--- title: "Vertical curve -0.7% and +0.5%, IP RL 650" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-neg07-pos05-rl-ip" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-neg07-pos05-rl-ip site: BCT Engineering --- # Vertical curve -0.7% and +0.5%, IP RL 650 Surveying II · Chapter 6: Curves - Asked in: 2075 Baisakh, 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-neg07-pos05-rl-ip - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question A grade of -0.7% is followed by another grade of +0.5%. The two ends of these portions are connected by a parabolic vertical curve. The chainage and RL of intersection point are 1000 and 650 m respectively. Calculate RLs of all the points on the curve. Take peg interval of 20 m and rate of change of grade is 0.1% per 20 m. ## Short answer Length 240 m based on grade change of 1.2% at rate 0.1% per 20 m. BVC at 880 m (RL 650.84 m), EVC at 1120 m (RL 650.60 m). Lowest point at chainage 1020 m, elevation 650.35 m. ## 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 $r = 0.1\%$ per 20 m, algebraic change in grade $= 0.5 - (-0.7) = 1.2\%$. $$ L = \frac{|g_2-g_1|}{r}\times 20 = \frac{1.2}{0.1}\times 20 = 240\ \text{m} $$ #### Step 2: BVC, EVC and their levels The curve is a sag (valley) curve with equal tangent lengths $L/2 = 120$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 1000 - 120 = 880\ \text{m}\\ \text{Ch. of EVC} &= 1000 + 120 = 1120\ \text{m}\\ \text{RL of BVC} &= 650 - \frac{(-0.7)}{100}\times 120 = 650.84\ \text{m}\\ \text{RL of EVC} &= 650 + \frac{(0.5)}{100}\times 120 = 650.6\ \text{m} \end{aligned} $$ #### 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 $g_1 = -0.7\%$, $g_2 = 0.5\%$, $L = 240$ m, so the offset is $y = \frac{(1.2)x^2}{200\times 240}$ and pegs are taken at multiples of 20 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 880 | 0 | 650.840 | 0.000 | 650.840 | | 900 | 20 | 650.700 | +0.010 | 650.710 | | 920 | 40 | 650.560 | +0.040 | 650.600 | | 940 | 60 | 650.420 | +0.090 | 650.510 | | 960 | 80 | 650.280 | +0.160 | 650.440 | | 980 | 100 | 650.140 | +0.250 | 650.390 | | 1000 | 120 | 650.000 | +0.360 | 650.360 | | 1020 | 140 | 649.860 | +0.490 | 650.350 | | 1040 | 160 | 649.720 | +0.640 | 650.360 | | 1060 | 180 | 649.580 | +0.810 | 650.390 | | 1080 | 200 | 649.440 | +1.000 | 650.440 | | 1100 | 220 | 649.300 | +1.210 | 650.510 | | 1120 | 240 | 649.160 | +1.440 | 650.600 | Check: the last curve RL (650.6 m) equals the RL of EVC (650.6 m), and the curve at the IP chainage lies 0.36 m above the IP, at RL 650.36 m. #### Step 4: Lowest point At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{0.7\times 240}{1.2} = 140$ m from BVC. - Chainage = 880 + 140 = 1020 m - RL = 650.84 + (-0.7/100)(140) + (1.2)(140)²/(200 × 240) = 650.35 m **Answer:** $L = 240$ m, BVC at 880 m (RL 650.84 m), EVC at 1120 m (RL 650.6 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With rate of change of grade 0.1% per 20 m and total change 1.2%, L is A. 240 m B. 120 m C. 24 m D. 60 m Answer: A. L = (1.2/0.1) x 20 = 240 m. (Key point: Length from rate of change) 2. How are BVC chainage and RL found? A. Chainage = IP - L = 760 m B. Chainage = IP - L/2 = 880 m; RL = RL of IP - g1 x L/2/100 = 650.84 m C. RL of BVC = 650 - 0.84 = 649.16 m D. Chainage = IP + L/2 = 1120 m Answer: B. The BVC is L/2 before the IP, and with a falling first grade it lies higher than the IP by 0.007 x 120. (Key point: BVC chainage and RL) 3. What is the offset of the curve from the tangent at distance x? A. y = (g2 - g1)x/(200L) B. y = (g2 + g1)x^2/(200L) C. y = (g2 - g1)x^2/(200L) D. y = g1 x^2/(100L) Answer: C. The parabolic offset is proportional to x^2: y = (g2 - g1)x^2/(200L) with grades in %. (Key point: Offset formula) 4. Where is the lowest point? A. x = 120 m, chainage 1000 m B. x = 100 m, chainage 980 m C. x = 240 m, chainage 1120 m D. x = -g1 L/(g2 - g1) = 140 m from BVC, chainage 1020 m, RL 650.35 m Answer: D. The grade becomes zero at x = 0.7 x 240/1.2 = 140 m. 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