--- title: "Anallatic tacheometer from O, RLs of A and B" subject: "Surveying II" chapter: 2 key: "semester_4/surveying_ii:tacheometer-o-distance-ab-rl-gradient" source: https://www.bctengineer.com/q/semester_4/surveying_ii/tacheometer-o-distance-ab-rl-gradient site: BCT Engineering --- # Anallatic tacheometer from O, RLs of A and B Surveying II · Chapter 2: Tacheometry - Asked in: 2065 Kartik (old course), 10 marks - Video: [KNEC ENGINEERING SURVEY TACHEOMETRY CALCULATION 2022 NOV SERIES](https://www.youtube.com/watch?v=at_6b9sV3iA) — Iwords Shots Empire - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/tacheometer-o-distance-ab-rl-gradient - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2 ## Question The following observations were made on a vertically held staff with a tacheometer fitted anallatic lens having multiplying constant of 100. | Instrument Station | Height of Instrument | Staff Station | Bearing | Zenith Angle | Hair Reading | Remarks | |---|---|---|---|---|---|---| | O | 1.55 | A | 30°30' | 85°30' | 1.155, 1.755, 2.355 | RL of O = 450.80 m | | | | B | 75°30' | 101°15' | 1.250, 2.000, 2.750 | | Calculate the distance AB and RLs of A and B. Find the gradient of the line AB. ## Short answer Anallatic lens (C=0, K=100). Zenith angles converted: A +4°30′, B −11°15′. Intercepts s=1.200 m, 1.500 m. D=119.261 m, 144.291 m; RL_A=459.981 m, RL_B=421.649 m. Angle at O 45° gives AB=103.47 m. Gradient falling 1 in 2.7 (37.05%). ## Answer Anallatic lens: $C=0$, $K=100$. Zenith angles are converted to vertical angles by $\theta=90^\circ-z$: for A, $90^\circ-85^\circ30'=+4^\circ30'$ (elevation); for B, $90^\circ-101^\circ15'=-11^\circ15'$ (depression). The RL of station O is taken as the ground RL, so the instrument axis is HI above it. Staff held vertical, line of sight inclined at angle $\theta$ (additive constant zero, which is the case for an anallatic lens or $C=0$): $$ D=Ks\cos^2\theta,\qquad V=\tfrac12Ks\sin2\theta $$ where $s$ is the staff intercept, $\theta$ the vertical angle and $K=100$, $C=0$. #### Step 1: Horizontal distance and vertical component of each sight **Point A** (bearing 30°30', angle of elevation $4^\circ 30'$): $$ \begin{aligned} s&=2.355-1.155=1.200\ \text{m}\\ D&=100\times1.200\times\cos^2 4^\circ 30'=119.261\ \text{m}\\ V&=\tfrac12\times100\times1.200\times\sin 2(4^\circ 30')=9.386\ \text{m} \end{aligned} $$ **Point B** (bearing 75°30', angle of depression $11^\circ 15'$): $$ \begin{aligned} s&=2.750-1.250=1.500\ \text{m}\\ D&=100\times1.500\times\cos^2 11^\circ 15'=144.291\ \text{m}\\ V&=\tfrac12\times100\times1.500\times\sin 2(11^\circ 15')=-28.701\ \text{m} \end{aligned} $$ #### Step 2: Reduced levels RL of instrument axis = RL of station + HI = 450.800 + 1.550 = 452.350 m. $$ \text{RL of staff point}=\text{RL of axis}+V-h $$ $$ \begin{aligned} \text{RL}_{A}&=452.350+9.386-1.755=459.981\ \text{m}\\ \text{RL}_{B}&=452.350-28.701-2.000=421.649\ \text{m} \end{aligned} $$ #### Step 3: Horizontal distance AB The angle at O between the two lines is the difference of the bearings, $45^\circ 00'$ (30°30' and 75°30'). $$ \begin{aligned} AB&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{119.261^2+144.291^2-2\times119.261\times144.291\cos 45^\circ 00'}=103.474\ \text{m} \end{aligned} $$ #### Step 4: Gradient $$ \text{Gradient}=\frac{\Delta H}{AB}=\frac{38.332}{103.474}=0.37045 $$ The line from A to B is falling: 1 in 2.7, i.e. 37.05 %, an angle of $20^\circ 19'38''$ with the horizontal. **Answer:** AB = 103.47 m; RL of A = 459.981 m, RL of B = 421.649 m; difference in level = -38.332 m; gradient falling 1 in 2.7 (37.05 %, 20°20'). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A zenith angle of 101°15' to point B is converted to a vertical angle of: A. +11°15' (elevation) B. -11°15' (depression) C. +101°15' D. -78°45' (depression) Answer: B. Vertical angle = 90° - z = 90° - 101°15' = -11°15', so it is a depression. (Key point: Zenith to vertical angle) 2. For a vertical staff with an anallatic lens (C = 0), the horizontal distance and vertical component are: A. D = Ks cos^2θ and V = (Ks/2) sin2θ B. D = Ks cosθ and V = Ks sinθ C. D = Ks sin^2θ and V = Ks cos2θ D. D = Ks cos^2θ and V = Ks sin^2θ Answer: A. With the staff vertical, D = Ks cos^2θ and V = (1/2)Ks sin2θ. The other forms are for a staff normal to the line of sight. (Key point: Formulae for inclined sight, vertical staff) 3. In the answer, the RL of a staff point is found from: A. RL of station + V + r, with HI ignored B. RL of axis - V + h C. RL of axis + V - h, where RL of axis = RL of station + HI D. RL of station + HI - V - h Answer: C. The axis is HI above the station, V is added (negative for depression), and the middle-hair reading h is subtracted. (Key point: RL of staff point) 4. How is the horizontal distance AB obtained from the distances D1 and D2 of A and B from O? A. AB = D1 + D2 B. AB = D1 - D2 C. By the cosine rule using the angle at O equal to the difference of the bearings D. By the sine rule using the difference of the zenith angles Answer: C. The angle AOB equals the difference of bearings (45°), then AB^2 = D1^2 + D2^2 - 2D1D2 cos45°. 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