--- title: "Vertical curve 180 m, -4.5% then +3.5%, IP 450" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-45down-35up-ip-450" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-45down-35up-ip-450 site: BCT Engineering --- # Vertical curve 180 m, -4.5% then +3.5%, IP 450 Surveying II · Chapter 6: Curves - Asked in: 2065 Kartik (old course), 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-45down-35up-ip-450 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question A down grade of 4.5% is followed by an upgrade of 3.5%. The reduced level and chainage of the point of intersection are 900.00 m and 450.00 m respectively. A vertical parabolic curve 180 m long is to be introduced to connect the two grades. The pegs are to be fixed at 20 m intervals. Calculate including lowest point also. ## Short answer Length 180 m; BVC chainage 360 m, RL 904.05 m; EVC chainage 540 m, RL 903.15 m. Lowest point at chainage 461.25 m, RL 901.772 m, computed by tangent-offset method using RL = RL_BVC + g1 x/100 + (g2-g1)x²/(200L). ## Answer A parabolic vertical curve of the given length is set out by the tangent-offset method. #### Step 1: Given data $g_1 = -4.5\%$, $g_2 = 3.5\%$, $L = 180$ m, change of grade $= 8\%$. Mid-ordinate $e = \dfrac{L|g_2-g_1|}{800} = 1.8$ m. #### Step 2: BVC, EVC and their levels The curve is a sag (valley) curve with equal tangent lengths $L/2 = 90$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 450 - 90 = 360\ \text{m}\\ \text{Ch. of EVC} &= 450 + 90 = 540\ \text{m}\\ \text{RL of BVC} &= 900 - \frac{(-4.5)}{100}\times 90 = 904.05\ \text{m}\\ \text{RL of EVC} &= 900 + \frac{(3.5)}{100}\times 90 = 903.15\ \text{m} \end{aligned} $$ #### 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 $g_1 = -4.5\%$, $g_2 = 3.5\%$, $L = 180$ m, so the offset is $y = \frac{(8)x^2}{200\times 180}$ and pegs are taken at multiples of 20 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 360 | 0 | 904.050 | 0.000 | 904.050 | | 380 | 20 | 903.150 | +0.089 | 903.239 | | 400 | 40 | 902.250 | +0.356 | 902.606 | | 420 | 60 | 901.350 | +0.800 | 902.150 | | 440 | 80 | 900.450 | +1.422 | 901.872 | | 460 | 100 | 899.550 | +2.222 | 901.772 | | 480 | 120 | 898.650 | +3.200 | 901.850 | | 500 | 140 | 897.750 | +4.356 | 902.106 | | 520 | 160 | 896.850 | +5.689 | 902.539 | | 540 | 180 | 895.950 | +7.200 | 903.150 | Check: the last curve RL (903.15 m) equals the RL of EVC (903.15 m), and the curve at the IP chainage lies 1.8 m above the IP, at RL 901.8 m. #### Step 4: Lowest point At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{4.5\times 180}{8} = 101.25$ m from BVC. - Chainage = 360 + 101.25 = 461.25 m - RL = 904.05 + (-4.5/100)(101.25) + (8)(101.25)²/(200 × 180) = 901.772 m **Answer:** $L = 180$ m, BVC at 360 m (RL 904.05 m), EVC at 540 m (RL 903.15 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is RL of BVC 904.05 m? A. 900 + 3.5% x 90 B. 900 - 4.05 C. 900 + 2 D. IP RL 900 + 4.5% x 90 m = 904.05 m, since the grade falls from BVC Answer: D. RL BVC = RL IP - g1 L/200 = 900 + 4.05. (Key point: RL of BVC) 2. What is the offset at the EVC (x = 180 m)? A. 0.08 m B. 7.2 m C. 3.6 m D. 1.8 m Answer: B. y = 8 x 180^2/(200 x 180) = 7.2 m, so tangent 895.95 + 7.2 = 903.15 = RL EVC. (Key point: Offsets) 3. What is the lowest point chainage? A. 450 m B. 461.25 m C. 360 + 90 D. 450 + 90 Answer: B. x = 4.5 x 180/8 = 101.25 m, chainage 360 + 101.25. (Key point: Lowest point) 4. What is the curve RL at the IP chainage? A. 901.8 m B. 898.2 m C. 902.0 m D. 900 m Answer: A. Sag: IP RL + e = 900 + 1.8. (Key point: Curve at IP) ## Close variants set in other exams - [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× ## Related in Surveying II - [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 +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 -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× - [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 -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 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