)\sin 2 \phi+\left(\frac{4}{45}n^{2}-\frac{32}{315}n^{3}\right)\sin 4\phi-\frac{524}{2835}n^{3} \sin 6\phi,\end{array}  \tag{37}\]  
\[ \begin{array}{rl}\varphi(\phi)&=\phi+\left(\frac{1}{24}e^{4}+\frac{5}{96}e^{ 6}\right)\sin 2\phi+\left(-\frac{1}{48}e^{4}-\frac{1}{60}e^{6}\right)\sin 4 \phi-\frac{1}{160}e^{6}\sin 6\phi\\ &=\phi+\left(\frac{2}{3}n^{2}+\frac{2}{3}n^{3}\right)\sin 2\phi+\left(-\frac{1}{ 3}n^{2}+\frac{4}{15}n^{3}\right)\sin 4\phi-\frac{2}{5}n^{3}\sin 6\phi, \end{array}  \tag{38}\]  
\[ \begin{array}{rl}u(\phi)&=\phi+\left(\frac{1}{4}e^{2}+\frac{1}{8}e^{4}+ \frac{5}{64}e^{6}\right)\sin 2\phi+\left(\frac{1}{32}e^{4}+\frac{1}{32}e^{6} \right)\sin 4\phi+\frac{1}{192}e^{6}\sin 6\phi\\ &=\phi+n\sin 2\phi+\frac{1}{12}n^{2}\sin 4\phi+\frac{1}{3}n^{3}\sin 6\phi, \end{array}  \tag{39}\]  
\[ \begin{array}{rl}B(\phi)&=\phi+\left(\frac{1}{2}e^{2}+\frac{1}{4}e^{4}+ \frac{1}{8}e^{6}\right)\sin 2\phi+\left(\frac{1}{8}e^{4}+\frac{1}{8}e^{6} \right)\sin 4\phi+\frac{1}{24}e^{6}\sin 6\phi\\ &=\phi+\left(2n-2n^{3}\right)\sin 2\phi+2n^{2}\sin 4\phi+\frac{8}{3}n^{3} \sin 6\phi.\end{array}  \tag{40}\]  
In conclusion, the use of Mathematica allows us to derive simplified, practical formulas for the common latitude, with geodetic latitude and geocentric latitude as variables. Equations (31) to (37) show that when the power series is expanded based on the third flattening, it is more compact, and its coefficients are simpler, which is consistent with the conclusions derived above.