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# Transformations between ECEF and ENU coordinates

Fundamentals
Title Transformations between ECEF and ENU coordinates
Author(s) J. Sanz Subirana, J.M. Juan Zornoza and M. Hernández-Pajares, Technical University of Catalonia, Spain.
Year of Publication 2011

The relation between the local East, North, Up (ENU) coordinates and the $(x,y,z)$ Earth Centred Earth Fixed (ECEF) coordinates is illustrated in the next figure:

Figure 2:: Transformations between ENU and ECEF coordinates.

From the figure 1 it follows that the ENU coordinates can be transformed to the $(x,y,z)$ ECEF by two rotations:

1. A clockwise rotation over east-axis by an angle $90-\varphi$ to align the up-axis with the $z$-axis. That is ${\mathbf R}_1[-(\pi/2-\varphi)]$.

2. A clockwise rotation over the $z$-axis by and angle $90+\lambda$ to align the east-axis with the $x$-axis. That is ${\mathbf R}_3[-(\pi/2+\lambda)]$.

That is:

$\left [ \begin{array}{l} x\\ y\\ z\\ \end{array} \right ] = {\mathbf R}_3[-(\pi/2+\lambda)]\,{\mathbf R}_1[-(\pi/2-\varphi)] \left [ \begin{array}{l} E\\ N\\ U\\ \end{array} \right ] \qquad \mbox{(1)}$

where, according to the expressions (2) (see Transformation between Terrestrial Frames)

$\begin{array}{l} \mathbb{\mathbf R}_1[\theta]=\left [ \begin{array}{ccc} 1 & 0 & 0\\ 0 & \cos \theta & \sin \theta \\ 0 & -\sin \theta & \cos \theta \\ \end{array} \right ] \;;\;\; \mathbb{\mathbf R}_2[\theta]=\left [ \begin{array}{ccc} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0\\ \sin \theta &0 & \cos \theta \\ \end{array} \right ]\\ \\ \mathbb{\mathbf R}_3[\theta]=\left [ \begin{array}{ccc} \cos \theta & \sin \theta &0\\ -\sin \theta & \cos \theta & 0\\ 0 & 0 & 1\\ \end{array} \right ] \end{array} \qquad \mbox{(2)}$

yields:

${\mathbf R}_3[-(\pi/2+\lambda)]\,{\mathbf R}_1[-(\pi/2-\varphi)]= \left ( \begin{array}{ccc} -\sin \lambda & -\cos \lambda \sin \varphi &\cos \lambda \cos \varphi\\ \cos \lambda & -\sin \lambda \sin \varphi & \sin \lambda \cos \varphi\\ 0 & \cos \varphi & \sin \varphi\\ \end{array} \right ) \qquad \mbox{(3)}$

The unit vectors in local East, North and Up directions as expressed in ECEF cartesian coordinates are given by the columns of matrix (3). That is:

$\begin{array}{l} \hat{\mathbf e}=\left ( -\sin \lambda \,,\,\cos \lambda\,,\, 0 \right )\\ \hat{\mathbf n}=\left ( - \cos \lambda \sin \varphi \,,\,- \sin \lambda \sin \varphi\,,\, \cos \varphi \right)\\ \hat{\mathbf u}=\left ( \cos \lambda \cos \varphi \,,\,\sin \lambda\ \cos \varphi\,,\, \sin \varphi \right) \qquad \mbox{(4)} \end{array}$

Note: If $(\lambda,\varphi)$ are ellipsoidal coordinates, thence, the vector $\hat{\mathbf u}$ is orthogonal to the tangent plane to the ellipsoid, which is defined by $(\hat{\mathbf e}, \hat{\mathbf n})$. If $(\lambda,\varphi)$ are taken as the spherical latitude and longitude, thence, the vector $\hat{\mathbf u}$ is in the radial direction and $(\hat{\mathbf e}, \hat{\mathbf n})$ defines the tangent plane to the sphere.

