Editing Geographic coordinate conversion (section)

=== From geodetic to ECEF coordinates ===
[[Image:Geodetic latitude and the length of Normal.svg|thumb|The length PQ, called the ''prime vertical radius'', is <math>N(\phi)</math>. The length IQ is equal to <math>\, e^2 N(\phi) </math>. <math>R = (X,\, Y,\, Z)</math>.]]

[[Geodetic coordinates]] (latitude <math>\ \phi</math>, longitude <math>\ \lambda</math>, height <math>h</math>) can be converted into [[ECEF]] coordinates using the following equation:<ref name="gps-chap10">{{cite book|title=GPS - theory and practice|author1=B. Hofmann-Wellenhof |author2=H. Lichtenegger |author3=J. Collins |isbn=3-211-82839-7|page=282|others=Section 10.2.1|year=1997 }}</ref>

: <math>\begin{align}
      X & = \left( N(\phi) + h\right)\cos{\phi}\cos{\lambda} \\
      Y & = \left( N(\phi) + h\right)\cos{\phi}\sin{\lambda} \\
      Z & = \left( \frac{b^2}{a^2} N(\phi) + h\right)\sin{\phi} \\
        & = \left( (1 - e^2)       N(\phi) + h\right)\sin{\phi} \\
        & = \left( (1 - f)^2       N(\phi) + h\right)\sin{\phi}
\end{align}</math>

where
: <math>
  N(\phi) = \frac{a^2}{\sqrt{a^2 \cos^2 \phi + b^2 \sin^2 \phi }}
          = \frac{a}{\sqrt{1 - e^2\sin^2\phi}} = \frac{a}{\sqrt{1 - \frac{e^2}{1 + \cot^2 \phi}}},
</math>

and <math>a</math> and <math>b</math> are the equatorial radius ([[semi-major axis]]) and the polar radius ([[semi-minor axis]]), respectively. <math>e^2 = 1 - \frac{b^2}{a^2}</math> is the square of the first numerical eccentricity of the ellipsoid. <math>f = 1 - \frac{b}{a}</math> is the flattening of the ellipsoid. The ''[[prime vertical radius of curvature]]'' <math>\, N(\phi) </math> is the distance from the surface to the Z-axis along the ellipsoid normal.

====Properties====
The following condition holds for the longitude in the same way as in the geocentric coordinates system:
:<math>\frac{X}{\cos\lambda} - \frac{Y}{\sin\lambda} = 0.</math>

And the following holds for the latitude:
:<math>\frac{p}{\cos\phi} - \frac{Z}{\sin\phi} - e^2 N(\phi) = 0,</math>

where <math>p = \sqrt{X^2 + Y^2}</math>, as the parameter <math>h</math> is eliminated by subtracting
:<math>\frac{p}{\cos\phi} = N + h</math>

and
:<math>\frac{Z}{\sin\phi} = \frac{b^2}{a^2}N + h.</math>

The following holds furthermore, derived from dividing above equations:
:<math>\frac{Z}{p} \cot \phi = 1 - \frac{e^2 N}{N + h}.</math>

====Orthogonality====
The [[Orthogonal coordinates|orthogonality]] of the coordinates is confirmed via differentiation:
:<math>\begin{align}
  \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}, \\[3pt]
  \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},
\end{align}</math>
<!--
: <math>\begin{align}
      & \big(dX,\, dY,\, dZ\big) \\[6pt]
    = & \big(-\sin\phi \cos\lambda,\, -\sin\phi \sin\lambda,\, \cos\phi\big) \left(M(\phi) + h\right)\, d\phi \\[6pt]
      &{}+ \big(-\sin\lambda,\, \cos\lambda,\, 0\big)\left(N(\phi) + h\right) \cos\phi\, d\lambda \\[6pt]
      &{}+ \big(\cos\lambda \cos\phi,\, \cos\phi \sin\lambda,\, \sin\phi\big)\, dh,
\end{align}</math>
-->

where
:<math>
 M(\phi) = \frac{a\left(1 - e^2\right)}{\left(1 - e^2 \sin^2 \phi\right)^\frac{3}{2}} = N(\phi) \frac{1 - e^2}{1 - e^2\sin^2\phi}
</math>

(see also "[[Meridian arc#Definition|Meridian arc on the ellipsoid]]").
<!--
The infinitesimal length caused by latitude and longitude is calculated as follows (see also "[[Meridian arc#Meridian distance on the ellipsoid|Meridian arc on the ellipsoid]]"):
: <math>
 ds^2 = \left(\frac{a\left(1 - e^2\right)}{\left(1 - e^2 \sin^2\phi\right)^\frac{3}{2}} + h\right)^2 d\phi^2 + \left(\frac{a}{\sqrt{1 - e^2 \sin^2\phi}} + h\right)^2 \cos^2\phi\, d\lambda^2 .
</math>
-->