traffic flow analysis - national chiao tung universityocw.nctu.edu.tw/course/ftf011/lec2-01.pdf ·...
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Traffic Flow Analysis: (traffic stream and shockwave)
Dr. Gang-Len Chang Professor and Director of Traffic Safety and
Operations Lab. University of Maryland, College Park, MD 20742
1
Q, K, V, Q = K. V (space mean speed)
K and V relation? Q and V relation? Kmax?
Qmax
Vmax
2
k
q
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
Shockwave
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Shock wave - a rapid change in traffic conditions (speed, density, and flow)
A moving boundary between two different traffic states
Forward, backward, and stationary shock waves
Shockwave Analysis A illustrative video: shockwave at signalized
intersections
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u1, k1
u2, k2
u3, k3
20
u1, k1 u2, k2
Low density High density
Shock wave speed
w
a b
12
12 )()(kk
kqkqws −−
=
Shockwave Analysis – General Equation
21
q2
k1
w=(q2-q1)/(k2-k1) 1 Low density
k
q Space
Time
0
w
u1
u2
k2
q1
2 High density
k 1
k 2
u2=q2/k2
u1=q1/k1
22
Traffic Interruption
kA kC=kJ/2 kJ
qC qA
wAC
k
q
A
C
J
wAJ
wCJ
uC uA
uJ= 0
Space
Time
0
C J A A
J
C
uC
wAC
wAJ
wCJ
uA
uJ=0
Traffic interaction between two density levels
23
u1, k1 u2, k2
Low density High density
Shock wave speed
w
a b
11 )( kwuqa −=
w u k u kk k
w q qk k
=−−
=−−
2 2 1 1
2 1
2 1
2 1
Since qa = qb, then (u1 - w)k1 = (u2 - w)k2
22 )( kwuqb −=
24
)1(
)1()1(
)1(and
and
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12
11
22
12
1122
12
12
jf
jf
jf
jf
kkkuw
kk
kkkuk
kku
w
kkuu
kkkukuw
ukqkkqqw
+−=
−
−−−
=
−=−−
=
=−−
=
Using the flow-density diagram, determine the speeds of the following shock waves: (1) Jam conditions change to capacity conditions (2) Density 3kj/4 rapidly decreases to kj/4
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k1= kA k2= kJ
J
AfAJ
J
JAfAJ
Jf
kku
wk
kkuw
kkkuw
−=+
−=
+−=
)1(
)1( 2112
Arrival conditions Jam conditions
Waves of Stopping
Wave of Starting
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k1 = kj k2 = kC = kj /2
2)2/1(
)1( 2112
fCJ
J
JJfCJ
Jf
uw
kkkuw
kkkuw
−=+
−=
+−=
Shockwave Theory- Congestion formulation and Dissipation Definition:
if the traffic stream is stationary over time and distance, then: k and n (and therefore λ) are independent of x (distance) and t
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tttxM
ttt
q ixix ii
∆∆
=∆
Φ−Φ=
),,()()( 0
)(]1)([lim
0t
tMP
xt
λ=∆
≥⋅→∆
xxtxN
xtt
k ixx ii
∆∆
=∆Φ−Φ
=),,()()(
)()],,([lim
0xk
xxtxNP
tx
=∆
∆→∆
29
Shockwave Theory- Congestion formulation and Dissipation
using a plane in n-x-t space, n(x,t) : number of cars at time t location x ------- assumed relation which satisfy:
akxt +−= λ
)tan(),(),( tconskx
txnconstt
txn−=
∂∂
===∂
∂ λ
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λ
v
∆λ∆x
u
Shockwave Theory- Congestion formulation and Dissipation
when vehicles or traffic flows change from (λ1, µ1) to (λ2, µ2) then, ⇒ along this intersection
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111111 axktn +−= λ
222222 axktn +−= λ
tn
tn
∂∂
=∂∂ 21
dtdxk
dtdxk 2211 −=− λλ⇒
kkkdtdx
∆∆
=−−
==λλλµ
21
21∴ (shockwave speed)
Shockwave Theory- Congestion formulation and Dissipation
32
t t
x
u3
u2
u1
Shockwave Theory- Congestion formulation and Dissipation
An Example on Freeway Incident Impact Analysis
On a section of freeway, an accident occurs at 10:00am at point B. First, both lanes are blocked, then after 15 minutes, one lane is cleared.
