Question

The following consecutive staff readings (in m) were taken with a dumpy level and a levelling staff at a common interval of 20 m: 0.385; 1.030; 1.925; 2.825; 3.730; 4.850; 1.045; 2.005; 3.330; 4.580. The dumpy level was shifted after taking the sixth reading. The gradient (in %) of the ground between the first and the last location of the readings is ______ (in integer).

Hint:

Reduce levels using the Height of Instrument method across the change point, then find the fall over the total distance to get the percentage gradient.

Answer 1

Correct Answer: -5

Step 1: Recognize the level book structure.
Ten staff readings are taken at 9 distinct points spaced 20 m apart along a line (the reading right after the instrument shift is a back sight to the SAME point where the last reading of the first setup was a fore sight, since that point is the "change point"). Using the Height of Instrument (HI) method: reading 1 is a back sight (BS) fixing the first HI, readings 2-5 are intermediate sights (IS), reading 6 is the fore sight (FS) closing setup 1, reading 7 is the BS of setup 2 (to the same change point as reading 6), readings 8-9 are IS, and reading 10 is the final FS.

Step 2: Take an arbitrary reduced level (RL) of 100.000 m for the first point, since only the difference in level matters for the gradient.
\[ HI_1 = RL_{P1}+BS_1 = 100.000+0.385=100.385 \]

Step 3: Reduce the levels of the intermediate points and the change point using \(RL = HI - \text{reading}\).
\(RL_{P2}=100.385-1.030=99.355\)
\(RL_{P3}=100.385-1.925=98.460\)
\(RL_{P4}=100.385-2.825=97.560\)
\(RL_{P5}=100.385-3.730=96.655\)
\(RL_{P6}=100.385-4.850=95.535\) (change point, closes setup 1)

Step 4: Start the second setup at the change point and reduce the remaining points.
\[ HI_2 = RL_{P6}+BS_2 = 95.535+1.045=96.580 \]
\(RL_{P7}=96.580-2.005=94.575\)
\(RL_{P8}=96.580-3.330=93.250\)
\(RL_{P9}=96.580-4.580=92.000\) (last point)

Step 5: Compute the gradient between the first and last points.
Fall in level \(= RL_{P1}-RL_{P9}=100.000-92.000=8.000\) m.
Horizontal distance \(=(9-1)\times20=160\) m (8 intervals of 20 m between 9 points).
\[ \text{Gradient} = \frac{-8.000}{160}\times100=-5\% \]

Final Answer:
The ground level falls steadily from the first point to the last, by 8 m over 160 m, giving a downward (negative) gradient.
\[ \boxed{\text{Gradient} = -5\%} \]