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Performance of BDS-3: satellite visibility and dilution of precision

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Abstract

We describe a method to assess the performance of the third-generation BeiDou navigation satellite system (BDS-3), in terms of satellite visibility and dilution of precision (DOP), on global and regional scales. Different from traditional methods, this method estimates the satellite visibility and DOP without requiring real or simulated ephemerides. Validated by the reference values derived from real ephemerides of GPS and GLONASS, the estimated number of visible satellites achieves an accuracy better than 0.15, and the estimated DOP values are lower than their reference values by less than 10% on average. Applying this method to BDS-3, with a 5° cutoff elevation angle, results show that the geostationary earth orbit (GEO) and inclined geosynchronous orbit (IGSO) satellites of BDS-3 together contribute 3–6 visible satellites in the area of 60°S–60°N and 50°E–170°E. In this area, the number of visible BDS-3 satellites is 11–14, which is more than GPS and Galileo by 1–3, and GLONASS by 3–7. With better satellite visibility, the average BDS-3 horizontal, vertical, and time DOPs over this area are 0.74, 1.08, and 0.67, which are, respectively, 5%, 9%, and 3% lower than those of GPS and Galileo, 14%, 16%, and 21% lower than those of GLONASS, and 16%, 19% and 14% lower than those of the 24-MEO-only BDS-3.

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Acknowledgements

This study is supported by National Natural Science Foundation of China (41604017 and 41674029). We would like to express our gratitude to editor Alfred Leick and anonymous reviewers for their constructive comments and suggestions. We thank MGEX and GFZ for providing multi-GNSS orbit products.

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Correspondence to Jiexian Wang.

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Appendix 1: retrieval of 0.5° × 0.5° grid cells along the figure-8-shape track of BDS-3 IGSO satellites

Appendix 1: retrieval of 0.5° × 0.5° grid cells along the figure-8-shape track of BDS-3 IGSO satellites

We define a \(\eta {\text{-}}\xi\) coordinate system on the circular orbit plane of a BDS-3 IGSO satellite, with the \(\eta {\text{-}}\)axis pointing to the ascending node and the \(\xi {\text{-}}\)axis pointing to the satellite on the argument of latitude of 90°. The satellite positions on the circular orbit with an interval of 0.25° are

$$\left\{ {\begin{array}{*{20}{c}} {{\eta _i}=R\cos \left( {\frac{{2\pi }}{{1440}}i} \right)} \\ {{\xi _i}=R\sin \left( {\frac{{2\pi }}{{1440}}i} \right)} \end{array}} \right.{\text{ (}}i=1,2,...,1440)$$
(20)

where R is the distance between the satellite and the earth center (42,157 km). Their corresponding ECEF Cartesian coordinates are

$$\left( {\begin{array}{*{20}{c}} {{X_i}} \\ {{Y_i}} \\ {{Z_i}} \end{array}} \right)={R_3}( - {\Omega _i}){R_1}( - {i_{{\text{orb}}}})\left( {\begin{array}{*{20}{c}} {{\eta _i}} \\ {{\xi _i}} \\ 0 \end{array}} \right){\text{ }}(i=1,2,...,1440)$$
(21)

where \({R_3}( - {\Omega _i})\) and \({R_1}( - {i_{{\text{orb}}}})\) are rotation matrices

$${R_3}( - {\Omega _i})=\left( {\begin{array}{*{20}{c}} {\cos {\Omega _i}}&{ - \sin {\Omega _i}}&0 \\ {\sin {\Omega _i}}&{\cos {\Omega _i}}&0 \\ 0&0&1 \end{array}} \right)$$
(22)
$${R_1}( - {i_{{\text{orb}}}})=\left( {\begin{array}{*{20}{c}} 1&0&0 \\ 0&{\cos {i_{{\text{orb}}}}}&{ - \sin {i_{{\text{orb}}}}} \\ 0&{\sin {i_{{\text{orb}}}}}&{\cos {i_{{\text{orb}}}}} \end{array}} \right)$$
(23)

where \({i_{{\text{orb}}}}\) is the orbit inclination (55°) and \({\Omega _i}\) is the right ascension of ascending node. Let the satellite be right above the intersection point of subsatellite track (0°, 118°E) at time \({t_0}=0\), and the 1440 satellite positions be corresponding to the positions at time \({t_i}=i\Delta t{\text{ (}}i=1,2,...,1440)\), where\(\Delta t=T/1440\) with \(T=86164\)s (orbit period). Then the \({\Omega _i}\) can be written as

$${\Omega _i}={118^ \circ }\frac{\pi }{{{{180}^ \circ }}} - \omega ({t_i} - {t_0})=({118^ \circ } - i \cdot {0.25^ \circ })\frac{\pi }{{{{180}^ \circ }}}$$
(24)

where \(\omega\) is the angular velocity of the satellite.

From the Cartesian coordinates \(({X_i},{Y_i},{Z_i})\), the geocentric latitude\({\varphi _i}\) and longitude \({\lambda _i}\) of the satellite can calculated as

$$\left\{ {\begin{array}{*{20}{c}} {{\varphi _i}=\arcsin (\frac{{{Z_i}}}{R})} \\ {{\lambda _i}=\arctan (\frac{{{Y_i}}}{{{X_i}}})} \end{array}} \right.$$
(25)

We divide the spherical orbit surface (with the radius of R) into 0.5°×0.5° grid cells by geocentric latitude and longitude and use the satellite positions \(({\varphi _i},{\lambda _i})\) \({\text{(}}i=1,2,...,1440)\) to determine which grid cells are on the figure-8-shape satellite track. The sampling interval of 0.25° for calculating the satellite positions ensures that all the 0.5° × 0.5° grid cells on the track are retrieved.

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Wang, M., Wang, J., Dong, D. et al. Performance of BDS-3: satellite visibility and dilution of precision. GPS Solut 23, 56 (2019). https://doi.org/10.1007/s10291-019-0847-x

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