Question

Determine the minimum radius of a horizontal circular curve for a route having a 70 mph design speed, super-elevation (e) 3%, and coefficient of side friction 0.12. (b) Determine the degree of curve using your radius value and the arc definition. (c.) Determine the length of curve (L) if the curve angle or central angel is 65 degrees. (d) If the PI is station 2827+45.50, determine the station of the PC and station of the PT.

Answer 1

Answer:

a) Rmin ≈ 52 m

b) D = 110.184°

c) Lc ≈ 59 m

d) PT Station = 2827+12.37

PT Station = 2827+78.63

Explanation:

a) Given

v = 70 mph = (70 mph)(1,609 m/1 mile)(1 h/3600 s) = 31.286 m/s

e = 3% = 0.03

f = 0.12

a) We can use the equation

Rmin = v²/(127*(e + f))

⇒ Rmin = (31.286)²/(127*(0.03 + 0.12))

⇒ Rmin = 51.38 m ≈ 52 m

b) We can use the equation

D = 5729.578/R  ⇒  D = 5729.578/52

⇒  D = 110.184°

c) We apply the formula

Lc = R*Δ/57.3

If  Δ = 65°  we have

Lc = 52*65/57.3

⇒ Lc = 58.98 m ≈ 59 m

d) If the PI is station 2827+45.50 we get the tangent length T as follows:

T = R*tan(Δ/2)

⇒ T = 52*tan(65/2) = 33.13 m

then, the station of the PC will be

PC Station = PI - T

⇒ PC Station = (2827+45.50) - (0+33.13) = 2827+12.37

and the station of the PT will be

PT Station = PI + T

⇒ PT Station = (2827+45.50) + (0+33.13) = 2827+78.63