Question

A signalized intersection approach has an upgrade of 4%. The total width of the cross street at this intersection is 60 feet. The average vehicle length of approaching traffic is 16 feet. The speed of approaching traffic is 40 mi/h. Determine the sum of the minimum necessary change and clearance intervals.

Answer 1

Answer:

change interval is 3.93 sec

clearance interval is 1.477 sec

Explanation:

Given data:

upgrade of intersection 4%

street total width at intersection is 60 ft

vehicle length of approaching traffic = 16 ft

speed of approaching traffic =40 mi/hr

85th percentile speed is calculated as

S_{85} = S +5

S_{ 85} = 40 + 5  = 45 mi/h

15th Percentile speed

$S_{15} =S-5$

          = 40 - 5 = 35 mi/hr

change in interval is calculated as

$y = t + \frac{1.47 S_{85}}{2a +(64.4\times 0.01 G}$

t is reaction time is 1.0,

deceleration rate is given as 10 ft/s^2

$y = 1.0 +\frac{1.47\times 45}{2a +(64.4\times 0.01\times 4}$

y = 3.93 s

clearance interval is calculated as

$a_r = \frac{W+ L}{1.47\times S_{15}}$

$a_r = \frac{60+16}{1.47\times 35} = 1.477 s$