Question

In levelling between two points P and Q on opposite banks of a river, the level was set up near P, and the staff readings on P and Q were 2.165 m and 3.810 m, respectively. The level was then moved and set up near Q and the respective staff readings on P and Q were 0.910 m and 2.355 m. The true difference of level between P and Q is ____________m (rounded off to 3 decimal places).

Hint:

In reciprocal leveling, average the differences in height from both setups to eliminate errors due to collimation, refraction, and curvature.

Answer 1

Correct Answer: 1.54

Let’s apply the principle of reciprocal leveling.

First setup near point P:
Backsight (BS) at P = 2.165 m
Foresight (FS) at Q = 3.810 m

\[ \Delta h_1 = \text{BS} - \text{FS} = 2.165 - 3.810 = -1.645~\text{m} \]

Second setup near point Q:
Backsight (BS) at Q = 2.355 m
Foresight (FS) at P = 0.910 m

\[ \Delta h_2 = \text{BS} - \text{FS} = 2.355 - 0.910 = 1.445~\text{m} \]

True difference of level:
\[ \Delta h_{\text{true}} = \frac{-1.645 + 1.445}{2} = \frac{-0.200}{2} = -0.100~\text{m} \]

This represents the collimation error correction.

Corrected difference of level:
\[ \text{Corrected difference} = -1.645 + 0.100 = -1.545~\text{m} \]

Hence, the true difference in level between P and Q is:
\[ \Delta h = 1.545~\text{m} \quad (\text{Q is lower than P}) \]

Final Answer: 1.545 m