Density and Spacing

k = \frac{1}{s} = \frac{5280}{s _{ft}} (veh/mi)

Macroscopic density k is the reciprocal of average spacing s (front-to-front). In US customary units, s in feet gives k = 5280/s in veh/mi. Metric: k (veh/km) = 1000/s_m. Headway h (s/veh) is the time analog: q = 3600/h. Flow, density, and space-mean speed remain tied by q = k v.

$k$ (veh/mi): Density
$s$ (ft/veh): Average spacing
$q$ (veh/h): Flow rate
$h$ (s/veh): Average headway

Step-by-step solved example

Average spacing is 88 ft/veh. Find density in veh/mi. If space-mean speed is 55 mph, find flow.

k = 1/s. Vehicles occupy space along the lane.

1. Density
k = 5280/88 = 60 veh/mi.
$k = 60 veh/mi$
2. Flow
q = k v = 60×55 = 3300 veh/h.
$q = 3300 veh/h$

Answer: k = 60 veh/mi, q = 3300 veh/h

10 practice questions

0/10 correct

1.Spacing 132 ft/veh corresponds to k =

2.Density k is the reciprocal of

3.Metric: s = 25 m/veh. k =

4.If k = 50 veh/mi, average spacing is

5.Headway h = 2.4 s implies q =

6.Doubling spacing (same speed) makes density

7.Jam spacing of 20 ft/veh is jam density

8.Spacing is measured

9.k = 40 veh/mi, v = 45 mph. q =

10.Time headway and spacing are related by s = v h when