Traffic Sensor Data: A: q = 2700 vph, s = 90km/h B: one-lane – q = 1500vph, s = 7.5km/h B: two-lane – q = 3600vph, s = 60km/h Kjam = 300veh/lane/km
Questions: 1. Where is the end of queue at 10:15am? 2. When will the queue stop to increase? 3. How long is the max queue length? 4. By what time must the second lane be cleared to avoid the impact from the accident extending to A? 5. What is the total incident impact period? 33
A B
Incident
14.17km
34
Q1: Where is the end of the queue at 10:15 AM? Speed of the shockwave:
hkmkkqq /10
30090
270002700
21
211 −=
−
−=
−−
=µ
(queue formulation speed) k2 = kmax , q2 = 0 After 15 min, X1(t = 15 min) = µ1×t = -10×(15/60) = -2.5 km (negative toward upstream)
35
Q2: When are vehicles last forced to stop by the queue?
After the partial clearning: Traffic flow q3 (one lane) = 1500 vph Traffic able changes again form a second shockwave k3 = 1500/7.5 = 200 , q2 = 0 µII = (queue dissipation speed) |µII| < |µ1| queue will be dissolved
The second wave meets the back of the first wave XI = X (t = 15) +µI×t = -2.5 - 10×t (Initial queue at 10 min) XII = µII×t = -15×t XI = XII -2.5-10×t = -15×t t* = 0.5 h = 30 min ⇒ 10:45 (30 min. after the start of the shockwave II)
hkmkkqq /15
20030015000
32
32 −=−
−=
−−
36
Q3: The maximum queue
by the time the congestion dissipation move commences at 10:15 AM, X1 = max. queue = 2.5 km
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Q4: What is the max. distance of the end of the queue
from the site of the accident? The max. distance at time 10:45 AM (30 min) ∴ XI (t = 30) = -2.5 - 10×(0.5 hr) = -7.5 km
At 10:45 AM, shock waves I and II meet and form a new shockwave: q1 = 2700 (coming flow), q2 = 1500 (outgoing) k1 = 30 k2 = 200 to reach X = 14.17 = µIII⋅t + X0 = -7.06t – 7.5 = 14.17 t = 57 min ⇒ the distribution should have reached A at 11:42 AM if the blockage is completely removed: ⇒ the 4th shockwave:
38
Q5: By what time must the second lane be cleared if the disturbance to the traffic flow resulting from the accident is not to extend to entrance A?
39
The 4th shockwave is a congestion dissipation wave: It takes 0.95h = 14.17/15 =57 min to reach A. Note: |µIV| > |µIII| ⇒ meet at A, not to disturb traffic at A This wave should begin 10:45 AM After wave IV meet wave III, a new shock wave will form: q1 = 2700 (coming flow), q5 = 3600 (outgoing) k1 = 30, k5 = 60 ∴
⇒ increasing speed
hkmkkqq
IV /1560/3600200
36001500
43
43 −=−−
=−−
=µ
hkmV /30603036002700
=−−
=µ
40
41
H
h
t∆
u1
x
1 2 nN
u2
v1
v2
t∆ta ∆tb
42
max1
11
0kk
q−−
=µmax2
2
2max
22
0kk
qkk
q−
=−
−=µ
1µ=∆th
2µ=−∆ Hth
)(21 Htt −∆=∆ µµ
21
2
µµµ−
−=∆
Ht ⇒
N = kmax ⋅ h = mile
nilevehicle
⋅ = vehicles Total vehicle
∴
Note: shockwave is usually negative
)()(
11
max21max12
max21
21
max
max
kkqkkqHkqq
kH
khN
−−−⋅⋅⋅
=
−
⋅=
⋅=
µµ
Shockwave use the negative sign to indicate the direction ⇒the actual queue is
)()( 2max11max2
max21
max
kkqkkqHkqq
khN
−−−⋅⋅⋅
=
⋅−=−
43
Homework (see p. 163): Please prove: The total stopped time of all N vehicles
−−
−⋅−
−⋅=2
2max
1
1max
max2)1(
qkk
qkk
kNNHNTH
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