Basic Navigational Mathematics, Reference Frames and the Earth’s Geometry

Abstract

Navigation algorithms involve various coordinate frames and the transformation of coordinates between them. For example, inertial sensors measure motion with respect to an inertial frame which is resolved in the host platform’s body frame. This information is further transformed to a navigation frame. A GPS receiver initially estimates the position and velocity of the satellite in an inertial orbital frame. Since the user wants the navigational information with respect to the Earth, the satellite’s position and velocity are transformed to an appropriate Earth-fixed frame. Since measured quantities are required to be transformed between various reference frames during the solution of navigation equations, it is important to know about the reference frames and the transformation of coordinates between them. But first we will review some of the basic mathematical techniques.

Appendices

Appendix A

2.1.1 Derivation of Meridian Radius and Normal Radius

For Earth ellipsoids, every meridian is an ellipse with equatorial radius $a$ (called the semimajor axis) and polar radius $b$ (called the semiminor axis). Figure 2.16 shows a meridian cross section of one such ellipse (Rogers 2007).

Fig. 2.16

A meridian cross section of the reference ellipsoid containing the projection of the point of interest P

This ellipse can be described by the equation

$$ \frac{{w^{2} }}{{a^{2} }} + \frac{{z^{2} }}{{b^{2} }} = 1 $$

(2.124)

and the slope of the tangent to point $P$ can be derived by differentiation

$$ \frac{2wdw}{{a^{2} }} + \frac{2zdz}{{b^{2} }} = 0 $$

(2.125)

$$ \frac{dz}{dw} = - \frac{{b^{2} w}}{{a^{2} z}} $$

(2.126)

An inspection of the Fig. 2.16 shows that the derivative of the curve at point P, which is equal to the slope of the tangent to the curve at that point, is

$$ \frac{dz}{dw} = \tan \left( {\frac{\pi }{2} + \varphi } \right) $$

(2.127)

$$ \frac{dz}{dw} = \frac{{\sin \left( {\frac{\pi }{2} + \varphi } \right)}}{{\cos \left( {\frac{\pi }{2} + \varphi } \right)}} = \frac{\cos \varphi }{ - \sin \varphi } = - \frac{1}{\tan \varphi } $$

(2.128)

Therefore

$$ \frac{1}{\tan \varphi } = \frac{{b^{2} w}}{{a^{2} z}} $$

(2.129)

From the definition of eccentricity, we have

$$ \begin{aligned} & e^{2} = 1 - \frac{{b^{2} }}{{a^{2} }} \\ & \frac{{b^{2} }}{{a^{2} }} = 1 - e^{2} \\ \end{aligned} $$

(2.130)

and Eq. (2.129) becomes

$$ z = w\left( {1 - e^{2} } \right)\tan \varphi $$

(2.131)

The ellipse described by Eq. (2.124) gives

$$ w^{2} = a^{2} \left( {1 - \frac{{z^{2} }}{{b^{2} }}} \right) $$

(2.132)

Substituting the value of $z$ from Eq. (2.131) yields

$$ \begin{aligned} & w^{2} = \left( {a^{2} - \frac{{a^{2} w^{2} \left( {1 - e^{2} } \right)^{2} \tan^{2} \varphi }}{{b^{2} }}} \right) \\ & w^{2} + \frac{{a^{2} w^{2} \left( {1 - e^{2} } \right)^{2} \tan^{2} \varphi }}{{b^{2} }} = a^{2} \\ & w^{2} \left( {\frac{{b^{2} + a^{2} \left( {1 - e^{2} } \right)^{2} \tan^{2} \varphi }}{{b^{2} }}} \right) = a^{2} \\ \end{aligned} $$

