--- title: "Gradient P to Q, constants 100 and 0" subject: "Surveying II" chapter: 2 key: "semester_4/surveying_ii:gradient-pq-tacheometric-constants-100-0" source: https://www.bctengineer.com/q/semester_4/surveying_ii/gradient-pq-tacheometric-constants-100-0 site: BCT Engineering --- # Gradient P to Q, constants 100 and 0 Surveying II · Chapter 2: Tacheometry - Asked in: 2078 Baisakh, 6 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/gradient-pq-tacheometric-constants-100-0 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-2 ## Question Compute the gradient between two instrument stations P and Q from the following observation having tacheometric constant 100 and 0. | Inst. Stn. | H.I | Sighted To | Bearing | Zenith Angle | Staff Reading | Remarks | |---|---|---|---|---|---|---| | P | 1.46 | X | 70°20' | 80°10' | 2.900, 2.565, 2.230 | RL of X is 1197.339 | | Q | 1.38 | | 335°0' | 95°20' | 1.440, 1.220, 1.000 | | ## Short answer Using K=100, C=0: from P to X, s=0.670, D=65.046 m, V=+11.274 m; from Q to X, s=0.440, D=43.620 m, V=−4.072 m. Axis RLs give RLP=1187.170 m and RLQ=1201.251 m. PQ=81.616 m at bearing 102°29′, so gradient from P to Q = +14.081/81.616 = 0.1725, rising 1 in 5.8 (17.25%). ## Answer **Reading of the data:** only one RL is given (RL of X = 1197.339 m), so the staff point sighted from Q is taken to be the **same point X** as sighted from P (the common-point layout). With $K=100$, $C=0$ and the staff vertical: $$ D=Ks\cos^2\theta,\qquad V=\tfrac12Ks\sin2\theta,\qquad \theta=90^\circ-z $$ Vertical angles: from P, $90^\circ-80^\circ10'=+9^\circ50'$ (elevation); from Q, $90^\circ-95^\circ20'=-5^\circ20'$ (depression). #### Step 1: Distances and vertical components **Point X from P** (bearing 70°20', angle of elevation $9^\circ 50'$): $$ \begin{aligned} s&=2.900-2.230=0.670\ \text{m}\\ D&=100\times0.670\times\cos^2 9^\circ 50'=65.046\ \text{m}\\ V&=\tfrac12\times100\times0.670\times\sin 2(9^\circ 50')=11.274\ \text{m} \end{aligned} $$ **Point X from Q** (bearing 335°00', angle of depression $5^\circ 20'$): $$ \begin{aligned} s&=1.440-1.000=0.440\ \text{m}\\ D&=100\times0.440\times\cos^2 5^\circ 20'=43.620\ \text{m}\\ V&=\tfrac12\times100\times0.440\times\sin 2(5^\circ 20')=-4.072\ \text{m} \end{aligned} $$ #### Step 2: RL of P and Q $$ \text{RL}_X=\text{RL of axis}+V-h\ \Rightarrow\ \text{RL of axis}=\text{RL}_X-V+h $$ $$ \begin{aligned} \text{Axis at P}&=1197.339-11.274+2.565=1188.630\ \text{m}\\ \text{RL}_P&=1188.630-1.46=1187.170\ \text{m}\\ \text{Axis at Q}&=1197.339-(-4.072)+1.220=1202.631\ \text{m}\\ \text{RL}_Q&=1202.631-1.38=1201.251\ \text{m} \end{aligned} $$ #### Step 3: Horizontal distance and bearing of PQ Take P as the origin. X is at $D_P=65.046$ m on bearing $70^\circ20'$, and Q is $D_Q=43.620$ m from X in the direction opposite to the bearing $335^\circ00'$ of the line from Q to X: $$ \begin{aligned} X&=(21.891,\ 61.251)\ \text{(N, E)}\\ Q&=X-D_Q(\cos335^\circ,\ \sin335^\circ)=(21.891-39.533,\ 61.251+18.435)=(-17.642,\ +79.686)\\ PQ&=\sqrt{17.642^2+79.686^2}=81.616\ \text{m},\qquad \theta_{PQ}=\tan^{-1}\frac{+79.686}{-17.642}=102^\circ 29'01'' \end{aligned} $$ #### Step 4: Gradient from P to Q $$ \text{Gradient}=\frac{\text{RL}_Q-\text{RL}_P}{PQ}=\frac{+14.081}{81.616}=0.17253 $$ **Answer:** RL