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

a) 400 meters

b) 67 meters

c) 124.790 meters

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 =
`(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(
`(g2-g1)/(L)` ) x + g1

∴ x =
`(-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

=
`(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