--- title: "Vertical curve +0.5% and -0.7%, IP RL 350.750" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-pos05-neg07-ip-500" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-pos05-neg07-ip-500 site: BCT Engineering --- # Vertical curve +0.5% and -0.7%, IP RL 350.750 Surveying II · Chapter 6: Curves - Asked in: 2069 Bhadra, 8 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-pos05-neg07-ip-500 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Calculate the RLs of pegs on a vertical curve connecting two grades of +0.5% and -0.7% at the point of intersection. The chainage and RL of intersection point are 500 m and 350.750 m respectively. The rate of change of grade is 0.1% per 30 m. ## Short answer Length L = 360 m from 1.2% grade change at 0.1% per 30 m. BVC at chainage 320 m, RL 349.850 m; EVC at 680 m, RL 349.490 m. Highest point at chainage 470 m, RL 350.225 m. Curve levels are obtained by adding tangent grade elevation and parabolic offset from BVC. ## 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 30 m, algebraic change in grade $= -0.7 - (0.5) = -1.2\%$. $$ L = \frac{|g_2-g_1|}{r}\times 30 = \frac{1.2}{0.1}\times 30 = 360\ \text{m} $$ #### Step 2: BVC, EVC and their levels The curve is a summit (crest) curve with equal tangent lengths $L/2 = 180$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 500 - 180 = 320\ \text{m}\\ \text{Ch. of EVC} &= 500 + 180 = 680\ \text{m}\\ \text{RL of BVC} &= 350.75 - \frac{(0.5)}{100}\times 180 = 349.85\ \text{m}\\ \text{RL of EVC} &= 350.75 + \frac{(-0.7)}{100}\times 180 = 349.49\ \text{m} \end{aligned} $$ #### 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 $g_1 = 0.5\%$, $g_2 = -0.7\%$, $L = 360$ m, so the offset is $y = \frac{(-1.2)x^2}{200\times 360}$ and pegs are taken at multiples of 30 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 320 | 0 | 349.850 | 0.000 | 349.850 | | 330 | 10 | 349.900 | -0.002 | 349.898 | | 360 | 40 | 350.050 | -0.027 | 350.023 | | 390 | 70 | 350.200 | -0.082 | 350.118 | | 420 | 100 | 350.350 | -0.167 | 350.183 | | 450 | 130 | 350.500 | -0.282 | 350.218 | | 480 | 160 | 350.650 | -0.427 | 350.223 | | 510 | 190 | 350.800 | -0.602 | 350.198 | | 540 | 220 | 350.950 | -0.807 | 350.143 | | 570 | 250 | 351.100 | -1.042 | 350.058 | | 600 | 280 | 351.250 | -1.307 | 349.943 | | 630 | 310 | 351.400 | -1.602 | 349.798 | | 660 | 340 | 351.550 | -1.927 | 349.623 | | 680 | 360 | 351.650 | -2.160 | 349.490 | Check: the last curve RL (349.49 m) equals the RL of EVC (349.49 m), and the curve at the IP chainage lies 0.54 m below the IP, at RL 350.21 m. #### Step 4: Highest point At the highest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{-0.5\times 360}{-1.2} = 150$ m from BVC. - Chainage = 320 + 150 = 470 m - RL = 349.85 + (0.5/100)(150) + (-1.2)(150)²/(200 × 360) = 350.225 m Peg interval is not stated, so it is taken as 30 m (equal to the length over which the rate of change of grade is specified). The BVC and EVC are included. **Answer:** $L = 360$ m, BVC at 320 m (RL 349.85 m), EVC at 680 m (RL 349.49 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A +0.5% grade followed by -0.7% gives which curve? A. Sag curve, offsets added B. Reverse curve C. Summit curve, offsets subtracted D. Transition curve Answer: C. Rising then falling makes a crest, so offsets are negative and the curve lies below the tangent. (Key point: Curve type) 2. What is L for 1.2% change at 0.1% per 30 m? A. 360 m B. 180 m C. 72 m D. 36 m Answer: A. L = (1.2/0.1) x 30 = 360 m. (Key point: Length) 3. Where is the highest point of the curve? A. At the IP chainage 500 m B. Where grade is zero, x = 150 m from BVC (chainage 470 m) C. At the BVC D. At the EVC Answer: B. x = -g1 L/(g2 - g1) = 0.5 x 360/1.2 = 150 m. (Key point: Highest point) 4. Where does the curve lie at the IP chainage? A. 0.54 m below the IP, at 350.21 m B. At the IP C. 2.16 m below the IP D. 0.54 m above the IP Answer: A. Mid-ordinate e = 360 x 1.2/800 = 0.54 m, subtracted for a summit curve. (Key point: Curve at the IP) ## Related in Surveying II - [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 -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× - [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 +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× ## 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, -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 - [Summit curve limited to 750 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-75bh-vc-summit-750) — Transportation Engineering I, chapter 3, asked 3×