Question

On a two-lane road with 12-ft lanes, a horizontal curve is designed for 50 mi/h with a superelevation of 10%. The PI of the curve is at 220+48 and the PC is at station 216+74. Determine the station of the PT and the middle ordinate for stopping sight distance.

Answer 1

Answer:

Detailed solution is given below:

Answer 2

Answer:

• 223 + 60 ( station of the PT )
• 32.27 ft ( middle ordinate for stopping distance )

Explanation:

lanes = 12-ft

speed = 50 mi/h

super elevation = 10%

PI of curve = 220 + 48

PC at the station = 216 + 74

PT = ?

Middle ordinate for SSD = ?

from the table of radius for a 10% super elevation and 50 mi/h

r = 694 + 6 = 700

Solution:

PT = PC + L  ----------------- equation 1

L = πr∅ / 180 ------------------ equation 2

∅ can be calculated from this equation

T = r * tan ∅ / 2 --------------- equation 3

while T = PI - PC

            = 220 + 48 - 216 + 74 = 3 + 74

            = 374 ft

back to equation 3

tan ∅ / 2 = T / r

tan ∅ / 2 = 374 / 700

∴ ∅ = 56.15 ⁰

back to equation 2

L = π * 700 * 56.15 / 180

L = 686.05 ft

back to equation 1 ( station of the PT )

= PC + L

= 216 + 74 + 6 +86

= 223 + 60

middle ordinate for stopping sight distance

for a speed of 50 mi/h the SSD ( stopping sight distance ) = 425 ft

middle ordinate ( M₀ ) = r ( $1-cos\frac{90*SSD}{\pi*r }$ )

                                    = 700 ( 1 - $cos\frac{90* 425}{\pi*700 }$ )

                                    = 32.27 ft