# Junction Eq

Topics: Orders of magnitude, Time, Dual carriageway Pages: 2 (430 words) Published: February 6, 2013
EQUATIONS FOR PREDICTING THE CAPACITY OF TURNING TRAFFIC STREAMS μba = D(627 + 14Wcr – Y(0.364qac + 0.144qab+ 0.229qca + 0.520qcb)) μbc = E(745 – Y( 0.364qac + 0.144qab )
μcb = F(745 – 0.364Y( qac + qab )
D = (1 + 0.094(wab – 3.65))(1 + 0.0009(Vrba – 120)) x (1 + 0.0006(Vlba – 150)) E = (1 + 0.094(wbc – 3.65))(1 + 0.0009(Vrbc – 120))
F = (1 + 0.094(wcb – 3.65))(1 + 0.0009(Vrcb – 120))
μ = stream capacity
qab = measured flow of stream a-b
qac = measured flow of stream a-c
qca = measured flow of stream c-a
qcb = measured flow of stream c-b
wba = avg. lane width over a dist. of 20m available to waiting vehicles in the stream b-a, meters wbc = avg. lane width over a dist. of 20m available to waiting vehicles in the stream b-c, meters wcb = avg. lane width available to waiting vehicles in the stream c-b, meters Vrba = Visibility to the right from a 10m back from the give-way line for vehicles making the b-a maneuver, meters Vrbc = Visibility to the right from a 10m back from the give-way line for vehicles making the b-c maneuver, meters Vrcb = Visibility to the right from a 10m back from the give-way line for vehicles making the c-b maneuver, meters wcr = width of the central reserve (only for dual carriageways), meters Y = (1 – 0.0345W)

W = total major road carriageway width, meters

L = ρ + C ρ²
(1 – ρ)

regular vehicle arrivals C = 0
random arrivals C = 1
ρ = flow(λ) ÷ capacity(μ)

L = ρ/ (1 – ρ)

L = (ρ – 1)μt

QUEUE LENGTH
L = 0.5 x ((A² + B) - A)

A = (1 – ρ)(μt)² + (1 – Lo)μt – 2(1 – C)(Lo + ρμt)
μt + (1 – C)

B = 4(Lo + ρμt)(μt – (1 – C)(Lo + ρμt))
μt + (1 – C)

DELAY PER UNIT TIME
Dt = 0.5 x ((F² + G) - F)

F = (1 – ρ)(μt)² – 2(Lo - 1)μt – 4(1 – C)(Lo + ρμt)
2(μt + 2(1 – C))
G = 2(2Lo + ρμt)(μt – (1 – C)(2Lo + ρμt))
μt + 2(1 – C)

DELAY PER ARRIVING VEHICLE
Dv = 0.5 x ((P² + Q) - P) + 1/μ
P = (0.5 x (1 – ρ)t) – (1/μ)(Lo – C)
Q = (2Ct/μ)(ρ + 2Lo/μt)
μ = capacity
ρ = ratio of...