9 间接平差的应用

测边网间接平差

以边长为观测值、以未知点纵横坐标为参数，可列立方程。

单测边误差方程

单未知点

• 参数：坐标 $X_k,Y_k$
• 近似值：坐标 $X_k^0,Y_k^0$，边长 $L_{jk}^0$

有观测方程

$${\hat{L}}_{jk}=\sqrt{({\hat{X}}_k-X_j{)}^2+({\hat{Y}}_k-Y_j{)}^2}$$

记 ${\hat{X}}_k=X_k^0+{\hat{x}}_k$，${\hat{Y}}_k=Y_k^0+{\hat{y}}_k$，有线性化

$$\begin{array}{rl}{\hat{L}}_{jk}& =L_{jk}^0+\frac{\partial {\hat{L}}_{jk}}{\partial {\hat{X}}_k}{\hat{x}}_k+\frac{\partial {\hat{L}}_{jk}}{\partial {\hat{Y}}_k}{\hat{y}}_k\\ & =L_{jk}^0+\frac{X_k^0-X_j}{L_{jk}^0}{\hat{x}}_k+\frac{Y_k^0-Y_j}{L_{jk}^0}{\hat{y}}_k\\ & =L_{jk}^0+{\hat{x}}_k\cos {\alpha }_{jk}^0+{\hat{y}}_k\sin {\alpha }_{jk}^0\end{array}$$

双未知点

• 参数：坐标 $X_j,Y_j,X_k,Y_k$
• 近似值：坐标 $X_j^0,Y_j^0,X_k^0,Y_k^0$，边长 $L_{jk}^0$

有观测方程

$${\hat{L}}_{jk}=\sqrt{({\hat{X}}_k-{\hat{X}}_j{)}^2+({\hat{Y}}_k-{\hat{Y}}_j{)}^2}$$

记 ${\hat{X}}_j=X_j^0+{\hat{x}}_j$，${\hat{Y}}_j=Y_j^0+{\hat{y}}_j$，${\hat{X}}_k=X_k^0+{\hat{x}}_k$，${\hat{Y}}_k=Y_k^0+{\hat{y}}_k$，有线性化

$$\begin{array}{rl}{\hat{L}}_{jk}& =L_{jk}^0+\frac{\partial {\hat{L}}_{jk}}{\partial X_j}{\hat{x}}_j+\frac{\partial {\hat{L}}_{jk}}{\partial {\hat{Y}}_j}{\hat{y}}_j+\frac{\partial {\hat{L}}_{jk}}{\partial {\hat{X}}_k}{\hat{x}}_k+\frac{\partial {\hat{L}}_{jk}}{\partial {\hat{Y}}_k}{\hat{y}}_k\\ & =L_{jk}^0+\frac{X_k^0-X_j^0}{L_{jk}^0}({\hat{x}}_k-{\hat{x}}_j)+\frac{Y_k^0-Y_j^0}{L_{jk}^0}({\hat{y}}_k-{\hat{y}}_j)\\ & =L_{jk}^0+({\hat{x}}_k-{\hat{x}}_j)\cos {\alpha }_{jk}^0+({\hat{x}}_k-{\hat{x}}_j)\sin {\alpha }_{jk}^0\end{array}$$

测边网求解

$$\begin{array}{rl}{\hat{L}}_1& =\sqrt{(X_A-{\hat{X}}_2{)}^2+(Y_A-{\hat{Y}}_2{)}^2}\\ {\hat{L}}_2& =\sqrt{(X_B-{\hat{X}}_2{)}^2+(Y_B-{\hat{Y}}_2{)}^2}\\ {\hat{L}}_3& =\sqrt{({\hat{X}}_1-{\hat{X}}_2{)}^2+({\hat{Y}}_1-{\hat{Y}}_2{)}^2}\\ {\hat{L}}_4& =\sqrt{(X_A-{\hat{X}}_1{)}^2+(Y_A-{\hat{Y}}_1{)}^2}\\ {\hat{L}}_5& =\sqrt{(X_B-{\hat{X}}_1{)}^2+(Y_B-{\hat{Y}}_1{)}^2}\end{array}$$

