RL of B for bridge axis AB (58.60 m)

The axis AB is crossed by reciprocal levelling: two sets of readings are taken, one with the instrument near A and one near B. The mean of the two apparent differences gives the true difference of level, free from collimation, curvature and refraction errors.

Check on the hair readings

For each pair of readings, the middle hair should equal the mean of the top and bottom hairs:

The hair readings are consistent, so the middle-hair readings are used.

Apparent differences (rise of B above A $=$ reading on A $-$ reading on B)

$$H_1 = 1.559 - 1.585 = -0.026\ \text{m} \quad (\text{instrument near A})$$ $$H_2 = 1.647 - 1.671 = -0.024\ \text{m} \quad (\text{instrument near B})$$

The two values agree within 2 mm, so the work is acceptable.

True difference and RL of B

$$H = \frac{H_1 + H_2}{2} = \frac{-0.026 + (-0.024)}{2} = -0.025\ \text{m}\ (\text{B is lower than A})$$ $$\text{RL of B} = 1295.000 - 0.025 = 1294.975\ \text{m}$$

Check: the error in the far reading is $e = (H_2 - H_1)/2 = +0.001$ m, which is only 1 mm in about 60 m, so the instrument is well adjusted. The mean of the top and bottom hairs gives $-0.0265$ and $-0.0245$, mean $-0.0255$, i.e. RL of B = 1294.9745 m, which agrees with 1294.975 m to 0.5 mm.

Answer: RL of B = 1294.975 m.