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Introduction to Transportation Engineering

Instructor Dr. Norman Garrick

Hamed Ahangari27th March 2014

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Traffic Stream Analysis

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Key equations

Flow headway (h) - (s/veh)

Flow rate (q) - (veh/h)

q=3600/(h) (1) T=3secT=0 sec

h1-2=3sec

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Key equations

Density Spacing (s) -(ft/veh)

density-concentration(k)- (veh/mi)

k=5280/(s) (2)

S1-2S2-3

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Key equations

Speed U(TMS)= 1/n ∑ vi (3)

U(SMS)= N

ivN 1

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Key equations

q=u.k (4)flow=u(SMS) * densityq = k u

(veh/hr) = (veh/mi) (mi/hr)

h = 1 / q(sec/veh) = 1 / (veh/hr) (3600)

s = 1 / k(ft/veh) = 1 / (veh/mi) (5280)

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Example 1Data obtained from aerial photography showed six vehicles on a 600 ft-long section of road. Traffic data collected at the same time indicated an average time headway of 4 sec. Determine

(a) the density on the highway, (b) the flow on the road, (c) the space mean speed.

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Solution• Given:

• h=4 sec, l=600 ft, n=6

• Part (a):– Density (k):K= (n)/(l)= 6/600= 0.01 veh/ft

k=0.01*5280= 52.8 veh/mile

• Part (b):– flow (q):q= 1/h= ¼= 0.25 veh/sec q = 0.25*3600= 900 veh/hour

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Solution

• Part (c):– Space Mean Speed (U(sms)):U(sms)= q/k

=900/52.8U(sms)= 17 miles/hour

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Traffic Flow CurvesMaximum Flow, Jam Concentration, Freeflow Speed

u

k

u

q

k

qqmax

qmaxkj

uf

kj - jam concentrationu = 0, k = kj

qmax - maximum flow

uf - free flow speedk = 0, u = uf

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Example 2Assume that :

u=57.5*(1-0.008 k)Find:

a) uf free flow speed

b) kj jam concentration

c) relationships q-u, d) relationships q-k,e) qmax capacity

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Solution

• Part a):free flow speed?i) when k=0 uf

ii) u=57.5*(1-0.008 k)

i+ii) u=57.5*(1-0.008 k)= 57.5*(1-0.008*0) uf =57.5 miles/hour

kj

uf

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Solution

• Part b): kj jam concentration?

i) when u=0 kj

ii) u=57.5*(1-0.008 k)

i+ii) 0=57.5*(1-0.008 k)---- 0.008k=1

kj = 125 veh/miles

kj

uf

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Solution

• Part c): relationships q-u?i) q=u.k ii) u=57.5*(1-0.008 k)---u/57.5=1-0.008k

1-u/57.5=0.008K-----K=125-2.17u) (iii)

i+iii) q= u.k= u.(125-2.17u)

q= 125u-2.17u^2u

q

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Solution

• Part d): relationships q-k?i) q=u.k u=q/kii) u=57.5*(1-0.008 k)

i+ii) q/k=57.5*(1-0.008k)

q=57.5k -0.46k^2k

q

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Solution• Part e): qmax capacity ?

i)

ii) q=57.5k -0.46k^2

d(q)/d(k)=0 57.5-0.92k=0 km=62.5

qmax =1796 veh/hour

k

q

:0,max dk

dqsetqfindTo

qmax

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Example 3• The data shown below were obtained on a highway.

Use linear regression analysis to fit these data and determine

– (a) the free speed, – (b) the jam density,– (c) the capacity, – d) the speed at maximum flow.

Speed (mi/h) Density (veh/mile)14.2 8524.1 7030.3 5536.8 4740.1 4150.6 2055.0 15

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Plot data

10 20 30 40 50 60 70 80 900

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20

30

40

50

60

f(x) = − 0.568553388627363 x + 62.9183254875588R² = 0.996125718732137

Density

Spee

d

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• from plot: u = -0.57k + 62.92

• Part a) uf=62.92 miles/hour

• Part b) kj = 110.8 veh/mi

• Part c) qmax = 1736 veh/hr

• @ qmax, u = 31.5 mph and k = 55.2 veh/mi

Solution

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Shockwave Analysis

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Example 4- Length of Queue Due to a Speed Reduction

• The volume at a section of a two-lane highway is 1500 veh/h in each direction and the density is about 25 veh/mi.

• A large dump truck from an adjacent construction site joins the traffic stream and travels at a speed of 10 mi/h for a length of 2.5 mi.

• Vehicles just behind the truck have to travel at the speed of the truck which results in the formation of a platoon having a density of 100 veh/mi and a flow of 1000 veh/h.

• Determine how many vehicles will be in the platoon by the time the truck leaves the highway.

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Solution

Approach Conditions(Case1)

q1=1500 veh/hrK1=25 veh/mi

Platoon Conditions (Case2)

q2=1000 veh/hrK2=100 veh/mi

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• -6.7 mile/hour

• Step 2: Growth rate of platoon:= 10+6.7=16.7 mile/hour

• Step 3: Time(truck)= distance/u= 2.5/10= 0.25 hour

• Step 4: Length of platoon= time*= 0.25*16.7= 4.2 mile

• Step 5: Queue length= density*distance=100*4.2= 420 vehicle

Solution

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Distance

Time

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Example 5- Length of Queue Due to Stop

• A vehicle stream is interrupted and stopped by policeman. • The traffic volume for the vehicle stream before the

interruption is 1500 veh/hr and the density is 50 veh/mi. • Assume that the jam density is 250 veh/mi. After four

minutes the policeman releases the traffic.• The flow condition for the release is a traffic volume of

1800 veh/hr and a speed of 18mph. Determine :

• the length of the queue • and the number of vehicles in the queue after five minutes. • how long it will take for the queue to dissipate after the

policeman releases the traffic.

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Solution

Shockwave 1 Shockwave 2

State 3q = 1800 veh/hru = 18 mi/hr

State 1q = 1500 veh/hrk = 50 veh/mi

State 2q = 0 veh/hr

k = 250 veh/mi

Approach conditions Platoon Conditions Release Conditions

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• -7.5 mi/hr

• Shockwave 1 is moving upstream at -7.5 mph

• Length of the queue after 4 minutesLength = u*t = 7.5 mph * 4/60 hr = 0.5 mile

• Vehicles are in the queue after 4 minutes# of vehicles = k * L = 250 *0.5 = 125 vehicles

Solution

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-12 mi/hr

• Shockwave 2 is moving upstream at 12 mph

• usw1 = - 7.5 mph usw2 = - 12 mph

• The queue will dissipate at rate of 4.5 mphTime to dissipate a 0.5 mile queue is L/speed=0.5 mile / 4.5 mph = 0.012 hr = 6.6 minutes

Solution