Reciprocal Levelling Solution with Collimation Error

This solution details the calculation of the corrected staff reading in a reciprocal levelling operation, accounting for collimation error.

Understanding the Problem

• The task is to find the correct staff reading on station B when the instrument is positioned near station A.
• The levelling operation involves reciprocal readings, meaning readings are taken from two different instrument stations.
• A collimation error is present in the instrument, which needs to be accounted for.
• All other errors are assumed to be negligible.

Data and Formulas

The observed staff readings are provided in the table:

Let:

• $r_{A1}$ = Reading on staff at A, instrument near A (BS) = 1.275 m
• $r_{B1}$ = Reading on staff at B, instrument near A (FS) = 2.005 m
• $r_{A2}$ = Reading on staff at A, instrument near B (FS) = 1.040 m
• $r_{B2}$ = Reading on staff at B, instrument near B (BS) = 1.660 m
• $h$ = True difference in level ($RL_B - RL_A$)
• $D$ = Distance between stations A and B.
• $c$ = Collimation error per unit distance.
• $X = c \times D$ = Total effect of collimation error over distance $D$. We assume distances from instrument to staff at the same station are negligible compared to the distance between stations.

The apparent differences in level are affected by collimation error ($X$):

• From position 1 (near A): The observed difference $r_{B1} - r_{A1}$ relates to $h$ as $r_{B1} - r_{A1} = h + X$.
• From position 2 (near B): The observed difference $r_{A2} - r_{B2}$ relates to $h$ as $r_{A2} - r_{B2} = -h + X$.

Calculations

1. Calculate the apparent differences in level for each position:
   • Position 1 (near A): $r_{B1} - r_{A1} = 2.005 \, \text{m} - 1.275 \, \text{m} = 0.730 \, \text{m}
   • Position 2 (near B): $r_{A2} - r_{B2} = 1.040 \, \text{m} - 1.660 \, \text{m} = -0.620 \, \text{m}
2. Set up the equations using the observed differences:
   1. $0.730 = h + X$
   2. $-0.620 = -h + X$
3. Solve for $X$ by adding the two equations: $(0.730) + (-0.620) = (h + X) + (-h + X)$ $0.110 = 2X$ $X = \frac{0.110}{2} = 0.055 \, \text{m}$
4. The question asks for the correct staff reading on station B when the instrument is near A. This is the observed reading ($r_{B1}$) corrected by subtracting the collimation error effect ($X$) relevant to that specific sight (to staff B from instrument near A). $ \text{Corrected reading} = r_{B1} - X $ $ \text{Corrected reading} = 2.005 \, \text{m} - 0.055 \, \text{m} $ $ \text{Corrected reading} = 1.950 \, \text{m} $

Final Answer

The correct staff reading over station B when the instrument is set very near to station A is 1.950 m.