A crest vertical curve connects a +4.90% grade and a –1.85% grade. The PVI is at station 77+00.00 at an elevation of 770.00 ft. The design speed is 60 mi/h. Determine: the following below
(a) The length of the vertical curve using the AASHTO method (“K” factors)
(b) The station of the BVC
(c) The elevation of the BVC
(d) The station of the EVC
(e) The elevation of the EVC
(f) The station of the high point
(g) The elevation of the high point
(h) The elevation of station 75+30.00
Also, provide a tabulation showing the elevations at each full station on the curve,
including the distance from the BVC for each full station, the elevation on initial
tangent, the offset, and the elevation on the vertical curve. This can be in the form of
a computer printout from Microsoft Excel or another program.
What will be an ideal response?
(a) From Table 15.4, K = 151L = kA = (151)|–4.90–1.85| = (151)(6.75) = 1019.25 ft
(b) Station of BVC = (77+00) – (1019.25)/2 = 71+90.38
(c) Elevation of BVC = 770.00 – (0.0490)(1019.25/2) = 745.03 ft
(d) Station of EVC = (77+00) + (1019.25)/2 = 82+09.63
(e) Elevation of EVC = 770.00 – (0.0185)(1019.25/2) = 760.57 ft
(f) Station of the high point = (71+90.38) + (1019.25)(4.90)/(6.75) = 79+30.28
(g) Elevation of the high point =
(h) Elevation of station 75+30 =
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