Collimation error with level mid-way and near B

Reading of the data (assumptions).

• "The distance to stations A and B was 60 m" is taken as 60 m from the mid-point level to each station, so AB = 120 m.
• With the level at the mid-point, A = 1.855 m and B = 1.625 m, so A reads 0.230 m more than B. In the second set-up the printed readings are A = 2.385 and B = 2.655, which would give A − B = −0.270 m, of opposite sign to 0.230 m. Taken literally, this means an error of −0.270 − 0.230 = −0.500 m in 120 m (1 in 240, inclined downward), far too large for a working level. The two readings are therefore taken as interchanged: A = 2.655 m and B = 2.385 m (apparent difference +0.270 m).
• "Near to station B" is taken as the level practically at B (distance from B about 0), so A is 120 m away.

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

With equal sight lengths the collimation error cancels, so

$$\text{True difference (A} - \text{B)} = 1.855 - 1.625 = 0.230\ \text{m}$$

Step 2: Apparent difference from the second setup

$$\text{Apparent difference (A} - \text{B)} = 2.655 - 2.385 = 0.270\ \text{m}$$

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 = 120 m and B = 0 m, so the difference of sight lengths is 120 m.

$$e = \text{apparent} - \text{true} = 0.270 - 0.230 = +0.040\ \text{m}$$ $$\text{Error per metre of sight} = \frac{0.040}{120} = 0.000333\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{3000}$$

This is 33.3 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 large and the error grows with distance (the far staff reads too much), so the line of collimation is inclined upward.

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

Error in the reading on A $= 0.000333 \times 120 = +0.040$ m; error in the reading on B $= 0.000333 \times 0 = 0$ m.

$$A = 2.655 - 0.040 = 2.615\ \text{m}, \qquad B = 2.385 - 0 = 2.385\ \text{m}$$

Check: A $-$ B $= 2.615 - 2.385 = 0.230$ m, equal to the true difference.

Answer: collimation error = 0.040 m in 120 m, i.e. 0.000333 m per metre (1 in 3000, 33 mm per 100 m); the sign is upward (the line of collimation is inclined upward, and the far staff A reads 0.040 m too much). Correct readings in the second set-up: A = 2.615 m, B = 2.385 m. (If the printed readings are used as they stand, the error is 0.500 m in 120 m, inclined downward.)