Collimation: level 5 m from B on AB produced

Data: A and B are 50 m apart. Setup 1 at the mid-point: A = 1.875 m, B = 1.790 m. Setup 2 at a point 5 m beyond B on the line AB produced (so B is 5 m and A is 55 m from the instrument): A = 1.630 m, B = 1.560 m.

Step 1: True difference of level (instrument at the mid-point)

With equal sight lengths the collimation error cancels, so

$$\text{True difference (A} - \text{B)} = 1.875 - 1.790 = 0.085\ \text{m}$$

Step 2: Apparent difference from the second setup

$$\text{Apparent difference (A} - \text{B)} = 1.630 - 1.560 = 0.070\ \text{m}$$

Step 3: Error and collimation precision

The reading error of a staff is proportional to its distance from the instrument. In the second setup the distances are A = 55 m and B = 5 m, so the difference of sight lengths is 50 m.

$$e = \text{apparent} - \text{true} = 0.070 - 0.085 = -0.015\ \text{m}$$ $$\text{Error per metre of sight} = \frac{0.015}{50} = 0.0003\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{3333}$$

This is 30.0 mm per 100 m of sight, which is more than the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs adjustment.

The readings are too small and the error grows with distance (the far staff reads too little), so the line of collimation is inclined downward.

Step 4: Correct readings for a horizontal line of sight (second setup)

Error in reading on A $= -0.0003 \times 55 = -0.0165$ m; error in reading on B $= -0.0003 \times 5 = -0.0015$ m.

$$A = 1.630 - (-0.0165) = 1.6465\ \text{m}, \qquad B = 1.560 - (-0.0015) = 1.5615\ \text{m}$$

Check: A $-$ B $= 1.6465 - 1.5615 = 0.085$ m, equal to the true difference.

Answer: with a horizontal line of sight the readings in the second setup are A = 1.6465 m and B = 1.5615 m; the line of collimation is inclined downward and the precision is 1 in 3333 (30 mm per 100 m), which needs adjustment.