--- title: "Tacheometer depression angles, distance A to B and RL" subject: "Surveying II" chapter: 2 key: "semester_4/surveying_ii:tachometer-depression-bm-distance-ab-rl" source: https://www.bctengineer.com/q/semester_4/surveying_ii/tachometer-depression-bm-distance-ab-rl site: BCT Engineering --- # Tacheometer depression angles, distance A to B and RL Surveying II · Chapter 2: Tacheometry - Asked in: 2066 Magh (old course), 9 marks - Video: [Solved Example of Case 2 of Stadia Tacheometer Method; Calculating horizontal distance and R.L.](https://www.youtube.com/watch?v=RqN00_Zajsg) — CIVIL Solved - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/tachometer-depression-bm-distance-ab-rl - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2 ## Question A tachometer is placed at a station A on a staff held upon a B.M. of R.L. = 1000.00 m and station B are 0.640, 2.200, 3.760 and 0.010, 2.120, 4.230 respectively. The angle of depression of the telescope in the first case is -6°19' and in the second case -7°42'. Find the horizontal distance from A to B and R.L. of the station B. (Constants are 100 and 0.3) ## Short answer Constants K=100, C=0.3. Staff intercepts: BM 3.120 m (−6°19′), B 4.220 m (−7°42′). D=Ks cos²θ+C cosθ, V=½Ks sin2θ+C sinθ. RL of instrument axis = 1036.352 m. Horizontal distance AB = 414.721 m, RL of B = 978.159 m. ## Answer **Reading of the data:** the tacheometer at A is first sighted on the staff held on the benchmark (BM, RL 1000.000 m) and then on the staff at B. Both sights are angles of depression. The constants are $K=100$ and $C=0.3$ m, and the staff is vertical, so $$ D=Ks\cos^2\theta+C\cos\theta,\qquad V=\tfrac12Ks\sin2\theta+C\sin\theta $$ with $\theta$ negative for depression. The horizontal distance "from A to B" is the distance from the instrument at A to the staff at B. #### Step 1: Sight on the BM (readings 0.640, 2.200, 3.760; angle $-6^\circ19'$) **Point BM** (angle of depression $6^\circ 19'$): $$ \begin{aligned} s&=3.760-0.640=3.120\ \text{m}\\ D&=100\times3.120\cos^2 6^\circ 19'+0.30\cos 6^\circ 19'=308.521\ \text{m}\\ V&=\tfrac12\times100\times3.120\sin 2(6^\circ 19')+0.30\sin 6^\circ 19'=-34.152\ \text{m} \end{aligned} $$ Since the BM is below the axis, RL of axis $=$ RL$_{BM}$ $+|V|+h$: $$ \text{RL}_{axis}=1000.000+34.152+2.200=1036.352\ \text{m} $$ #### Step 2: Sight on B (readings 0.010, 2.120, 4.230; angle $-7^\circ42'$) **Point B** (angle of depression $7^\circ 42'$): $$ \begin{aligned} s&=4.230-0.010=4.220\ \text{m}\\ D&=100\times4.220\cos^2 7^\circ 42'+0.30\cos 7^\circ 42'=414.721\ \text{m}\\ V&=\tfrac12\times100\times4.220\sin 2(7^\circ 42')+0.30\sin 7^\circ 42'=-56.073\ \text{m} \end{aligned} $$ $$ \text{RL}_B=\text{RL}_{axis}-|V|-h=1036.352-56.073-2.120=978.159\ \text{m} $$ **Answer:** horizontal distance from A to B = **414.721 m**; R.L. of B = **978.159 m**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which formulae apply (vertical staff, constants 100, 0.3)? A. D = Ks cos^2θ + C cosθ, V = (1/2)Ks sin2θ + C sinθ B. D = Ks cos^2θ, V = (1/2)Ks sin2θ C. D = Ks + C, V = 0 D. D = Ks cosθ + C, V = Ks sinθ Answer: A. Non-zero additive constant adds C cosθ and C sinθ terms. (Key point: Formulae with C = 0.3) 2. For a BM below the axis, how is RL of axis found? A. RL_BM - |V| - h B. RL_BM + |V| + h C. RL_BM + |V| - h D. RL_BM + h only Answer: B. The axis lies |V| above the staff reading point, and the point is h below the sighted hair; with BM below, add both. (Key point: RL of axis from BM) 3. How is RL of B found? A. RL_axis + |V| - h B. RL_axis - |V| + h C. RL_axis - |V| - h D. RL_BM - |V| Answer: C. B is below the axis by |V| plus the axial reading h. (Key point: RL of B) 4. Which sight gives the horizontal distance A to B? A. The sight on BM (308.521 m) B. Sum of both C. Difference of both D. The sight on B (414.721 m) Answer: D. Distance from instrument A to staff at B uses B's readings. (Key point: Distance A to B) ## Related in Surveying II - [Anallatic tacheometer distance CD and gradient](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-staff-cd-gradient-abcd) — Surveying II, chapter 2 - [Anallatic tacheometer A to B distance, gradient](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-bm-b-distance-gradient-970) — Surveying II, chapter 2 - [RL of hill station P (A-B 120 m, RL 1280)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-top-rl-2074b-ab-120) — Surveying II, chapter 3, asked 2× - [Subtense bar and tacheometer, A-B distance and level](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Asubtense-bar-2m-theodolite-c-ab) — 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× - [Anallatic tacheometer distance XY, gradient](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-xy-gradient-p-2081c) — Surveying II, chapter 2, asked 3× ## Related in other subjects - [Initial line design for 200 MW over 120 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aline-design-200mw-120km) — Transmission and Distribution System Design, chapter 2 - [Mean angles and VCR, station O (A-D)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-o-abcd-vcr-120-15-10-ls10) — Surveying I, chapter 7, asked 3× - [Mean angles and VCR, station O (reading 179°59′40″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-o-abcd-first-reversed-179-59-40) — Surveying I, chapter 7, asked 3× - [Road earthwork, rising 1 in 200, falling 1 in 120](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-i-rising-falling) — Estimating and Costing, chapter 6