Question

1. A vertical curve joins a -1.2% grade to a +0.8% grade. The two grades intersect at station 75 + 00 and elevation 50.90 m above sea level. The centerline of the roadway must clear a pipe located at station 75 + 40 by 0.80 m. The elevation of the top of the pipe is 51.10 m above sea level. What is the minimum length of the vertical curve that can be used?

Answer 1

Answer:

The minimum length of the vertical curve that can be used is 416.63 m

Explanation:

g₁ = -1.2% , g₂ = 0.8%

Station (Sta) = 75 + 00

Elevation (Ele) = 50.90 m

x₀ = Sta(VPC) = Sta(VPI) - L/2 ⇒ 7500 - L/2

y₀ = Ele(VPC) = Ele(VPI) - g₁L/2 == 50.90 - (-0.012)L/2 ⇒ 50.90 + 0.012L/2

at Sta 75+40, x = 7540 -x₀ = 40 + L/2

at Ele, y = 51.10 + C = 51.10 + 0.80 ⇒ y = 51.90 m

Equation of vertical curve is

$y = y_0 + g_1x + \frac{g_2-g_1}{2L}x^2$

Substituting the values for x and y

$51.90 = (50.90 + \frac{0.012}{2}L)+(-0.012)(40+\frac{L}{2)}+\frac{(0.8+1.2)}{100*2L}*(40+\frac{L}{2})^2$

$0 = -1.48 + \frac{0.02}{2L}(1600+\frac{L^2}{4}+40L)$

$0 = -148L + 1600 + \frac{L^2}{4} + 40L$

Solving : L = 416.63 m