経緯度
( d x d y d z ) = ( − sin  λ − sin  φ cos  λ cos  φ cos  λ cos  λ − sin  φ sin  λ cos  φ sin  λ 0 cos  φ sin  φ ) ( d E d N d U ) , ( d E d N d U ) = ( ( N ( φ ) + h ) cos  φ 0 0 0 M ( φ ) + h 0 0 0 1 ) ( d λ d φ d h ) , N ( φ ) ≜ a 1 − e 2 sin 2  φ , M ( φ ) ≜ a ( 1 − e 2 ) ( 1 − e 2 sin 2  φ ) 3 / 2 . {\displaystyle {\begin{aligned}{\begin{pmatrix}dx\\dy\\dz\\\end{pmatrix}}&={\begin{pmatrix}-\sin \lambda &-\sin \phi \cos \lambda &\cos \phi \cos \lambda \\\cos \lambda &-\sin \phi \sin \lambda &\cos \phi \sin \lambda \\0&\cos \phi &\sin \phi \\\end{pmatrix}}{\begin{pmatrix}dE\\dN\\dU\\\end{pmatrix}},\\{\begin{pmatrix}dE\\dN\\dU\\\end{pmatrix}}&={\begin{pmatrix}\left(N(\phi )+h\right)\cos \phi &0&0\\0&M(\phi )+h&0\\0&0&1\\\end{pmatrix}}{\begin{pmatrix}d\lambda \\d\phi \\dh\\\end{pmatrix}},\\N(\phi )&\triangleq {\frac {a}{\sqrt {1-e^{2}\sin ^{2}\phi }}},\\M(\phi )&\triangleq {\frac {a(1-e^{2})}{\left(1-e^{2}\sin ^{2}\phi \right)^{3/2}}}.\end{aligned}}}