建筑4等测量题
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发布时间:2024-03-02 03:30
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时间:2024-03-21 16:08
求高差闭合差f=1.24-1.42+1.787-1.582
=0.025
测站数n=5+4+6+10
=25
高差改正数v1=-(0.025/25)*5
=-0.005
v2=-(0.025/25)*4
=-0.004
v3=-(0.025/25)*6
=-0.006
v4=-(0.025/25)*10
=-0.010
改正后高差h1=1.24-0.005
=1.235
h2=-1.42-0.004
=-1.424
h3=1.787-0.006
=1.781
h4=-1.582-0.01
=-1.592
求得高程H1=12.648+1.235
=13.883
H2=13.883-1.424
=12.459
H3=12.459+1.781
=14.24
H4=14.24-1.592
=12.648=H已知