## From ECEF to ENU coordinates

Taking into account the properties of the rotation matrices ${\mathbf R}_i(\alpha)$,i.e., ${\mathbf R}_i^{-1}(\alpha)= {\mathbf R}_i(-\alpha)={\mathbf R}_i^T(\alpha)$, thence, the inverse transformation of (1) is given by:

$\left [ \begin{array}{l} E\\ N\\ U\\ \end{array} \right ] = {\mathbf R}_1[\pi/2-\varphi]\,{\mathbf R}_3[\pi/2+\lambda] \left [ \begin{array}{l} x\\ y\\ z\\ \end{array} \right ] \qquad \mbox{(5)}$

where the transformation matrix of (5) is the transpose of matrix (3):

${\mathbf R}_1[\pi/2-\varphi]\,{\mathbf R}_3[\pi/2+\lambda]= \left ( \begin{array}{ccc} -\sin \lambda & \cos \lambda &0\\ - \cos \lambda \sin \varphi & -\sin \lambda \sin \varphi & \cos \varphi\\ \cos \lambda \cos \varphi & \sin \lambda \cos \varphi & \sin \varphi\\ \end{array} \right ) \qquad \mbox{(6)}$

The unit vectors in the ECEF $\hat{\mathbf x}$, $\hat{\mathbf y}$ and $\hat{\mathbf z}$ directions, as expressed in ENU coordinates, are given by the columns of matrix (6). That is:

$\begin{array}{l} \hat{\mathbf x}=\left ( -\sin \lambda \,,\,-\cos \lambda \sin \varphi\,,\, \cos \lambda \cos \varphi \right )\\ \hat{\mathbf y}=\left (\cos \lambda\,,\,- \sin \lambda \sin \varphi\,,\, \sin \lambda \cos \varphi \right)\\ \hat{\mathbf z}=\left ( 0 \,,\, \cos \varphi \,,\,\sin \varphi \right) \end{array} \qquad \mbox{(7)}$

## Elevation and azimuth computation

Given the line of sight unit vector

$\hat {\boldsymbol \rho}=\displaystyle \frac{{\mathbf r}^{sat}-{\mathbf r}_{rcv}}{\| {\mathbf r}^{sat}-{\mathbf r}_{rcv} \|} \qquad \mbox{(8)}$

where ${\mathbf r}^{sat}$ and ${\mathbf r}_{rcv}$ are the geocentric position of the satellite and receiver, respectively, the elevation and azimuth in the local system coordinates (ENU), defined by the unit vectors $\hat{\mathbf e}$, $\hat{\mathbf n}$ and $\hat{\mathbf u}$ can be computed from (see figure 2):

$\begin{array}{l} \hat {\boldsymbol \rho}\cdot \hat{\mathbf e} =\cos E \sin A\\ \hat {\boldsymbol \rho}\cdot \hat{\mathbf n}=\cos E \cos A\\ \hat {\boldsymbol \rho}\cdot \hat{\mathbf u} = \sin E \end{array} \qquad \mbox{(9)}$

Thence the elevation and azimuth of satellite in the local coordinates system are given by:

$E=\arcsin(\hat {\boldsymbol \rho}\cdot \hat{\mathbf u}) \qquad \mbox{(10)}$
$A=\arctan \left (\frac{\hat {\boldsymbol \rho}\cdot \hat{\mathbf e}}{\hat {\boldsymbol \rho}\cdot \hat{\mathbf n}}\right ) \qquad \mbox{(11)}$

Figure 2:: Local coordinate frame showing the elevation ($E$) and azimuth ($A$).

Note: If $(\lambda,\varphi)$ are ellipsoidal coordinates, thence, the vector $\hat{\mathbf u}$ is orthogonal to the tangent plane to the ellipsoid, which is defined by $(\hat{\mathbf e}, \hat{\mathbf n})$. If $(\lambda,\varphi)$ are taken as the spherical latitude and longitude, thence, the vector $\hat{\mathbf u}$ is in the radial direction and $(\hat{\mathbf e}, \hat{\mathbf n})$ defines the tangent plane to the sphere.