--- title: "RL, distance, gradient P to Q, instrument RL 1325.750" subject: "Surveying II" chapter: 2 key: "semester_4/surveying_ii:anallatic-rl-distance-gradient-pq-1325" source: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-rl-distance-gradient-pq-1325 site: BCT Engineering --- # RL, distance, gradient P to Q, instrument RL 1325.750 Surveying II · Chapter 2: Tacheometry - Asked in: 2079 Chaitra, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/anallatic-rl-distance-gradient-pq-1325 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2 ## Question Find out RL, distance and gradient between P and Q by using below data given in table, the staff held vertical to the line of sight by using anallatic lens. Take RL of instrument station is 1325.750 m. | Instrument St'n | Staff | Line | Bearing | Vertical angle | Hair Reading (m) | |---|---|---|---|---|---| | A | P | AP | 84°36' | 3°30' | 1.35, 2.10, 2.85 | | A | Q | AQ | 142°24' | 2°45' | 1.955, 2.875, 3.795 | ## Short answer Using stadia formula with K=100, C=0 for staff normal to line of sight: RL of P=1332.811m, RL of Q=1331.706m, horizontal distance PQ=164.04m, gradient falling 1 in 148.5 (0.67%). ## Answer **Reading of the data:** the staff is held at right angles (normal) to the line of sight; the anallatic lens gives $C=0$ with $K=100$. No height of instrument is given, so 1325.750 m is taken as the RL of the instrument axis. Staff normal to the line of sight, angle of elevation $\theta$ (anallatic lens, $C=0$): $$ D=Ks\cos\theta+h\sin\theta,\qquad V=Ks\sin\theta,\qquad \text{RL}=\text{RL of axis}+V-h\cos\theta $$ where $s$ is the staff intercept, $h$ the axial (middle) hair reading, $\theta$ the vertical angle and $K=100$. #### Step 1: Horizontal distance and vertical component of each sight **Point P** (bearing 84°36', angle of elevation $3^\circ 30'$, $h=2.100$ m): $$ \begin{aligned} s&=2.850-1.350=1.500\ \text{m}\\ D&=(100\times1.500)\cos 3^\circ 30'+2.100\sin 3^\circ 30'=149.720+0.128=149.848\ \text{m}\\ V&=(100\times1.500)\sin 3^\circ 30'=9.157\ \text{m} \end{aligned} $$ **Point Q** (bearing 142°24', angle of elevation $2^\circ 45'$, $h=2.875$ m): $$ \begin{aligned} s&=3.795-1.955=1.840\ \text{m}\\ D&=(100\times1.840)\cos 2^\circ 45'+2.875\sin 2^\circ 45'=183.788+0.138=183.926\ \text{m}\\ V&=(100\times1.840)\sin 2^\circ 45'=8.828\ \text{m} \end{aligned} $$ #### Step 2: Reduced levels RL of instrument axis = 1325.750 m. $$ \begin{aligned} \text{RL}_{P}&=1325.750+9.157-2.100\cos3^\circ30'=1325.750+9.157-2.096=1332.811\ \text{m}\\ \text{RL}_{Q}&=1325.750+8.828-2.875\cos2^\circ45'=1325.750+8.828-2.872=1331.706\ \text{m} \end{aligned} $$ #### Step 3: Horizontal distance PQ The angle at A between the two lines is the difference of the bearings, $57^\circ 48'$ (84°36' and 142°24'). $$ \begin{aligned} PQ&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{149.848^2+183.926^2-2\times149.848\times183.926\cos 57^\circ 48'}=164.043\ \text{m} \end{aligned} $$ #### Step 4: Gradient $$ \text{Gradient}=\frac{\Delta H}{PQ}=\frac{1.105}{164.043}=0.00674 $$ The line from P to Q is falling: 1 in 148.5, i.e. 0.67 %, an angle of $0^\circ 23'09''$ with the horizontal. **Answer:** PQ = 164.04 m; RL of P = 1332.811 m, RL of Q = 1331.706 m; difference in level = -1.105 m; gradient falling 1 in 148.5 (0.67 %, 0°23'). If the staff is taken as vertical instead, the results are RL of P = 1332.790 m, RL of Q = 1331.693 m, PQ = 163.69 m and a gradient falling 1 in 149.2, very close to the above because the vertical angles are small. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When the staff is held normal to the line of sight with C = 0: A. D = Ks cos^2θ, V = (Ks/2) sin2θ B. D = Ks cosθ, V = Ks sinθ C. D = Ks sinθ, V = Ks cosθ D. D = Ks, V = 0 Answer: B. For a normal staff, D = Ks cosθ and V = Ks sinθ. (Key point: Staff normal to line of sight) 2. The staff intercept for P is: A. Middle minus top reading B. Top minus bottom reading, 2.85 - 1.35 = 1.50 C. Sum of the hair readings D. The middle reading 2.10 Answer: B. s = top - bottom. (Key point: Staff intercept) 3. The angle at A between AP and AQ is: A. Difference of bearings, 57°48' B. Sum of bearings C. Difference of vertical angles D. 90° Answer: A. 142°24' - 84°36' = 57°48'. 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