5.6.1单导线的平差解算和精度评定 纵坐标条件式:∑△x1=xc-x ∑1+=O ∑Ax ii+I 根据误差传播率:Ax1= S coST coST Vn=v4+v2+………+v 代入上式,得: ∑[cosT·]-∑[4y·(v+va+…+v)+=05.6.1 单导线的平差解算和精度评定 纵坐标条件式: c n i x A i i x n i x C A n i i i w x x x v w x x x i i = + − + = = − = + = = + + 1 , 1 1 1 , 1 0 , 1 i i i T i i i T i i s i i x i i i i v v v v v T v y v x S T = + + + = − = + + + 2 0 , 1 0 , 1 , 1 cos 根据误差传播率: cos 代入上式,得: [cos ] [ ( )] 0 1 0 , 1 1 0 1 2 − + + + + = = + = x n i i i n i T i v s y v v v w i i