Asignalized intersection approach has three lanes with no exclusive left or right turning lanes. the approach has a 40-second green out of a 75-second cycle. the yellow plus all-red intervals for phase total 4.0 seconds. if the start-up lost time is 2.3s/phase, the clearance lost time is 1.1s/phase, and thesaturation headway is 2.48 s/veh under prevailing conditions, what is the capacity of the intersection approach?

786.01

Explanation:

Calculate the saturation flow rate by using the following equation

S = 3600/h, where h is the saturation headway

S = 3600/2.48 = 1452 veh/hg

Calculate the effective green time

G(i) = G(i) + Y(i) – t(Li), where G(i) is the actual green time for movement, Y(i) is the sum of yellow and all red intervals for movement, and t(Li), is the total lost time for movement

G(i) = 40, Y(i) = 4, t(Li) = 2.3 + 1.1 = 3.4

G(i) = 40 + 4 – 3.4 = 40.6

Calculate the capacity of an intersection approach by using the following equation

C(i) = s(i) [g(i)/C], where s(i) the saturation flow rate, g(i) is the effective green time, and C is the cycle length

s(i) = 1452, g(i) = 40.6, C = 75

C(i) = 1452 x (40.6/75) = 786.01 veh/h/ln

The capacity of intersection approach is 786.01

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )