--- title: "Vertical curve 5% and 3%, 1474.50 m" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-5-3-ip-1475" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-5-3-ip-1475 site: BCT Engineering --- # Vertical curve 5% and 3%, 1474.50 m Surveying II · Chapter 6: Curves - Asked in: 2075 Bhadra, 6 marks - Video: [SURVEYING- CURVES (LECTURE-5)( VERTICAL CURVES - SUMMIT & VALLEY)](https://www.youtube.com/watch?v=hh6iRE6lYVQ) — Ekta Dwivedi - 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-5-3-ip-1475 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question A grade of 5% meets another grade of 3%. The elevation and chainage of IP are 1475.0 m and 3500 respectively. Field condition require that the vertical curve should pass through a point of elevation 1474.50 m at chainage 3500.00 m compute a suitable equal tangent vertical curve including full station elevations take peg interval 30 m. ## Short answer The vertical gap at IP is e = –0.5 m. For equal-tangent parabola, L = 800e/|g2–g1| = 200 m. BVC at chainage 3400 m (RL 1470 m), EVC at 3600 m (RL 1478 m). Curve elevations at 30 m intervals are computed using tangent offsets y = –(g2–g1)x²/(200L). ## Answer 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 summit curve, so the curve lies below the IP: $$ e = 1474.5 - 1475 = -0.5\ \text{m}\ \Rightarrow\ |e| = 0.5\ \text{m} $$ For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence $$ L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 0.5}{|3-(5)|} = \frac{400}{2} = 200\ \text{m} $$ #### Step 2: BVC, EVC and their levels The curve is a summit (crest) curve with equal tangent lengths $L/2 = 100$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 3500 - 100 = 3400\ \text{m}\\ \text{Ch. of EVC} &= 3500 + 100 = 3600\ \text{m}\\ \text{RL of BVC} &= 1475 - \frac{(5)}{100}\times 100 = 1470\ \text{m}\\ \text{RL of EVC} &= 1475 + \frac{(3)}{100}\times 100 = 1478\ \text{m} \end{aligned} $$ #### 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 $g_1 = 5\%$, $g_2 = 3\%$, $L = 200$ m, so the offset is $y = \frac{(-2)x^2}{200\times 200}$ and pegs are taken at multiples of 30 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 3400 | 0 | 1470.000 | 0.000 | 1470.000 | | 3420 | 20 | 1471.000 | -0.020 | 1470.980 | | 3450 | 50 | 1472.500 | -0.125 | 1472.375 | | 3480 | 80 | 1474.000 | -0.320 | 1473.680 | | 3510 | 110 | 1475.500 | -0.605 | 1474.895 | | 3540 | 140 | 1477.000 | -0.980 | 1476.020 | | 3570 | 170 | 1478.500 | -1.445 | 1477.055 | | 3600 | 200 | 1480.000 | -2.000 | 1478.000 | Check: the last curve RL (1478 m) equals the RL of EVC (1478 m), and the curve at the IP chainage lies 0.5 m below the IP, at RL 1474.5 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 1470 m) and the highest level is at the EVC (RL 1478 m). **Answer:** $L = 200$ m, BVC at 3400 m (RL 1470 m), EVC at 3600 m (RL 1478 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 passes through RL 1474.5 m at the IP (RL 1475 m). What does this tell you? A. It is a sag curve with e = 0.5 m above the IP B. It is a summit curve, the curve lying 0.5 m below the IP C. The grades are equal D. L cannot be found Answer: B. The curve is below the IP, so it is a summit curve with |e| = 0.5 m. (Key point: Summit curve and sign of e) 2. What is L for grades 5% and 3% with e = 0.5 m? A. 100 m B. 400 m C. 160 m D. 200 m Answer: D. L = 800e/|g2-g1| = 800 x 0.5/2 = 200 m. (Key point: Length of curve) 3. With L = 200 m and IP RL 1475 m at chainage 3500 m, what are RL of BVC and EVC? A. 1475 m and 1475 m B. 1470 m and 1478 m C. 1470 m and 1480 m D. 1480 m and 1478 m Answer: B. RL BVC = 1475 - 0.05 x 100 = 1470 m; RL EVC = 1475 + 0.03 x 100 = 1478 m. (Key point: BVC and EVC levels) ## Close variants set in other exams - [Vertical curve (-)3% and (-)35%, IP 2477 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg3-neg35-ip-2477) — Surveying II, chapter 6, asked 6× - [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 -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 +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 -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 +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× ## Related in Surveying II - [Vertical curve -3.5% and +0.5%, IP 1300 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg35-up-05-1300) — Surveying II, chapter 6, asked 5× - [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 -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 - [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× - [Vertical curve +1.25% and +4.75%, 1267.70 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-125-475-126770) — Surveying II, chapter 6, asked 3× - [Highway curve through M 16 m from IP](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Ahighway-curve-pqr-140-point-m) — 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 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 5% and 4% at 50+050](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-78p-vc-5-4) — 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