$$ \begin{aligned} & w^{2} = \frac{{a^{2} b^{2} }}{{b^{2} + a^{2} \left( {1 - e^{2} } \right)^{2} \tan^{2} \varphi }} \\ & w^{2} = \frac{{a^{2} \left[ {a^{2} \left( {1 - e^{2} } \right)} \right]}}{{a^{2} \left( {1 - e^{2} } \right) + a^{2} \left( {1 - e^{2} } \right)^{2} \tan^{2} \varphi }} \\ & w^{2} = \frac{{a^{2} }}{{1 + \left( {1 - e^{2} } \right)\tan^{2} \varphi }} \\ & w^{2} = \frac{{a^{2} }}{{1 + \left( {1 - e^{2} } \right)\frac{{\sin^{2} \varphi }}{{\cos^{2} \varphi }}}} \\ & w^{2} = \frac{{a^{2} \cos^{2} \varphi }}{{\cos^{2} \varphi + \left( {1 - e^{2} } \right)\sin^{2} \varphi }} \\ & w^{2} = \frac{{a^{2} \cos^{2} \varphi }}{{1 - e^{2} \sin^{2} \varphi }} \\ \end{aligned} $$

(2.133)

$$ w = \frac{a\cos \varphi }{{\left( {1 - e^{2} \sin^{2} \varphi } \right)^{\frac{1}{2}} }} $$

(2.134)

Substituting this expression for w in Eq. (2.131) produces

$$ z = \frac{{a\left( {1 - e^{2} } \right)\sin \varphi }}{{\left( {1 - e^{2} \sin^{2} \varphi } \right)^{{^{\frac{1}{2}} }} }} $$

(2.135)

which will be used later to derive the meridian radius.

It can easily be proved from Fig. 2.16 that

$$ w = R_{N} \cos \varphi $$

(2.136)

From this and Eq. (2.134) we have the expression for the normal radius, $R_{N} ,$ also known as the radius of curvature in the prime vertical

$$ R_{N} = \frac{a}{{\left( {1 - e^{2} \sin^{2} \varphi } \right)^{{^{\frac{1}{2}} }} }} $$

(2.137)

The radius of curvature of an arc of constant longitude is

$$ R_{M} = \frac{{\left[ {1 + \left( \frac{dz}{dw} \right)^{2} } \right]}}{{ \pm \frac{{d^{2} z}}{{dw^{2} }}}}^{\frac{3}{2}} $$

(2.138)

The second derivative of Eq. (2.126) provides

$$ \begin{aligned} & \frac{dz}{dw} = - \frac{{b^{2} w}}{{a^{2} z}} = - \frac{{b^{2} w}}{{a^{2} }}\left( {b^{2} - \frac{{b^{2} w^{2} }}{{a^{2} }}} \right)^{ - 1/2} \\ & \frac{{d^{2} z}}{{dw^{2} }} = - \frac{{b^{2} }}{{a^{2} }}\left( {b^{2} - \frac{{b^{2} w^{2} }}{{a^{2} }}} \right)^{ - 1/2} - \frac{{b^{2} w}}{{a^{2} }}\left( { - \frac{1}{2}} \right)\left( {b^{2} - \frac{{b^{2} w^{2} }}{{a^{2} }}} \right)^{ - 3/2} \left( { - \frac{{b^{2} }}{{a^{2} }}} \right)2w \\ & \frac{{d^{2} z}}{{dw^{2} }} = - \frac{{b^{2} }}{{a^{2} z}} - \frac{{b^{4} w^{2} }}{{a^{4} z^{3} }} \\ & \frac{{d^{2} z}}{{dw^{2} }} = \frac{{ - b^{2} a^{2} z^{2} - b^{4} w^{2} }}{{a^{4} z^{3} }} = \frac{{ - b^{2} a^{2} \left( {b^{2} - \frac{{b^{2} w^{2} }}{{a^{2} }}} \right) - b^{4} w^{2} }}{{a^{4} z^{3} }} \\ & \frac{{d^{2} z}}{{dw^{2} }} = \frac{{ - b^{4} a^{2} + b^{4} w^{2} - b^{4} w^{2} }}{{a^{4} z^{3} }} \\ \end{aligned} $$

which simplifies to

$$ \frac{{d^{2} z}}{{dw^{2} }} = - \frac{{b^{4} }}{{a^{2} z^{3} }} $$

(2.139)

