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5 Different Numericals on Rise and Fall Method : Engineering Surveying

 

The rise and fall method calculates the difference in elevation between consecutive points in leveling work. Some of the points you have to know before starting numerical are:

 

a. Backsight

The first reading after seeing the instrument is called Backsight. It is taken at a known point, such as Benchmark.

It is denoted as B.S.

 

b. Intermediate Sight

All readings taken between B.S and F.S are known as intermediate sights.

It is denoted as I.S.

 

c. Foresight

The last reading after setting the instrument is known by foresight.

It is denoted as F.S.

 

d. Benchmark

A benchmark is a point with known elevation. 1

It is denoted as B.M.

 


 

1. Steps in Numerical Calculation  

 

a. Determine  rise or fall using:

B.S – I.S  or I.S – F.S

 

b. If:

B.S – I.S  or I.S – F.S = +ve →Rise

B.S – I.S  or I.S – F.S = -ve →Fall

 

c. Determine new RL:

New R.L = Old R.L – Fall = Old R.L + Rise

 

d. Arithmetic checking:

∑B.S – ∑F.S = ∑Rise – ∑Fall = Last R.L – First R.L

 


 

2. Numerical Examples of the Rise and Fall Method  

 

Question no. 1:

The following readings were observed successively with a leveling instrument. The instrument was shifted after the fifth and eleventh readings.

0.485, 1.210, 1.635, 3.395, 3.775, 0.650, 1.400, 1.795, 2.575, 3.375, 3.895, 1.735, 0.635, 1.605 m

Determine the R.L. of various points and show the entries in a level book if the R.L. of the first point is 100m, using the rise and fall method.

 

Solution,

( Note: We request mobile users to see numerical on desktop view. )

S.N. B.S I.S F.S Rise Fall RL Remarks
1 0.485 100 B.M
2 1.210 0.725 99.275
3 1.635 0.425 98.850
4 3.395 1.76 97.09
5 0.650 3.775 0.38 96.71 C.P.1
6 1.400 0.75 95.96
7 1.795 0.395 95.565
8 2.575 0.78 94.785
9 3.375 0.80 93.985
10 1.735 3.895 0.52 93.465 C.P.2
11 0.635 1.10 94.565
12 1.605 0.97 93.595
Total 2.87 9.275 1.10 7.505

( Note: C.P. denotes Changing Point)

 

Arithmetic Check

∑B.S – ∑F.S = ∑Rise – ∑Fall = Last R.L – First R.L

2.870 – 9.275 = 1.1 – 7.505 = 93.595 – 100.00

– 6.405 = -6.404 = – 6.405

Hence checked.

 

 

Question no. 2:

The data from the survey are shown below. Use the Rise and Fall method to reduce the data. Use arithmetic checks to support your answer.

Station Point B.S I.S F.S Rise Fall R.L Remarks
1 TBM 0.771
1,2 A 0.802 1.552
2 B 2.311
2,3 C 3.580 1.990
3 D 1.220
3 E 3.675
3,4 F 2.408 4.02
4 G 0.339
4 H 0.157

 

Solution;

( Note: We request mobile users to see numerical on desktop view. )

Station Point B.S I.S F.S Rise Fall R.L Remarks
1 TBM 0.771 43.000 B.M
1,2 A 0.802 1.552 (0.771-1.552)
0.781
(43.00-0.781)
42.219
C.P.1
2 B 2.311 (0.802-2.311)
1.509
(42.219-1.509)
40.710
2,3 C 3.580 1.990 (2.311-1.990)
0.321
(40.710 +0.321)
41.031
C.P.2
3 D 1.220 (3.580-1.220)
2.360
(41.031 +2.360)
43.391
3 E 3.675 (1.220-3.675)
2.455
(43.391-2.455)
40.936
3,4 F 2.408 4.02 (3.675-4.020)
0.345
(40.936-0.345)
40.591
C.P.3
4 G 0.339 (2.408-0.339)
2.069
(40.591 +2.069)
42.660
4 H 0.157 (0.339-0.157)
0.182
(42.660 +0.182)
42.842
Total 7.561 7.719 4.932 5.090

 

Arithmetic check:

1. Last R.L – First R.L = 42.842 – 43.000 = -0.158 m

2. ∑B.S – ∑F.S = 7.561 – 7.719 = -0.158 m

3. ∑Rise – ∑Fall = 4.932 – 5.090 = -0.158 m

Hence checked.

