Collimation: pegs 60 m apart, Q near B

Data: A and B are 60 m apart; with the level at the mid-point the true difference of level is 0.320 m (reading on A is greater than on B by 0.320 m, since the apparent difference is also positive). Setup 2: level at Q on AB, 6 m from B (so 54 m from A): readings A = 1.387 m, B = 1.069 m.

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

With equal sight lengths the collimation error cancels, so

$$\text{True difference of readings (A} - \text{B)} = 0.320\ \text{m}$$

Step 2: Apparent difference from the second setup

$$\text{Apparent difference} = 1.387 - 1.069 = 0.318\ \text{m}$$

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 = 54 m and B = 6 m, so the difference of sight lengths is 48 m.

$$e = \text{apparent} - \text{true} = 0.318 - 0.320 = -0.002\ \text{m}$$ $$\text{Error per metre of sight} = \frac{0.002}{48} = 0.000042\ \text{m/m}, \qquad \text{collimation precision} = \frac{1}{24000}$$

This is 4.2 mm per 100 m of sight, which is within the usual permissible limit of about 10 mm per 100 m (1 in 10,000) for ordinary levelling, so the instrument needs no 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 $= -0.000042 \times 54 = -0.00225$ m; error in reading on B $= -0.000042 \times 6 = -0.00025$ m.

$$A = 1.387 - (-0.00225) = 1.38925\ \text{m}, \qquad B = 1.069 - (-0.00025) = 1.06925\ \text{m}$$

Check: A $-$ B $= 1.38925 - 1.06925 = 0.320$ m, equal to the true difference.

Answer: correct readings A = 1.389 m (1.38925), B = 1.069 m (1.06925); collimation precision 1 in 24,000 (the line of collimation is inclined downward, but the error is small, 4.2 mm per 100 m).