Substituting Eqs. (2.139) and (2.126) into (2.138) yields

$$ R_{M} = \frac{{\left[ {1 + \frac{{b^{4} w^{2} }}{{a^{4} z^{2} }}} \right]}}{{\frac{{b^{4} }}{{a^{2} z^{3} }}}}^{\frac{3}{2}} $$

(2.140)

and since $\frac{{b^{2} }}{{a^{2} }} = 1\,-\,e^{2}$

$$ \begin{aligned} & R_{M} = \frac{{\left[ {1 + \frac{{\left( {1\,-\,e^{2} } \right)^{2} w^{2} }}{{z^{2} }}} \right]}}{{\frac{{b^{2} \left( {1\,-\,e^{2} } \right)}}{{z^{3} }}}}^{\frac{3}{2}} \\ & R_{M} = \frac{{\left[ {\frac{{z^{2} + \left( {1\,-\,e^{2} } \right)^{2} w^{2} }}{{z^{2} }}} \right]}}{{\frac{{b^{2} \left( {1\,-\,e^{2} } \right)}}{{z^{3} }}}}^{\frac{3}{2}} \\ \end{aligned} $$

$$ R_{M} = \frac{{\left[ {z^{2} + \left( {1\,-\,e^{2} } \right)^{2} w^{2} } \right]}}{{b^{2} \left( {1\,-\,e^{2} } \right)}}^{\frac{3}{2}} $$

(2.141)

Substituting the value of $w$ from Eq. (2.134) and $z$ from (2.135) gives

$$ \begin{aligned} & R_{M} = \frac{{\left[ {\frac{{a^{2} \left( {1\,-\,e^{2} } \right)^{2} \sin^{2} \varphi }}{{1\,-\,e^{2} \sin^{2} \varphi }} + \left( {1\,-\,e^{2} } \right)^{2} \frac{{a^{2} \cos^{2} \varphi }}{{1\,-\,e^{2} \sin^{2} \varphi }}} \right]}}{{b^{2} \left( {1\,-\,e^{2} } \right)}}^{\frac{3}{2}} \\ & R_{M} = \frac{{\left[ {\frac{{a^{2} \left( {1\,-\,e^{2} } \right)^{2} \sin^{2} \varphi + \left( {1\,-\,e^{2} } \right)^{2} a^{2} \cos^{2} \varphi }}{{1\,-\,e^{2} \sin^{2} \varphi }}} \right]}}{{b^{2} \left( {1\,-\,e^{2} } \right)}}^{\frac{3}{2}} \\ & R_{M} = \frac{{\left[ {a^{2} \left( {1\,-\,e^{2} } \right)^{2} \left( {\sin^{2} \varphi + \cos^{2} \varphi } \right)} \right]}}{{b^{2} \left( {1\,-\,e^{2} } \right)\left( {1\,-\,e^{2} \sin^{2} \varphi } \right)^{\frac{3}{2}} }}^{\frac{3}{2}} \\ & R_{M} = \frac{{a^{3} \left( {1\,-\,e^{2} } \right)^{3} }}{{b^{2} \left( {1\,-\,e^{2} } \right)\left( {1\,-\,e^{2} \sin^{2} \varphi } \right)^{\frac{3}{2}} }} \\ & R_{M} = \frac{{a\left( {1\,-\,e^{2} } \right)^{3} }}{{\left( {1\,-\,e^{2} } \right)\left( {1\,-\,e^{2} } \right)\left( {1\,-\,e^{2} \sin^{2} \varphi } \right)^{\frac{3}{2}} }} \\ \end{aligned} $$

leading to the meridian radius of curvature

$$ R_{M} = \frac{{a\left( {1 - e^{2} } \right)}}{{\left( {1 - e^{2} \sin^{2} \varphi } \right)^{\frac{3}{2}} }} $$

(2.142)

Appendix B

2.2.1 Derivation of the Conversion Equations From Geodetic to Rectangular Coordinates in the ECEF Frame

Figure 2.17 shows the relationship between geodetic and rectangular coordinates in the ECEF frame.

