--- title: "Anallatic tacheometer AB gradient from P" subject: "Surveying II" chapter: 2 key: "semester_4/surveying_ii:anallatic-ab-gradient-percent-angle" source: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-ab-gradient-percent-angle site: BCT Engineering --- # Anallatic tacheometer AB gradient from P Surveying II · Chapter 2: Tacheometry - Asked in: 2078 Chaitra, 6 marks - Video: [Determination of gradient between two points in tacheometric surveying- part 1](https://www.youtube.com/watch?v=FJos96jcQPQ) — Gayathri VG - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-ab-gradient-percent-angle - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2 ## Question A tacheometer fitted with anallatic lens and having multiplying constant of 100 was setup at station 'P'. The following readings were taken with the staff held vertically. | S.N. | Staff Station | Bearing | Vertical Angle | Staff Intercept | Axial hair reading | |---|---|---|---|---|---| | 1 | A | 345° | +15°10'10" | 1.370 | 1.435 | | 2 | B | 250° | +09°50'40" | 2.425 | 1.835 | Calculate the distance AB and the gradient between AB. Also express the gradient as decimal percentage and angle. ## Short answer AB = 277.38 m, difference in level = +5.850 m (B higher). Gradient from A to B is rising 1 in 47.4 (2.11%, 1°12'30"). ## Answer Anallatic lens: $C=0$, $K=100$. The intercept $s$ is given directly, and the height of instrument is not needed because only the **difference** in level between A and B is required. With the staff held vertical and the line of sight inclined at angle $\theta$ (additive constant zero): $$ D=Ks\cos^2\theta,\qquad V=\tfrac12Ks\sin2\theta $$ where $s$ is the staff intercept and $\theta$ the vertical angle. #### Step 1: Horizontal distance and vertical component of each sight **Point A** (bearing 345°, angle of elevation $15^\circ 10'10''$): $$ \begin{aligned} s&=1.370\ \text{m}\\ D&=100\times1.370\times\cos^2 15^\circ 10'10''=127.619\ \text{m}\\ V&=\tfrac12\times100\times1.370\times\sin 2(15^\circ 10'10'')=34.600\ \text{m} \end{aligned} $$ **Point B** (bearing 250°, angle of elevation $9^\circ 50'40''$): $$ \begin{aligned} s&=2.425\ \text{m}\\ D&=100\times2.425\times\cos^2 9^\circ 50'40''=235.411\ \text{m}\\ V&=\tfrac12\times100\times2.425\times\sin 2(9^\circ 50'40'')=40.851\ \text{m} \end{aligned} $$ #### Step 2: Difference in level The height of instrument is the same for both sights, so it cancels (RL of point $=$ RL of axis $+V-h$): $$ \begin{aligned} \Delta H&=(V_{B}-h_{B})-(V_{A}-h_{A})\\ &=(+40.851-1.835)-(+34.600-1.435)=+5.850\ \text{m} \end{aligned} $$ #### Step 3: Horizontal distance AB The angle at P between the two lines is the difference of the bearings: $345^\circ00'-250^\circ00'=95^\circ 00'$. $$ \begin{aligned} AB&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{127.619^2+235.411^2-2\times127.619\times235.411\cos 95^\circ 00'}=277.384\ \text{m} \end{aligned} $$ #### Step 4: Gradient $$ \text{Gradient}=\frac{\Delta H}{AB}=\frac{5.850}{277.384}=0.02109 $$ The line from A to B is rising: 1 in 47.4, i.e. 2.11 %, an angle of $1^\circ 12'30''$ with the horizontal. **Answer:** AB = 277.38 m; difference in level = +5.850 m; gradient rising 1 in 47.4 (2.11 %, 1°12'30"). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the height of instrument not needed to find the level difference between A and B? A. Both sights are from the same axis, so HI cancels in the difference B. HI is always zero for an anallatic lens C. Staff intercept replaces HI D. HI is included in the bearing Answer: A. Both points are sighted from the same instrument axis, so only V - h differences matter. (Key point: Height of instrument cancels) 2. How is AB obtained from the two distances D1 and D2 from P? A. D1 + D2 B. D1 - D2 C. Cosine rule with the angle between bearings (95°) D. Pythagoras with D1 and D2 Answer: C. Angle at P = 345° - 250° = 95°, and AB^2 = D1^2 + D2^2 - 2 D1 D2 cos(95°). (Key point: Horizontal distance AB) 3. Which expression gives the level difference of B over A? A. (V_B - h_B) - (V_A - h_A) B. (V_B + h_B) - (V_A + h_A) C. V_B - V_A + HI D. (h_B - h_A) only Answer: A. RL = axis + V - h, so the difference is (V_B - h_B) - (V_A - h_A) = +5.850 m. (Key point: Difference in level) 4. Delta H = 5.850 m, AB = 277.384 m. What is the gradient? A. Rising 1 in 4.7 (21.1 %) B. Falling 1 in 47.4 C. Rising 1 in 47.4 (2.11 %) D. Rising 1 in 474 (0.21 %) Answer: C. 5.850/277.384 = 0.0211, a rise of 1 in 47.4 or 2.11 %. (Key point: Gradient forms) ## Close variants set in other exams - [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 Surveying II - [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 - [Anallatic tacheometer from O, RLs of A and B](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atacheometer-o-distance-ab-rl-gradient) — Surveying II, chapter 2 - [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 - [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 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 - [Length and bearing of CD from stations A, B](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atacheometer-length-bearing-cd-from-a) — Surveying II, chapter 2