关于平差第二题的公式及线性化角度就是按照题目给的顺序L1^=α12−α31^−π \widehat{L_{1}}=\alpha_{12}-\widehat{\alpha_{31}}-{\pi}L1=α12−α31−πL2^=π−(α12−α24^) \widehat{L_{2}}={\pi}-(\alpha_{12}-\widehat{\alpha_{24}})L2=π−(α12−α24)L3^=π−(α43^−α31^) \widehat{L_{3}}={\pi}-(\widehat{\alpha_{43}}-\widehat{\alpha_{31}})L3=π−(α43−α31)L4^=α43^−α53^ \widehat{L_{4}}=\widehat{\alpha_{43}}-\widehat{\alpha_{53}}L4=α43−α53L5^=α46^−α43^ \widehat{L_{5}}=\widehat{\alpha_{46}}-\widehat{\alpha_{43}}L5=α46−α43L6^=π−(α24^−α43^) \widehat{L_{6}}={\pi}-(\widehat{\alpha_{24}}-\widehat{\alpha_{43}})L6=π−(α24−α43)L7^=π−(α65^−α53^) \widehat{L_{7}}={\pi}-(\widehat{\alpha_{65}}-\widehat{\alpha_{53}})L7=π−(α65−α53)L8^=π−(α46^−α65^) \widehat{L_{8}}={\pi}-(\widehat{\alpha_{46}}-\widehat{\alpha_{65}})L8=π−(α46−α65)即L1^=α12−α31^−180L2^=α24^−α12+180L3^=α31^−α43^+180L4^=α43^−α53^L5^=α46^−α43^L6^=α43^−α24^+180L7^=α53^−α65^+180L8^=α65^−α46^+180 \begin{gathered} \widehat{L_1}=\alpha_{12}-\widehat{\alpha_{31}}-180\\ \widehat{L_{2}}=\widehat{\alpha_{24}}-\alpha_{12}+180\\ \widehat{L_3}=\widehat{\alpha_{31}}-\widehat{\alpha_{43}}+180\\ \widehat{L_4}=\widehat{\alpha_{43}}-\widehat{\alpha_{53}}\\ \widehat{L_5}=\widehat{\alpha_{46}}-\widehat{\alpha_{43}}\\ \widehat{L_6}=\widehat{\alpha_{43}}-\widehat{\alpha_{24}}+180\\ \widehat{L_7}=\widehat{\alpha_{53}}-\widehat{\alpha_{65}}+180\\ \widehat{L_8}=\widehat{\alpha_{65}}-\widehat{\alpha_{46}}+180 \end{gathered}L1=α12−α31−180L2=α24−α12+180L3=α31−α43+180L4=α43−α53L5=α46−α43L6=α43−α24+180L7=α53−α65+180L
