--- title: "Anallatic tacheometer distance CD and gradient" subject: "Surveying II" chapter: 2 key: "semester_4/surveying_ii:anallatic-staff-cd-gradient-abcd" source: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-staff-cd-gradient-abcd site: BCT Engineering --- # Anallatic tacheometer distance CD and gradient Surveying II · Chapter 2: Tacheometry - Asked in: 2079 Jestha, 8 marks - Video: [Calculate the horizontal distance between stations and RL of B when inst. constant is 100](https://www.youtube.com/watch?v=5BOguNaD3ZA) — SBK Concept - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-staff-cd-gradient-abcd - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2 ## Question To determine the distance between two points C and D and their elevations, the following observations were taken upon a vertically held staff from two traverse stations A and B. The tachometer was fitted with an anallatic lens, the constant of the instrument being 100. Bearing of AC and BD are 330°20' and 20°36' respectively. | Station | HI | Northing | Easting | Staff Station | Vertical angle | Staff reading (m) | |---|---|---|---|---|---|---| | A | 1.58 | 218.3 | 164.7 | C | +12°12' | 1.255, 1.86, 2.456 | | B | 1.50 | 518.2 | 207.6 | D | +10°36' | 1.3, 1.885, 2.47 | Calculate: a) The distance CD b) RL of C and D given that those of A to B are 432.550 and 425.5. c) The gradient from C to D. ## Short answer CD = 336.30 m; RL of C = 457.08 m, RL of D = 446.27 m. Gradient from C to D is falling at 1 in 31.1 (3.21%, -0.03213). ## Answer Anallatic lens: additive constant $C=0$, $K=100$. With the staff vertical and the line of sight inclined at $\theta$: $$ D=Ks\cos^2\theta,\qquad V=\tfrac12Ks\sin2\theta,\qquad \text{RL}=\text{RL}_{inst}+\text{HI}+V-h $$ #### Step 1: Distances AC and BD and vertical components **Point C** (bearing 330°20', angle of elevation $12^\circ 12'$): $$ \begin{aligned} s&=2.456-1.255=1.201\ \text{m}\\ D&=100\times1.201\times\cos^2 12^\circ 12'=114.737\ \text{m}\\ V&=\tfrac12\times100\times1.201\times\sin 2(12^\circ 12')=24.807\ \text{m} \end{aligned} $$ **Point D** (bearing 20°36', angle of elevation $10^\circ 36'$): $$ \begin{aligned} s&=2.470-1.300=1.170\ \text{m}\\ D&=100\times1.170\times\cos^2 10^\circ 36'=113.041\ \text{m}\\ V&=\tfrac12\times100\times1.170\times\sin 2(10^\circ 36')=21.155\ \text{m} \end{aligned} $$ #### Step 2: Coordinates of C and D Coordinates of the instrument stations: A (218.3 N, 164.7 E) and B (518.2 N, 207.6 E). $$ \begin{aligned} N_C&=218.3+114.737\cos330^\circ20'=317.997,\quad E_C=164.7+114.737\sin330^\circ20'=107.911\\ N_D&=518.2+113.041\cos20^\circ36'=624.013,\quad E_D=207.6+113.041\sin20^\circ36'=247.373 \end{aligned} $$ #### (a) Distance CD $$ CD=\sqrt{(N_D-N_C)^2+(E_D-E_C)^2}=\sqrt{(+306.016)^2+(+139.462)^2}=336.297\ \text{m} $$ (Bearing of CD = 24°30'.) #### (b) Reduced levels of C and D $$ \begin{aligned} \text{RL}_C&=432.550+1.58+24.807-1.86=457.077\ \text{m}\\ \text{RL}_D&=425.500+1.50+21.155-1.885=446.270\ \text{m} \end{aligned} $$ #### (c) Gradient from C to D $$ \text{Gradient}=\frac{\text{RL}_D-\text{RL}_C}{CD}=\frac{-10.807}{336.297}=-0.03213 $$ The ground **falls** from C to D at 1 in 31.1 (3.21 %). **Answer:** CD = 336.297 m; RL of C = 457.077 m, RL of D = 446.270 m; gradient 1 in 31.1 falling from C to D. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are the instrument constants with an anallatic lens? A. K = 100, C = 0.3 B. K = 100, C = 0 C. K = 0, C = 100 D. K = 50, C = 0 Answer: B. An anallatic lens makes the additive constant zero; K = 100 as given. (Key point: Anallatic lens constants) 2. For a vertical staff and inclined sight with C = 0, which formulas give D and V? A. D = Ks sin^2(theta), V = Ks sin(theta) B. D = Ks cos^2(theta), V = (1/2) Ks sin(2 theta) C. D = Ks cos(theta), V = Ks sin(theta) D. D = Ks, V = Ks tan(theta) Answer: B. D = Ks cos^2(theta) and V = (1/2) Ks sin 2theta for a vertical staff. (Key point: Formulas for vertical staff) 3. How is CD found from the two tacheometric sights? A. From the coordinates of C and D computed from stations A and B using bearing and D B. By subtracting the two horizontal distances C. By adding the two horizontal distances D. From the difference in staff readings Answer: A. Coordinates of C and D are found by N = N0 + D cos(bearing), E = E0 + D sin(bearing), then CD = sqrt(dN^2 + dE^2). (Key point: Distance CD) 4. How is RL of C computed? A. RL_A + HI + V - axial reading B. RL_A - HI + V + axial reading C. RL_A + V + s D. RL_A + HI - V + axial reading Answer: A. RL = RL of station + HI + V - h (axial hair), e.g. 432.550 + 1.58 + 24.807 - 1.86 = 457.077. (Key point: RL of staff point) 5. RL_C = 457.077, RL_D = 446.270, CD = 336.297 m. What is the gradient C to D? A. Rising about 1 in 31 B. Falling about 1 in 31 C. Falling about 1 in 3.1 D. Rising about 1 in 3.1 Answer: B. Fall of 10.807 m over 336.297 m is 0.0321, i.e. falling 1 in 31.1. 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