Question

A level rural two-lane highway has a 10-degree curve section as it passes by a large building. The obstruction corner of the building is 24.5 ft. away from the center line of the highway. PRT: 2.5 s, f= 0.5.

A. If the lanes of the highway are 11 ft wide, what is the required SSD on this section and speed?
B. What is the safe operating speed on the section and required SSD in case the corner is only 17 ft away from the centerline of the highway?

Answer 1

A. 336.3 ft.

B. 279.83 ft.

Step-by-step explanation:

The formula for Stopping Site Distance SSD is given s follows;

`SSD=(2 * R * \theta * \pi)/(180)`

`R = (M)/(1 - cos\theta)`

R = 5730/D = 5730/10 = 573 ft

`\theta = (SSD * D)/(200)`

D = Degree of curvature

θ = (SSD×D)/20 = (28.65×SSD)/573

Where:

`(M)/(R) = 1 - cos \left((28.65 \cdot SSD)/(R) \right)`

D = 10°

M is the distance of the obstruction from the highway = 24.5 ft

`(24.5)/(573) = 1 - cos \left((28.65 \cdot SSD)/(573) \right)`

`cos \left((28.65 \cdot SSD)/(573) \right) = 1 - (24.5)/(573) =(1097)/(1146)`

`cos^(-1) \left((1097)/(1146)\right) = (28.65 \cdot S)/(573) =16.815`

573×16.815= 28.65×SSD

SSD = 9635.13/28.65 = 336.3 ft

B. When M = 17, we have

`(17)/(573) = 1 - cos \left((28.65 \cdot SSD)/(573) \right)`

`cos \left((28.65 \cdot SSD)/(573) \right) = 1 - (17)/(573) = (556)/(573)`

`cos^(-1) \left( (556)/(573)\right) = (28.65 \cdot SSD)/(573) = 13.99 \approx 14`

SSD = 573×14/28.65 = 279.83 ft.