Three-Dimensional Coordinate Transformation Model and Its Robust Estimation Method Under Gauss-Helmert Model
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摘要: 对三维坐标转换的高斯-赫尔默特（Gauss-Helmert，GH）模型，采用牛顿-高斯（Newton-Gauss）迭代算法构建了该模型的拉格朗日目标函数，推导了其解算方法，并给出了具体的计算步骤。在此基础上，考虑到可能出现的粗差对观测空间与结构空间的综合影响，基于标准化残差构造权因子函数，推导了该模型的抗差解法。仿真实验结果表明，GH模型用于三维坐标转换时不受旋转角度大小和其他附加条件限制，解算结果与现有算法一致，且估计参数的维数大大降低，计算效率有一定程度的提高；所提出的抗差解法效果良好，与现有基于整体最小二乘的三维坐标转换的抗差解法相比，表现出了更好的稳健性。Abstract: For three-dimensional coordinate transformation, it's impossible using the Gauss-Markov model to obtain optimal parameter estimation from the functional model with error in its coefficient matrix. On the other hand, errors-in-variables model has difficulty expressing the functional model, and partial errors-in-variables model is complex as well as too much parameters to be estimated for the quasi-observation method. Therefore, Gauss-Helmert model is employed for three-dimensional coor-dinate transformation. The target function of the proposed model is established based on Newton-Gauss iterative algorithm, and the estimated method and its derivation procedure also are presented in this paper. Beyond the above process, we proposed a new robust estimation method for the proposed model, which is based on the normalized residual error and takes the influence of gross error on both observation and structure spaces into consideration. Meanwhile, derivational process of statistical tests and iterative algorithm are presented. The simulation experiment results show that the proposed estimation method has the same accuracy as the traditional method, which has robust with angular dimension and other additional conditions, but less estimation parameters. In addition, the new robust estimation method has effective robustness when comparing with the other existing robust total least square methods for the coordinate transformation.
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图 1 方案5和6估值与真值的差值序列(3个粗差)
Figure 1. Difference Values Between Truth Values and Estimated Results of Schemes 5 and 6 with Three Outliers
表 1 存在1个粗差时各方案的均方根误差对比
Table 1 Root Mean Square Errors of Estimated Parameters for Different Schemes with One Outlier
方案 X0 /m Y0 /m Z0 /m μ/(10-5 rad) α1 /(10-5 rad) α2 /(10-5 rad) α3 /(10-5 rad) 方案1 19.302 1 10.579 6 15.580 1 2.600 1.510 1.160 1.750 方案2 3.922 1 2.666 3 3.613 2 0.632 0.405 0.351 0.366 方案3 3.92 21 2.666 3 3.613 2 0.632 0.405 0.351 0.366 方案4 6.470 3 3.981 9 5.938 8 1.010 0.644 0.557 0.570 方案5 4.502 9 3.163 6 4.381 6 0.696 0.453 0.410 0.452 方案6 4.194 9 2.907 2 4.106 6 0.666 0.435 0.382 0.407 
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表 2 存在3个粗差时各方案的均方根误差对比
Table 2 Root Mean Square Errors of Estimated Parameters for Different Schemes with Three Outliers
