The document provides various examples related to traffic flow analysis, including calculations for flow, density, time mean speed, and space mean speed based on vehicle speeds and intervals. It outlines the methods for calculating these metrics using formulas and real data, such as vehicle times and counts on highways. It also introduces models like Greenshield's and Greenberg's to derive relationships between speed and density in traffic scenarios.

1. Examples
1. Fundamentals of traffic Flow
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2. Examples
1. The figure shown below that the vehicles traveling at constant speed
on two lane highway between section X and Y with their positions and
speeds obtained at an instant of time by photogrammetry. An observer
located at point X observes four the vehicles passing point X during the
period of 2minutes. The velocities of the vehicles measured as 45, 45, 40
and 30 km/hr respectively. Therefore, calculate the flow, density, space
mean speed and time mean speed.
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Mohammed A.
2

3. Examples
Given data
 number of vehicles n = 4
The velocities of the vehicles
vehicle A = 45 km/hr
vehicle B = 45 km/hr
vehicle C = 40 km/hr
vehicle D = 30 km/hr
The time period T = 2minutes
 Required
calculate:
a) flow,
b) density,
c) time mean speed and
d) space mean speed
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4. Examples
Solution
a) Flow
The flow can be expressed by the
formula:
𝑞 =
4∗3600
2∗60
= 120 veh/hr
b) Density
It can be given by the formula:
𝐾 =
4∗1000
300
= 13.3 veh/km
c) time mean speed
It can be found by the equations:
𝑈𝑡 =
45+45+40+30
4
= 40 km/hr
d) space mean speed
It can be also found by:
L is the distance between X and Y.
Now, the time ti takes the ith vehicle
to travel from X toY at speed ui can
be found by:𝒕𝒊 =
𝑳
𝑼𝒊
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5. Examples
𝑡𝐴 =
300
0.278∗45
= 24 sec
𝑡𝐵 =
300
0.278∗45
= 24 sec
𝑡𝐶 =
300
0.278∗40
= 27 sec
 𝑡𝐷 =
300
0.278∗30
= 36 sec
Therefore, the space mean speed becomes:
=
4∗300
24+24+27+36
= 10.81 m/sec
= 38.92km/hr
Note:
the time mean speed is always higher
than the space mean speed.
≥
i.e 40 km/hr > 𝟑𝟖. 𝟗𝟐𝒌𝒎/𝒉𝒓 ……ok!!!
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6. Examples
2. Two set of vehicles are timed over a kilometer, and the flows are also recorded.
In the first sets, four vehicles are take 52,56,63 and 69 seconds when the flow is
1500 veh/hr. In the second set, four vehicles are take 70,77,74 and 79 seconds
when the flow is 1920 veh/hr. These vehicles have been timed over space, then
calculate the density, space mean speed, free flow speed, jam density and capacity
for the given two sets of vehicles?
Given:
Data for the first set of vehicles: Data for the 2nd set of vehicles:
Number of vehicles, n = 4 Number of vehicles, n = 4
Timed over a kilometer, ti Timed over a kilometer, ti
52,56,63 and 69 seconds 70,77,74 and 79 seconds
The flow, q = 1500 veh/hr. The flow, q = 1920 veh/hr.
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7. Examples
Required:
1. the density,
2. space mean speed,
3. free flow speed,
4. jam density and
5. capacity
Solution:
1. the density,
It can be found by:
 =
52+56+63+69
4
= 60 sec. for the 1st set of vehicles
 =
70+77+74+79
4
= 75 sec. for 2nd set of vehicles
Now, the density for both sets of vehicles can be:
𝑘 = 1500 ∗
60
3600
= 25 veh/km for 1st set of vehicles
𝑘 = 1920 ∗
75
3600
= 40 veh/km for 2nd set of vehicles
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8. Examples
2. space mean speed
It can be given by the relations:
=
𝑞
𝑘
=
1500
25
= 60 km/hr 1st set
=
1920
40
= 48 km/hr 2nd set of vehicle
3. free flow speed
In order to determine the free flow speed we
Need to develop the equation using Green shield
model:
Now, substituting the values of
speed and density into the equation
and get the required values.
60Kj = Uf*Kj- 25Uf……..1
48Kj = Uf*Kj- 40Uf……..2
Multiplying equation 1 by 8 and
equation 2 by -5 and we get:
480Kj = 8UfKj – 200Uf….1
-240Kj = -5UfKj + 200Uf….2
Add the two equations and we get:
240Kj = 3UfKj
 Uf = 80 km/hr.
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









j
f
k
k
u
u 1

9. Examples
4. jam density
To determine jam density substitute
the value of free flow speed into
equation 1 and we get:
60Kj = UfKj- 25Uf…1
Kj = 100 veh/km
5. Capacity
It can be calculated by:
𝑞𝑐𝑎𝑝 =
100 ∗ 80
4
qcap = 2000 veh/hr.
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4
f
j
cap
cap
cap
u
k
u
k
q 


10. Examples
3. The speeds of five vehicles were measured (with radar) at the
midpoint of a 0.5-mile section of roadway. The speeds for vehicles 1,
2, 3, 4, and 5 were 44, 42, 51, 49, and 46 mi/h, respectively.
Assuming all vehicles were traveling at constant speed over this
roadway section, calculate the time-mean and space-mean speeds.
Given data
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Vehicles Speeds (mi/hr)
1 44
2 42
3 51
4 49
5 46

11. Examples
Required:
a) Time mean speed
b) Space mean speed
Solution:
The time mean speed is given by:
=
44+42+51+49+46
5
= 46.4 mi/hr
 the space mean speed is given by:
ok!!!
=
𝟓
𝟏
𝟒𝟒
+
𝟏
𝟒𝟐
+
𝟏
𝟓𝟏
+
𝟏
𝟒𝟗
+
𝟏
𝟒𝟔
= 46.17 mi/hr
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s
t v
v 

