A complete closed-form method for transformation from Cartesian to geodetic coordinates

Abstract

By introducing the auxiliary variable with respect to the reduced latitude, a new closed-form method for transforming Cartesian to geodetic coordinates has been proposed based on the solution of a special constructed unary quartic equation. The algorithm comes with rigorous and concise procedure of root-finding. Moreover, through theoretical analysis, different approaches with respective pros and cons to determine the geodetic latitude and height have been explored. Besides fast computation, numerical experiments covering the magnitude of the geodetic height from $- 6.33 \times 10^{6} {\text{m}}$ to $10^{10} {\text{m}}$ have also shown that the new method can be operational with high precision at almost any point including the region near or at the pole, the equator and the center of the reference ellipsoid. Considering the accuracy, efficiency and adaptability simultaneously, it is prospective to be applied into computation and inspection on critical occasions in comparison to existing methods.

Appendices

Appendix 1: Steps of the new algorithm

Step 1: Prepare the initialized constants and variables

$$ \left\{ \begin{gathered} l = \left( {ae^{2} } \right)^{2} \hfill \\ m = W^{2} = X^{2} + Y^{2} \hfill \\ n = Z^{2} \hfill \\ n_{c} = \left( {1 - e^{2} } \right)Z^{2} \hfill \\ p = m + n_{c} - l \hfill \\ q = 27mn_{c} l \hfill \\ \end{gathered} \right. $$

Step 2: Distinguish the applicable zone

If $p^{3} + q\geqslant 0,$

$$ t = p + \sqrt[3]{{\left( {\sqrt {p^{3} + q} + \sqrt q } \right)^{2} }} + \sqrt[3]{{\left( {\sqrt {p^{3} + q} - \sqrt q } \right)^{2} }} $$

And unless near the astroid, it can be accelerated with only one cubic root operation as

$$ t = p + \sqrt[3]{{p^{3} + 2q + 2\sqrt {q\left( {q + p^{3} } \right)} }} + \frac{{p^{2} }}{{\sqrt[3]{{p^{3} + 2q + 2\sqrt {q\left( {q + p^{3} } \right)} }}}} $$

Otherwise,

$$ t = \frac{{\sqrt { - \frac{q}{p}} }}{{\cos \left( {\frac{{\arccos \sqrt { - \frac{q}{{p^{3} }}} }}{3}} \right)}} $$

To reduce operations of one division and one square root, use

$$ t = - p\frac{{\sqrt { - \frac{q}{{p^{3} }}} }}{{\cos \left( {\frac{{\arccos \sqrt { - \frac{q}{{p^{3} }}} }}{3}} \right)}} $$

Step 3: Compute the intermediate variables

$$ \left\{ \begin{gathered} u_{m} = \sqrt {36ml + t^{2} } \hfill \\ u_{{n_{c} }} = \sqrt {36n_{c} l + t^{2} } \hfill \\ v = u_{m} + u_{{n_{c} }} \hfill \\ w = 2t + 6l + v \hfill \\ I = kW = \frac{{2\left( {t + u_{{n_{c} }} } \right)}}{{w + \sqrt {6l\left[ {w + v + 6\left( {m + n_{c} } \right)} \right]} }} \hfill \\ S = \sqrt {I^{2} + n} \hfill \\ \end{gathered} \right. $$

Step 4: Compute geodetic coordinates $\left( {\lambda ,\;\varphi ,\;h} \right)$

If $t > 0$ or $n > 0,$

$$ \left\{ \begin{gathered} \lambda = {\text{msign}}\left( Y \right) \cdot \left( {\frac{\pi }{2} - 2\arctan \frac{X}{W + \left| Y \right|}} \right) \hfill \\ \varphi = 2\arctan \frac{Z}{I + S} \hfill \\ h = \frac{{WI + n - a\sqrt {I^{2} + n_{c} } }}{S} \hfill \\ \end{gathered} \right. $$

where the function $y = {\text{msign}}\left( x \right)$ returns $1$ when the argument is greater than or equal to zero, and $- 1$ when the argument is negative.

Unless on a sphere ($e^{2} \ne 0$), the computation for $h$ can be accelerated with one square root operation reduced via

$$ h = \frac{{k + e^{2} - 1}}{{e^{2} k}}S $$

Step 5: Compute geodetic coordinates $\left( {\lambda ,\;\varphi ,\;h} \right)$ on Singular Line ($t = n = 0$)

$$ \left\{ \begin{gathered} \lambda = {\text{msign}}\left( Y \right) \cdot \left( {\frac{\pi }{2} - 2\arctan \frac{X}{W + \left| Y \right|}} \right) \hfill \\ \varphi = \pm 2\arctan \left( {\frac{{\sqrt {l - m} }}{{\sqrt {l - e^{2} m} + \sqrt {\left( {1 - e^{2} } \right)m} }}} \right) \hfill \\ h = - \sqrt {1 - e^{2} } \sqrt {a^{2} - m/e^{2} } \hfill \\ \end{gathered} \right. $$

Appendix 2: Particular deficiency of Vermeille’s algorithm

In the computational procedure of Vermeille (2011), for points $\left( {0, \pm {{ae^{2} } \mathord{\left/ {\vphantom {{ae^{2} } {\sqrt {1 - e^{2} } }}} \right. \kern-0pt} {\sqrt {1 - e^{2} } }}} \right)$, which are two vertexes of the astroid resulting in $r=p=8r^{3} + e^{4} pq=0$, the equation chosen to compute $u$ should be the one applied inside the evolute. To ensure the consistent correctness, ${\text{atan2}}\left( {\sqrt {e^{4} pq} ,\;\sqrt { - 8r^{3} - e^{4} pq} + \sqrt { - 8r^{3} } } \right)$ (the sequence of arguments may be opposite in some programming environment) ought to be adopted rather than ${\text{atan}}\left( {{{\sqrt {e^{4} pq} } \mathord{\left/ {\vphantom {{\sqrt {e^{4} pq} } {\left( {\sqrt { - 8r^{3} - e^{4} pq} + \sqrt { - 8r^{3} } } \right)}}} \right. \kern-0pt} {\left( {\sqrt { - 8r^{3} - e^{4} pq} + \sqrt { - 8r^{3} } } \right)}}} \right)$, which will make the operating speed of computing other points decreased meaninglessly. Crucially, the result of arctangent function with two parameters both equal to zero is still indeterminate in some programming environment such as ${\text{ArcTan}}\left[ {0,\;0} \right]$ in Mathematica.

Appendix 3: Usability problems of the algorithm by Guo and Shen

The main variable in the closed-form method by Guo and Shen (2023) is also relative to the reduced latitude, but some hidden troubles should be noticed except that the speed is slower than the method by Vermeille (2011) which was confirmed in their paper (2023, pp. 173 ~ 174). Primarily, the first eccentricity squared $e^{2}$ at denominators of some equations (2023, Eqs. (A10) and (A11)) makes that algorithm lose effect when the geodetic model is a sphere ($e = 0$). More seriously, significant digit loss will happen at the time of the subtraction between two close values (2023, e.g., the first and fourth items of equation group (A11)), which can be easily checked by two numerical examples:

(a) when φ = 89.999992°, h = 3.02 × 10⁷m, the error of recovered latitude will reach 1.58 × 10⁻⁹rad (about 0.003″). This makes about 0.06m bias in the equatorial direction;

(b) when φ = 47, h = − 6.34681246356 × 10⁶m, it will increase to 1.81 × 10⁻⁶rad (about 0.37″).

These degrees of errors cannot be accepted, and the method should be used prudentially.