\(\displaystyle{ \varphi=\tan^{-1}\left(\frac{1+n}{1-n}t_\psi\right)\, ,\:\: \lambda=\lambda_0+\tan^{-1}\left(\frac{\sinh\eta^{\prime}}{\cos\xi^{\prime}}\right) }\)
\(\displaystyle{ \gamma=\tan^{-1}\left(\frac{\tau^{\prime}+\sigma^{\prime}\tan\xi^{\prime}\tanh\eta^{\prime}}{\sigma^{\prime}-\tau^{\prime}\tan\xi^{\prime}\tanh\eta^{\prime}}\right)\, ,\:\: m=\frac{\bar A}{a}\sqrt{\frac{\cos^2\xi^{\prime}+\sinh^2\eta^{\prime}}{{\sigma^{\prime}}^2+{\tau^{\prime}}^2}\left(1+t_\psi^2\right)} }\)
ただし、
\(x, y\)： 新点の\(\,X\,\)座標及び\(\,Y\,\)座標
\(\varphi_0, \lambda_0\)： 平面直角座標系原点の緯度【ラジアン換算必須】及び経度
\(a, F\)： 地球楕円体の長半径及び逆扁平率\(\,(1/f)\)
\(m_0\)： 平面直角座標系の\(\,X\,\)軸上における縮尺係数 (0.9999)
\(\displaystyle{ n=\frac{1}{2F-1}\, ,\:\: \xi=\frac{x+\bar S_{\varphi_0}}{\bar A}\, ,\:\: \eta=\frac{y}{\bar A} }\)
\(\displaystyle{ \xi^{\prime}=\xi-\sum_{j=1}^{5}\beta_j\sin 2j\xi\cosh 2j\eta\, ,\:\: \eta^{\prime}=\eta-\sum_{j=1}^{5}\beta_j\cos 2j\xi\sinh 2j\eta }\)
\(\displaystyle{ \sigma^{\prime}=1-\sum_{j=1}^{5}2j\beta_j\cos 2j\xi\cosh 2j\eta\, ,\:\: \tau^{\prime}=\sum_{j=1}^{5}2j\beta_j\sin 2j\xi\sinh 2j\eta }\)
\(\displaystyle{ \beta_1=\frac{1}{2}n-\frac{2}{3}n^2+\frac{37}{96}n^3-\frac{1}{360}n^4-\frac{81}{512}n^5\, ,\:\: \beta_2=\frac{1}{48}n^2+\frac{1}{15}n^3-\frac{437}{1440}n^4+\frac{46}{105}n^5\, ,\:\: }\)
\(\displaystyle{ \beta_3=\frac{17}{480}n^3-\frac{37}{840}n^4-\frac{209}{4480}n^5\, ,\:\: \beta_4=\frac{4\,397}{161\,280}n^4-\frac{11}{504}n^5\, ,\:\: \beta_5=\frac{4\,583}{161\,280}n^5 }\)
\(\displaystyle{ \chi=\sin^{-1}\left(\frac{\sin\xi^{\prime}}{\cosh\eta^{\prime}}\right)\, ,\:\: t_\psi=\frac{1+n}{1-n}\tan\left(\chi+\sum_{j=1}^5 \delta_j\sin 2j\chi\right) }\)
\(\displaystyle{ \delta_1=-\frac{2}{3}n^2-\frac{2}{3}n^3+\frac{4}{9}n^4+\frac{2}{9}n^5\, ,\:\: \delta_2=\frac{1}{3}n^2-\frac{4}{15}n^3-\frac{23}{45}n^4+\frac{68}{45}n^5\, ,\:\: }\)
\(\displaystyle{ \delta_3=\frac{2}{5}n^3-\frac{24}{35}n^4-\frac{46}{35}n^5\, ,\:\: \delta_4=\frac{83}{126}n^4-\frac{80}{63}n^5\, ,\:\: \delta_5=\frac{52}{45}n^5 }\)
\(\displaystyle{ \bar S_{\varphi_0}=\bar A\varphi_0+\frac{m_0 a}{1+n}\sum_{j=1}^{5}A_j\sin 2j\varphi_0\, ,\:\: \bar A=\frac{m_0 a}{1+n}A_0 }\)
\(\displaystyle{ A_0=1+\frac{n^2}{4}+\frac{n^4}{64}\, ,\:\: A_1=-\frac{3}{2}\left(n-\frac{n^3}{8}-\frac{n^5}{64}\right)\, ,\:\: A_2=\frac{15}{16}\left(n^2-\frac{n^4}{4}\right) }\)
\(\displaystyle{ A_3=-\frac{35}{48}\left(n^3-\frac{5}{16}n^5\right)\, ,\:\: A_4=\frac{315}{512}n^4\, ,\:\: A_5=-\frac{693}{1280}n^5 }\)