Subtense bar, RL difference, gradient, bearing AB

Assumptions: the "vertical reading" is the zenith angle; the horizontal circle reading $0^\circ00'00''$ is on CA, which is taken as the reference direction (bearing of CA = 0°, since no bearing is given). The height of instrument at C is not given, but it cancels in the RL difference.

Distance CB by the subtense bar (b = 2 m, θ = 2°40'20")

$$D_{CB} = \frac{b}{2}\cot\frac{\theta}{2} = 1\times\cot(1°20'10") = 42.875\ \text{m}$$

Distance CA by stadia ($k = 100$, $c = 0$)

• $s = 1.839 - 1.459 = 0.380$ m, $z = 95^\circ10'$, $\theta = -5°10'$ (depression)
• $H_{CA} = 100\,s\sin^2 z = 37.692$ m
• $V_A = 50\,s\sin 2\theta = -3.408$ m

RL of A and B relative to the axis of C

$$RL_A - RL_{axis} = V_A - r_A = -3.408 - 1.649 = -5.057\ \text{m}$$

For B, $z = 70^\circ 23'$ so $\theta = +19°37'$ and $V_B = D_{CB}\tan\theta = 15.281$ m:

$$RL_B - RL_{axis} = 15.281 - 1.235 = 14.046\ \text{m}$$

Difference in RL

$$RL_B - RL_A = 14.046 - (-5.057) = 19.103\ \text{m}$$

B is higher than A by 19.103 m.

Length and gradient of AB

Angle ACB = $80^\circ24'20''$.

$$AB = \sqrt{37.692^2 + 42.875^2 - 2(37.692)(42.875)\cos 80^\circ24'20''} = 52.156\ \text{m}$$ $$\text{Gradient} = \frac{19.103}{52.156} = 1\text{ in }2.73\ (\text{rising from A to B})$$

Bearing of AB

Taking C as origin, CA along north: A = (0, 37.692), B = (42.275 E, 7.146 N).

$$\tan\beta = \frac{\Delta E}{\Delta N} = \frac{42.275}{-30.546}, \qquad \text{bearing of AB} = 125°51'00"$$

Answer: RL of B − RL of A = 19.103 m; gradient = 1 in 2.73; AB = 52.16 m; bearing of AB = 125°51'00" (if CA is taken as 0°; add the actual bearing of CA otherwise).