An equaltangent crest vertical curve that is 1200 feet long

An equal-tangent crest vertical curve that is 1200 feet long connects tangents that intersect at Station 118 + 00 and elevation 1000 ft. The initial grade is +3.0% and the final grade is 2.0%.

a) Determine the elevation and stationing of the high point, PVC, and PVT.

b) what is the elevation of the curve at Station 120+00?

Solution

Solution:-

(1)

D₁ = L/2 =1200/2 = 600 feet

Station of PVC = station of PVI – D₁/100

                          = 118+00 – 600/100

                         = 112 station    Answer

Elevation of PVC = elevation of PVI – (G_i*N₁)

= 1000 – (3*6) = 982 feet   Answer

D₂ = 1200/2 = 600 feet

Station of PVT = station of PVI + D₂/100

= 118+00 + 600/100

= 124 station Answer

Elevation of PVT = elevation of PVI + (G_f *N₂)

= 1000 + (-2*6)

= 988 feet Answer

(2) Elevation of curve station at 120 +00

Change in station = 120 – 118 = 2

Elevation 0f station 120+00 = elevation of PVI + (G_f*N)

= 1000 + (-2 *2)

= 996 feet    Answer