--- title: "Subtense bar, RL difference, gradient, bearing AB" subject: "Surveying II" chapter: 2 key: "semester_4/surveying_ii:subtense-bar-c-difference-rl-gradient" source: https://www.bctengineer.com/q/semester_4/surveying_ii/subtense-bar-c-difference-rl-gradient site: BCT Engineering --- # Subtense bar, RL difference, gradient, bearing AB Surveying II · Chapter 2: Tacheometry - Asked in: 2076 Bhadra, 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-c-difference-rl-gradient - 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 C was 2°40'20". Intermediate level information was later recorded using a theodolite having constant 100 and 0 at station C and staff held vertical to the line of sight. The following data was recorded on two stations A and B. What is the difference in RL between A and B, gradient between A and B and bearing of AB. | Sighting | Horizontal angle | Vertical Reading | Staff Reading (m) | |---|---|---|---| | C-A | 0°0'0" | 95°10' | 1.459, 1.649, 1.839 | | C-B | 80°24'20" | 70°23' | ......, 1.235, ...... | ## Short answer Subtense bar gives CB=42.875m, stadia gives CA=37.692m. RL difference B-A=19.103m (B higher), AB=52.16m, gradient 1 in 2.73 rising, bearing of AB=125°51' (assuming CA bearing 0°). ## Answer Assumptions: the "vertical reading" is the zenith angle; the horizontal circle reading $0^\circ00'00''$ is on CA, which is taken as the reference direction (bearing of CA = 0°, since no bearing is given). The height of instrument at C is not given, but it cancels in the RL difference. #### Distance CB by the subtense bar (b = 2 m, θ = 2°40'20") $$ D_{CB} = \frac{b}{2}\cot\frac{\theta}{2} = 1\times\cot(1°20'10") = 42.875\ \text{m} $$ #### Distance CA by stadia ($k = 100$, $c = 0$) - $s = 1.839 - 1.459 = 0.380$ m, $z = 95^\circ10'$, $\theta = -5°10'$ (depression) - $H_{CA} = 100\,s\sin^2 z = 37.692$ m - $V_A = 50\,s\sin 2\theta = -3.408$ m #### RL of A and B relative to the axis of C $$ RL_A - RL_{axis} = V_A - r_A = -3.408 - 1.649 = -5.057\ \text{m} $$ For B, $z = 70^\circ 23'$ so $\theta = +19°37'$ and $V_B = D_{CB}\tan\theta = 15.281$ m: $$ RL_B - RL_{axis} = 15.281 - 1.235 = 14.046\ \text{m} $$ #### Difference in RL $$ RL_B - RL_A = 14.046 - (-5.057) = 19.103\ \text{m} $$ B is higher than A by 19.103 m. #### Length and gradient of AB Angle ACB = $80^\circ24'20''$. $$ AB = \sqrt{37.692^2 + 42.875^2 - 2(37.692)(42.875)\cos 80^\circ24'20''} = 52.156\ \text{m} $$ $$ \text{Gradient} = \frac{19.103}{52.156} = 1\text{ in }2.73\ (\text{rising from A to B}) $$ #### Bearing of AB Taking C as origin, CA along north: A = (0, 37.692), B = (42.275 E, 7.146 N). $$ \tan\beta = \frac{\Delta E}{\Delta N} = \frac{42.275}{-30.546}, \qquad \text{bearing of AB} = 125°51'00" $$ **Answer: RL of B − RL of A = 19.103 m; gradient = 1 in 2.73; AB = 52.16 m; bearing of AB = 125°51'00" (if CA is taken as 0°; add the actual bearing of CA otherwise).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A 2 m bar subtending θ gives distance CB by: A. (2/2) cot(θ/2) B. 2 tanθ C. Ks cos^2θ D. 2/θ with θ in degrees Answer: A. D = (b/2) cot(θ/2) = 1 x cot(1°20'10 sec). (Key point: Distance CB) 2. Distance CA is by stadia since: A. A has a staff with three hair readings B. A has a bar C. The bearing is zero D. It is a BM Answer: A. The staff intercept gives H = 100 s sin^2 z. (Key point: Distance CA) 3. The HI at C is not needed because: A. It is zero B. It cancels in RL_B - RL_A C. It is in K D. It is the bearing Answer: B. Both points are referred to the same axis RL. (Key point: Height of instrument) 4. Bearing of AB is found by: A. Coordinates of A and B from C (origin) taking CA as reference B. Averaging the zenith angles C. Subtracting heights D. The cosine rule only Answer: A. Using distances and horizontal angles from C as coordinates gives ΔE/ΔN of AB. 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