Fig. 2.17

The relationship between geodetic and rectangular coordinates in the ECEF frame

It is evident that

$$ {\mathbf{r}}_{P} = {\mathbf{r}}_{Q} + h{\mathbf{n}} $$

(2.143)

where

$$ {\mathbf{r}}_{Q} = \left[ {\begin{array}{*{20}c} {R_{N} \cos \varphi \cos \lambda } \\ {R_{N} \cos \varphi \sin \lambda } \\ {R_{N} \sin \varphi - R_{N} e^{2} \sin \varphi } \\ \end{array} } \right] = R_{N} \left[ {\begin{array}{*{20}c} {\cos \varphi \cos \lambda } \\ {\cos \varphi \sin \lambda } \\ {\sin \varphi - R_{N} e^{2} \sin \varphi } \\ \end{array} } \right] $$

(2.144)

Also, the unit vector along the ellipsoidal normal is

$$ {\mathbf{n}} = \left[ {\begin{array}{*{20}c} {\cos \varphi \cos \lambda } \\ {\cos \varphi \sin \lambda } \\ {\sin \varphi } \\ \end{array} } \right] $$

(2.145)

Substituting Eqs. (2.144) and (2.145) into (2.143) gives

$$ {\mathbf{r}}_{P} = R_{N} \left[ {\begin{array}{*{20}c} {\cos \varphi \cos \lambda } \\ {\cos \varphi \sin \lambda } \\ {\sin \varphi - R_{N} e^{2} \sin \varphi } \\ \end{array} } \right] + h\left[ {\begin{array}{*{20}c} {\cos \varphi \cos \lambda } \\ {\cos \varphi \sin \lambda } \\ {\sin \varphi } \\ \end{array} } \right] $$

(2.146)

$${\mathbf{r}}_{P} = \left[ {\begin{array}{*{20}l} {\left( {R_{N} + h} \right)\cos \varphi \cos \lambda } \\ {\left( {R_{N} + h} \right)\cos \varphi \sin \lambda } \\ {\left\{ {R_{N} \left( {1 - e^{2} } \right) + h} \right\}\sin \varphi } \\ \end{array} } \right] $$

(2.147)

$$ \left[ \begin{gathered} x^{e} \hfill \\ y^{e} \hfill \\ z^{e} \hfill \\ \end{gathered} \right] = \left[ {\begin{array}{*{20}c} {\left( {R_{N} + h} \right)\cos \varphi \cos \lambda } \\ {\left( {R_{N} + h} \right)\cos \varphi \sin \lambda } \\ {\left\{ {R_{N} \left( {1 - e^{2} } \right) + h} \right\}\sin \varphi } \\ \end{array} } \right] $$

(2.148)

Appendix C

2.3.1 Derivation of Closed Form Equations From Rectangular to Geodetic Coordinates in the ECEF Frame

Here we derive a closed form algorithm which uses a series expansion (Schwarz and Wei Jan 1999) to compute e-frame geodetic coordinates directly from e-frame rectangular coordinates.

From Fig. 2.18a it can is evident that, for given rectangular coordinates, the calculation of geocentric coordinates $\left( {\lambda ,\varphi^{\prime},r} \right)$ is simply

$$ {\mathbf{r}} = \sqrt {\left( {x^{e} } \right)^{2} + \left( {y^{e} } \right)^{2} + \left( {z^{e} } \right)^{2} } $$

(2.149)

$$ \lambda = \tan^{ - 1} \left( {\frac{{y^{e} }}{{x^{e} }}} \right) $$

(2.150)

$$ \varphi^{\prime} = \sin^{ - 1} \left( {\frac{{z^{e} }}{r}} \right) $$

(2.151)

Fig. 2.18

Reference diagram for conversion of rectangular coordinates to geodetic coordinates in the ECEF frame through a closed-form method

