Question

The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the road if (a) the road is level and (b) the road has a maximum grade of 4%? Assume the perception-reaction time = 2.5 sec

Answer 1

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

• SSD is the stopping sight distance which is to be calculated.
• u is the speed which is given as 60 mi/hr
• t is the perception-reaction time given as 2.5 sec.
• a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
• G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              $SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

• SSD is the stopping sight distance which is to be calculated.
• u is the speed which is given as 60 mi/hr
• t is the perception-reaction time given as 2.5 sec.
• a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
• G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              $SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.

For downgrade of 4%, Substituting values

                              $SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.