--- title: "Gradient and bearing of PQ from R" subject: "Surveying II" chapter: 2 key: "semester_4/surveying_ii:tacheometer-gradient-bearing-pq-s60e" source: https://www.bctengineer.com/q/semester_4/surveying_ii/tacheometer-gradient-bearing-pq-s60e site: BCT Engineering --- # Gradient and bearing of PQ from R Surveying II · Chapter 2: Tacheometry - Asked in: 2072 Magh, 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/tacheometer-gradient-bearing-pq-s60e - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2 ## Question Determine gradient and bearing of PQ. K = 100, C = 0. The staff was held vertical. | Inst. St. | Staff St. | Bearing | Zenithal Angle | Staff Readings T | M | B | |---|---|---|---|---|---|---| | R | P | S60°E | 79°28' | 2.36 | 1.81 | 1.25 | | | Q | S30°W | 95°06' | 2.94 | 2.12 | 1.30 | ## Short answer Using D=Ks cos²θ and V=½Ks sin2θ, then cosine rule for PQ. PQ = 194.89 m, ΔH = -34.781 m, gradient falling 1 in 5.6 (17.85%), bearing 243°24' (S63°24'W). ## Answer $K=100$, $C=0$. The zenith angles are converted to vertical angles by $\theta=90^\circ-z$: for P, $90^\circ-79^\circ28'=+10^\circ32'$ (elevation); for Q, $90^\circ-95^\circ06'=-5^\circ06'$ (depression). The middle reading is the axial hair reading $h$. The height of instrument is not given and cancels in the difference of levels. 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 P** (bearing S60°E = 120°00', angle of elevation $10^\circ 32'$): $$ \begin{aligned} s&=2.360-1.250=1.110\ \text{m}\\ D&=100\times1.110\times\cos^2 10^\circ 32'=107.291\ \text{m}\\ V&=\tfrac12\times100\times1.110\times\sin 2(10^\circ 32')=19.950\ \text{m} \end{aligned} $$ **Point Q** (bearing S30°W = 210°00', angle of depression $5^\circ 06'$): $$ \begin{aligned} s&=2.940-1.300=1.640\ \text{m}\\ D&=100\times1.640\times\cos^2 5^\circ 06'=162.704\ \text{m}\\ V&=\tfrac12\times100\times1.640\times\sin 2(5^\circ 06')=-14.521\ \text{m} \end{aligned} $$ #### Step 2: Difference in level The height of instrument is the same for both sights, so it cancels (RL of point $=$ RL of axis $+V-h$): $$ \begin{aligned} \Delta H&=(V_{Q}-h_{Q})-(V_{P}-h_{P})\\ &=(-14.521-2.120)-(+19.950-1.810)=-34.781\ \text{m} \end{aligned} $$ #### Step 3: Horizontal distance PQ The angle at R between the two lines is the difference of the bearings, $90^\circ 00'$ (120°00' and 210°00'). $$ \begin{aligned} PQ&=\sqrt{D_1^2+D_2^2-2D_1D_2\cos\gamma}\\ &=\sqrt{107.291^2+162.704^2-2\times107.291\times162.704\cos 90^\circ 00'}=194.894\ \text{m} \end{aligned} $$ #### Step 4: Gradient $$ \text{Gradient}=\frac{\Delta H}{PQ}=\frac{34.781}{194.894}=0.17846 $$ The line from P to Q is falling: 1 in 5.6, i.e. 17.85 %, an angle of $10^\circ 07'06''$ with the horizontal. #### Step 5: Bearing of PQ Taking the instrument station as the origin: $$ \begin{aligned} P&=(-53.645,\ +92.916),\qquad Q=(-140.906,\ -81.352)\\ \tan\theta&=\frac{\Delta E}{\Delta N}=\frac{-174.268}{-87.261}\ \Rightarrow\ \theta_{PQ}=243^\circ 24'06'' \end{aligned} $$ Bearing of PQ = 243°24' (S 63°24' W). **Answer:** PQ = 194.89 m; difference in level = -34.781 m; gradient falling 1 in 5.6 (17.85 %, 10°07'); bearing of PQ = 243°24'. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How are the zenith angles converted to vertical angles? A. θ = 90° - z, so P is +10°32' and Q is -5°06' B. θ = z - 90°, so P is -10°32' C. θ = z, so P is 79°28' D. θ = 180° - z Answer: A. The vertical angle is the complement of the zenith angle; z > 90° means depression. (Key point: Zenith angle conversion) 2. What are the whole circle bearings of S60°E and S30°W? A. 60° and 30° B. 120° and 210° C. 240° and 150° D. 300° and 330° Answer: B. S60°E = 180° - 60° = 120°; S30°W = 180° + 30° = 210°. (Key point: Whole circle bearings) 3. Why is HI not needed here? A. The instrument is at RL zero B. HI equals the axial reading C. Same axis for both sights, so it cancels in the difference of levels D. Because K = 100 Answer: C. ΔH = (VQ - hQ) - (VP - hP); the axis RL appears in both. (Key point: Height of instrument) 4. How is the bearing of PQ obtained? A. By averaging the bearings of RP and RQ B. By subtracting the bearings of RP and RQ C. From the gradient D. From the coordinate differences ΔE/ΔN of P and Q taken from the station as origin Answer: D. Coordinates of P and Q are computed from D and bearing; tan^-1(ΔE/ΔN) gives the bearing, here 243°24'. 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