方案 X0 /m Y0 /m Z0 /m μ/(10-5 rad) α1 /(10-5 rad) α2 /(10-5 rad) α3 /(10-5 rad) 方案1 10.307 9 12.340 7 18.474 0 1.460 1.390 1.730 1.130 方案2 3.752 9 2.956 4 2.844 9 0.507 0.289 0.311 0.417 方案3 3.752 9 2.956 4 2.844 9 0.507 0.289 0.311 0.417 方案4 8.797 8 6.880 5 6.709 9 1.230 0.664 0.770 0.952 方案5 7.068 0 5.744 2 5.810 6 0.978 0.592 0.628 0.798 方案6 4.695 8 3.896 1 3.931 2 0.671 0.404 0.413 0.518
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表 3 存在5个粗差时各方案的均方根误差对比
Table 3 Root Mean Square Errors of Estimated Parameters for Different Schemes with Five Outliers
方案 X0 /m Y0 /m Z0 /m μ/(10-5 rad) α1 /(10-5 rad) α2 /(10-5 rad) α3 /(10-5 rad) 方案1 12.748 6 11.219 6 15.764 9 2.250 1.790 1.150 1.400 方案2 3.414 0 3.380 1 4.576 3 0.610 0.469 0.380 0.424 方案3 3.414 0 3.380 1 4.576 3 0.610 0.469 0.380 0.424 方案4 10.173 6 10.600 6 14.184 5 1.960 1.480 1.120 1.320 方案5 8.900 8 8.467 6 11.919 8 1.580 1.290 0.956 1.110 方案6 6.064 6 6.039 3 7.999 3 1.090 0.818 0.639 0.757
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表 4 方案5和方案6计算检查点的均方根误差(1个粗差)/m
Table 4 Root Mean Square Errors of Check Points for Schemes 5 and 6 with One Outlier/m
方向 方案 点位1 点位2 点位3 点位4 点位5 点位6 点位7 X 方案5 0.079 2 0.081 3 0.072 9 0.077 6 0.081 5 0.080 3 0.080 3 方案6 0.078 4 0.080 3 0.072 4 0.077 1 0.080 8 0.079 0 0.078 7 Y 方案5 0.096 2 0.040 7 0.076 9 0.039 8 0.065 6 0.068 2 0.041 0 方案6 0.095 9 0.039 1 0.077 0 0.038 6 0.065 3 0.068 2 0.040 1 Z 方案5 0.084 8 0.052 2 0.067 5 0.059 8 0.077 6 0.082 8 0.057 3 方案6 0.084 7 0.052 2 0.066 8 0.058 3 0.076 4 0.081 9 0.056 1
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表 5 方案5和方案6计算检查点的均方根误差(3个粗差)/m
Table 5 Root Mean Square Errors of Check Points for Schemes 5 and 6 with Three Outliers/m
方向 方案 点位1 点位2 点位3 点位4 点位5 点位6 点位7 X 方案5 0.078 5 0.081 1 0.073 9 0.082 2 0.082 7 0.080 3 0.082 9 方案6 0.077 8 0.078 5 0.071 7 0.078 2 0.080 4 0.078 5 0.078 0 Y 方案5 0.098 7 0.048 1 0.082 7 0.044 6 0.065 1 0.073 0 0.044 9 方案6 0.094 6 0.041 9 0.078 4 0.038 7 0.063 9 0.070 7 0.039 2 Z 方案5 0.089 3 0.056 9 0.068 3 0.061 8 0.076 8 0.083 4 0.059 3 方案6 0.087 1 0.052 9 0.066 2 0.057 6 0.076 2 0.082 1 0.057 2
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表 6 方案5和方案6计算检查点的均方根误差(5个粗差)/m
Table 6 Root Mean Square Errors of Check Points for Schemes 5 and 6 with Five Outliers/m
方向 方案 点位1 点位2 点位3 点位4 点位5 点位6 点位7 X 方案5 0.084 4 0.09 22 0.080 6 0.089 3 0.087 3 0.090 7 0.097 6 方案6 0.079 7 0.083 5 0.075 0 0.082 1 0.081 3 0.081 6 0.085 5 Y 方案5 0.104 0 0.057 7 0.087 7 0.052 2 0.072 8 0.079 4 0.057 6 方案6 0.097 9 0.047 5 0.083 1 0.041 9 0.069 5 0.072 1 0.046 8 Z 方案5 0.099 5 0.066 1 0.078 1 0.073 4 0.085 4 0.087 9 0.068 1 方案6 0.091 9 0.060 1 0.070 8 0.062 1 0.078 9 0.083 5 0.060 0
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