--- title: "Vertical curve, -3.5% and +0.5%, 3268 m" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-parabolic-why-3267-1780" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-parabolic-why-3267-1780 site: BCT Engineering --- # Vertical curve, -3.5% and +0.5%, 3268 m Surveying II · Chapter 6: Curves - Asked in: 2081 Chaitra, 1+5 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-parabolic-why-3267-1780 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Why vertical curves are generally used parabolic curve? The elevation and chainage of intersection point are 3267 and 1+780 km. A grade of -3.5% meets another grade of 0.5%. Field condition requires that the vertical curve should pass through a point of elevation 3268 m at chainage 1+780 km. Compute a suitable equal tangent vertical curve and full station elevations. Use parabolic equation. The peg interval is 30 m. ## Short answer A parabola gives constant rate of grade change for smooth riding and simple offsets. For e = 1 m, L = 200 m; BVC at 1680 m (RL 3270.5 m), EVC at 1880 m (RL 3267.5 m). Curve RLs computed via tangent-offset equation; at IP chainage 1780 m, RL = 3268 m. ## Answer #### Why a parabola? A parabola is used for vertical curves because it gives a constant rate of change of grade along the curve, so the change in slope is uniform and the ride is smooth. Its equation is simple ($y = ax^2 + bx$), offsets from the tangents vary as the square of the distance and are very easy to compute, the curve is symmetrical about the IP for equal tangents, and a circle and a parabola are almost identical for the very flat curves used on roads. #### Computation Chainage of IP $= 1+780$ km $= 1780$ m. 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: $$ e = 3268 - 3267 = 1\ \text{m}\ \Rightarrow\ |e| = 1\ \text{m} $$ 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. $$ \begin{aligned} \text{Ch. of BVC} &= 1780 - 100 = 1680\ \text{m}\\ \text{Ch. of EVC} &= 1780 + 100 = 1880\ \text{m}\\ \text{RL of BVC} &= 3267 - \frac{(-3.5)}{100}\times 100 = 3270.5\ \text{m}\\ \text{RL of EVC} &= 3267 + \frac{(0.5)}{100}\times 100 = 3267.5\ \text{m} \end{aligned} $$ #### 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) | |---|---|---|---|---| | 1680 | 0 | 3270.500 | 0.000 | 3270.500 | | 1710 | 30 | 3269.450 | +0.090 | 3269.540 | | 1740 | 60 | 3268.400 | +0.360 | 3268.760 | | 1770 | 90 | 3267.350 | +0.810 | 3268.160 | | 1800 | 120 | 3266.300 | +1.440 | 3267.740 | | 1830 | 150 | 3265.250 | +2.250 | 3267.500 | | 1860 | 180 | 3264.200 | +3.240 | 3267.440 | | 1880 | 200 | 3263.500 | +4.000 | 3267.500 | Check: the last curve RL (3267.5 m) equals the RL of EVC (3267.5 m), and the curve at the IP chainage lies 1 m above the IP, at RL 3268 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 = 1680 + 175 = 1855 m - RL = 3270.5 + (-3.5/100)(175) + (4)(175)²/(200 × 200) = 3267.438 m **Answer:** $L = 200$ m, BVC at 1680 m (RL 3270.5 m), EVC at 1880 m (RL 3267.5 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 a parabola commonly used for vertical curves? A. It has constant radius B. It needs no tangents C. It gives a constant rate of change of grade and simple offsets proportional to x^2 D. It has the shortest length Answer: C. A parabola changes grade uniformly, giving smooth riding and easy offset computation. (Key point: Why parabola) 2. For equal tangents, e = L|g2 - g1|/800. With e = 1 m and change of grade 4%, L is A. 400 m B. 100 m C. 3200 m D. 200 m Answer: D. L = 800 x 1/4 = 200 m. (Key point: Length from mid-ordinate) 3. The curve passes 1 m above the IP. What does this tell us? A. It is a sag (valley) curve B. The grades are equal C. It is a summit curve D. It is straight Answer: A. A curve lying above the IP is a sag curve. (Key point: Sag vs summit) 4. Where is the lowest point? A. at the EVC B. x = 175 m from BVC, chainage 1855 m C. x = 100 m from BVC D. at the IP, 1780 m Answer: B. x = -g1L/(g2 - g1) = 3.5 x 200/4 = 175 m, chainage 1680 + 175 = 1855 m. 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