线性化得到 $\hat{\mathbf{L}}=\mathbf{B}\hat{\mathbf{X}}+\mathbf{d}$

$$\left[\begin{array}{c}{\hat{L}}_1\\ {\hat{L}}_2\\ {\hat{L}}_3\\ {\hat{L}}_4\\ {\hat{L}}_5\end{array}\right]=\left[\begin{array}{cccc}0& 0& \cos {\alpha }_{A2}^0& \sin {\alpha }_{A2}^0\\ 0& 0& \cos {\alpha }_{B2}^0& \sin {\alpha }_{B2}^0\\ \cos {\alpha }_{21}^0& \sin {\alpha }_{21}^0& -\cos {\alpha }_{21}^0& -\sin {\alpha }_{21}^0\\ \cos {\alpha }_{A1}^0& \sin {\alpha }_{A1}^0& 0& 0\\ \cos {\alpha }_{B1}^0& \sin {\alpha }_{B1}^0& 0& 0\end{array}\right]\left[\begin{array}{c}{\hat{x}}_1\\ {\hat{y}}_1\\ {\hat{x}}_2\\ {\hat{y}}_2\end{array}\right]+\left[\begin{array}{c}{L}_{A2}^0\\ L_{B2}^0\\ L_{21}^0\\ L_{A1}^0\\ L_{B1}^0\end{array}\right]$$

边长误差方程的特点：

• 两待定点的坐标参数的系数等值反号
• 已知点坐标系数不出现在设计矩阵中
• $j,k$ 均为已知点时，无误差方程
• $j,k$ 边的误差方程，$j\to k$ 或 $k\to j$ 列立，其结果相同

精度评定

$${\hat{\sigma }}_0^2=\frac{{\mathbf{V}}^{\mathrm{T}}\mathbf{PV}}{n-t}=\frac{{\mathbf{V}}^{\mathrm{T}}\mathbf{PV}}{r}{\mathbf{Q}}_{\hat{x}\hat{x}}={\mathbf{N}}^{-1}$$

• 函数式

  $$\hat{F}={\hat{L}}_{jk}=\sqrt{({\hat{X}}_k-{\hat{X}}_j{)}^2+({\hat{Y}}_k-{\hat{Y}}_j{)}^2}$$

• 权函数式

  $$\begin{array}{c}\mathrm{d}\hat{F}=-\frac{\Delta X_{jk}^0}{L_{jk}^0}\mathrm{d}{\hat{x}}_j-\frac{\Delta Y_{jk}^0}{L_{jk}^0}\mathrm{d}{\hat{y}}_j+\frac{\Delta X_{jk}^0}{L_{jk}^0}\mathrm{d}{\hat{x}}_k+\frac{\Delta Y_{jk}^0}{L_{jk}^0}\mathrm{d}{\hat{y}}_k={\mathbf{f}}^T\mathrm{d}\hat{\mathbf{x}}\\ Q_{\hat{F}\hat{F}}={\mathbf{f}}^T{\mathbf{Q}}_{\hat{x}\hat{x}}\mathbf{f}\\ {\hat{\sigma }}_{\hat{F}}={\hat{\sigma }}_0\sqrt{Q_{\hat{F}\hat{F}}}\end{array}$$

• 相对精度

  $$\frac{{\hat{\sigma }}_{\hat{F}}}{\hat{F}}=\frac{{\hat{\sigma }}_0\sqrt{Q_{\hat{F}\hat{F}}}}{{\hat{L}}_{jk}}$$

方向观测网

陀螺经纬仪可以直接观测方位角。

设未知点的坐标 $C(X_1,Y_1)$，$D(X_2,Y_2)$ 为参数，对于点 $j$ 与点 $k$ 组成的边 $jk$ 有观测方程

$${\hat{\alpha }}_{jk}=\arctan \frac{{\hat{Y}}_k-{\hat{Y}}_j}{{\hat{X}}_k-{\hat{X}}_j}$$

进行线性化

$$\begin{array}{rl}{\alpha }_{jk}^0& =\arctan \frac{Y_k^0-Y_j^0}{X_k^0-X_j^0}\\ a_{jk}& ={\left(\frac{\partial {\hat{\alpha }}_{jk}}{\partial {\hat{X}}_j}\right)}_0=\frac{\sin {\alpha }_{jk}^0}{L_{jk}^0}{\rho }^{\prime \prime }\\ b_{jk}& ={\left(\frac{\partial {\hat{\alpha }}_{jk}}{\partial {\hat{X}}_j}\right)}_0=-\frac{\cos {\alpha }_{jk}^0}{L_{jk}^0}{\rho }^{\prime \prime }\end{array}$$

得到

$${\hat{\alpha }}_{jk}={\alpha }_{jk}^0+a_{jk}{\hat{x}}_j+b_{jk}{\hat{y}}_j-a_{jk}{\hat{y}}_k-b_{jk}{\hat{y}}_k$$

