--- title: "Vertical curve -3.5% and +5%, 1268 m at 780" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-35down-5up-1268-at" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-35down-5up-1268-at site: BCT Engineering --- # Vertical curve -3.5% and +5%, 1268 m at 780 Surveying II · Chapter 6: Curves - Asked in: 2073 Bhadra, 6 marks - Video: [Horizontal Curves || Vertical Curves](https://www.youtube.com/watch?v=f9k__IXcvYo) — Civil Engineer - 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-35down-5up-1268-at - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question A grade of -3.5% meets another grade of +0.5%. The elevation and chainage of IP are 1267.00 m and 780 m respectively. Field condition requires that the vertical curve should pass through a point of elevation 1268 m at chainage 780 m. Compute a suitable equal tangent vertical curve and full stations elevations when normal chord = 30 m. ## Short answer Equal-tangent parabolic vertical curve. L = 200 m from mid-ordinate e = 1 m. BVC chainage 680 m, RL 1270.5 m; EVC chainage 880 m, RL 1267.5 m. Peg RLs computed with y = RL_BVC + g1 x/100 + (g2-g1)x²/(200L). Lowest point at chainage 855 m, RL 1267.438 m. ## Answer An equal-tangent parabolic vertical curve is set out. Because the curve passes through the given level at the IP chainage, the vertical gap between the IP and the curve there is the mid-ordinate $e$ of the curve. #### Step 1: Length of the curve The curve is a sag curve, so the curve lies above the IP: $$ e = 1268 - 1267 = 1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m} $$ For an equal-tangent parabola, $e = \dfrac{L\,|g_2-g_1|}{800}$ (grades in %), hence $$ L = \frac{800\,e}{|g_2-g_1|} = \frac{800\times 1}{|0.5-(-3.5)|} = \frac{800}{4} = 200\ \text{m} $$ #### Step 2: BVC, EVC and their levels The curve is a sag (valley) curve with equal tangent lengths $L/2 = 100$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 780 - 100 = 680\ \text{m}\\ \text{Ch. of EVC} &= 780 + 100 = 880\ \text{m}\\ \text{RL of BVC} &= 1267 - \frac{(-3.5)}{100}\times 100 = 1270.5\ \text{m}\\ \text{RL of EVC} &= 1267 + \frac{(0.5)}{100}\times 100 = 1267.5\ \text{m} \end{aligned} $$ #### 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 $g_1 = -3.5\%$, $g_2 = 0.5\%$, $L = 200$ m, so the offset is $y = \frac{(4)x^2}{200\times 200}$ and pegs are taken at multiples of 30 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 680 | 0 | 1270.500 | 0.000 | 1270.500 | | 690 | 10 | 1270.150 | +0.010 | 1270.160 | | 720 | 40 | 1269.100 | +0.160 | 1269.260 | | 750 | 70 | 1268.050 | +0.490 | 1268.540 | | 780 | 100 | 1267.000 | +1.000 | 1268.000 | | 810 | 130 | 1265.950 | +1.690 | 1267.640 | | 840 | 160 | 1264.900 | +2.560 | 1267.460 | | 870 | 190 | 1263.850 | +3.610 | 1267.460 | | 880 | 200 | 1263.500 | +4.000 | 1267.500 | Check: the last curve RL (1267.5 m) equals the RL of EVC (1267.5 m), and the curve at the IP chainage lies 1 m above the IP, at RL 1268 m. #### Step 4: Lowest point At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{3.5\times 200}{4} = 175$ m from BVC. - Chainage = 680 + 175 = 855 m - RL = 1270.5 + (-3.5/100)(175) + (4)(175)²/(200 × 200) = 1267.438 m **Answer:** $L = 200$ m, BVC at 680 m (RL 1270.5 m), EVC at 880 m (RL 1267.5 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Grades -3.5% then +0.5%, curve above the IP: it is a A. sag curve B. summit curve C. crest curve D. compound curve Answer: A. Falling then rising grade. (Key point: Curve type) 2. L from e = 1 m and grade change 4% is A. 100 m B. 200 m C. 400 m D. 800 m Answer: B. L = 800e/4 = 200 m. (Key point: Length) 3. RL of BVC is A. 1263.5 m B. 1267.5 m C. 1267 + 0.035 x 100 = 1270.5 m D. 1267 m Answer: C. First grade falls so the BVC is higher. (Key point: RL of BVC) 4. Lowest point is A. x = 100 m, chainage 780 B. x = 200 m, chainage 880 C. x = 150 m D. x = 175 m, chainage 855 m, RL 1267.438 m Answer: D. x = 3.5 x 200/4. (Key point: Lowest point) ## Close variants set in other exams - [Vertical curve -2.5% and +3.5%, 1268.50 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-25down-35up-12685) — Surveying II, chapter 6, asked 9× - [Vertical curve +0.5% and -3.5%, 1266 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-05up-35down-1266-at) — Surveying II, chapter 6, asked 9× ## Related in Surveying II - [Vertical curve +3.5% and -0.5%, 1266 m at 780](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-35down-05-1266-780) — Surveying II, chapter 6, asked 8× - [Vertical curve +1.25% and +4.75%, 1267.70 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-125-475-126770) — Surveying II, chapter 6, asked 3× - [Vertical curve -3.5% and -0.5%, 1268 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg35-neg05-1268-lowest) — Surveying II, chapter 6 - [Vertical curve (-)3% and (-)35%, IP 2477 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg3-neg35-ip-2477) — Surveying II, chapter 6, asked 6× - [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× - [Vertical curve -3.5% and +0.5%, IP 1300 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-neg35-up-05-1300) — Surveying II, chapter 6, asked 5× ## Related in other subjects - [Vertical curve +4% and -3%, through given point](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-70m-vc-4-3) — 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 - [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