--- title: "Vertical curve table, -4.5% and +3.5%, 200 m" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-table-neg45-pos35-200m" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-table-neg45-pos35-200m site: BCT Engineering --- # Vertical curve table, -4.5% and +3.5%, 200 m Surveying II · Chapter 6: Curves - Asked in: 2070 Bhadra, 10 marks - Video: [How To Solve Vertical Curves | Vertical Curves Calculations And Formulas | Vertical Curves Examples](https://www.youtube.com/watch?v=ksazfhyqq1M) — Civil 3d Pro - 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-table-neg45-pos35-200m - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Prepare a table giving all necessary data for setting out a vertical curve. In a road alignment a grade of -4.5% followed by +3.5%, R.L. of I.P. = 1000 m, chainage of IP = 1500 m connect the two grade by a parabolic curve 200 m long. Take peg interval = 20 m. ## Short answer Curve length 200 m; BVC at 1400 m (RL 1004.5 m), EVC at 1600 m (RL 1003.5 m). Lowest point at chainage 1512.5 m, RL 1001.969 m. Levels at 20 m intervals computed by tangent-offset method are provided in the table. ## Answer A parabolic vertical curve of the given length is set out by the tangent-offset method. #### Step 1: Given data $g_1 = -4.5\%$, $g_2 = 3.5\%$, $L = 200$ m, change of grade $= 8\%$. Mid-ordinate $e = \dfrac{L|g_2-g_1|}{800} = 2$ m. #### Step 2: BVC, EVC and their levels The curve is a sag (valley) curve with equal tangent lengths $L/2 = 100$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 1500 - 100 = 1400\ \text{m}\\ \text{Ch. of EVC} &= 1500 + 100 = 1600\ \text{m}\\ \text{RL of BVC} &= 1000 - \frac{(-4.5)}{100}\times 100 = 1004.5\ \text{m}\\ \text{RL of EVC} &= 1000 + \frac{(3.5)}{100}\times 100 = 1003.5\ \text{m} \end{aligned} $$ #### 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 $g_1 = -4.5\%$, $g_2 = 3.5\%$, $L = 200$ m, so the offset is $y = \frac{(8)x^2}{200\times 200}$ and pegs are taken at multiples of 20 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 1400 | 0 | 1004.500 | 0.000 | 1004.500 | | 1420 | 20 | 1003.600 | +0.080 | 1003.680 | | 1440 | 40 | 1002.700 | +0.320 | 1003.020 | | 1460 | 60 | 1001.800 | +0.720 | 1002.520 | | 1480 | 80 | 1000.900 | +1.280 | 1002.180 | | 1500 | 100 | 1000.000 | +2.000 | 1002.000 | | 1520 | 120 | 999.100 | +2.880 | 1001.980 | | 1540 | 140 | 998.200 | +3.920 | 1002.120 | | 1560 | 160 | 997.300 | +5.120 | 1002.420 | | 1580 | 180 | 996.400 | +6.480 | 1002.880 | | 1600 | 200 | 995.500 | +8.000 | 1003.500 | Check: the last curve RL (1003.5 m) equals the RL of EVC (1003.5 m), and the curve at the IP chainage lies 2 m above the IP, at RL 1002 m. #### Step 4: Lowest point At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{4.5\times 200}{8} = 112.5$ m from BVC. - Chainage = 1400 + 112.5 = 1512.5 m - RL = 1004.5 + (-4.5/100)(112.5) + (8)(112.5)²/(200 × 200) = 1001.969 m **Answer:** $L = 200$ m, BVC at 1400 m (RL 1004.5 m), EVC at 1600 m (RL 1003.5 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the mid-ordinate e at the IP for L = 200 m and change of grade 8%? A. e = 8/200 = 0.04 m B. e = L(g2 - g1)/800 = 2 m C. e = L(g2 - g1)/200 = 8 m D. e = 4 m Answer: B. e = 200 x 8/800 = 2 m, which equals the offset at x = L/2. (Key point: Mid-ordinate) 2. Why is the RL of BVC 1004.5 m when IP RL is 1000 m? A. The curve is a summit B. BVC is 4.5 m below the IP C. The offset 2 m is added D. Grade of -4.5% falls for 100 m after BVC, so BVC is 4.5 m higher than IP Answer: D. RL BVC = RL IP - g1 x L/200 = 1000 + 4.5 = 1004.5 m. (Key point: RL of BVC) 3. What is the curve RL at the IP chainage? A. 1000 m, same as IP B. 1002 m, 2 m above the IP (sag curve) C. 1004.5 m D. 998 m Answer: B. For a sag curve the curve lies above the IP tangent by the mid-ordinate e = 2 m. (Key point: Curve RL at the IP) 4. What is the chainage of the lowest point? A. 1500 m B. 1512.5 m C. 1400 m D. 1487.5 m Answer: B. x = 4.5 x 200/8 = 112.5 m from BVC, so chainage 1512.5 m. (Key point: Lowest point) ## Related in Surveying II - [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 -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 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 -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, -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× - [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 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 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 - [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 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