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

Let’s apply the principle of reciprocal leveling.

First setup near point P:
\[ \text{Backsight (BS) at P} = 2.165\, \text{m} \] \[ \text{Foresight (FS) at Q} = 3.810\, \text{m} \]
\[ \Delta h_1 = \text{BS} - \text{FS} = 2.165 - 3.810 = -1.645\, \text{m} \]

Second setup near point Q:
\[ \text{BS at Q} = 2.355\, \text{m} \] \[ \text{FS at P} = 0.910\, \text{m} \]
\[ \Delta h_2 = \text{BS} - \text{FS} = 2.355 - 0.910 = 1.445\, \text{m} \]

Now, the true difference of level is the average of the two:
\[ \Delta h_{\text{true}} = \frac{-1.645 + 1.445}{2} = \frac{-0.200}{2} = -0.100\, \text{m} \]

This is the correction due to collimation error. Now, calculate the corrected difference of level from either setup by applying half the error:
\[ \text{Corrected difference (from first setup)} = -1.645 + 0.100 = -1.545\, \text{m} \]

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

Final Answer: \(\boxed{1.545\, \text{m}}\)