 

 

Question no. 3:

The following readings were taken with a level and 4m staff having initial B.M=100 m.

0.578, 0.933, 1.768, 2.450, 2.005, 0.567, 1.888, 1.181, 3.679, 0.612, 0.705, 1.810

The instrument is shifted after the 5th and 9th reading. Determine the R.L. of various points.

Solution,

( Note: We request mobile users to see numerical on desktop view. )

S.N. B.S I.S F.S Rise Fall R.L Remarks
1 0.578 100 B.M
2 0.933 0.355 99.645
3 1.768 0.835 98.81
4 2.450 0.682 98.128
5 0.567 2.005 0.445 98.573 C.P.1
6 1.888 1.321 97.252
7 1.181 0.707 97.959
8 0.612 3.679 2.498 95.461 C.P.2
9 0.705 0.093 95.368
10 1.810 1.105 94.263
Total 1.7578 7.494 1.152 6.889

 

Arithmetic check:

∑B.S – ∑F.S = ∑Rise – ∑Fall = Last R.L – First R.L

1.7578 – 7.494 = 1.152 – 6.889 = 94.263 – 100

-5.737 = -5.737 = -5.737

Hence checked.

 

 

Question no. 4:

The following readings (in meters) were taken with a leveling instrument using a staff of 4m long.

2.0, 2.5, 3.1, 3.9, 0.6, 1.3, 2.0, 2.8, 2.1, 3.1, 0.9, 3.0, 1.2, 2.3, 0.8

The instrument was shifted after the 4th, 9th, 11th, and 13th reading. The first reading was taken on BM of RL 100m. Calculate the RL of all points.

Solution,

( Note: We request mobile users to see numerical on desktop view. )

S.N.  B.S I.S F.S Rise Fall R.L Remarks
A 2.0 100 B.M
B 2.5 0.5 99.5
C 3.1 0.6 98.9
D 0.6 3.9 0.8 98.1 C.P.1
E 1.3 0.7 97.4
F 2.0 0.7 96.7
G 2.8 0.8 95.9
H 3.1 2.1 0.7 96.6 C.P.2
I 3.0 0.9 2.2 98.8 C.P.3
J 2.3 1.2 1.8 100.6 C.P.4
K 0.8 1.5 102.1
Total 8.9 6.2 4.1

 

Arithmetic check:

∑B.S – ∑F.S = ∑Rise – ∑Fall = Last R.L – First R.L

11 – 8.9 = 6.2 – 4.1 = 102.1 – 100

2.1 = 2.1 = 2.1

Hence checked.

Read Also: 5R Principle of Solid Waste Management

 

 

Question no. 5 ( With Gradient Calculation):

The following consecutive readings were taken with a level and 5m leveling staff on continuously sloping ground at a common interval of 20m. 0.385, 1.030, 1.925, 2.825, 3.730,4.685, 0.625, 2.005, 3.110, 4.485. The reduced level of the first point was 208.125m. Rule out a page of the level field book and enter the above readings. Calculate the reduced levels of the points by the rise and fall method. Also find the gradient of the line joining the first and last point.

Solution,

( Note: We request mobile users to see numerical on desktop view. )

S.N. B.S I.S F.S Rise Fall R.L Remarks
1 0.385 208.125 B.M
2 1.030 0.645 207.480
3 1.925 0.895 206.585
4 2.825 0.900 205.685
5 3.730 0.905 204.780
6 0.625 4.685 0.955 203.825 C.P.
7 2.005 1.380 202.445
8 3.110 1.105 201.340
9 4.485 1.375 199.965
Total 1.010 9.17 0 8.160

 

Arithmetic check:

∑B.S – ∑F.S = ∑Rise – ∑Fall = Last R.L – First R.L

-8.160 = -8.160 = -8.160

Hence checked.

 

Horizontal Distance = (S.N. – 1) * Interval = (9-1) * 20 = 160 m

Gradient = ( Last RL- First RL ) / Horizontal Distance

= ( 199.965-208.125) / 160

= -8.16/160

= -1 / (160/8.16)

= -1/19.60

= 1 in 19.60 ( Falling )

Note:

If a gradient is positive = Rising

If a gradient is negative = Falling

 

Written By Er. Madhu Krishna Poudel
Verified By Er. Surakchya Gynawali
References
  1. Benchmark []
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