---
title: "Gradient between P and Q (constants 100, 0)"
subject: "Surveying II"
chapter: 2
key: "semester_4/surveying_ii:tacheometer-gradient-pq-from-r-100"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/tacheometer-gradient-pq-from-r-100
site: BCT Engineering
---

# Gradient between P and Q (constants 100, 0)

Surveying II · Chapter 2: Tacheometry

- Asked in: 2070 Bhadra, 8 marks
- Video: [Calculate the horizontal distance between stations and RL of B when inst. constant is 100](https://www.youtube.com/watch?v=5BOguNaD3ZA) — SBK Concept
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/tacheometer-gradient-pq-from-r-100
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

The following observation were made with a tacheometer. The staff was held vertical (constants are 100 and 0).

| Inst. St | Staff St | Bearing | Vertical angle | Staff Readings |
|---|---|---|---|---|
| R | P | 100° | +8°20' | 2.60, 1.85, 1.10 |
| R | Q | 200° | -2°30' | 2.50, 1.91, 1.32 |

Find the gradient between P and Q.

## Short answer

Using D=Ks cos²θ and V=½Ks sin2θ, then cosine rule for PQ. PQ = 203.57 m, ΔH = -26.712 m, gradient falling 1 in 7.6 (13.12%).

## Answer

$K=100$, $C=0$. The height of instrument is not given and cancels in the difference of levels. The middle reading is the axial hair reading $h$.

Staff held vertical, line of sight inclined at angle $\theta$ (additive constant zero, which is the case for an anallatic lens or $C=0$):

$$
D=Ks\cos^2\theta,\qquad V=\tfrac12Ks\sin2\theta
$$

where $s$ is the staff intercept, $\theta$ the vertical angle and $K=100$, $C=0$.

#### Step 1: Horizontal distance and vertical component of each sight
**Point P** (bearing 100°00', angle of elevation $8^\circ 20'$):

$$
\begin{aligned}
s&=2.600-1.100=1.500\ \text{m}\\
D&=100\times1.500\times\cos^2 8^\circ 20'=146.849\ \text{m}\\
V&=\tfrac12\times100\times1.500\times\sin 2(8^\circ 20')=21.510\ \text{m}
\end{aligned}
$$

**Point Q** (bearing 200°00', angle of depression $2^\circ 30'$):

$$
\begin{aligned}
s&=2.500-1.320=1.180\ \text{m}\\
D&=100\times1.180\times\cos^2 2^\circ 30'=117.775\ \text{m}\\
V&=\tfrac12\times100\times1.180\times\sin 2(2^\circ 30')=-5.142\ \text{m}
\end{aligned}
$$

#### Step 2: Difference in level
The height of instrument is the same for both sights, so it cancels (RL of point $=$ RL of axis $+V-h$):

$$
\begin{aligned}
\Delta H&=(V_{Q}-h_{Q})-(V_{P}-h_{P})\\
&=(-5.142-1.910)-(+21.510-1.850)=-26.712\ \text{m}
\end{aligned}
$$

#### Step 3: Horizontal distance PQ
The angle at R between the two lines is the difference of the bearings, $100^\circ 00'$ (100°00' and 200°00').

$$
\begin{aligned}
PQ&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\
&=\sqrt{146.849^2+117.775^2-2\times146.849\times117.775\cos 100^\circ 00'}=203.574\ \text{m}
\end{aligned}
$$

#### Step 4: Gradient
$$
\text{Gradient}=\frac{\Delta H}{PQ}=\frac{26.712}{203.574}=0.13122
$$

The line from P to Q is falling: 1 in 7.6, i.e. 13.12 %, an angle of $7^\circ 28'32''$ with the horizontal.

**Answer:** PQ = 203.57 m; difference in level = -26.712 m; gradient falling 1 in 7.6 (13.12 %, 7°29').

## Check yourself

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

1. The readings are 2.60, 1.85, 1.10. What is the intercept s and axial reading h?
   A. s = 1.500 m, h = 1.85 m
   B. s = 1.85 m, h = 1.500 m
   C. s = 3.70 m, h = 1.10 m
   D. s = 0.75 m, h = 2.60 m
   Answer: A. s = top - bottom and the middle hair is the axial reading. (Key point: Stadia hairs order)

2. Why is V negative for Q?
   A. The staff reading is smaller
   B. Q is sighted at an angle of depression of 2°30'
   C. The bearing exceeds 180°
   D. The intercept is smaller than at P
   Answer: B. V = (1/2)Ks sin2θ takes the sign of θ. (Key point: Sign of V)

3. What angle is used in the cosine rule?
   A. 300°
   B. 80°
   C. 100° (200° - 100°)
   D. 200°
   Answer: C. The included angle at R is the bearing difference. (Key point: Angle between PQ lines)

4. What is the direction and gradient of PQ?
   A. Rising, about 1 in 7.6
   B. Rising, about 1 in 131
   C. Falling, about 1 in 39
   D. Falling, about 1 in 7.6
   Answer: D. ΔH = -26.712 m over 203.57 m gives 13.12 % falling. (Key point: Gradient result)

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