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:
( XYZ ) GCRS = [ P ] [ N ] [ R ] [ W ] ( XYZ ) ITRS \left(\begin{array}{l} X \\ Y \\ Z \ end{array}\right)_{\mathrm{GCRS}}=[\boldsymbol{P}][\boldsymbol{N}][\boldsymbol{R}][\boldsymbol{W}]\left(\begin{ array}{l} X \\ Y \\ Z \end{array}\right)_{\mathrm{ITRS}}⎝
⎛XYZ⎠
⎞GCRS=[P][N][R][W]⎝
⎛XYZ⎠
⎞ITRS
( X Y Z ) ITRS = [ W ] − 1 [ R ] − 1 [ N ] − 1 [ P ] − 1 ( X Y Z ) G C R S \left(\begin{array}{l} X \\ Y \\ Z \end{array}\right)_{\text {ITRS }}=[\boldsymbol{W}]^{-1}[\boldsymbol{R}]^{-1}[\boldsymbol{N}]^{-1}[\boldsymbol{P}]^{-1}\left(\begin{array}{l} X \\ Y \\ Z \end{array}\right)_{\mathrm{GCRS}} ⎝
⎛XYZ⎠
⎞ITRS =[W]−1[R]−1[N]−1[P]−1⎝
⎛XYZ⎠
⎞GCRS
where, [ P ] [\boldsymbol{P}][ P ] is the precession matrix;[ N ] [\boldsymbol{N}][ N ] is the nutation matrix;[ R ] [\boldsymbol{R}][ R ] is the earth rotation matrix;[ W ] [\boldsymbol{W}][ W ] 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 ti t_iti
We know that GCRS is the reference time t 0 = J 2000.0 t_0=\mathrm{J} 2000.0t0=The flat celestial coordinate system at J 2000.0 , it should be transformed into the observation time ti t_itiWhen the flat celestial coordinate system, just consider [ t 0 − ti ] \left[t_0-t_i\right][t0−ti] time period precession correction, that is, multiplied by[ P ] − 1 [\boldsymbol{P}]^{-1}[P]− 1 matrix is enough.
(2)把 t i t_i tiThe flat celestial coordinate system at that time is transformed into the true celestial coordinate system at the same time.
The observation time ti t_itiTo 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 [ N ] − 1 [\boldsymbol{N}]^{-1}[N]− 1 matrix is enough.
(3)把 t i t_i tiThe 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 XXThe x- axis is pointing to the true equinox at that momentγ \gammaγ , andXXThe 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:
GAST = 36 0 ∘ 2 4 h ( UT 1 + 6 h 41 m 50.54841 s + 8640184.812866 s ⋅ t + 0.093104 s ⋅ t 2 − 6.2 s × 1 0 − 6 ⋅ t 3 ) + Δ Ψ cos ( ε ˉ + Δ ε ) \begin{aligned} \text { GAST }=& \frac{360^{\circ}}{24^{\mathrm{h}}}\left(\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 \right.\\ &\left.-6.2 \mathrm{~s} \times 10^{-6} \cdot t^3\right)+\Delta \Psi_{\cos }(\bar{\varepsilon}+ \Delta \varepsilon) \end{aligned} GAST =24h360∘(UT1+6 h41 m50.54841 s+8640184.812866 s⋅t+0.093104 s⋅t2−6.2 s×10−6⋅t3)+D Pscos(eˉ+D e ).
where, ttt is 2000.0 \mathrm{J} 2000.0fromJulian century number of J 2000.0 ; ε ˉ \bar{\varepsilon}eˉ is the oblique angle when only precession is considered,ε ˉ = 2 3 ∘ 2 6 ′ 21.44 8 ′ ′ − \bar{\varepsilon}=23^{\circ} 26^{\prime} 21.448^{\prime \ prime}-eˉ=23∘26′21.448′′− 46.81 5 ′ ′ ⋅ t − 0.0005 9 ′ ′ ⋅ t 2 + 0.00181 3 ′ ′ ⋅ t 3 ; Δ ψ 46.815^{\prime \prime} \cdot t-0.00059^{\prime \prime} \cdot t^2+0.001813^{\prime \prime} \cdot t^3 ; \Delta \psi 46.815′′⋅t−0.00059′′⋅t2+0.001813′′⋅t3;Δψ 46.81 5 ′ ′ ⋅ t − 0.0005 9 ′ ′ ⋅ t 2 + 0.00181 3 ′ ′ ⋅ t 3 ; Δ Ψ 46.815^{\prime \prime} \cdot t-0.00059^{\prime \prime} \cdot t^2+0.001813^{\prime \prime} \cdot t^3 ; \Delta \Psi 46.815′′⋅t−0.00059′′⋅t2+0.001813′′⋅t3;ΔΨ is nutation of Huang Jing;Δ ε \Delta \varepsilonΔ ε is angular nutation;UT 1 \mathrm{UT1}UT1 can be obtained according to the UTC and (UTC-UT1) values at the time of observation.
