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摘要:
高超声速滑翔飞行器滑翔飞行高度在30 km以上,大气极其稀薄,传统采用气压高度计的地形匹配辅助导航方式将无法正常工作。为实现高精度地形匹配,在分析匹配算法对地形常值误差不敏感的基础上,详细论证了基于惯性系统解算绝对高度方案,并对比分析了将短时滑翔段弹道简化为等高飞行方案。在捷联惯性导航系统(SINS)误差模型基础上,结合高度通道方块图,通过拉普拉斯变换,建立了惯性系统高度通道短时稳定性解析模型,并以CAV-H为研究对象建立数值仿真环境。仿真结果表明,解析模型精度较高,基于SINS解算绝对高度能够满足地形匹配辅助导航系统精度要求,优于气压高度计正常工作时的精度。
Abstract:The height of the hypersonic vehicle gliding flight is over 30 km, so the atmosphere is extremely thin and the traditional terrain match aided navigation with barometric altimeter cannot work properly. In order to improve the accuracy of terrain match, on the basis of analyzing the insensitivity of the matching algorithm to the terrain constant value error, the scheme of using inertial system to solve absolute height is demonstrated in detail, and the scheme that simplifies the short-time gliding trajectory to equal high flight is compared and analyzed. Based on the strap-down inertial navigation system (SINS) error model, the analytic model of the short-time stability of the inertial system is established by the height channel block diagram and Laplace transformation. In addition, the numerical simulation environment is established with CAV-H as study object. The simulation results show that the analytical model has high accuracy and the scheme that the absolute height is solved by SINS can meet the accuracy requirements of terrain match aided navigation system, which is better than the accuracy of the normal operation of the barometric altimeter.
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表 1 不同升阻比下2000s时的δh
Table 1. δh at 2000s with different lift-drag ratios
K δh/m 2.30 -32.78 2.45 -24.03 2.60 -19.07 2.75 -15.88 2.90 -13.65 3.05 -12.00 3.20 -10.74 3.35 -9.73 3.50 -8.91 表 2 不同初始高度下2000s时的δh
Table 2. δh at 2000s with different initial heights
H0/km δh/m 50 -64.64 51 -39.16 52 -27.92 53 -21.63 54 -17.64 55 -14.91 60 -8.72 70 -5.86 80 -5.31 表 3 ΔT=1s时δh的大小
Table 3. Value of δh when ΔT=1s
t0/s δh/m 1 000 5.10 1 300 8.80 1 500 12.57 1 800 21.36 2 000 30.36 表 4 不同加速度计误差系数偏差下t0=2000s时δh的大小
Table 4. Value of δh when t0=2000s with different deviations of accelerometer error coefficient
误差系数偏差 δh/m 0.3×10-5 0.65 1.5×10-5 3.23 3×10-5 6.46 6×10-5 12.92 9×10-5 19.38 12×10-5 25.84 15×10-5 32.30 30×10-5 64.60 表 5 不同ϕ′m0下t0=2000s时δh的大小
Table 5. Value of δh when t0=2000s with different ϕ′m0
ϕ′m0/(°) δh/m 0 16.80 5 17.79 10 18.13 15 18.68 20 19.16 25 19.56 30 19.87 35 20.09 40 20.23 45 20.28 表 6 工具误差系数精度
Table 6. Instrumental error coefficient accuracy
误差系数偏差 系数精度 ΔD0x/((°)·h-1) 0.01 ΔD0y/((°)·h-1) 0.01 ΔD0z/((°)·h-1) 0.01 ΔK0x/g0 3×10-5 ΔK0y/g0 3×10-5 ΔK0z/g0 3×10-5 ΔK1x 3×10-5 ΔK1y 3×10-5 ΔK1z 3×10-5 -
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