This method consists in determining the difference of level between consecutive points by comparing each point, after the first, with that immediately preceding it. The difference between their staff readings indicates rise or fall depending on whether the staff reading at the point is smaller or greater than that at the preceding point. The reduced level of each point is then found by adding the rise to, or subtracting the fall from, the reduced level of the preceding point.
It is to be noted that the terms ‘rise’ and ‘fall’ always refer to rise or fall from the first point to the second point, second point to the third point and not conversely.
The steps involved are as follows:
- It consists in determining the difference of levels between the consecutive points by comparing their staff readings.
- Obtain the rise or fall by calculating the difference between the consecutive staff readings. Rise is indicated if the back sight is more than the fore sight, and a fall if the back sight is less than the fore sight.
- Find out the reduced levels of each point by adding the rise to, or by subtracting the fall from, the reduced level of the preceding point.
Check: ΣBS – ΣFS = ΣRise – ΣFall = Last RL – First RL
Example-2
The following readings were extracted from a level field book.
ΣBS = 5.430; ΣFS = 4.235; ΣRise = 3.955; ΣFall = 2.760
Here also the reduced level of the bench mark is assumed as 100.00 and the back sight to the bench mark is taken. Then the staff is held at B and reading recorded as intermediate sight.
RL of B = BM – Fall
(As the intermediate sight at B is greater than the back sight at A)
Then RL of C = RL of B – Fall of C
(Fall of C = IS at C – IS at B)
Now there is a change point and so the back sight and fore sight are to be taken, and as the intermediate sight (i.e., 2.540) is greater than fore sight (i.e., 0.385) there is a rise.
Rise = 2.540 – 0.385 = 2.155
RL of D = RL of C + Rise = 98.125 + 2.155 = 100.280
Now there is another change point and here also, as the back sight (3.200) is greater than the F.S (1.400), there is a rise.
Rise = 3.200 – 1.400 = 1.800
RL of E = RL of D + Rise = 100.280 + 1.800 = 102.080
Intermediate sight is taken with the staff at F and here as the intermediate sight (i.e., 2.000) is greater than back sight (i.e., 1.565) there is a fall.
Fall = 2.000 – 1.565 = 0.435
RL of F = RL of E – Fall = 102.080 – 0.435 = 101.645
This procedure is repeated till the last point is reached and the usual checks are applied to assess the correctness of the calculations.
Check: ΣBS – ΣFS = ΣRise – ΣFall = Last RL – First RL = 1.195
Problem-1
Given are the levels recorded in a levelling book. Complete the levels using rise and fall method.
Solution: The levels are computed as below using rise and fall method.
Check: ΣBS – ΣFS = Last RL – First RL = ΣRise – ΣFall
6.44 – 7.91 = –1.47, 192.15 – 193.62 = –1.47, 7.27 – 8.74 = –1.47
Hence checked.
Problem-2
Given are the levels recorded in a field book. Compute the levels using rise and fall method.
Solution: The levels are computed as below using the rise and fall method.
Check: ΣBS – ΣFS = Last RL – First RL = ΣRise – ΣFall
Σ6.00 – Σ3.42 = 2.58, 25.68 – 23.10 = 2.58, 2.58 – 0.00 = 2.58
Hence checked.
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