Coordinate transformation between ITRS and GCRS

1. Coordinate transformation between ITRS and GCRS

Due to the rotation of the earth, the earth coordinate system is not an inertial coordinate system, and the orbit calculation is based on Newtonian mechanics, so the orbit determination work cannot be carried out in the earth coordinate system. As mentioned above, GCRS is a pretty good quasi-inertial coordinate system, and the orbit determination work is generally carried out in this coordinate system, but the user uses the satellite navigation and positioning system to finally obtain the position and velocity in the earth coordinate system, Therefore, it is also necessary to transform the satellite orbit (satellite position and velocity) obtained in GCRS into the earth coordinate system ITRS (WGS 84).

There is the following conversion relationship between ITRS and GCRS:
$${\left(\begin{array}{l}X\\ Y\\ Z\end{array}\right)}_{\mathrm{GCRS}}=[\mathbf{P}][\mathbf{N}][\mathbf{R}][\mathbf{W}]{\left(\begin{array}{l}X\\ Y\\ Z\end{array}\right)}_{}$$
$${\left(\begin{array}{l}X\\ Y\\ Z\end{array}\right)}_{}=[\mathbf{W}{]}^{-1}[\mathbf{R}{]}^{-1}[\mathbf{N}{]}^{-1}[\mathbf{P}{]}^{-1}{\left(\begin{array}{l}X\\ Y\\ Z\end{array}\right)}_{\mathrm{GCRS}}$$
where, $\left[P\right]$ is the precession matrix;$\left[N\right]$ is the nutation matrix;$\left[R\right]$ is the earth rotation matrix;$\left[W\right]$ is the pole shift matrix. Considering that the IGS has completed the coordinate conversion work, the position and velocity of the satellite center of mass in the ITRS are directly given in the precise ephemeris, while the accuracy of the broadcast ephemeris is limited, some approximate conversion methods are allowed, so the following coordinates In the conversion, we still use the classic conversion method and terms (basically consistent with the methods given in IS-GPS-200D and IS-GPS-705). Rigorous methods with high precision can be found in IAU resolution documents and references such as space geodesy.

(1) Convert GCRS to observation time $t_i$
We know that GCRS is the reference time t $t_0=$The flat celestial coordinate system at$J$$2000.0$$t_i$When the flat celestial coordinate system, just consider $\left[t_0-t_i\right]$ time period precession correction, that is, multiplied by$[\mathbf{P}{]}^{-1}$ matrix is ​​enough.

（2）把 $t_i$The flat celestial coordinate system at that time is transformed into the true celestial coordinate system at the same time.
The observation time $t_i$To convert the flat celestial coordinate system at , to the true celestial coordinate system, we only need to take into account the nutation at this moment, that is, we only need to multiply $[\mathbf{N}{]}^{-1}$ matrix is ​​enough.

（3）把 $t_i$The true celestial coordinate system at time is transformed into the true terrestrial coordinate system at the same moment.
We know that the true celestial coordinate system $The\; x-$ axis is pointing to the true equinox at that moment$\gamma$ , and$The\; X-$ axis points to the intersection of the origin meridian and the equator, and the angle between the two is called Greenwich True Sidereal Time (GAST). The calculation formula is as follows:
$$\begin{array}{rl}\text{ GAST }=& \frac{360^{\circ }}{24^{\mathrm{h}}}(\mathrm{UT}1+6\mathrm{h}41\mathrm{m}50.54841\mathrm{s}+8640184.812866\mathrm{s}\cdot \mathrm{t}+0.093104\mathrm{s}\cdot t^2\\ & -6.2\mathrm{s}\times 10^{-6}\cdot t^3)+D{Ps}_{\cos }(\bar{e}+De\end{array}$$
where, $t$ is 2000.0 \mathrm{J} 2000.0fromJulian century number of$J$$2000.0$$\bar{e}$ is the oblique angle when only precession is considered,$\bar{e}=23^{\circ }26^{\prime }21.448^{\prime \prime }-$ $46.815^{\prime \prime }\cdot t-0.00059^{\prime \prime }\cdot t^2+0.001813^{\prime \prime }\cdot t^3;\Delta \psi$$46.815^{\prime \prime }\cdot t-0.00059^{\prime \prime }\cdot t^2+0.001813^{\prime \prime }\cdot t^3;\Delta \Psi$ is nutation of Huang Jing;$\Delta \epsilon$ is angular nutation;$UT1$ can be obtained according to the UTC and (UTC-UT1) values ​​at the time of observation.

Turn the true celestial coordinate system around $After\; the\; Z$ axis is rotated by the GAST angle, it can be converted to the real earth coordinate system, and the rotation matrix$\mathbf{R}$ 为:
$$\mathbf{R}=\left(\begin{array}{ccc}\cos GAST& \mathrm{SinGAST}& 0\\ -\mathrm{SinGAST}& \cos GAST& 0\\ 0& 0& 1\end{array}\right)$$
（4）把 $t_i$Time's True Earth Coordinate System Conversion to ITRS (WGS 84)

As can be seen from the above figure, only $t_i$The true earth coordinate system at time revolves around $y-$ axis rotation$\left(-X_p\right)$ corner, and then around$x-$ axis rotation$\left(-Y_p\right)$ angle, we can put the straight earth coordinate system$O-xyz$ converts to$\mathrm{ITRS}\left(WGS84\right)$ coordinate system$O-XY{Z}_{\text{。 }}$Right now

$$\left(\begin{array}{l}X\\ Y\\ Z\end{array}\right)={\mathbf{R}}_x\left(-Y_p\right)R_y\left(-X_p\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& \cos Y_p& -\sin Y_p\\ 0& \sin Y_p& \cos Y_p\end{array}\right)\left(\begin{array}{ccc}\cos X_p& 0& \sin X_p\\ 0& 1& 0\\ -\sin X_p& 0& \cos X_p\end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)$$
Due to the extreme shift value $X_p$和$Y_p$are all less than $0.5^{\prime \prime }$ , so$\cos X_p=\cos Y_p=1,\sin X_p=X_p$, $\sin Y_p=Y_p$, so we have $:$
$$\left(\begin{array}{l}X\\ Y\\ Z\end{array}\right)=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& -Y_p\\ 0& Y_p& 1\end{array}\right)\left(\begin{array}{ccc}1& 0& X_p\\ 0& 1& 0\\ -X_p& 0& 1\end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}1& 0& X_p\\ 0& 1& -Y_p\\ -X_p& Y_p& 1\end{array}\right)\left(\begin{array}{l}x\\ y\\ z\end{array}\right)=[\mathbf{W}]\left(\begin{array}{l}x\\ y\\ z\end{array}\right)$$

From: Chapter Two of "GPS Measurement and Data Processing".