A single lane highway has a horizontal curve. The curve has a super elevation of 4% and a design speed of 45 mph. The PC station is 105+00 and the PI is at 108+75.

What is the station of the PT?

Answer: 112 + 19.27

Explanation:

Super elevation is an inward transverse slope provided through out the length of the horizontal curve which ends up serving as a counteract to the centrifugal force and checks tendency of overturning. It changes from infinite radius to radius of a transition curve.

Super curve elevation (e) = 4%

4/100 = 0.04

e = V^2/gR

Make R the subject of the formula.

egR = V^2

R = V^2/eg

V = 45mph

=45 * 0.44704m/s

=20.1168m/s

g (force due to gravity) = 9.81

Therefore,

R = (20.1168) ^2/9.81 * 0.04

= 1031.31m

Tangent Length (T) = PI - PC

Tangent Length = 10875 - 10500

=375m

T = R Tan (I/2)

375 = 1031.31 * Tan (I/2)

I = 39.96

Also,

L = πRI/180

= 719.27m

Station PT = Stat PC + L

10500 + 719.27

=11219.27

=112 + 19.27