SME
Transportation Engineering · Problem type
Shock wave at a queue
Use this for “Traffic at 40 pc/km/ln and 2667 pc/h/ln meets a queue at 100 pc/km/ln and 1667 pc/h/ln. How fast does the back of the queue move?”
Chart
q(40) ≈ 2,667
q(k) = shock(k)
The curves cross at k = 40 and k = 100.
This is an example. Type your own values to start a new problem.
uw (km/h)
=
(
q₂ (pc/h/ln)
−
q₁ (pc/h/ln)
)
÷
(
k₂ (pc/km/ln)
−
k₁ (pc/km/ln)
)
k₁ and q₁ are the density and flow upstream; k₂ and q₂ are those in the queue.
v₁
Speed upstream
calculated · km/h
v₂
Speed in the queue
calculated · km/h
vf
Free-flow speed
calculated · km/h
kⱼ
Jam density
calculated · pc/km/ln
Formulas
q₁
=
k₁
×
v₁
2,666.6667 = 40 × 66.6667
q₂
=
k₂
×
v₂
1,666.6667 = 100 × 16.6667
uw
=
q₂
−
q₁
k₂
−
k₁
−16.6667
=
1,666.6667 − 2,666.6667
100 − 40
vf
=
v₁
×
k₂
−
v₂
×
k₁
k₂
−
k₁
100 = (66.6667 × 100 − 16.6667 × 40) ÷ (100 − 40)
v₁
=
vf
×
(1
−
k₁
kⱼ
)
66.6667
=
100
×
(1
−
40
120
)
v₂
=
vf
×
(1
−
k₂
kⱼ
)
16.6667
=
100
×
(1
−
100
120
)
Assumptions
•
A shock wave is the moving boundary between two traffic states: here free flow behind a queue.
•
uw < 0: the boundary moves upstream, against the traffic, so the queue grows backward.
•
Both states lie on one Greenshields line; typed q₁ and q₂ fit vf and kⱼ through them.
•
Values are per lane; each vehicle counts as one passenger car (pc).
Step-by-step
WE KNOW
k₁ = 40 pc/km/ln, q₁ = 2,666.6667 pc/h/ln, k₂ = 100 pc/km/ln, q₂ = 1,666.6667 pc/h/ln
FIND
speed upstream (v₁), speed in the queue (v₂), wave speed (uw), free-flow speed (vf), jam density (kⱼ)
Step 1 · Find speed upstream (v₁)
Use q₁ = k₁ × v₁
Divide the flow by the density: the space-mean speed.
v₁
=
q₁
k₁
v₁
=
2,666.6667
40
v₁
=
66.6667
km/h
Step 2 · Find speed in the queue (v₂)
Use q₂ = k₂ × v₂
Divide the flow by the density: the space-mean speed.
v₂
=
q₂
k₂
v₂
=
1,666.6667
100
v₂
=
16.6667
km/h
Step 3 · Find wave speed (uw)
Use uw = (q₂ − q₁) ÷ (k₂ − k₁)
The jump in flow over the jump in density: how fast the boundary between them moves.
uw
=
1,666.6667 − 2,666.6667
100 − 40
uw
=
−
1,000
60
uw = −16.6667 km/h
Step 4 · Find free-flow speed (vf)
Use vf = (v₁ × k₂ − v₂ × k₁) ÷ (k₂ − k₁)
The straight speed–density line through both states meets k = 0 at vf.
vf = (66.6667 × 100 − 16.6667 × 40) ÷ (100 − 40)
vf
=
6,666.6667 − 666.6667
60
vf
=
6,000
60
vf = 100 km/h
Step 5 · Find jam density (kⱼ)
Use v₁ = vf × (1 − k₁ ÷ kⱼ)
Divide the density by the share of free-flow speed lost.
kⱼ
=
k₁
1 −
v₁
÷ vf
kⱼ
=
40
1 − 66.6667 ÷ 100
kⱼ
=
40
1 − 0.6667
kⱼ
=
40
0.3333
kⱼ = 120 pc/km/ln
Check
Put the numbers back in. Both sides should match.
2,666.6667 = 40 × 66.666667 ✓
1,666.6667 = 100 × 16.666667 ✓
−16.6667
=
1,666.6667 − 2,666.6667
100 − 40
✓
100 = (66.6667 × 100 − 16.6667 × 40) ÷ (100 − 40) ✓
66.6667
=
100
×
(1
−
40
120
)
✓
16.6667
=
100
×
(1
−
100
120
)
✓
Topic check
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