of P = 1187.170 m, RL of Q = 1201.251 m, PQ = 81.616 m (bearing 102°29'). The gradient from P to Q is **rising 1 in 5.8 (17.25 %)**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Zenith angle 80°10' corresponds to what vertical angle? A. +9°50' (elevation) B. -9°50' (depression) C. +80°10' D. +170°10' Answer: A. theta = 90° - z = 90° - 80°10' = +9°50'; a zenith angle above 90° gives a depression. (Key point: Vertical angle from zenith) 2. How is the missing relation between P and Q established? A. Both sight the same point X whose RL is given B. P and Q are assumed to be at the same level C. The gradient is read from the table D. Q is assumed to be north of P Answer: A. Only one RL is given, so the staff point from Q is taken as the same point X. (Key point: Reading of data) 3. How is the axis RL at P obtained? A. RL_X - V + h B. RL_X + V + h C. RL_X + V - h D. RL_X - V - h Answer: A. RL_X = axis + V - h, so axis = RL_X - V + h. (Key point: RL of axis) 4. How is PQ found? A. D_P + D_Q B. D_P - D_Q C. By coordinates of P (origin), X from D_P and bearing, and Q from X using D_Q and the reverse bearing D. From the vertical angles only Answer: C. X is plotted from P, then Q is located from X by going back along the Q-to-X direction; PQ is the resulting distance. (Key point: Horizontal distance PQ) 5. RL_P = 1187.170, RL_Q = 1201.251, PQ = 81.616 m. Gradient P to Q? A. Falling 1 in 5.8 B. Rising 1 in 58 C. Rising 1 in 5.8 (17.25 %) D. Rising 1 in 0.17 Answer: C. 14.081/81.616 = 0.1725, rising about 1 in 5.8. (Key point: Gradient) ## Close variants set in other exams - [Gradient A to B, RL of P = 1234.560](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Agradient-ab-tacheometer-100-0-p) — Surveying II, chapter 2, asked 2× ## Related in Surveying II - [Gradient P to Q from common point R](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Agradient-pq-common-point-r) — Surveying II, chapter 2 - [Tacheometric RL of B, C, gradient BC](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-rl-b-c-gradient-bc) — Surveying II, chapter 2, asked 2× - [Gradient between P and Q (constants 100, 0)](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Atacheometer-gradient-pq-from-r-100) — 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 - [RL of hill station P from stations R and Q](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahill-station-signal-r-q-p) — Surveying II, chapter 3 - [Tacheometric RL of B, C, gradient BC](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aanallatic-rl-b-c-gradient-bc-2) — Surveying II, chapter 2, asked 2× ## Related in other subjects - [Two peg test, AB 100 m, Q 10 m](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-ab-100m-q-10m-inside) — Surveying I, chapter 5 - [Level book, benchmark RL 100.00 at 3rd reading](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Alevel-book-instrument-shifted-after-3rd-6th) — Surveying I, chapter 5 - [Two peg test, AB = 40 m, near B](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Atwo-peg-ab-40m-inner-side-b) — Surveying I, chapter 5 - [Mean angles and VCR, station O (reading 179°59′40″)](https://www.bctengineer.com/question.md?key=semester_3%2Fsurveying_i%3Amean-direction-o-abcd-first-reversed-179-59-40) — Surveying I, chapter 7, asked 3×