ECEF From Curvilinear Conversion

The meridian (or the north-south motion) radius of curvature, \(R_{N}\), varies with latitude and is smallest at the equator, where the geocentric radius is the largest, and largest at the poles:

\[ R_{N}(L) = \frac{R_0 (1 - e^2)}{(1 - e^2 \text{sin}^2(L))^{3/2}}, \]

where \(L\) is the geodetic latitude in radians, \(e\) is the eccentricity of the reference ellipsoid (WGS 84 - 0.0818191908425), \(R_0\) is the equatorial radius (WGS 84 - 6,378,137.0m).

The radius of curvature for east-west motion, \(R_E\) is the vertical plane perpendicular to the meridian plane and is not the plane of constant latitude. It varies with latitude and is smallest at the equator:

\[ R_{E}(L) = \frac{R_0}{\sqrt{1 - e^2 \text{sin}^2(L)}}. \]

The Cartesian ECEF position can be obtained from curvilinear position by:

\[ \begin{align} \mathbf{r}^{e}_{eb} = \left[ \begin{array}{c} x^{e}_{eb} \\ y^{e}_{eb} \\ z^{e}_{eb} \end{array} \right] = \left[ \begin{array}{c} \left( R_E (L_b) + h_b \right) \text{cos}L_b \text{cos} \lambda_b \\ \left( R_E (L_b) + h_b \right) \text{cos}L_b \text{sin} \lambda_b \\ \left[ (1 - e^2) R_E (L_b) + h_b \right] \text{sin} L_b \end{array} \right], \end{align} \]

where \(h_b\) is the geodetic height or altitude (distance from a body to the ellipsoid surface along the normal to that ellipsoid).

Curvilinear From ECEF Conversion

Conversion from ECEF position to LLA is given by:

\[ \begin{align} \text{tan}L_b &= \frac{z^{e}_{eb} \left[R_E(L_b) + h_b \right]}{\sqrt{(x^e_{eb})^2 + (y^e_{eb})^2} \left[(1 - e^2)R_E(L_b) + h_b \right]} \\ \text{tan}\lambda_b &= \frac{y^e_{eb}}{x^{e}_{eb}} \\ h_b &= \frac{\sqrt{(x^e_{eb})^2 + (y^{e}_{eb})^2}}{\text{cos} L_b} - R_E(L_b), \end{align} \]

where a four-quadrant arctangent function must be used for longitude. Also, since \(R_E\) is a function of latitude, the latitude and heigh must be solved iteratively. The approximate closed-form latitude solution (accurate to within 1cm for positions close to the Earth's surface) is given by:

\[ \text{tan}L_b \approx \frac{z^e_{eb} \sqrt{1 - e^2} + e^2 R_0 sin^3 \zeta_b}{\sqrt{1 - e^2} (\sqrt{(x^e_{eb})^2 + (y^e_{eb})^2} - e^2 R_0 \text{cos}^3 \zeta_b)}, \]

where

\[ \text{tan}\zeta_b = \frac{z^e_{eb}}{\sqrt{1-e^2} \sqrt{(x^e_{eb})^2 + (y^e_{eb})^2}}. \]

\(1^o\) of longitude is about 110 km (60 nautical miles) at the equator, and 80 km at \(45^o\) latitude.