Question

Question 3 (10 marks) A 1,500ft long sag vertical curve (equal tangent) has a PC at stenine 1,500f. The initial grade is -3% and the fina on 3+70 and elevation I grade is +6%. Determine the elevation and station 3+70 and a PVC at stationing of the low point, PVI, and PVT

Answer 1

Given,

Length of vertical curve, L = 1500 ft = 15 stations

G1 = -3%

G2 = +6%

PVC station = (3+70)

PVC elevation = 1500 ft

a) Station of PVC = Station of PVI - L/2

Station of PVI = Station of PVC + L/2 = (3+70) + (7+50) = (11+20)

Station of PVT = Station of PVI + L/2 = (11+20) + (7+50) = (18+80)

b) Elevation of PVC = Elevation of PVI - G1 * L/2

Elevation of PVI = Elevation of PVC + G1 * L/2 = 1500 + (-0.03) * 750 = 1477.5 ft

Elevation of PVT = Elevation of PVI + G2 * L/2 = 1477.5 + (0.06) * 750 = 1522.5 ft

c) Equation of the curve: $y = ax^{2}+bx+c$

a = (G2-G1)/2L = (0.06 - (-0.03))/(2*1500) = -0.00003

b = G1 = -0.03

c = elevation of PVC = 1500 ft

Lowest point of curve:

X = (-b/2a) = (-(-0.03)/2*(0.00003)) = 500 ft = (5+00) stations

Station of lowest point = station of PVC + X = (3+70) + (5+00) = (8+70)

Elevation of lowest point = (0.00003 * 500 * 500) + (-0.03 * 500) + 1500 = 1492.5 ft