--- title: "Vertical curve -1% and +0.5% by parabolic method" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-parabolic-neg1-pos05-ip" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-parabolic-neg1-pos05-ip site: BCT Engineering --- # Vertical curve -1% and +0.5% by parabolic method Surveying II · Chapter 6: Curves - Asked in: 2073 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/vertical-curve-parabolic-neg1-pos05-ip - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question In a road alignment a grade of (-)1% is followed by another grade of 0.5%. The chainage and RL of intersection pt are 1500 m and 1250 m respectively. The rate of change of grade is 0.1% /20 m. Calculate the necessary data required for setting out of vertical curve by parabolic equation method take peg interval = 30 m. ## Short answer The rate of change of grade 0.1%/20 m and algebraic change 1.5% give length L = (1.5/0.1)×20 = 300 m. BVC at chainage 1350 m (RL 1251.5 m), EVC at 1650 m (RL 1250.75 m). The lowest point is at chainage 1550 m, RL 1250.5 m. ## Answer A parabolic vertical curve with a constant rate of change of grade is set out using the tangent-offset method. #### Step 1: Length of the curve Rate of change of grade $r = 0.1\%$ per 20 m, algebraic change in grade $= 0.5 - (-1) = 1.5\%$. $$ L = \frac{|g_2-g_1|}{r}\times 20 = \frac{1.5}{0.1}\times 20 = 300\ \text{m} $$ #### Step 2: BVC, EVC and their levels The curve is a sag (valley) curve with equal tangent lengths $L/2 = 150$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 1500 - 150 = 1350\ \text{m}\\ \text{Ch. of EVC} &= 1500 + 150 = 1650\ \text{m}\\ \text{RL of BVC} &= 1250 - \frac{(-1)}{100}\times 150 = 1251.5\ \text{m}\\ \text{RL of EVC} &= 1250 + \frac{(0.5)}{100}\times 150 = 1250.75\ \text{m} \end{aligned} $$ #### 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 $g_1 = -1\%$, $g_2 = 0.5\%$ and $L = 300$ m, the offset is $y = \frac{(1.5)x^2}{200\times 300}$. The pegs are taken at multiples of 30 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 1350 | 0 | 1251.500 | 0.000 | 1251.500 | | 1380 | 30 | 1251.200 | +0.023 | 1251.223 | | 1410 | 60 | 1250.900 | +0.090 | 1250.990 | | 1440 | 90 | 1250.600 | +0.203 | 1250.803 | | 1470 | 120 | 1250.300 | +0.360 | 1250.660 | | 1500 | 150 | 1250.000 | +0.563 | 1250.563 | | 1530 | 180 | 1249.700 | +0.810 | 1250.510 | | 1560 | 210 | 1249.400 | +1.103 | 1250.503 | | 1590 | 240 | 1249.100 | +1.440 | 1250.540 | | 1620 | 270 | 1248.800 | +1.823 | 1250.623 | | 1650 | 300 | 1248.500 | +2.250 | 1250.750 | Check: the last curve RL (1250.75 m) equals the RL of EVC (1250.75 m), and at the IP chainage the curve lies 0.563 m above the IP, at RL 1250.563 m. #### Step 4: Lowest point At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{1\times 300}{1.5} = 200$ m from BVC. - Chainage = 1350 + 200 = 1550 m - RL = 1251.5 + (-1/100)(200) + (1.5)(200)²/(200 × 300) = 1250.5 m **Answer:** $L = 300$ m, BVC at 1350 m (RL 1251.5 m), EVC at 1650 m (RL 1250.75 m), lowest point at chainage 1550 m (RL 1250.5 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Rate 0.1% per 20 m and grade change 1.5% give L = A. 300 m B. 150 m C. 30 m D. 600 m Answer: A. L = (1.5/0.1) x 20 = 300 m. (Key point: Length) 2. With L = 300 m and change 1.5%, offset y at x is A. 1.5 x/(200 x 300) B. 1.5 x^2/(200 x 300) C. 1.5 x^2/300 D. 1.5 x^2/(100 x 300) Answer: B. y = (g2 - g1)x^2/(200L). (Key point: Offset formula) 3. RL of BVC is A. 1250 - 1.5 = 1248.5 m B. 1250.75 m C. 1250 + 0.01 x 150 = 1251.5 m D. 1250 m Answer: C. The first grade is falling, so BVC is higher than the IP. (Key point: BVC RL) 4. Lowest point from BVC is at A. x = 150 m, chainage 1500 m B. x = 100 m C. x = 300 m D. x = 200 m, chainage 1550 m, RL 1250.5 m Answer: D. x = 1 x 300/1.5 = 200 m. (Key point: Lowest point) ## Close variants set in other exams - [Vertical curve -1% and +0.5%, IP 500 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-falling-1-rising-05) — Surveying II, chapter 6, asked 2× ## Related in Surveying II - [Vertical curve -0.7% and +0.5%, IP RL 650](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg07-pos05-rl-ip) — Surveying II, chapter 6 - [Vertical curve table, -4.5% and +3.5%, 200 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-table-neg45-pos35-200m) — Surveying II, chapter 6 - [Vertical curve -2% and +3%, IP 2600, 0.5% per 20 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-2down-3up-ip-2600) — Surveying II, chapter 6 - [Derive RL on parabolic vertical curve](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Aparabolic-vertical-curve-rl-expression) — Surveying II, chapter 6 - [Vertical curve +0.5% and -0.7%, IP RL 350.750](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-pos05-neg07-ip-500) — Surveying II, chapter 6 - [Short note: Setting out vertical curve](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-setting-out-note) — Surveying II, chapter 6 ## Related in other subjects - [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 35 m headlight and momentum grade](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-80c-vc-headlight-35m) — 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 - [Vertical curve with 300 m grade sections](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-65k-vc-300) — Transportation Engineering I, chapter 3