--- title: "Vertical curve -2% and +3%, IP 2600, 0.5% per 20 m" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-2down-3up-ip-2600" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-2down-3up-ip-2600 site: BCT Engineering --- # Vertical curve -2% and +3%, IP 2600, 0.5% per 20 m Surveying II · Chapter 6: Curves - Asked in: 2065 Chaitra (old course), 10 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-2down-3up-ip-2600 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question A 2% down gradient meets a 3% up gradient at a chainage of 2600 m, the RL of the point of intersection being 1200.00 m. A vertical parabolic curve is to be set out to connect two grades with pegs at 20 m interval. The rate of change of grade is 0.5% per 20 m chain. Tabulate the chainages and RLs of the station pegs including lowest point on the curve. ## Short answer Length 200 m. BVC chainage 2500 m, RL 1202.00 m; EVC chainage 2700 m, RL 1203.00 m. Lowest point at chainage 2580 m, RL 1201.20 m, found by x = -g1 L/(g2-g1) = 80 m. RL computed from tangent-offset formula RL = RL_BVC + g1 x/100 + (g2-g1)x²/(200L). ## 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 $r = 0.5\%$ per 20 m, algebraic change in grade $= 3 - (-2) = 5\%$. $$ L = \frac{|g_2-g_1|}{r}\times 20 = \frac{5}{0.5}\times 20 = 200\ \text{m} $$ #### Step 2: BVC, EVC and their levels The curve is a sag (valley) curve with equal tangent lengths $L/2 = 100$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 2600 - 100 = 2500\ \text{m}\\ \text{Ch. of EVC} &= 2600 + 100 = 2700\ \text{m}\\ \text{RL of BVC} &= 1200 - \frac{(-2)}{100}\times 100 = 1202\ \text{m}\\ \text{RL of EVC} &= 1200 + \frac{(3)}{100}\times 100 = 1203\ \text{m} \end{aligned} $$ #### 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 $g_1 = -2\%$, $g_2 = 3\%$, $L = 200$ m, so the offset is $y = \frac{(5)x^2}{200\times 200}$ and pegs are taken at multiples of 20 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 2500 | 0 | 1202.000 | 0.000 | 1202.000 | | 2520 | 20 | 1201.600 | +0.050 | 1201.650 | | 2540 | 40 | 1201.200 | +0.200 | 1201.400 | | 2560 | 60 | 1200.800 | +0.450 | 1201.250 | | 2580 | 80 | 1200.400 | +0.800 | 1201.200 | | 2600 | 100 | 1200.000 | +1.250 | 1201.250 | | 2620 | 120 | 1199.600 | +1.800 | 1201.400 | | 2640 | 140 | 1199.200 | +2.450 | 1201.650 | | 2660 | 160 | 1198.800 | +3.200 | 1202.000 | | 2680 | 180 | 1198.400 | +4.050 | 1202.450 | | 2700 | 200 | 1198.000 | +5.000 | 1203.000 | Check: the last curve RL (1203 m) equals the RL of EVC (1203 m), and the curve at the IP chainage lies 1.25 m above the IP, at RL 1201.25 m. #### Step 4: Lowest point At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{2\times 200}{5} = 80$ m from BVC. - Chainage = 2500 + 80 = 2580 m - RL = 1202 + (-2/100)(80) + (5)(80)²/(200 × 200) = 1201.2 m **Answer:** $L = 200$ m, BVC at 2500 m (RL 1202 m), EVC at 2700 m (RL 1203 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The change in grade is 5% and the rate is 0.5% per 20 m chain. What is L? A. 50 m B. 200 m C. 400 m D. 100 m Answer: B. L = (5/0.5) x 20 = 200 m. (Key point: Length from rate of change) 2. With IP at 2600 m and L = 200 m, the BVC and EVC chainages are A. 2550 m and 2650 m B. 2600 m and 2800 m C. 2500 m and 2700 m D. 2400 m and 2800 m Answer: C. Equal tangents of L/2 = 100 m on both sides of the IP. (Key point: BVC and EVC) 3. How is the lowest point on the sag curve found? A. Where the offset is maximum B. At the IP chainage C. At the mid-point of the curve D. Where the grade becomes zero: x = -g1 L/(g2 - g1) = 80 m from BVC Answer: D. At the minimum the slope is zero, giving x = 2 x 200/5 = 80 m, chainage 2580 m. (Key point: Lowest point) 4. How does the offset from the tangent vary along the parabola? A. As the square of x B. Linearly with x C. As the cube of x D. It is constant Answer: A. y = (g2 - g1)x^2/(200L), so offsets grow as x^2; e.g., 0.05 m at 20 m and 0.20 m at 40 m. (Key point: Offsets) 5. What is the RL of the lowest point? A. 1202.000 m B. 1201.200 m C. 1200.000 m D. 1201.250 m Answer: B. RL = 1202 - 0.02 x 80 + 5 x 80^2/(200 x 200) = 1202 - 1.6 + 0.8 = 1201.2 m. 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