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Keywords: Levelling examples, levelling, rise and fall method, missing entries problem
Since we have already studied about levelling, now let’s move on to a specific problem in this section. The problem of missing entries in the field book is an application level problem, often asked in question papers. An example on this type of problem is illustrated as follows:
Problem: The following level readings have been taken from a page of a level book. Some of the readings are missing. It is required to fill all the entries in the page. Fill up the missing readings and apply the arithmetic checks.
Station | BS | IS | FS | Rise | Fall | RL | Remarks |
A | 3.125 | BM | |||||
B | × | × | 1.325 | 125.505 | TP | ||
C | 2.320 | 0.055 | |||||
D | × | 123.850 | |||||
E | × | 2.655 | TP | ||||
F | 1.620 | 3.205 | 2.165 | TP | |||
G | 3.625 | ||||||
H | × | 123.090 | BM |
Solution:
The steps in the solution are as follows:
FS of station B= 3.125-1.325=1.800
BS of station B= 2.320-0.055=2.265
RL of BM= 125.505-1.325=124.180M
Fall of station E= 125.850-125.115=0.735
IS of station D = 2.655-0.735=1.920
BS of station E= 3.205-2.165=1.040
Rise of station H=123.090-120.945=2.145
FS of station H= 3.625-2.145=1.480
The missing entries are filled and presented in the following table:
Station | BS | IS | FS | Rise | Fall | RL | Remarks |
A | 3.125 | 124.180 | BM | ||||
B | 2.265 | 1.800 | 1.325 | 125.505 | TP | ||
C | 2.320 | 0.055 | 125.450 | ||||
D | 1.920 | 0.400 | 123.850 | ||||
E | 1.040 | 2.655 | 0.735 | 125.115 | TP | ||
F | 1.620 | 3.205 | 2.165 | 122.950 | TP | ||
G | 3.625 | 2.005 | 120.945 | ||||
H | 1.480 | 2.145 | 123.090 | BM | |||
SUM | 8.050 | 9.140 | 3.870 | 4.960 |
Arithmetic checks:
ΣBS-ΣFS=8.050-9.140=-1.09
ΣRise-ΣFall=3.87-4.96=-1.09
LRL-FRL=123.09-124.180=-1.09
Since, ΣBS-ΣFS=LRL – FRL=ΣRise-ΣFall
Therefore, the calculations are correct.
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