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Introduction to Transportation Engineering. Instructor Dr. Norman Garrick Hamed Ahangari 27th March 2014. Traffic Stream Analysis. Key equations. T=3sec. T=0 sec. h 1-2 =3sec. Key equations. S 2-3. S 1-2. Key equations. Key equations. q= u.k (4) flow=u(SMS) * density. - PowerPoint PPT Presentation
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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 speedb) kj jam concentrationc) 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 dkdqsetqfindTo
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