Turn the true celestial coordinate system around ZZAfter the Z axis is rotated by the GAST angle, it can be converted to the real earth coordinate system, and the rotation matrixR \boldsymbol{R}R 为:
R = ( cos G A S T sinGAST 0 − sinGAST cos G A S T 0 0 0 1 ) \boldsymbol{R}=\left(\begin{array}{ccc} \cos G A S T & \operatorname{sinGAST} & 0 \\ -\operatorname{sinGAST} & \cos G A S T & 0 \\ 0 & 0 & 1 \end{array}\right) R=⎝
⎛cosGAST−SinGAST0SinGASTcosGAST0001⎠
⎞
(4)把 t i t_i tiTime's True Earth Coordinate System Conversion to ITRS (WGS 84)
As can be seen from the above figure, only ti t_itiThe true earth coordinate system at time revolves around yyy- axis rotation( − X p ) \left(-X_p\right)(−Xp) corner, and then aroundxxx- axis rotation( − Y p ) \left(-Y_p\right)(−Yp) angle, we can put the straight earth coordinate systemO − xyz Ox yzO−x yz converts toITRS ( WGS 84 ) \operatorname{ITRS}(\mathrm{WGS} 84)ITRS ( WGS 84 ) coordinate systemO − XYZ. OX Y Z_{\text {. }}O−XYZ。 Right now
( X Y Z ) = R x ( − Y p ) R y ( − X p ) ( x y z ) = ( 1 0 0 0 cos Y p − sin Y p 0 sin Y p cos Y p ) ( cos X p 0 sin X p 0 1 0 − sin X p 0 cos X p ) ( x y z ) \left(\begin{array}{l} X \\ Y \\ Z \end{array}\right)=\boldsymbol{R}_x\left(-Y_p\right) \boldsymbol{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) ⎝
⎛XYZ⎠
⎞=Rx(−Yp)Ry(−Xp)⎝
⎛xyz⎠
⎞=⎝
⎛1000cosYpsinYp0−sinYpcosYp⎠
⎞⎝
⎛cosXp0−sinXp010sinXp0cosXp⎠
⎞⎝
⎛xyz⎠
⎞
Due to the extreme shift value X p X_pXp和Y p Y_pYpare all less than 0. 5 ′ ′ 0.5^{\prime \prime}0.5′′ , socos X p = cos Y p = 1 , sin X p = X p \cos X_p=\cos Y_p=1, \sin X_p=X_pcosXp=cosYp=1,sinXp=Xp, sin Y p = Y p \sin Y_p=Y_psinYp=Yp, so we have : ::
( X Y Z ) = ( 1 0 0 0 1 − Y p 0 Y p 1 ) ( 1 0 X p 0 1 0 − X p 0 1 ) ( x y z ) = ( 1 0 X p 0 1 − Y p − X p Y p 1 ) ( x y z ) = [ W ] ( x y z ) \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)=[\boldsymbol{W}]\left(\begin{array}{l} x \\ y \\ z \end{array}\right) ⎝
⎛XYZ⎠
⎞=⎝
⎛10001Yp0−Yp1⎠
⎞⎝
⎛10−Xp010Xp01⎠
⎞⎝
⎛xyz⎠
⎞=⎝
⎛10−Xp01YpXp−Yp1⎠
⎞⎝
⎛xyz⎠
⎞=[W]⎝
⎛xyz⎠
⎞
From: Chapter Two of "GPS Measurement and Data Processing".