--- title: "Vertical curve 200 m, +1.50% then -2.5%" subject: "Surveying II" chapter: 6 key: "semester_4/surveying_ii:vertical-curve-200m-upward-15-down" source: https://www.bctengineer.com/q/semester_4/surveying_ii/vertical-curve-200m-upward-15-down site: BCT Engineering --- # Vertical curve 200 m, +1.50% then -2.5% Surveying II · Chapter 6: Curves - Asked in: 2079 Jestha, 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-200m-upward-15-down - Whole chapter: https://www.bctengineer.com/md/semester_4/surveying_ii/chapter-6 ## Question A 200 meter equal tangent parabolic vertical curve is to be placed to negotiate a upward grade of 1.50% followed by a downward grade at 2.5% intersecting at a station having elevation 1185.795 m above mean sea level. Calculate elevations at even 20 m stations on the vertical curve and determine the station and elevation of the highest point on the vertical curve. ## Short answer The 200 m equal-tangent parabolic summit curve has its highest point at chainage 975.00 m and RL 1184.858 m. Elevations at 20 m stations are computed from PVC (RL 1184.295 m) using rate r = –0.0002 per m, as tabulated. ## Answer Data: $g_1 = +1.5\% = +0.015$, $g_2 = -2.5\% = -0.025$, $L = 200$ m (equal tangents of 100 m), RL of the point of intersection (IP) = 1185.795 m. The chainage of the IP is not given; it is assumed to be 1000.000 m (0+1000), so that the PVC is at 900 m and the curve pegs at even 20 m stations fall at 900, 920, ... 1100 m. It is a summit (crest) curve. ```text g1=+1.5% IP g2=-2.5% ___.-'`'-.___ PVC./ (crest) \.PVT 900 1100 ``` #### Elements - Rate of change of grade: $r = \dfrac{g_2 - g_1}{L} = \dfrac{-0.025 - 0.015}{200} = -0.000200$ per m - PVC chainage $= 1000 - 100 = 900.00$ m; PVT chainage $= 1100.00$ m - RL of PVC $= 1185.795 - 0.015\times 100 = 1184.295$ m - RL of PVT $= 1185.795 - 0.025\times 100 = 1183.295$ m #### Level on the curve (parabola) $$ RL_x = RL_{PVC} + g_1x + \frac{r}{2}x^2 $$ where $x$ is the distance from the PVC. | Chainage (m) | x from PVC (m) | Tangent level (m) | Offset $y=\frac{r}{2}x^2$ (m) | Curve level (m) | |---|---|---|---|---| | 900.00 | 0.00 | 1184.295 | 0.000 | 1184.295 | | 920.00 | 20.00 | 1184.595 | -0.040 | 1184.555 | | 940.00 | 40.00 | 1184.895 | -0.160 | 1184.735 | | 960.00 | 60.00 | 1185.195 | -0.360 | 1184.835 | | 980.00 | 80.00 | 1185.495 | -0.640 | 1184.855 | | 1000.00 | 100.00 | 1185.795 | -1.000 | 1184.795 | | 1020.00 | 120.00 | 1186.095 | -1.440 | 1184.655 | | 1040.00 | 140.00 | 1186.395 | -1.960 | 1184.435 | | 1060.00 | 160.00 | 1186.695 | -2.560 | 1184.135 | | 1080.00 | 180.00 | 1186.995 | -3.240 | 1183.755 | | 1100.00 | 200.00 | 1187.295 | -4.000 | 1183.295 | #### Highest point The curve is highest where the grade is zero: $g_1 + r x = 0$. $$ x = \frac{-g_1}{r} = \frac{0.015}{0.0002} = 75.00\ \text{m} $$ Chainage $= 900.00 + 75.00 = 975.00$ m and $$ RL = 1184.295 + 0.015\times 75.00 - 0.0001\times 75.00^2 = 1184.858\ \text{m} $$ Check: the level at mid-curve (x = 100 m) is 1184.795 m, which is midway between the IP and the mid-point of the chord PVC-PVT (1183.795 m), i.e. $e = \dfrac{(g_1-g_2)L}{8} = 1.00$ m below the IP. **Answer: highest point at chainage 975.00 m with RL 1184.858 m; the levels at the 20 m stations are given in the table.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With L = 200 m, IP at assumed chainage 1000 and RL 1185.795, g1 = +1.5%, what are the PVC chainage and RL? A. Chainage 900.00, RL 1184.295 B. Chainage 900.00, RL 1187.295 C. Chainage 1100.00, RL 1183.295 D. Chainage 1000.00, RL 1185.795 Answer: A. PVC is L/2 = 100 m before the IP, and its level is 1185.795 - 0.015 × 100 = 1184.295 m. (Key point: PVC chainage and level) 2. What is r = (g2 - g1)/L for this curve? A. +0.0002 per m B. -0.0200 per m C. -0.0002 per m D. -0.0001 per m Answer: C. r = (-0.025 - 0.015)/200 = -0.0002 per m; the negative sign shows a summit curve. (Key point: Rate of change of grade) 3. How is the highest point located? A. At x = g1/L from the PVC B. At the IP chainage C. At x = -g1/r = 75 m from PVC, where the grade is zero (chainage 975) D. At the PVT Answer: C. The curve is highest where g1 + r·x = 0, so x = 0.015/0.0002 = 75 m, chainage 975.00, RL 1184.858 m. (Key point: Highest point) 4. What is the offset e of the curve below the IP? A. e = (g1 - g2)L/8 = 1.00 m B. e = (g1 - g2)L/2 = 4.00 m C. e = (g1 + g2)L/8 = 0.20 m D. e = (g1 - g2)L/4 = 2.00 m Answer: A. e = 0.04 × 200/8 = 1.00 m, so the curve at mid-point is 1184.795 m, 1 m below the IP. 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