|
Staff Reading |
Height of Instrument (m)
|
R.L. (m)
|
Points |
B.S (m) |
F.S.(m) |
B.M.1 |
2.125 |
|
|
|
T.P.1 |
1.830 |
2.945 |
|
|
T.P.2 |
2.100 |
3.225 |
|
|
T.P.3 |
1.650 |
3.605 |
|
|
B.M.2 |
2.365 |
2.805 |
|
|
T.P.4 |
2.885 |
2.530 |
|
|
T.P.5 |
3.065 |
2.350 |
|
|
B.M.3 |
3.855 |
1.100 |
|
|
T.P.6 |
3.270 |
1.660 |
|
|
T.P.7 |
3.865 |
2.110 |
|
|
B.M.1 |
|
3.455 |
|
|
Solution :
|
Staff Reading |
Height of Instrument (m)
|
R.L. (m)
|
Points |
B.S (m) |
F.S.(m) |
B.M.1 |
2.125 |
|
102.125 |
100.000 |
T.P.1 |
1.830 |
2.945 |
101.010 |
99.18 |
T.P.2 |
2.100 |
3.225 |
99.885 |
97.785 |
T.P.3 |
1.650 |
3.605 |
97.93 |
96.280 |
B.M.2 |
2.365 |
2.805 |
97.49 |
95.125 |
T.P.4 |
2.885 |
2.530 |
97.845 |
94.960 |
T.P.5 |
3.065 |
2.350 |
98.56 |
95.495 |
B.M.3 |
3.855 |
1.100 |
101.315 |
97.46 |
T.P.6 |
3.270 |
1.660 |
102.925 |
99.655 |
T.P.7 |
3.865 |
2.110 |
104.680 |
100.815 |
B.M.1 |
|
3.455 |
|
101.225 |
Error of closure = 101.225 - 100 = + 1.225 m
There are ten (10) set up for the instrument. Thus for each set up, there is an error of 0.1225 m.
Therefore correction for each set up = - 0.1225 m
Adjusted elevation of B.M.2 = 95.125 - 4 x .1225 = 94.635 m
Adjusted elevation of B.M. 3 = 97.46 - 7 x .1225 = 96.603 m