Two peg test: collimation inclined up or down

Given: staffs at C and D, 100 m apart. Setup 1: instrument at the mid-point F, readings C = 2.000 m, D = 3.000 m. Setup 2: instrument on the line DC produced, near C, with DF = 120 m, so CF = 120 − 100 = 20 m; readings C = 1.500 m, D = 2.750 m.

Step 1: True difference of level (mid-point setup)

Equal sight lengths remove the collimation error, so

$$\text{True difference (D below C)} = 3.000 - 2.000 = 1.000\ \text{m}$$

Step 2: Apparent difference from setup 2

$$\text{Apparent difference} = 2.750 - 1.500 = 1.250\ \text{m}$$

Step 3: Error and direction

$$e = 1.250 - 1.000 = 0.250\ \text{m for a difference of sight lengths } (120 - 20) = 100\ \text{m}$$

The error per metre of sight is $0.250/100 = 0.0025$ m/m (1 in 400; angle about 8′36″). The far staff D reads more than it should, so the line of collimation is inclined upward.

Step 4: Readings for a horizontal line of sight (instrument near C)

Error at C (20 m): $0.0025 \times 20 = 0.050$ m. Error at D (120 m): $0.0025 \times 120 = 0.300$ m.

$$C = 1.500 - 0.050 = 1.450\ \text{m}, \qquad D = 2.750 - 0.300 = 2.450\ \text{m}$$

Check: $2.450 - 1.450 = 1.000$ m, equal to the true difference.

Answer: the line of collimation is inclined upward; the readings for a horizontal line of sight are C = 1.450 m and D = 2.450 m. The cross-hair is then adjusted to read 1.450 on C (or 2.450 on D) with the bubble centred.