测角网

设未知点的坐标 $C(X_1,Y_1)$，$D(X_2,Y_2)$ 为参数。有观测方程

$$\begin{array}{rl}{\hat{L}}_1& ={\hat{\alpha }}_{AB}-{\hat{\alpha }}_{AC}={\alpha }_{AB}-\arctan \frac{{\hat{Y}}_1-Y_A}{{\hat{X}}_1-X_A}\\ {\hat{L}}_2& ={\hat{\alpha }}_{AC}-{\hat{\alpha }}_{AD}=\arctan \frac{{\hat{Y}}_1-Y_A}{{\hat{X}}_1-X_A}-\arctan \frac{{\hat{Y}}_2-Y_A}{{\hat{X}}_2-X_A}\end{array}$$

单角线性化

对于 $L_i=\angle hjk$ 的线性化：

$${\hat{L}}_i={\hat{\alpha }}_{jk}-{\hat{\alpha }}_{jh}$$$$\begin{array}{rl}{\alpha }_{jk}^0& =\arctan \frac{Y_k^0-Y_j^0}{X_k^0-X_j^0}\\ a_{jk}& ={\left(\frac{\partial {\hat{\alpha }}_{jk}}{\partial {\hat{X}}_j}\right)}_0=\frac{\sin {\alpha }_{jk}^0}{L_{jk}^0}{\rho }^{\prime \prime }\\ b_{jk}& ={\left(\frac{\partial {\hat{\alpha }}_{jk}}{\partial {\hat{X}}_j}\right)}_0=-\frac{\cos {\alpha }_{jk}^0}{L_{jk}^0}{\rho }^{\prime \prime }\end{array}$$$$\begin{array}{rl}{\hat{\alpha }}_{jk}=\arctan \frac{{\hat{Y}}_k-{\hat{Y}}_j}{{\hat{X}}_k-{\hat{X}}_j}& ={\alpha }_{jk}^0+a_{jk}{\hat{x}}_j+b_{jk}{\hat{y}}_j-a_{jk}{\hat{x}}_k-b_{jk}{\hat{y}}_k\\ {\hat{\alpha }}_{jh}=\arctan \frac{{\hat{Y}}_h-{\hat{Y}}_j}{{\hat{X}}_h-{\hat{X}}_j}& ={\alpha }_{jh}^0+a_{jh}{\hat{x}}_j+b_{jh}{\hat{y}}_j-a_{jh}{\hat{x}}_h-b_{jh}{\hat{y}}_h\end{array}$$

代入 ${\hat{L}}_i={\hat{\alpha }}_{jk}-{\hat{\alpha }}_{jh}$ 即得到该角的线性化测量方程

$$\begin{array}{rl}{\hat{L}}_i=& {\alpha }_{jk}^0-{\alpha }_{jh}^0\\ & +(a_{jk}-a_{jh}){\hat{x}}_j+(b_{jk}-b_{jh}){\hat{y}}_j\\ & -a_{jk}{\hat{x}}_k-b_{jk}{\hat{y}}_k+a_{jh}{\hat{x}}_h+b_{jh}{\hat{y}}_h\end{array}$$

角度相关权

考虑在 O 点观测了 3 个独立等精度方向值 ${\alpha }_1,{\alpha }_2,{\alpha }_3$，相减得到角度值 ${\beta }_1,{\beta }_2$。

此时角度值 ${\beta }_1,{\beta }_2$ 相关，若视为独立等权，则平差结果不严密。

此例中取 ${\mathbf{Q}}_{\alpha \alpha }=\mathbf{I}=\mathrm{diag}(1,1,1)$，有 $\mathbf{\beta }=\mathbf{A\alpha }$

$$\left[\begin{array}{c}{\beta }_1\\ {\beta }_2\end{array}\right]=\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\end{array}\right]\left[\begin{array}{c}{\alpha }_1\\ {\alpha }_2\\ {\alpha }_3\end{array}\right]$$

故有

$${\mathbf{Q}}_{\beta \beta }={\mathbf{A}}^{\mathrm{T}}{\mathbf{Q}}_{\alpha \alpha }\mathbf{A}={\mathbf{A}}^{\mathrm{T}}\mathbf{A}=\left[\begin{array}{cc}2& -1\\ -1& 2\end{array}\right]$$$${\mathbf{P}}_{\beta \beta }={\mathbf{Q}}_{\beta \beta }^{-1}=\frac{1}{3}\left[\begin{array}{cc}2& 1\\ 1& 2\end{array}\right]$$

同理可求更多角组合的情况。中点多边形的中点也可采用此方法。