Question

A vertical parabolic sag curve has a grade of -0.4% followed by a grade of 2% intersecting at station 12 150.60 at elevation 124.80 m above sea level. The change of grade of the sag curve is restricted to 0.6%. (a) Compute the length of the curve (b) Compute the elevation of the lowest point of the curve (c) Compute the elevation at Sta. 12 125.60

Answer 1

Answer:

a) 400 meters

b) 67 meters

c) 124.790 meters

Step-by-step explanation:

Given data :

g1 = - 0.4 %

g2 = +2 %

change of grade of sag curve ( a ) = 0.6 %

Elevation of PVI ( h2 ) = 124.80 m

a) compute the length of the curve using the relation below

a = $\frac{g2-g1}{L}$ .  hence ; L = ( g2 - g1 ) / a

L = ( 2 -(-0.4 ) / 0.6

  = 2.4 / 0.6 = 4  

 =  400 meters

b) compute the elevation of the lowest point of the curve

slope of the vertical curve = 0

0 = 2ax l + b

 = 2( $\frac{g2-g1}{L}$ ) x  + g1

∴ x = $\frac{-g1L}{(g2-g1)}$  = ( 0.4 * 4 ) / ( 2 + 0.4 ) = 1.6 / 2.4 =  0.67 stations

therefore x ( elevation of the lowest point ) = 67 meters

C) compute the elevation at Sta. 12 + 125.60

Given that the rate of change of elevation is the same at all points

= $\frac{h2 - h1 }{PVI -P0}$  = 0.4 -------- ( 1 )

where h2 = 124.80

           h1 = ? ( elevation of p0 ) at 12 + 125.60

           PVI = 12 + 150.06

            p0 =  12 + 125.60

back to the above equation

- h1 = 0.4 ( PVI - P0 ) - h2

     = 0.4 ( 12.150 - 12.1256 ) - 124.80

-h1   = -124.790

hence h1 = 124.790 m