--- title: "Vertical curve -1% and +0.5%, IP 500 m" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-falling-1-rising-05" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-falling-1-rising-05 site: BCT Engineering --- # Vertical curve -1% and +0.5%, IP 500 m Surveying II · Chapter 6: Curves - Asked in: 2071 Magh, 10 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-falling-1-rising-05 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question In a road alignment a falling grade of 1% is followed by rising grade of 0.5%. The chainage and RL of the intersection point are 500 and 350 m respectively. The rate of change of grade is 0.1% per 20 m. Calculate the necessary data required for setting out the vertical curve, take peg interval of 30 m. ## Short answer Curve length 300 m; BVC at 350 m (RL 351.5 m), EVC at 650 m (RL 350.75 m). Lowest point at chainage 550 m, RL 350.5 m. Tangent-offset method yields levels at 30 m intervals as tabulated. ## 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. $$ \begin{aligned} \text{Ch. of BVC} &= 500 - 150 = 350\ \text{m}\\ \text{Ch. of EVC} &= 500 + 150 = 650\ \text{m}\\ \text{RL of BVC} &= 350 - \frac{(-1)}{100}\times 150 = 351.5\ \text{m}\\ \text{RL of EVC} &= 350 + \frac{(0.5)}{100}\times 150 = 350.75\ \text{m} \end{aligned} $$ #### 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) | |---|---|---|---|---| | 350 | 0 | 351.500 | 0.000 | 351.500 | | 360 | 10 | 351.400 | +0.003 | 351.403 | | 390 | 40 | 351.100 | +0.040 | 351.140 | | 420 | 70 | 350.800 | +0.123 | 350.923 | | 450 | 100 | 350.500 | +0.250 | 350.750 | | 480 | 130 | 350.200 | +0.423 | 350.623 | | 510 | 160 | 349.900 | +0.640 | 350.540 | | 540 | 190 | 349.600 | +0.903 | 350.503 | | 570 | 220 | 349.300 | +1.210 | 350.510 | | 600 | 250 | 349.000 | +1.563 | 350.563 | | 630 | 280 | 348.700 | +1.960 | 350.660 | | 650 | 300 | 348.500 | +2.250 | 350.750 | Check: the last curve RL (350.75 m) equals the RL of EVC (350.75 m), and at the IP chainage the curve lies 0.563 m above the IP, at RL 350.563 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 = 350 + 200 = 550 m - RL = 351.5 + (-1/100)(200) + (1.5)(200)²/(200 × 300) = 350.5 m The pegs are at multiples of 30 m; the BVC (350 m) and EVC (650 m) are added as the end points of the curve. **Answer:** $L = 300$ m, BVC at 350 m (RL 351.5 m), EVC at 650 m (RL 350.75 m), lowest point at chainage 550 m (RL 350.5 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Grade change is 1.5% and rate is 0.1% per 20 m. What is L? A. 30 m B. 75 m C. 150 m D. 300 m Answer: D. L = (1.5/0.1) x 20 = 300 m. (Key point: Length of curve) 2. With IP at chainage 500 m and L = 300 m, what are the chainages of BVC and EVC? A. 450 m and 550 m B. 200 m and 800 m C. 350 m and 650 m D. 500 m and 800 m Answer: C. BVC = IP - L/2 = 350 m and EVC = IP + L/2 = 650 m for an equal tangent curve. (Key point: BVC and EVC) 3. How does the offset y from the tangent vary along the curve? A. It is constant B. Linearly with x C. As the cube of x D. As the square of x, y = (g2 - g1)x^2/(200L) Answer: D. The curve is a parabola with constant rate of change of grade, so the offset is proportional to x^2. (Key point: Offsets) 4. Where is the lowest point, and what is the check for it? A. At the IP, chainage 500 m B. x = 150 m from BVC at the mid-point C. At the EVC D. x = 200 m from BVC (chainage 550 m) where grade is zero Answer: D. x = -g1 L/(g2 - g1) = 1 x 300/1.5 = 200 m, giving chainage 550 m and RL 350.5 m. (Key point: Lowest point) ## Close variants set in other exams - [Vertical curve -1% and +0.5% by parabolic method](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-parabolic-neg1-pos05-ip) — Surveying II, chapter 6, asked 2× ## Related in Surveying II - [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 - [Vertical curve -0.5% and +0.7%, IP RL 500](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg05-pos07-ip-1000) — 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 - [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 180 m, -3.5% then +4.5%, IP 2450](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35down-45up-ip-2450) — Surveying II, chapter 6, asked 2× - [Vertical curve 180 m, -4.5% then +3.5%, IP 450](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-45down-35up-ip-450) — Surveying II, chapter 6, asked 2× ## Related in other subjects - [Road earthwork, rising 1 in 200, falling 1 in 120](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aroad-earthwork-i-rising-falling) — Estimating and Costing, chapter 6 - [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 - [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