12. Examples
4. Vehicle time headways and spacing were measured at a point along a
highway, from a single lane, over the course of an hour. The average values
were calculated as 2.5 s/veh for headway and 200 ft/veh (61 m/veh) for
spacing. Calculate the, flow, density and speed of the traffic.
Given Data:
Average headway = 2.5 sec/veh
Average spacing = 200 ft/veh or 61m/veh
Required:
I. Flow
II. Density and
III. speed
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13. Examples
Solution:
I. Flow
It can be found by the relation:
𝑞 =
1
2.5 𝑠𝑒𝑐/𝑣𝑒ℎ
= 0.4 veh/sec
because the data were collected for an
hour: 𝑞 = 0.4
𝑣𝑒ℎ
𝑠𝑒𝑐
∗ 3600 sec/h = 1440veh/h
II. Density
It also calculated by using the formula:
𝑘 =
1
61 𝑚/𝑣𝑒ℎ
= 0.0164 veh/m
 applying this spacing over the
course of one kilometer:
𝑘 = 0.0164
𝑣𝑒ℎ
𝑚
∗ 1000𝑚/𝑘𝑚
K = 16.4 veh/km
III. Speed
It is given by the formula:
𝒖 =
𝟏𝟒𝟒𝟎 𝒗𝒆𝒉/𝒉
𝟏𝟔.𝟒 𝒗𝒆𝒉/𝒌𝒎
= 87.8 km/h
It can be also computed by:
𝑢 =
61 𝑚/𝑣𝑒ℎ
2.5 𝑠𝑒𝑐/𝑣𝑒ℎ
= 24.4 m/s
u = 24.4 m/s*3.6 = 87.8km/h
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h
q
1

d
k
1


14. Examples
5. A section of highway is known to have a free-flow speed of 55 mi/h and a
capacity of 3300 veh/h. In a given hour, 2100 vehicles were counted at a
specified point along this highway section. If the linear speed-density
relationship applies, what would you estimate the space-mean speed of these
2100 vehicles to be?
Given data:
Uf = 55 mi/hr
qcap = 3300 veh/h
q = 2100 veh in a given hour
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15. Examples
Required:
the space-mean speed
Solution:
linear speed-density relationship
is given by:
Rearranging the above equation
to solve for u,
Now, determine the jam density
using:
Substitute the values of the given data in
to the equation and we get:
By applying quadratic equation we get
the required space mean speed as:
u = 44.08 mi/h or 10.92 mi/h
From the results both of these speeds
are feasible.
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16. Examples
6. Consider Greenshield’s linear speed-density model U= 55 – 0.45K , then drive
flow-density relationships, and calculate Free flow speed, Jam density and
Capacity.(where speed is in mi/hr, density veh/mi and flow veh/hr).
Given data
Linear speed- density model, U = 55 – 0.45K
Required
Drive flow- density relationships
Free flow speed
Jam density
capacity
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17. Examples
Solution
Derivation of flow- density relationships
U= 55 – 0.45K but q = u*k, u = q/k substitute
(q/k) = 55 – 0.45K finally we get,
q = 55k – 0.45k^2 it is parabolic relationship
Free flow speed
Free flow speed occurs when the density is zero.
U= 55 – 0.45K, k=0
Uf = 55 – 0.45 *0
Uf = 55 mi/hr
Jam density
Jam density occurs when speed is
zero.
U= 55 – 0.45K, u = 0
0 = 55 – 0.45*Kj
Kj = 55/0.45 = 122.22 veh/mi
capacity
It is computed by:
𝑞 =
55 ∗122.22
4
=1680.53 veh/hr/ln
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4
f
j
cap
cap
cap
u
k
u
k
q 


18. Examples
8.A section of highway has the following speed-flow relationship:
q+100u = 300u -4u^2, then calculate the following:
a) free flow speed
b) density at capacity
c) capacity
d) the flow rate corresponding to a density of 80vpm.
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19. Examples
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20. 9. The data shown below were obtained on a highway. Use regression analysis to
develop Greenshields model/equation based on the given data shown below in
the table.
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Examples

21. Solution:
Based on Greenshield model there is a linear relation ship between traffic stream
parameters. Therefore, in order to determine the equation we use linear regression
with the general form shown below.
Where, a and b are regression constant which is given by the formula:
The computation is shown below in the table
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Examples

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Examples

23.  we see that the speed in the Greenshields expression is represented by y in the
estimated regression function, the mean free speed uf is represented by a, and
the value of the mean free speed uf divided by the jam density kj is represented
by -b. We therefore obtain:
Finally the required regression equation becomes: y = 62.68 – 0.53x
Where, y represents space mean speed and x represents density and we get
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Examples

24. 10. Redo example 9 using Greenberg model.
Solution:
Based on Greenberg there is logarithmic relationship between traffic stream
parameters which can be given by the expression
e see that in the Greenberg expression is represented by y in the estimated
regression function, c ln kj is represented by a, c is represented by -b, and ln k
is represented by x. We therefore obtain
The computation is shown below in the table
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Examples

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Examples
 Since a =145.06 and b = -28.68, the speed for maximum flow is c = 28.68 mi/h. Finally,
Then we get the equation

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Examples
11. A driver takes 3.5 s to react a complex situation while traveling at a speed of
60 km/hr. how far does the vehicle travel before the driver initiates a physical
response to the situation?
Solution:
The vehicle will travel : dr = 0.278vt, where
v = 60km/h
t = 3.5s
Therefore, the distance traveled by the vehicle cab be:
dr = 0.278*60*3.5
= 58.4 m

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Examples

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Examples

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