---
title: "Gradient P to Q from station A"
subject: "Surveying II"
chapter: 2
key: "semester_4/surveying_ii:gradient-p-to-q-from-a"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/gradient-p-to-q-from-a
site: BCT Engineering
---

# Gradient P to Q from station A

Surveying II · Chapter 2: Tacheometry

- Asked in: 2073 Magh, 6 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-p-to-q-from-a
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

Find the gradient from P to Q using data below,

| Instrument at | Staff at | Line | Bearing | Vertical Angle | Staff readings (m) |
|---|---|---|---|---|---|
| A | P | AP | 84°36' | 3°30' | 1.35, 2.10, 2.85 |
| A | Q | AQ | 142°24' | 2°45' | 1.955, 2.860, 3.765 |

The staff was held vertical to the line of sight in both cases.

## Short answer

With staff normal to line of sight, use D=Ks cosθ+h sinθ, V=Ks sinθ and RL=RL axis+V−h cosθ. PQ=162.16 m, level difference −1.234 m, so gradient from P to Q is falling 1 in 131.4 (0.76%).

## Answer

**Reading of the data:** the staff is held at right angles (normal) to the line of sight at both points, and no instrument constants are given, so $K=100$ and $C=0$ are taken. No heights of instrument are given, so only the **difference in level** between P and Q is found; the height of instrument cancels.

Staff normal to the line of sight, angle of elevation $\theta$, axial hair reading $h$ ($C=0$). The axial reading lies along the staff, which is tilted by $\theta$ from the vertical, so it adds $h\sin\theta$ to the horizontal distance and $h\cos\theta$ vertically:

$$
D=Ks\cos\theta+h\sin\theta,\qquad V=Ks\sin\theta,\qquad \text{RL of staff point}=\text{RL of axis}+V-h\cos\theta
$$

where $s$ is the staff intercept, $\theta$ the vertical angle and $K=100$.

#### Step 1: Horizontal distance and vertical component of each sight
**Point P** (bearing 84°36', angle of elevation $3^\circ 30'$, $h=2.100$ m):

$$
\begin{aligned}
s&=2.850-1.350=1.500\ \text{m}\\
D&=(100\times1.500)\cos 3^\circ 30'+2.100\sin3^\circ30'=149.720+0.128=149.848\ \text{m}\\
V&=(100\times1.500)\sin 3^\circ 30'=9.157\ \text{m},\qquad h\cos\theta=2.096\ \text{m}
\end{aligned}
$$

**Point Q** (bearing 142°24', angle of elevation $2^\circ 45'$, $h=2.860$ m):

$$
\begin{aligned}
s&=3.765-1.955=1.810\ \text{m}\\
D&=(100\times1.810)\cos 2^\circ 45'+2.860\sin2^\circ45'=180.792+0.137=180.929\ \text{m}\\
V&=(100\times1.810)\sin 2^\circ 45'=8.684\ \text{m},\qquad h\cos\theta=2.857\ \text{m}
\end{aligned}
$$

#### Step 2: Difference in level
The height of instrument is the same for both sights, so it cancels:

$$
\begin{aligned}
\Delta H&=(V_{Q}-h_{Q}\cos\theta_Q)-(V_{P}-h_{P}\cos\theta_P)\\
&=(8.684-2.857)-(9.157-2.096)=5.827-7.061=-1.234\ \text{m}
\end{aligned}
$$

#### Step 3: Horizontal distance PQ
The angle at A between the two lines is the difference of the bearings: $142^\circ24'-84^\circ36'=57^\circ 48'$.

$$
\begin{aligned}
PQ&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\
&=\sqrt{149.848^2+180.929^2-2\times149.848\times180.929\cos 57^\circ 48'}=162.158\ \text{m}
\end{aligned}
$$

#### Step 4: Gradient
$$
\text{Gradient}=\frac{\Delta H}{PQ}=\frac{1.234}{162.158}=0.00761
$$

The line from P to Q is falling: 1 in 131.4, i.e. 0.76 %, an angle of $0^\circ 26'09''$ with the horizontal.

**Answer:** PQ = 162.16 m; difference in level = -1.234 m; gradient falling 1 in 131.4 (0.76 %, 0°26').

If the staff is instead taken as vertical, the corresponding result is $PQ=161.81$ m and gradient 1 in 132.0 (difference in level $-1.226$ m), which agrees closely because the vertical angles are small.

## Check yourself

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

1. When the staff is held normal to the line of sight (C = 0), which formulae apply?
   A. D = Ks cosθ, V = Ks sinθ
   B. D = Ks cos^2θ, V = (1/2)Ks sin2θ
   C. D = Ks sinθ, V = Ks cosθ
   D. D = Ks, V = Ks tanθ
   Answer: A. With the staff normal, no cosθ correction to the intercept is needed, so D = Ks cosθ and V = Ks sinθ. (Key point: Staff normal formulae)

2. Why is the height of instrument not needed to find the level difference between P and Q?
   A. It is always zero
   B. It is the same for both sights, so it cancels in the difference
   C. It is replaced by the bearing
   D. It is equal to the middle hair reading
   Answer: B. Both points are sighted from the same axis, so RL of axis appears in both and cancels. (Key point: Height of instrument cancels)

3. How is the horizontal distance PQ found from D1 and D2?
   A. Add D1 and D2
   B. Subtract D2 from D1
   C. Cosine rule using the angle between bearings (57°48')
   D. Pythagoras with D1 and D2 regardless of angle
   Answer: C. A, P, Q form a triangle with included angle equal to the bearing difference, so the cosine rule applies. (Key point: Horizontal distance PQ)

4. How is the gradient of PQ computed?
   A. Difference in level divided by D1
   B. Horizontal distance PQ divided by difference in level
   C. Vertical angle difference
   D. Difference in level divided by horizontal distance PQ
   Answer: D. Gradient = ΔH / PQ, here 1.233/162.030, falling about 1 in 131. (Key point: Gradient)

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