---
title: "Subtense bar, level difference and gradient A-B"
subject: "Surveying II"
chapter: 2
key: "semester_4/surveying_ii:subtense-bar-tacheometer-level-diff-gradient"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/subtense-bar-tacheometer-level-diff-gradient
site: BCT Engineering
---

# Subtense bar, level difference and gradient A-B

Surveying II · Chapter 2: Tacheometry

- Asked in: 2071 Magh, 8 marks
- Video: [Tacheometer problem Solution | Calculate Gradient of distance AB and the RLs of A and B](https://www.youtube.com/watch?v=tQDd7sLym-E) — Civil Guys
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/surveying_ii/subtense-bar-tacheometer-level-diff-gradient
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

Calculate the elevation difference and gradient between stations A and B from the given data which are observed by a tacheometer from station R. Staff was vertically held at A and subtense bar at B. The subtended angle between the instrument and 2 m long subtense bar was 00°42'15".

| Instrument Station | Sighted to | Bearing | Zenith angle | Staff readings (m) | Subtense bar height |
|---|---|---|---|---|---|
| R | A | 345°00' | 96°30' | 0.650, 1.250, 1.850 | X |
| R | B | 225°00' | 85°00' | X | 1.180 m |

## Short answer

For A: D=Ks cos²θ, V=½Ks sin2θ. For B: D = (l/2) cot(β/2) from subtense bar, V = D tanθ. Then cosine rule for AB. AB = 244.525 m, B is 27.804 m higher, gradient rising 1 in 8.8 (11.37%).

## Answer

**Reading of the data:** the tacheometer is taken as $K=100$, $C=0$ (not stated). The bar at B is horizontal and normal to the line of sight. The theodolite measures the angle it subtends in the horizontal plane, so the distance found from this angle is the horizontal distance, whatever the inclination of the sight. The height of instrument is the same for both sights and cancels in the difference of level. The bar height 1.180 m is the height of the bar's sighting target above B.

#### Step 1: Staff point A (stadia)
Zenith angle $96^\circ30'$ gives $\theta=-6^\circ30'$ (depression).

**Point A** (bearing 345°00', angle of depression $6^\circ 30'$):

$$
\begin{aligned}
s&=1.850-0.650=1.200\ \text{m}\\
D&=100\times1.200\times\cos^2 6^\circ 30'=118.462\ \text{m}\\
V&=\tfrac12\times100\times1.200\times\sin 2(6^\circ 30')=-13.497\ \text{m}
\end{aligned}
$$

#### Step 2: Bar point B (subtense)
Horizontal distance from the bar angle $\beta=0^\circ42'15''$ and bar length $l=2$ m:

$$
D_B=\frac{l}{2}\cot\frac{\beta}{2}=\cot\frac{0^\circ42'15''}{2}=162.732\ \text{m}
$$

Zenith angle $85^\circ00'$ gives $\theta=+5^\circ00'$ (elevation):

$$
V_B=D_B\tan5^\circ=162.732\times0.08749=+14.237\ \text{m}
$$

#### Step 3: Difference in level
$$
\begin{aligned}
\text{RL}_B-\text{RL}_A&=(V_B-1.180)-(V_A-1.250)\\
&=(+14.237-1.180)-(-13.497-1.250)=+27.804\ \text{m}
\end{aligned}
$$

#### Step 4: Horizontal distance AB
Bearings from R: A $=345^\circ$, B $=225^\circ$, so angle ARB $=120^\circ$.

$$
AB=\sqrt{118.462^2+162.732^2-2\times118.462\times162.732\cos120^\circ}=244.525\ \text{m}
$$

#### Step 5: Gradient
$$
\text{Gradient}=\frac{+27.804}{244.525}=0.11371
$$

**Answer:** B is **27.804 m higher** than A; AB = 244.525 m; gradient from A to B **rising 1 in 8.8 (11.37 %)**.

## Check yourself

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

1. How is the inclined distance computed from the bar angle β and bar length l?
   A. S = (l/2) cot(β/2)
   B. S = (l/2) tan(β/2)
   C. S = l tanβ
   D. S = l cosβ
   Answer: A. In the isosceles triangle, l/2 = S tan(β/2). (Key point: Subtense bar distance)

2. With the bar normal to the sight at elevation 5°, how are D and V found?
   A. D = S cos^2 5°, V = (1/2)S sin10°
   B. D = S cos5°, V = S sin5°
   C. D = S sin5°, V = S cos5°
   D. D = S, V = 0
   Answer: B. S is already the inclined distance, so it is only resolved by cos and sin. (Key point: Horizontal and vertical from the bar)

3. What is the vertical angle for the zenith angle 96°30' at A?
   A. +6°30' (elevation)
   B. +96°30'
   C. -6°30' (depression)
   D. -96°30'
   Answer: C. θ = 90° - 96°30' = -6°30'. (Key point: Zenith angle sign)

4. How is RL_B - RL_A formed?
   A. (VB + 1.180) - (VA + 1.250)
   B. VB - VA only
   C. (VB - 1.250) - (VA - 1.180)
   D. (VB - 1.180) - (VA - 1.250); here +27.750 m
   Answer: D. Bar target height 1.180 m at B and axial reading 1.250 m at A are the lowered heights; HI cancels. (Key point: Level difference)

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