Subtense bar, elevation difference and gradient P-Q

Assumptions: the height of instrument at A is not given, so it cancels in the RL difference. "Subtense bar height 1.75 m" is taken as the height of the bar (the point sighted) above the ground at P. Staff is vertical; $k = 100$, $c = 0$.

Station P (subtense bar, b = 3 m, θ = 17'40")

$$D_{AP} = \frac{b}{2}\cot\frac{\theta}{2} = 1.5\cot(0°08'50") = 583.767\ \text{m}$$

$V_P = D_{AP}\tan(-10^\circ31') = -108.370$ m

Level of P above the axis of instrument: $RL_P - RL_{axis} = V_P - 1.75 = -110.120$ m

Station Q (staff, vertical angle +7°15')

• $s = 2.66 - 0.56 = 2.100$ m
• $H_{AQ} = k s\cos^2\theta = 100\times 2.10\times\cos^2 7^\circ15' = 206.656$ m
• $V_Q = \frac{ks}{2}\sin 2\theta = 50\times 2.10\times\sin 14^\circ30' = 26.290$ m
• $RL_Q - RL_{axis} = V_Q - 1.61 = 24.680$ m

Elevation difference

$$RL_Q - RL_P = 24.680 - (-110.120) = 134.800\ \text{m}$$

Q is higher than P by 134.800 m.

Distance PQ

Angle PAQ = $112^\circ15' - 37^\circ45' = 74^\circ30'$.

$$PQ = \sqrt{583.767^2 + 206.656^2 - 2(583.767)(206.656)\cos 74^\circ30'} = 564.811\ \text{m}$$

Gradient

$$\frac{134.800}{564.811} = 1\text{ in }4.19$$

Answer: elevation difference = 134.80 m (Q higher than P); PQ = 564.81 m; gradient = 1 in 4.19 rising from P to Q.