Editing Geographic coordinate conversion (section)

==== Ferrari's solution ====
The quartic equation of <math>\kappa</math>, derived from the above, <!--for this transformation--> can be solved by [[Quartic equation#Ferrari.27s solution|Ferrari's solution]]<ref>{{cite journal|first1=H. |last1=Vermeille, H.|title=Direct Transformation from Geocentric to Geodetic Coordinates|journal= J. Geod.|volume=76|number=8|pages=451–454
|year= 2002|doi=10.1007/s00190-002-0273-6|s2cid=120075409 }}</ref><ref>{{cite journal|first1=Laureano|last1=Gonzalez-Vega|first2=Irene|last2=PoloBlanco|title=A symbolic analysis of Vermeille and Borkowski polynomials for transforming 3D Cartesian to geodetic coordinates|journal=J. Geod.|volume=83|number=11|pages=1071–1081|doi=10.1007/s00190-009-0325-2|year=2009|bibcode=2009JGeod..83.1071G |s2cid=120864969 }}</ref> to yield:
: <math>
\begin{align}
   \zeta &= \left(1 - e^2\right)\frac{z^2}{a^2} ,\\[4pt]
    \rho &= \frac{1}{6}\left(\frac{p^2}{a^2} + \zeta - e^4\right) ,\\[4pt]
       s &= \frac{e^4 \zeta p^2}{4\rho^3 a^2} ,\\[4pt]
       t &= \sqrt[3]{1 + s + \sqrt{s(s + 2)}} ,\\[4pt]
       u &= \rho \left(t + 1 + \frac{1}{t}\right) ,\\[4pt]
       v &= \sqrt{u^2 + e^4 \zeta} ,\\[4pt]
       w &= e^2 \frac{u + v - \zeta}{2v} ,\\[4pt]
  \kappa &= 1 + e^2 \frac{\sqrt{u + v + w^2} + w}{u + v}.
\end{align}
</math>

===== The application of Ferrari's solution =====
A number of techniques and algorithms are available but the most accurate, according to Zhu,<ref>{{cite journal|first1=J.|last1=Zhu|title=Conversion of Earth-centered Earth-fixed coordinates to geodetic coordinates|journal=IEEE Transactions on Aerospace and Electronic Systems|volume=30|issue=3|year=1994|pages=957–961|doi=10.1109/7.303772|bibcode=1994ITAES..30..957Z }}</ref> is the following procedure established by Heikkinen,<ref>{{cite journal|first1=M.|last1=Heikkinen|title=Geschlossene formeln zur berechnung räumlicher geodätischer koordinaten aus rechtwinkligen koordinaten.|journal=Z. Vermess.|volume=107|year=1982|pages=207–211|language=de}}</ref> as cited by Zhu. This overlaps with above. It is assumed that geodetic parameters <math>\{a,\, b,\, e\}</math> are known

: <math>\begin{align}
        a &= 6378137.0 \text{     m. Earth Equatorial Radius} \\[3pt]
        b &= 6356752.3142 \text{      m. Earth Polar Radius} \\[3pt]
      e^2 &= \frac{a^2-b^2}{a^2} \\[3pt]
     e'^2 &= \frac{a^2 - b^2}{b^2} \\[3pt]
        p &= \sqrt{X^2 + Y^2} \\[3pt]
        F &= 54b^2 Z^2 \\[3pt]
        G &= p^2 + \left(1 - e^2\right)Z^2 - e^2\left(a^2 - b^2\right) \\[3pt]
        c &= \frac{e^4 Fp^2}{G^3} \\[3pt]
        s &= \sqrt[3]{1 + c + \sqrt{c^2 + 2c}} \\[3pt]
        k &= s + 1 + \frac{1}{s}\\[3pt]
        P &= \frac{F}{3 k^2 G^2} \\[3pt]
        Q &= \sqrt{1 + 2e^4 P} \\[3pt]
      r_0 &= \frac{-Pe^2 p}{1 + Q} + \sqrt{\frac{1}{2} a^2\left(1 + \frac{1}{Q}\right) - \frac{P\left(1 - e^2\right)Z^2}{Q(1 + Q)} - \frac{1}{2}Pp^2} \\[3pt]
        U &= \sqrt{\left(p - e^2 r_0\right)^2 + Z^2} \\[3pt]
        V &= \sqrt{\left(p - e^2 r_0\right)^2 + \left(1 - e^2\right)Z^2} \\[3pt]
      z_0 &= \frac{b^2 Z}{aV} \\[3pt]
        h &= U\left(1 - \frac{b^2}{aV}\right) \\[3pt]
     \phi &= \arctan\left[\frac{Z + e'^2 z_0}{p}\right] \\[3pt]
  \lambda &= \operatorname{arctan2}[Y,\, X]
\end{align}</math>

Note: [[atan2|arctan2]][Y, X] is the four-quadrant inverse tangent function.