---
title: "Gradient P to Q from common point R"
subject: "Surveying II"
chapter: 2
key: "semester_4/surveying_ii:gradient-pq-common-point-r"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/gradient-pq-common-point-r
site: BCT Engineering
---

# Gradient P to Q from common point R

Surveying II · Chapter 2: Tacheometry

- Asked in: 2080 Chaitra, 2+2+2 marks
- Video: [Determination of gradient between two points in tacheometric surveying- part 1](https://www.youtube.com/watch?v=FJos96jcQPQ) — Gayathri VG
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/gradient-pq-common-point-r
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

From two unknown instrument stations P and Q tacheometric observations are taken for common point 'R' holding the staff vertical and following observations are noted.

| Inst. stn. | Sighted to | hi. (m) | Zenith angle | Bearing | Staff Reading |
|---|---|---|---|---|---|
| P | R | 1.380 | 81°00' | 60°30' | 1.000, 2.000, 3.000 |
| Q | R | 1.420 | 85°00' | 335°45' | 1.300, 2.450, 3.600 |

If RL of 'R' is 1200.00 m, compute the gradient between P and Q. Take K = 100 and C = 0.00.

## Short answer

Using D=Ks cos^2θ, V=1/2 Ks sin2θ, RL_P=1169.718 m, RL_Q=1181.060 m, PQ=286.384 m. Gradient from P to Q is rising 1 in 25.2 (3.96%).

## Answer

$K=100$, $C=0$, staff vertical: $D=Ks\cos^2\theta$, $V=\tfrac12Ks\sin2\theta$. The zenith angles give the vertical angles $\theta=90^\circ-z$: from P, $90^\circ-81^\circ=+9^\circ$; from Q, $90^\circ-85^\circ=+5^\circ$. The middle hair reading is the axial reading $h$.

#### Step 1: Distances and vertical components
**Point R from P** (bearing 60°30', angle of elevation $9^\circ 00'$):

$$
\begin{aligned}
s&=3.000-1.000=2.000\ \text{m}\\
D&=100\times2.000\times\cos^2 9^\circ 00'=195.106\ \text{m}\\
V&=\tfrac12\times100\times2.000\times\sin 2(9^\circ 00')=30.902\ \text{m}
\end{aligned}
$$

**Point R from Q** (bearing 335°45', angle of elevation $5^\circ 00'$):

$$
\begin{aligned}
s&=3.600-1.300=2.300\ \text{m}\\
D&=100\times2.300\times\cos^2 5^\circ 00'=228.253\ \text{m}\\
V&=\tfrac12\times100\times2.300\times\sin 2(5^\circ 00')=19.970\ \text{m}
\end{aligned}
$$

#### Step 2: RL of P and Q (RL of R = 1200.000 m)
$$
\text{RL}_R=\text{RL}_{station}+HI+V-h\ \Rightarrow\ \text{RL}_{station}=\text{RL}_R-HI-V+h
$$

$$
\begin{aligned}
\text{RL}_P&=1200.000-1.380-30.902+2.000=1169.718\ \text{m}\\
\text{RL}_Q&=1200.000-1.420-19.970+2.450=1181.060\ \text{m}
\end{aligned}
$$

#### Step 3: Horizontal distance PQ
Take P as the origin, with coordinates written as (N, E). R is $D_P=195.106$ m from P on a bearing of $60^\circ30'$. The bearing $335^\circ45'$ is from Q to R, so Q is found by going back from R a distance $D_Q=228.253$ m along that bearing:

$$
\begin{aligned}
R&=(96.075,\ 169.811)\\
Q&=R-D_Q(\cos335^\circ45',\ \sin335^\circ45')=(96.075-208.112,\ 169.811+93.748)=(-112.038,\ 263.559)\\
PQ&=\sqrt{(-112.038)^2+263.559^2}=286.384\ \text{m},\qquad \text{bearing of PQ}=113^\circ 01'48''
\end{aligned}
$$

#### Step 4: Gradient from P to Q
$$
\text{Gradient}=\frac{\text{RL}_Q-\text{RL}_P}{PQ}=\frac{+11.342}{286.384}=0.03960
$$

**Answer:** RL of P = 1169.718 m, RL of Q = 1181.060 m, PQ = 286.384 m; the gradient from P to Q is **rising 1 in 25.2 (3.96 %)**.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. From P the zenith angle is 81°00'. The vertical angle is:
   A. -9°00'
   B. +9°00' (elevation)
   C. +81°00'
   D. -81°00'
   Answer: B. θ = 90° - 81° = +9°. (Key point: Zenith to vertical)

2. RL of R is known, so the RL of each station is found from:
   A. RL_R + HI + V - h
   B. RL_R - HI - V + h
   C. RL_R + HI - V + h
   D. RL_R - HI + V - h
   Answer: B. RL_R = RL_station + HI + V - h, rearranged to RL_station = RL_R - HI - V + h. (Key point: RL of instrument station)

3. Why is a coordinate method used to find PQ?
   A. P and Q are on the same line as R
   B. The bearings are measured from P to R and from Q to R, so Q is offset from R in the reverse direction of the bearing Q to R
   C. The zenith angles are equal
   D. K is unknown
   Answer: B. Taking P as origin, R is plotted from P, then Q = R minus D_Q along the bearing Q to R. (Key point: Locating Q)

4. Gradient between P and Q is:
   A. (RL_Q - RL_P) / horizontal PQ
   B. (RL_R - RL_P) / D_P
   C. (V_P - V_Q) / PQ
   D. (HI_P - HI_Q) / PQ
   Answer: A. ΔH between stations divided by horizontal distance PQ gives the gradient. (Key point: Gradient)

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