RL of hill station P from stations R and Q

Assumptions: the RL of the BM is not given, so it is taken as $RL_{BM}$; the final answer is also given for $RL_{BM} = 1000.000$ m. The stations are in line in the order R, Q, P with $QR = 125$ m (horizontal). The staff reading 1.250 m on the BM is a horizontal back-sight from Q.

Axis of Q and axis of R

• Axis of Q $= RL_{BM} + 1.250$
• R sights the vane at Q (rod height at Q = 1.75 m): depression $= 98^\circ15' - 90^\circ = 8^\circ15'$. Ground at Q $= (RL_{BM} + 1.250) - 1.420$. The vane is at ground Q + 1.75 $= RL_{BM} + 1.580$. Vane level $=$ axis R $- 125\tan 8^\circ15'$, so axis R $= RL_{BM} + 1.580 + 125\tan 8^\circ15' = RL_{BM} + 19.704$ m

Elevation angles of P

• From Q: $90^\circ - 62^\circ30' = 27^\circ30'$ ($\tan = 0.52057$)
• From R: $90^\circ - 70^\circ45' = 19^\circ15'$ ($\tan = 0.34922$)

Horizontal distance and RL of P

Let $x$ = horizontal distance from R to P (Q is between R and P):

$$x = \frac{(axis_Q - axis_R) - 125\tan 27^\circ30'}{\tan 19^\circ15' - \tan 27^\circ30'} = \frac{(1.250 - 19.704) - 125\times 0.52057}{0.34922 - 0.52057} = 487.449\ \text{m}$$

(these use $RL_{BM} = 0$ as the datum, i.e. levels are measured from the BM)

$$RL_P = RL_{BM} + 19.704 + 487.449\times 0.34922 = RL_{BM} + 189.929\ \text{m}$$

Check from Q: $RL_{BM} + 1.250 + (487.449 - 125)\times 0.52057 = RL_{BM} + 189.929$ m.

Answer: RL of the top of the hill station P = RL of BM + 189.929 m. For RL of BM = 1000.000 m, RL of P = 1189.929 m.