--- title: "Vertical curve -0.5% and +0.7%, IP RL 500" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-neg05-pos07-ip-1000" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-neg05-pos07-ip-1000 site: BCT Engineering --- # Vertical curve -0.5% and +0.7%, IP RL 500 Surveying II · Chapter 6: Curves - Asked in: 2072 Magh, 6 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-neg05-pos07-ip-1000 - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question Calculate the R.L.s of pegs on a vertical curve connecting two grades of -0.5% and +0.7% at the intersection point which has chainage = 1000 m and R.L. = 500 m. The rate of change of grade is 0.1% per 30 m. Take peg interval = 20. ## Short answer Rate of grade change 0.1%/30 m gives L=360 m. Equal-tangent curve: BVC chainage 820 m (RL 500.9 m), EVC chainage 1180 m (RL 501.26 m). Curve RLs from y = RL_BVC + g1 x/100 + (g2-g1)x²/(200L) at 20 m intervals; lowest point at 970 m, RL 500.525 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 30 m, algebraic change in grade $= 0.7 - (-0.5) = 1.2\%$. $$ L = \frac{|g_2-g_1|}{r}\times 30 = \frac{1.2}{0.1}\times 30 = 360\ \text{m} $$ #### Step 2: BVC, EVC and their levels The curve is a sag (valley) curve with equal tangent lengths $L/2 = 180$ m. $$ \begin{aligned} \text{Ch. of BVC} &= 1000 - 180 = 820\ \text{m}\\ \text{Ch. of EVC} &= 1000 + 180 = 1180\ \text{m}\\ \text{RL of BVC} &= 500 - \frac{(-0.5)}{100}\times 180 = 500.9\ \text{m}\\ \text{RL of EVC} &= 500 + \frac{(0.7)}{100}\times 180 = 501.26\ \text{m} \end{aligned} $$ #### 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 $g_1 = -0.5\%$, $g_2 = 0.7\%$, $L = 360$ m, so the offset is $y = \frac{(1.2)x^2}{200\times 360}$ and pegs are taken at multiples of 20 m. | Chainage (m) | x from BVC (m) | Tangent RL (m) | Offset y (m) | Curve RL (m) | |---|---|---|---|---| | 820 | 0 | 500.900 | 0.000 | 500.900 | | 840 | 20 | 500.800 | +0.007 | 500.807 | | 860 | 40 | 500.700 | +0.027 | 500.727 | | 880 | 60 | 500.600 | +0.060 | 500.660 | | 900 | 80 | 500.500 | +0.107 | 500.607 | | 920 | 100 | 500.400 | +0.167 | 500.567 | | 940 | 120 | 500.300 | +0.240 | 500.540 | | 960 | 140 | 500.200 | +0.327 | 500.527 | | 980 | 160 | 500.100 | +0.427 | 500.527 | | 1000 | 180 | 500.000 | +0.540 | 500.540 | | 1020 | 200 | 499.900 | +0.667 | 500.567 | | 1040 | 220 | 499.800 | +0.807 | 500.607 | | 1060 | 240 | 499.700 | +0.960 | 500.660 | | 1080 | 260 | 499.600 | +1.127 | 500.727 | | 1100 | 280 | 499.500 | +1.307 | 500.807 | | 1120 | 300 | 499.400 | +1.500 | 500.900 | | 1140 | 320 | 499.300 | +1.707 | 501.007 | | 1160 | 340 | 499.200 | +1.927 | 501.127 | | 1180 | 360 | 499.100 | +2.160 | 501.260 | Check: the last curve RL (501.26 m) equals the RL of EVC (501.26 m), and the curve at the IP chainage lies 0.54 m above the IP, at RL 500.54 m. #### Step 4: Lowest point At the lowest point the grade is zero, so $x = \dfrac{-g_1 L}{g_2-g_1} = \dfrac{0.5\times 360}{1.2} = 150$ m from BVC. - Chainage = 820 + 150 = 970 m - RL = 500.9 + (-0.5/100)(150) + (1.2)(150)²/(200 × 360) = 500.525 m **Answer:** $L = 360$ m, BVC at 820 m (RL 500.9 m), EVC at 1180 m (RL 501.26 m); curve levels as tabulated. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The grade changes by 1.2% and the rate of change of grade is 0.1% per 30 m. What is the curve length L? A. 360 m B. 36 m C. 120 m D. 3600 m Answer: A. L = (1.2/0.1) x 30 = 360 m. (Key point: Length of curve) 2. Why is the curve from -0.5% to +0.7% a sag (valley) curve? A. Because the chainage of IP is 1000 m B. Because the curve is a parabola C. Because the grade changes from falling to rising D. Because the grade changes from rising to falling Answer: C. A falling grade followed by a rising grade makes a valley, so the curve lies above the IP tangent offsets (offsets are added). (Key point: Curve type) 3. Why is the RL of BVC (500.9 m) higher than the RL of IP (500 m)? A. Because the first grade is falling, so ground rises backwards from IP to BVC B. Because L/2 is added to the chainage C. Because the offset is added at BVC D. Because the curve is a summit curve Answer: A. BVC is 180 m before the IP on a -0.5% grade, so it is 0.5% x 180 = 0.9 m higher than the IP. (Key point: BVC level) 4. How is the lowest point located? A. At the mid-point x = L/2 = 180 m B. Where the offset is largest, at the EVC C. Where the grade is zero, x = -g1 L/(g2 - g1) from BVC, here 150 m D. At the IP chainage Answer: C. The lowest point has zero grade; x = 0.5 x 360/1.2 = 150 m, chainage 970 m, which is not at the IP. 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