From the triangle POC (elaborated in Fig. 2.18b) it is apparent that the difference between the geocentric latitude $\varphi^{\prime}$ and the geodetic latitude $\varphi$ is

$$ \begin{aligned} & D + \varphi^{\prime} + \frac{\pi }{2} + \left( {\frac{\pi }{2} - \varphi } \right) = \pi \\ & D + \varphi^{\prime} + \pi - \varphi = \pi \\ & D = \varphi - \varphi^{\prime} \\ \end{aligned} $$

(2.152)

where $D$ is the angle between the ellipsoidal normal at $Q$ and the normal to the sphere at $P.$

Applying the law of sines to the triangle in Fig. 2.18b provides

$$ \begin{aligned} & \frac{\sin D}{{R_{N} e^{2} \sin \varphi }} = \frac{{\sin \left( {\frac{\pi }{2} - \varphi } \right)}}{r} \\ & \frac{\sin D}{{R_{N} e^{2} \sin \varphi }} = \frac{\cos \varphi }{r} \\ & \sin D = \frac{{R_{N} e^{2} \sin \varphi \cos \varphi }}{r} \\ & D = \sin^{ - 1} \left( {\frac{{R_{N} e^{2} \sin \varphi \cos \varphi }}{r}} \right) \\ \end{aligned} $$

(2.153)

$$ D = \sin^{ - 1} \left( {\frac{{R_{N} e^{2} \frac{1}{2}\sin 2\varphi }}{r}} \right) $$

(2.154)

Substituting the definition of the normal radius $R_{N}$ given in Eq. (2.137)

$$ R_{N} = \frac{a}{{\left( {1 - e^{2} \sin^{2} \varphi } \right)^{1/2} }} $$

(2.155)

into Eq. (2.154) gives

$$ D = \sin^{ - 1} \left( {\frac{{\frac{a}{{\left( {1 - e^{2} \sin^{2} \varphi } \right)^{1/2} }}e^{2} \frac{1}{2}\sin 2\varphi }}{r}} \right) $$

(2.156)

$$ D = \sin^{ - 1} \left( {\frac{k\sin 2\varphi }{{\left( {1 - e^{2} \sin^{2} \varphi } \right)^{1/2} }}} \right) $$

(2.157)

where

$$ k = \frac{{e^{2} a}}{2} $$

(2.158)

To achieve a first approximation, if $\varphi = \varphi^{\prime}$ then $D$ can be computed by using the geocentric latitude

$$ D_{c} = \sin^{ - 1} \left( {\frac{{k\sin 2\varphi^{\prime}}}{{\left( {1 - e^{2} \sin^{2} \varphi^{\prime}} \right)^{1/2} }}} \right) $$

(2.159)

Expanding Eq. (2.157) for this approximation, $D_{c} ,$ gives

$$ D = D\left( {\varphi = \varphi^{\prime}} \right) + \left. {D^{\prime}\left( \varphi \right)} \right|_{{\varphi = \varphi^{\prime}}} \left( {\varphi - \varphi^{\prime}} \right) + \left. {D^{\prime\prime}\left( \varphi \right)} \right|_{{\varphi = \varphi^{\prime}}} \frac{{\left( {\varphi - \varphi^{\prime}} \right)^{2} }}{2!} + \left. {D^{\prime\prime\prime}\left( \varphi \right)} \right|_{{\varphi = \varphi^{\prime}}} \frac{{\left( {\varphi - \varphi^{\prime}} \right)^{3} }}{3!} + \ldots $$

(2.160)

$$ D = D_{c} + D^{\prime}\left( {\varphi^{\prime}} \right)D + \frac{1}{2!}D^{\prime\prime}\left( {\varphi^{\prime}} \right)D^{2} + \frac{1}{3!}D^{\prime\prime\prime}\left( {\varphi^{\prime}} \right)D^{3} + \ldots $$

(2.161)

