CivilSolve

Transportation & Surveying

Greenshields Traffic Flow

Greenshields' linear speed-density model: v = vf(1 − k/kj) and q = k·v. Give the free-flow speed and jam density plus ONE of current density, speed, or flow, and it returns the rest — with the capacity qmax = vf·kj/4 and (for a given flow) both the uncongested and congested density solutions. Units: densities in veh per the SAME distance unit as the speeds (km or miles). Use for: 'calculate the flow rate using Greenshields', speed-density-flow conversions, capacity questions. Not for: other models (Greenberg, Underwood) or signalized intersections.

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Inputs

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Worked examples

These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.

Example 1

flow from density (the motivating exam problem)

Given

free flow speed
80 km/h
jam density
120
density
30

Assumptions

  • Greenshields' linear speed–density relation; densities are per km; flows are vehicles per hour on one lane.

Solution steps

  1. Apply the linear model

    Speed falls linearly from vf at empty road to zero at jam density; flow is density times speed.

    v = vf·(1 − k/kj); q = k·v

    k = 30 veh per kilometre; v = 60 km/h; q = 1800 veh/h

    = 1800

  2. Capacity check

    The model's maximum flow occurs at half the jam density and half the free-flow speed.

    qmax = vf·kj/4

    qmax = 2400 veh/h

Results

Traffic flow q (veh/h)

1800

Space-mean speed v

16.67m/s

Density k (veh/km)

30

Capacity qmax (veh/h)

2400

Where this answer was checked
source
v = vf(1 − k/kj); q = k·v — vf = 80 km/h, kj = 120 veh/km, k = 30
verified by
hand-recomputed
derivation
v = 80·(1 − 0.25) = 60 km/h; q = 30·60 = 1800 veh/h; qmax = 80·120/4 = 2400 veh/h.

Example 2

density from speed

Given

free flow speed
100 km/h
jam density
100
speed
75 km/h

Assumptions

  • Greenshields' linear speed–density relation; densities are per km; flows are vehicles per hour on one lane.

Solution steps

  1. Apply the linear model

    Speed falls linearly from vf at empty road to zero at jam density; flow is density times speed.

    v = vf·(1 − k/kj); q = k·v

    k = 25 veh per kilometre; v = 75 km/h; q = 1875 veh/h

    = 1875

  2. Capacity check

    The model's maximum flow occurs at half the jam density and half the free-flow speed.

    qmax = vf·kj/4

    qmax = 2500 veh/h

Results

Traffic flow q (veh/h)

1875

Space-mean speed v

20.83m/s

Density k (veh/km)

25

Capacity qmax (veh/h)

2500

Where this answer was checked
source
k = kj(1 − v/vf) — vf = 100, kj = 100, current v = 75 km/h
verified by
hand-recomputed
derivation
k = 100·(1 − 0.75) = 25 veh/km; q = 25·75 = 1875 veh/h.

Example 3

two states for one flow (uncongested and congested branches)

Given

free flow speed
80 km/h
jam density
120
flow
1800

Assumptions

  • Greenshields' linear speed–density relation; densities are per km; flows are vehicles per hour on one lane.

Solution steps

  1. Apply the linear model

    Speed falls linearly from vf at empty road to zero at jam density; flow is density times speed.

    v = vf·(1 − k/kj); q = k·v

    k = 30 veh per kilometre; v = 60 km/h; q = 1800 veh/h

    = 1800

  2. Capacity check

    The model's maximum flow occurs at half the jam density and half the free-flow speed.

    qmax = vf·kj/4

    qmax = 2400 veh/h

Results

Traffic flow q (veh/h)

1800

Space-mean speed v

16.67m/s

Density k (veh/km)

30

Capacity qmax (veh/h)

2400

Congested-branch density (veh/km)

90

Notes

  • A given flow corresponds to TWO traffic states — a fast/uncongested one and a slow/congested one; both densities are reported.
Where this answer was checked
source
quadratic (vf/kj)k² − vf·k + q = 0 — vf = 80, kj = 120, q = 1800
verified by
hand-recomputed
derivation
0.6667k² − 80k + 1800 = 0 → k = (80 ± 40)/1.3333 → k = 30 (fast) and k = 90 (congested).