Two peg test, AB = 40 m, near B

Data: AB = 40 m. Setup 1: instrument at the mid-point, A = 1.464, B = 1.154. Setup 2: instrument near B on the inner side (between A and B), A = 1.642, B = 1.321.

Assumption: the distance of the setup from B is not stated, so the instrument is taken as practically at B (error at the near staff B negligible). The whole error then appears on the far staff A (distance 40 m).

Step 1: True difference of level

$$\text{True difference} = 1.464 - 1.154 = 0.310\ \text{m}\ (\text{B higher than A})$$

Step 2: Apparent difference from the second setup

$$1.642 - 1.321 = 0.321\ \text{m}$$

Step 3: Collimation error and precision

$$e = 0.321 - 0.310 = 0.011\ \text{m in 40 m}$$ $$\text{Collimation precision} = \frac{0.011}{40} = 0.000275 = \frac{1}{3636}\ \ (\text{i.e. } 27.5\ \text{mm per 100 m})$$

The usual permissible limit for ordinary levelling is about 10 mm per 100 m (1 in 10,000). The error is larger, so the instrument needs adjustment.

The far staff A reads too high (the apparent difference is greater than the true one), so the line of collimation is inclined upward.

Step 4: Correct reading on A for the second setup

$$A = 1.642 - 0.011 = 1.631\ \text{m}, \qquad B = 1.321\ \text{m}$$

Check: $1.631 - 1.321 = 0.310$ m = true difference.

Answer: collimation precision = 1 in 3636 (error 0.011 m in 40 m, line of collimation inclined upward); the staff reading on A for a horizontal line of sight is 1.631 m (B = 1.321 m).

If the setup is a distance $x$ from B, the correction is $0.011 \times (40 - x)/(40 - 2x)$ at A and $0.011\, x/(40 - 2x)$ at B.