For a very small value of $D$ (less than 0.005), it can be assumed that $R_{N} \simeq r,$ and hence

$$ D = \frac{{e^{2} }}{2}\sin 2\varphi $$

(2.162)

So for this situation, let $k = \frac{{e^{2} }}{2}$ so that

$$ D = k\sin 2\varphi $$

(2.163)

The series used above can therefore be truncated after the fourth-order term and be considered as a polynomial equation. Solving this polynomial provides

$$ D = \frac{{D_{c} }}{{1 - 2k\cos 2\varphi^{\prime} + 2k^{2} \sin^{2} \varphi^{\prime}}} $$

(2.164)

The geodetic latitude can be given as

$$ \varphi = \varphi^{\prime} + D = \sin^{ - 1} \frac{{z^{e} }}{r} + D $$

(2.165)

The geodetic longitude is the same as shown earlier

$$ \lambda = \tan^{ - 1} \left( {\frac{{y^{e} }}{{x^{e} }}} \right) $$

(2.166)

The ellipsoidal height is calculated from the triangle PRC of Fig. 2.18a

$$ P^{e} = \left( {R_{N} + h} \right)\cos \varphi $$

(2.167)

$$ h = \frac{{P^{e} }}{\cos \varphi } - R_{N} $$

(2.168)

$$ h = \frac{{\sqrt {\left( {x^{e} } \right)^{2} + \left( {y^{e} } \right)^{2} } }}{\cos \varphi } - R_{N} $$

(2.169)

Equations (2.165), (2.166) and (2.169) are therefore closed-form expressions to convert from rectangular coordinates to geodetic coordinates in the e-frame.

Appendix D

2.4.1 Derivation of the Iterative Equations From Rectangular to Geodetic Coordinates in the ECEF Frame

From Eq. (2.148), which relates geodetic and rectangular coordinates in the e-frame, we see that

$$ x^{e} = \left( {R_{N} + h} \right)\cos \varphi \cos \lambda $$

(2.170)

$$ y^{e} = \left( {R_{N} + h} \right)\cos \varphi \sin \lambda $$

(2.171)

$$ z^{e} = \left\{ {R_{N} \left( {1 - e^{2} } \right) + h} \right\}\sin \varphi $$

(2.172)

Given the rectangular coordinates and these equations, the geodetic longitude is

$$ \lambda = \tan^{ - 1} \left( {\frac{{y^{e} }}{{x^{e} }}} \right) $$

(2.173)

Equation (2.169) specifies the relationship for the altitude as

$$ h = \frac{{\sqrt {\left( {x^{e} } \right)^{2} + \left( {y^{e} } \right)^{2} } }}{\cos \varphi } - R_{N} $$

(2.174)

Also, from Eqs. (2.170) and (2.171) it can be shown that

$$ \left( {x^{e} } \right)^{2} + \left( {y^{e} } \right)^{2} = \left( {R_{N} + h} \right)^{2} \cos \varphi^{2} \left[ {\cos^{2} \lambda + \sin^{2} \lambda } \right] $$

(2.175)

$$ \sqrt {\left( {x^{e} } \right)^{2} + \left( {y^{e} } \right)^{2} } = \left( {R_{N} + h} \right)\cos \varphi $$

(2.176)

And finally, dividing Eq. (2.172) by Eq. (2.176) yields

$$ \frac{{z^{e} }}{{\sqrt {\left( {x^{e} } \right)^{2} + \left( {y^{e} } \right)^{2} } }} = \frac{{\left[ {R_{N} \left( {1 - e^{2} } \right) + h} \right]}}{{\left( {R_{N} + h} \right)}}\tan \varphi $$

$$ \varphi = \tan^{ - 1} \left\{ {\frac{{z^{e} \left( {R_{N} + h} \right)}}{{\left[ {R_{N} \left( {1 - e^{2} } \right) + h} \right]\sqrt {\left( {x^{e} } \right)^{2} + \left( {y^{e} } \right)^{2} } }}} \right\} $$

(2.177)