--- title: "Vertical curve +1.25% and +4.75%, 1267.70 m" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-125-475-126770" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-125-475-126770 site: BCT Engineering --- # Vertical curve +1.25% and +4.75%, 1267.70 m Surveying II · Chapter 6: Curves - Asked in: 2076 Bhadra, 6 marks - Video: [Stations and Elevations of PVC, PVT and High point of Vertical Curve|Vertical Curve Fundamentals](https://www.youtube.com/watch?v=KLN9Oe0EJc4) — Master Civil Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-125-475-126770 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question A grade of 1.25% meets another grade of 4.75%. The elevation and chainage of intersection point are 1267 m and 1+780 km. Field condition requires that the vertical curve should pass through a point of elevation 1267.70 m at chainage (1+780) km. Compute a suitable equal tangent vertical curve and full station elevation. Use parabolic equation. ## Short answer Using mid-ordinate e = 0.7 m, L = 800×0.7/|4.75−1.25| = 160 m. Equal-tangent sag curve: BVC at 1700 m (1266 m), EVC at 1860 m (1270.8 m). Lowest point at BVC, highest at EVC. ## Answer Chainage of IP $= 1+780$ km $= 1780$ m. Peg interval is not stated; the usual 30 m is used. An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve. #### Step 1: Length of the curve The curve is a sag curve, so the curve lies above the IP: $$ e = 1267.7 - 1267 = 0.7\ \text{m}\ \Rightarrow\ |e| = 0.7\ \text{m} $$ For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence $$ L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 0.7}{|4.75-(1.25)|} = \frac{560}{3.5} = 160\ \text{m} $$ #### Step 2: BVC, EVC and their levels The curve is a sag (valley) curve with equal tangent lengths $L/2 = 80$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 1780 - 80 = 1700\ \text{m}\\ \text{Ch. of EVC} &= 1780 + 80 = 1860\ \text{m}\\ \text{RL of BVC} &= 1267 - \frac{(1.25)}{100}\times 80 = 1266\ \text{m}\\ \text{RL of EVC} &= 1267 + \frac{(4.75)}{100}\times 80 = 1270.8\ \text{m} \end{aligned} $$ #### Step 3: RL of curve points Using the tangent-offset form of the parabola ($x$ measured from BVC, grades in %): $$ y_x = \text{RL}_{BVC} + \frac{g_1 x}{100} + \frac{(g_2-g_1)x^2}{200L} $$ with $g_1 = 1.25\%$, $g_2 = 4.75\%$, $L = 160$ m, so the offset is $y = \frac{(3.5)x^2}{200\times 160}$ and pegs are taken at multiples of 30 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 1700 | 0 | 1266.000 | 0.000 | 1266.000 | | 1710 | 10 | 1266.125 | +0.011 | 1266.136 | | 1740 | 40 | 1266.500 | +0.175 | 1266.675 | | 1770 | 70 | 1266.875 | +0.536 | 1267.411 | | 1800 | 100 | 1267.250 | +1.094 | 1268.344 | | 1830 | 130 | 1267.625 | +1.848 | 1269.473 | | 1860 | 160 | 1268.000 | +2.800 | 1270.800 | Check: the last curve RL (1270.8 m) equals the RL of EVC (1270.8 m), and the curve at the IP chainage lies 0.7 m above the IP, at RL 1267.7 m. #### Step 4: Highest and lowest points Both grades are rising, so the grade never becomes zero within the curve and there is no turning point; the lowest level is at the BVC (RL 1266 m) and the highest level is at the EVC (RL 1270.8 m). **Answer:** $L = 160$ m, BVC at 1700 m (RL 1266 m), EVC at 1860 m (RL 1270.8 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The curve must pass through RL 1267.70 m at the IP chainage, where the IP is at RL 1267 m. What does the 0.7 m difference represent? A. The tangent length B. The mid-ordinate e of the curve C. The rate of change of grade D. The RL difference between BVC and EVC Answer: B. At the IP chainage the vertical gap between the IP and the curve is the mid-ordinate e, here 0.7 m. (Key point: Mid-ordinate from given level) 2. For an equal-tangent parabolic curve with grades 1.25% and 4.75% and e = 0.7 m, which expression gives L? A. L = 400e/|g2-g1| B. L = 800e/(g1+g2) C. L = e|g2-g1|/800 x 2 D. L = 800e/|g2-g1| Answer: D. Since e = L|g2-g1|/800, L = 800 x 0.7/3.5 = 160 m. (Key point: Length of curve formula) 3. With L = 160 m and IP at chainage 1780 m, what are the chainages of BVC and EVC? A. 1620 m and 1940 m B. 1700 m and 1860 m C. 1780 m and 1940 m D. 1700 m and 1780 m Answer: B. Each tangent length is L/2 = 80 m, so BVC = 1780 - 80 = 1700 m and EVC = 1780 + 80 = 1860 m. (Key point: Sag curve and BVC/EVC) 4. Why is there no highest/lowest turning point inside this curve? A. Because the curve is too short B. Because the curve is a summit curve C. Because the offsets are all positive D. Because both grades are rising, so the grade never becomes zero Answer: D. Grades +1.25% and +4.75% are both positive, so the slope never reaches zero; lowest level is at BVC and highest at EVC. (Key point: Turning point) ## Close variants set in other exams - [Vertical curve +1.25% and +4.75%, 1518.30 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-125-475-151760-2030) — Surveying II, chapter 6, asked 7× - [Vertical curve, -3.5% and +0.5%, 3268 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-parabolic-why-3267-1780) — Surveying II, chapter 6, asked 3× ## Related in Surveying II - [Vertical curve +3.5% and -0.5%, 1266 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35down-05-1266-780) — Surveying II, chapter 6, asked 8× - [Vertical curve +0.5% and -3.5%, 1266 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-05up-35down-1266-at) — Surveying II, chapter 6, asked 9× - [Vertical curve -3.5% and +5%, 1268 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35down-5up-1268-at) — Surveying II, chapter 6, asked 3× - [Vertical curve +3.5% and -2.75%, 30 m pegs](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35-down-275-through) — Surveying II, chapter 6, asked 7× - [Vertical curve -2.5% and +3.5%, 1268.50 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-25down-35up-12685) — Surveying II, chapter 6, asked 9× - [Vertical curve -3.5% and -0.5%, 1268 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg35-neg05-1268-lowest) — Surveying II, chapter 6 ## Related in other subjects - [Vertical curve +4% and -3%, through given point](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70m-vc-4-3) — Transportation Engineering I, chapter 3 - [Vertical curve -3.25% and +3.75%, 225 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-73m-vc-225) — Transportation Engineering I, chapter 3 - [Vertical curve chainage and elevations, R 8000 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-69bh-vc-chainage) — Transportation Engineering I, chapter 3 - [Vertical curve, -3% and 1.5% at chainage 2000](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-79a-vc-2000) — Transportation Engineering I, chapter 3