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

# Subtense bar and tacheometer, A-B distance and level

Surveying II · Chapter 2: Tacheometry

- Asked in: 2070 Magh, 10 marks
- Video: [Distance Measurement (Tacheometry by Subtense Bar)](https://www.youtube.com/watch?v=QcAo13TQ6OY) — Surveying With MamJ
- 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-2m-theodolite-c-ab
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

A 2 m long subtense bar was placed above station B and the angle subtended at station A was read as 02°40'20". Intermediate level information was later recorded using a theodolite with tachemetric constants 100 and 0 at station C and the staff was held vertical. The following data were recorded on to stations A and B.

| Inst st'n | Sighted to | Horizontal circle | Vertical angle | Staff readings (m) |
|---|---|---|---|---|
| C (hi = 1.55 m) | A | 00°00'00" | -05°10'00" | 1.459, 1.649, 1.839 |
| | B | 80°24'20" | +10°23'30" | -, 1.235, - |

Find the difference in elevation and distance between A and B, the horizontal angle ACB was 60°00'00".

## Short answer

Subtense bar gives AB = 42.875 m. Stadia gives CA = 37.692 m, V_A = -3.408 m. Cosine rule with angle 60° gives CB = 46.644 m, V_B = 8.554 m. ΔH = +12.376 m, B higher.

## Answer

**Reading of the data:** the distance AB comes from the subtense bar (angle measured at A, bar at B); the levels come from the theodolite at C ($K=100$, $C=0$, staff vertical). The distance CB is not read directly (only the middle hair is read on B), so it is found from triangle CAB with the stated angle $ACB=60^\circ00'$. (The circle reading $80^\circ24'20''$ for B in the table does not agree with this; the stated angle is used.) The height of instrument (1.55 m) is the same for both sights and cancels in the difference of level.

#### Step 1: Distance AB from the subtense bar
Bar length $l=2$ m, angle $\beta=2^\circ40'20''$:

$$
AB=\frac{l}{2}\cot\frac{\beta}{2}=\cot(1^\circ20'10'')=42.875\ \text{m}
$$

#### Step 2: Distance CA and vertical component from the stadia readings
Vertical angle $-5^\circ10'$ (depression):

**Point A** (angle of depression $5^\circ 10'$):

$$
\begin{aligned}
s&=1.839-1.459=0.380\ \text{m}\\
D&=100\times0.380\times\cos^2 5^\circ 10'=37.692\ \text{m}\\
V&=\tfrac12\times100\times0.380\times\sin 2(5^\circ 10')=-3.408\ \text{m}
\end{aligned}
$$

#### Step 3: Distance CB from triangle ACB
$$
AB^2=CA^2+CB^2-2\,CA\,CB\cos60^\circ
$$

$$
CB^2-37.692\,CB+(37.692^2-42.875^2)=0\ \Rightarrow\ CB=46.644\ \text{m}
$$

(the other root is negative).

#### Step 4: Vertical component for B and difference in level
Vertical angle $+10^\circ23'30''$ and the middle hair reading 1.235 m:

$$
V_B=CB\tan10^\circ23'30''=46.644\times0.18338=+8.554\ \text{m}
$$

$$
\begin{aligned}
\text{RL}_B-\text{RL}_A&=(V_B-1.235)-(V_A-1.649)\\
&=(+8.554-1.235)-(-3.408-1.649)=+12.376\ \text{m}
\end{aligned}
$$

**Answer:** horizontal distance AB = **42.875 m** (from the bar); B is **12.376 m higher** than A.

## Check yourself

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

1. How is AB found?
   A. From the subtense bar, AB = (l/2)cot(β/2) = 42.875 m
   B. From stadia readings at B
   C. From the circle readings at C
   D. As CA + CB
   Answer: A. The bar at B with angle measured at A gives AB directly. (Key point: Distance AB)

2. Why must CB be found by solving triangle ACB?
   A. B is below the instrument
   B. Only the middle hair is read at B, so no stadia distance is available
   C. The bar gives CB
   D. CB equals CA
   Answer: B. Without top and bottom hairs, CB comes from AB, CA and the angle ACB via the cosine rule. (Key point: Why CB needs triangle)

3. How is VB found at B?
   A. VB = (1/2)Ks sin2θ with the intercept
   B. VB = CB cos(10°23'30")
   C. VB = AB tanθ
   D. VB = CB tan(10°23'30") = +8.554 m
   Answer: D. With D = CB known, V = D tanθ. (Key point: Vertical component at B)

## Related in Surveying II

- [Subtense bar and tacheometer gradient A-B](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asubtense-bar-3m-level-difference-gradient) — Surveying II, chapter 2
- [Subtense bar, RL difference, gradient, bearing AB](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asubtense-bar-c-difference-rl-gradient) — Surveying II, chapter 2
- [Subtense bar, level difference and gradient A-B](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asubtense-bar-tacheometer-level-diff-gradient) — Surveying II, chapter 2
- [Subtense bar, elevation difference and gradient P-Q](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asubtense-staff-elevation-difference-gradient-pq) — Surveying II, chapter 2
- [Anallatic tacheometer distance XY, gradient](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-xy-gradient-p) — Surveying II, chapter 2, asked 3×
- [Tacheometer depression angles, distance A to B and RL](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atachometer-depression-bm-distance-ab-rl) — Surveying II, chapter 2

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