Sunday, 8 December 2013

Height of Instrument Method(Differential Levelling)


  • In any particular set up of an instrument height of instrument, which is the elevation of the line of sight, is constant. The elevation of unknown points can be obtained by subtracting the staff readings at the desired points from the height of instrument. This is the basic behind the height of instrument method for reduction of level.


  • With reference to Figure 13.2 and Table 13.2, when the instrument is at I1, the staff reading observed at A is 2.365m. The elevation of the line of sight i.e., the height of instrument is 102.365m obtained by adding the elevation of A (100.0m) with the staff reading observed at A (2.365m). The elevation of S1 (101.130m) is determined by subtracting its foresight reading (1.235m) from the the height of instrument (102.365m) when the instrument is at I1 . Next, the instrument is set up at I2. S1 is considered as a point of known elevation and backsight reading ( 0.685m) is taken . The height of the instrument (101.815 m) is then calculated by adding backsight reading ( 0.685m) with the elevation (R.L.) of point S1 (101.130m). Foresight is taken at S2 and its elevation (98.245m) is determined by subtracting the foresight (3.570m) from the height of the instrument (101.815 m). In this way, elevation of points are calculated by Height of instrument method.


Table 13.2 Level book note for Height of instrument method


Staff Reading
Height of Instrument (m)
R.L. (m)
Remarks
Points
B.S (m)
F.S.(m)
A
2.365
 
102.365
100.000
B.M.
S 1
0.685
1.235
101.815
101.130
T.P.1
S2
 
3.570
 
98.245
T.P.2
B
 
2.340
 
97.650

EXAMPLE PROBLEM: If you understood go through the following examples.

Example
  • Ex13-1 Data from a differential leveling have been found in the order of B.S., F.S..... etc. starting with the initial reading on B.M. (elevation 150.485 m) are as follows : 1.205, 1.860, 0.125, 1.915, 0.395, 2.615, 0.880, 1.760, 1.960, 0.920, 2.595, 0.915, 2.255, 0.515, 2.305, 1.170. The final reading closes on B.M.. Put the data in a complete field note form and carry out reduction of level by Rise and Fall method. All units are in meters.
    Solution :
    B.S. (m)
    F.S. (m)
    Rise (m)
    Fall (m)
    Elevation (m)
    Remark
    1.205
    		   
    150.485
    B.M.
    0.125
    1.860
     
    0.655
    149.830
     
    0.395
    1.915
     
    1.7290
    148.040
     
    0.880
    2.615
     
    2.220
    145.820
     
    1.960
    1.760
     
    0.880
    144.940
     
    2.595
    0.920
    1.040
    		 
    145.980
     
    2.255
    0.915
    1.680
    		 
    147.660
     
    2.305
    0.515
    1.740
    		 
    149.450
     
     
    1.170
    1.135
    		 
    150.535
    B.M.
    << Back | Next >>

    Arithmetic Check for Reduction of Level
    In case of Rise and Fall method for Reduction of level, following arithmetic checks are applied to verify calculations.
    S B.S. - S F.S. = S Rise - S Fall = Last R.L. - First R.L.
    With reference to Table 13.3:
    S B.S. - S F.S. = 4.795 - 7.145 = - 2.350
    S Rise - S Fall. = 1.130 - 3.480 = - 2.350
    Last R.L. - First R.L.= 97.650 - 100.000 = -2.350
    Table 13.3 Field book for Reduction of level

    Staff Reading (m)
    Difference in elevation (m)
    H.I (m)
    R.L. (m)
    Remarks
    Points
    B.S.
    I.S.
    F.S.
    Rise
    Fall
    A
    2.365
    		 
     
     
    		 
    102.365
    100.000
    B.M.
    S 1
    0.685
    		 
    1.235
    1.130
    		 
    101.815
    101.130
    T.P.1
    S2
    1.745 
    		 
    3.570
     
    2.885
    99.990
    98.245
    T.P.2
    B
     
     
    2.340
     
    0.595
    102.365
    97.650
     
    S
     4.795
    		 
    7.145
     
    3.480
    101.815
     
    		 
Example
  • Ex13-2 Carry out the arithmetic checks for Reduction of level of Ex13-1.
    Solution :
    S B.S. = 11.720 m; S F.S. = 11.670 m
    Therefore S B.S - S F.S. = 0.050 m
    S Rise = 5.595 m; S Fall = 5.545 m
    Therefore S Rise - S Fall = 0.050 m
    Last R.L. - First R.L. = 150.535 - 150.485 = 0.050 m.
    S B.S - S F.S. = S Rise - S Fall = Last R.L. - First R.L.
    Example
    Ex13-3 Complete the differential-level notes and determine the error of closure of the level circuit and adjust the elevations of B.M.2 and B.M.3 assuming that the error is constant per set up.
    Level book note for Level Net

    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

Posted by at 2 comments:
Subscribe to: Posts (Atom)