---
title: "Subtense bar and tacheometer gradient A-B"
subject: "Surveying II"
chapter: 2
key: "semester_4/surveying_ii:subtense-bar-3m-level-difference-gradient"
source: https://www.bctengineer.com/q/semester_4/surveying_ii/subtense-bar-3m-level-difference-gradient
site: BCT Engineering
---

# Subtense bar and tacheometer gradient A-B

Surveying II · Chapter 2: Tacheometry

- Asked in: 2077 Chaitra, 6 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-3m-level-difference-gradient
- Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2

## Question

A 3 m long subtense bar was placed above the station B and angle subtended in the instrument placed at station A was read out to be 0°50'20" but the bar was deviated 4° from being normal to the line joining the instrument and bar station. By using a tacheometer with constant 100 and 0 at station R following observations were produced with staff held vertical. Calculate the level difference and gradient between A and B.

| | Sighted to | Bearing | Zenithal Angle | Staff Readings |
|---|---|---|---|---|
| R | A | 315° | 98°45' | -1.100- |
| | B | 210° | 83°45' | 0.65, 1.25, 1.85 |

## Short answer

Correct the bar length for deviation: l′=3cos4°=2.9927 m, giving AB=(l′/2)cot(25′10″)=204.396 m. From R to B: s=1.200, D=118.578 m, V=+12.986 m; RA from triangle RAB is 138.599 m, with V_A=138.599tan(−8°45′)=−21.332 m. Level difference B−A = +34.169 m; gradient = 0.1672, rising 1 in 6.0 (16.72%).

## Answer

**Reading of the data:** the subtense bar at B, observed from A, gives the horizontal distance AB. The tacheometer at R ($K=100$, $C=0$, staff vertical) gives the distance RB and the heights. For A only the axial reading (1.100 m) is recorded, so RA cannot be found from stadia readings; it is found from the triangle RAB (AB and RB known, angle ARB known from the bearings). The instrument axis at R is the same for both sights, so only the readings and vertical components are needed for the **difference** in level.

#### Step 1: Distance AB from the subtense bar (angle at A)
The bar is out of normal by $4^\circ$, so its effective length is $l\cos4^\circ=3\cos4^\circ=2.9927$ m.

$$
AB=\frac{l'}{2}\cot\frac{\beta}{2}=\frac{2.9927}{2}\cot\frac{0^\circ50'20''}{2}=204.396\ \text{m}
$$

#### Step 2: Distance RB and vertical component from R
Zenith angle $83^\circ45'$ gives $\theta=+6^\circ15'$.

**Point B** (bearing 210°00', angle of elevation $6^\circ 15'$):

$$
\begin{aligned}
s&=1.850-0.650=1.200\ \text{m}\\
D&=100\times1.200\times\cos^2 6^\circ 15'=118.578\ \text{m}\\
V&=\tfrac12\times100\times1.200\times\sin 2(6^\circ 15')=12.986\ \text{m}
\end{aligned}
$$

#### Step 3: Distance RA from the triangle RAB
Bearings from R: RA $=315^\circ$ and RB $=210^\circ$, so angle ARB $=105^\circ$. By the cosine rule:

$$
AB^2=RA^2+RB^2-2\,RA\,RB\cos105^\circ
$$

$$
RA^2-2(118.578)\cos105^\circ\,RA+(118.578^2-204.396^2)=0\ \Rightarrow\ RA=138.599\ \text{m}
$$

(the other root is negative).

#### Step 4: Vertical component for A
Zenith angle $98^\circ45'$ gives $\theta=-8^\circ45'$ (depression):

$$
V_A=RA\tan\theta=138.599\times\tan(-8^\circ45')=-21.332\ \text{m}
$$

#### Step 5: Difference in level and gradient
$$
\begin{aligned}
\text{RL}_A&=\text{RL}_{axis}+V_A-1.100=\text{RL}_{axis}-21.332-1.100\\
\text{RL}_B&=\text{RL}_{axis}+V_B-1.250=\text{RL}_{axis}+12.986-1.250\\
\Delta H&=\text{RL}_B-\text{RL}_A=(12.986-1.250)-(-21.332-1.100)=+34.169\ \text{m}
\end{aligned}
$$

$$
\text{Gradient}=\frac{\Delta H}{AB}=\frac{+34.169}{204.396}=0.16717
$$

**Answer:** B is **34.169 m higher** than A; AB = 204.396 m; the gradient from A to B is **rising 1 in 6.0 (16.72 %)**.

## Check yourself

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

1. The 3 m bar is 4° off normal. What effective length is used?
   A. 3 / cos 4°
   B. 3 cos 4° = 2.9927 m
   C. 3 sin 4°
   D. 3 m unchanged
   Answer: B. The projection normal to the line of sight is l cos(delta). (Key point: Effective bar length)

2. Which formula gives AB from the subtense angle?
   A. (l'/2) cot(beta/2)
   B. l' tan(beta)
   C. (l'/2) tan(beta/2)
   D. l' sin(beta)
   Answer: A. AB = (l'/2) cot(beta/2) with l' = 2.9927 m, giving 204.396 m. (Key point: Distance AB)

3. How is RA found when only the axial reading is recorded for A?
   A. Cosine rule in triangle RAB with RB and AB known and angle ARB = 105°
   B. From the stadia constant
   C. Equal to RB
   D. From the zenith angle only
   Answer: A. No stadia intercept for A, so RA comes from the triangle R-A-B. (Key point: Finding RA)

4. Which formula gives V_A?
   A. RA tan(theta)
   B. RA cos(theta)
   C. RA sin(theta)/2
   D. Ks
   Answer: A. V = horizontal distance x tan(theta); theta = -8°45' gives -21.332 m. (Key point: Vertical component for A)

5. Which expression gives RL_B - RL_A?
   A. (V_B - 1.250) - (V_A - 1.100)
   B. (V_B + 1.250) - (V_A + 1.100)
   C. V_B + V_A
   D. (V_B - V_A) / AB
   Answer: A. RL = axis + V - h, so the axis cancels: (12.986 - 1.250) - (-21.332 - 1.100) = +34.169 m. (Key point: Level difference)

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