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- Lat Long To Ecef
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I am studying pArk Apple sample code, and how it is works. Anyone knows how convert the latitude and longitude to ECEF coordinates, and Covert ECEF to ENU coordinates centered at given lat, lon fun.
| 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. |
| Level | Advanced |
| Year of Publication | 2011 |
The relation between the local East, North, Up (ENU) coordinates and the [math](x,y,z)[/math] Earth Centred Earth Fixed (ECEF) coordinates is illustrated in the next figure:
From the figure 1 it follows that the ENU coordinates can be transformed to the [math](x,y,z)[/math] ECEF by two rotations:
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- 1. A clockwise rotation over east-axis by an angle [math]90-varphi[/math] to align the up-axis with the [math]z[/math]-axis. That is [math]{mathbf R}_1[-(pi/2-varphi)][/math].
- 2. A clockwise rotation over the [math]z[/math]-axis by and angle [math]90+lambda[/math] to align the east-axis with the [math]x[/math]-axis. That is [math]{mathbf R}_3[-(pi/2+lambda)][/math].
That is:
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- [math]left [begin{array}{l}xyzend{array}right ] ={mathbf R}_3[-(pi/2+lambda)],{mathbf R}_1[-(pi/2-varphi)]left [begin{array}{l}ENUend{array}right ]qquad mbox{(1)}[/math]
where, according to the expressions (2) (see Transformation between Terrestrial Frames)
- [math]begin{array}{l}mathbb{mathbf R}_1[theta]=left [begin{array}{ccc}1 & 0 & 00 & 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 & 0sin 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 & 00 & 0 & 1end{array}right ]end{array}qquad mbox{(2)}[/math]
yields:
- [math]{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 varphicos lambda & -sin lambda sin varphi & sin lambda cos varphi0 & cos varphi & sin varphiend{array}right )qquad mbox{(3)}[/math]
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:
- [math]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}[/math]
Note: If [math](lambda,varphi)[/math] are ellipsoidal coordinates, thence, the vector [math]hat{mathbf u}[/math] is orthogonal to the tangent plane to the ellipsoid, which is defined by [math](hat{mathbf e}, hat{mathbf n})[/math]. If [math](lambda,varphi)[/math] are taken as the spherical latitude and longitude, thence, the vector [math]hat{mathbf u}[/math] is in the radial direction and [math](hat{mathbf e}, hat{mathbf n})[/math] defines the tangent plane to the sphere.
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From ECEF to ENU coordinates
Taking into account the properties of the rotation matrices [math]{mathbf R}_i(alpha)[/math],i.e., [math]{mathbf R}_i^{-1}(alpha)= {mathbf R}_i(-alpha)={mathbf R}_i^T(alpha)[/math], thence, the inverse transformation of (1) is given by:
- [math]left [begin{array}{l}ENUend{array}right ] ={mathbf R}_1[pi/2-varphi],{mathbf R}_3[pi/2+lambda]left [begin{array}{l}xyzend{array}right ]qquad mbox{(5)}[/math]
where the transformation matrix of (5) is the transpose of matrix (3):
- [math]{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 varphicos lambda cos varphi & sin lambda cos varphi & sin varphiend{array}right )qquad mbox{(6)}[/math]
The unit vectors in the ECEF [math]hat{mathbf x}[/math], [math]hat{mathbf y}[/math] and [math]hat{mathbf z}[/math] directions, as expressed in ENU coordinates, are given by the columns of matrix (6). That is:
- [math]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)}[/math]
Elevation and azimuth computation
Given the line of sight unit vector
- [math]hat {boldsymbol rho}=displaystyle frac{{mathbf r}^{sat}-{mathbf r}_{rcv}}{| {mathbf r}^{sat}-{mathbf r}_{rcv} |}qquad mbox{(8)}[/math]
where [math]{mathbf r}^{sat}[/math] and [math]{mathbf r}_{rcv}[/math] 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 [math]hat{mathbf e}[/math], [math]hat{mathbf n}[/math] and [math]hat{mathbf u}[/math] can be computed from (see figure 2):
- [math]begin{array}{l}hat {boldsymbol rho}cdot hat{mathbf e} =cos E sin Ahat {boldsymbol rho}cdot hat{mathbf n}=cos E cos Ahat {boldsymbol rho}cdot hat{mathbf u} = sin Eend{array}qquad mbox{(9)}[/math]
Thence the elevation and azimuth of satellite in the local coordinates system are given by:
- [math]E=arcsin(hat {boldsymbol rho}cdot hat{mathbf u})qquad mbox{(10)}[/math]
- [math]A=arctan left (frac{hat {boldsymbol rho}cdot hat{mathbf e}}{hat {boldsymbol rho}cdot hat{mathbf n}}right )qquad mbox{(11)}[/math]
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Note: If [math](lambda,varphi)[/math] are ellipsoidal coordinates, thence, the vector [math]hat{mathbf u}[/math] is orthogonal to the tangent plane to the ellipsoid, which is defined by [math](hat{mathbf e}, hat{mathbf n})[/math]. If [math](lambda,varphi)[/math] are taken as the spherical latitude and longitude, thence, the vector [math]hat{mathbf u}[/math] is in the radial direction and [math](hat{mathbf e}, hat{mathbf n})[/math] defines the tangent plane to the sphere.