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Two sets of students are collecting traffic data at
t wo sections A and B of a highway 200 m apart.
Observation at A shows that 4 vehicles pass that
section at intervals of 8.18 sec, 9.09 sec, 10.
sec, and 11.68 sec.
The data below shows the result of the flow
of traffic at certain part of the highway by
observing the arrival times for four vehicles
at t wo sections A and B that are 150 m
apart. Compute the space mean speed of the
vehicles.
FAHAD SINSUAT
FUNDAMENTAL OF TRAFFIC FLOW
speed
→
Total distance^ traveled^ over^ total^ time^
of
travel
.
TWO TYPES^ OF SPEED
Time Mean Speed
MK
)
speed taking
into account a
point (^) along
a
roadway
arithmetic mean^ of the^ speed
.
Mk
=
EV
n
space
Mean
speed / Ms)
speed taking
into account a
segment (^) of the^ solution
:
roadway
.
Ms
= ? Ms
=
n
> Harmonic mean
of
the speeds
. Ev
"
-4=1/-1-7-
I__^ 7- 24sec
Ms
=
n
gy
or
^
tz=Tt
Itt (^) 3)
= (^) 6.37sec
E
'T
tz=Tt
12.47 -
(Ft 6sec)=^ 6.47sec
sample
Problem
1-4=171-21.
V
= (^) 150m
7. 24s
= 20 -92m/ s
Vz= 150m
= (^) 23.55m
/ s
6- (^) 37s
V
,
=
150m
= 23.18mV
a.) (^) Mk=? 6.47s
V. =^ 200m
g.1g
= 24.45m / s
V
= 150m
9- (^) 78s
=/ (^) 5.34m
/ s
Vz
= (^) 200m
g. og
= 22m
Mk=24 -451-221-19.551-17.
Ms
= n
Sv
V3 =^ 200M^
= 19.55m / s
Nk
= 20 - 78m
Ms =^
V
= 200m = 17.12m
/ s
MK
= 207813-6)
(20.721-11-123.55511-123.18)
'
Uk
kph
μ,^
= 20.09 Mls
Ms
= 20.09 (^) /3.^ 6)
b.) Ms
= ?
2078m "
/ I}ffm)¥f)
Ms
72.34 kph
Ms
= (^) n =
kph
EV
"
Ms =^4
't
122511-119-555't^ /17.
'
Ms
Mls
Ms
= 20.42 (^) Mls (3.
Us
kph
3.) From the following data of a freeway
surveillance, there are 5 vehicles under
observation and the following distances are the
distance each vehicle had traveled when
observed every 2 sec.
a.) Compute the time and space mean speed
b.) Standard deviation of the space distribution
speed.
The spot speeds in kph observed at a road
section are 66, 62, 45, 79, 32, 51, 56, 60, 53,
and 49. Evaluate the median speed in kph.
1.) There are 4 vehicles passing through
an intersection of highway in a period of
20 seconds.
a.) Determine the space mean speed in
kph.
b.) Computethe flow of traffic if the
traffic density is 60 veh/km.
FAHAD SINSUAT
Reg
'd :^ median^ speed
= 54-
2
a.) (^) Ms =^ ? ;Mss=
b.) standard deviation
DENSITY AND FLOW
Us =^ n
EV
"
Traffic
Flow
/g)
N' =^ 24.4m
= 12.2m, ,
number
of
vehicles
passing through^
a
2s
point
V2 = 25.8m^
=/ 2.
amps
> Veh
hr
2s
V3 = 247m
= (^) 12.35mi,
Traffic
Density (^) / K )
2s
(^) number
of
vehicles
traveling per^
length
V
= 26.7m Of
the roadway
.
= 13.45 Mls
2s > (^) Veh / km
g-
=
22.gg#=11-45m1s
Speed
Ms =^
space mean^ speed
't
/12.9^ )
't
/12.
't (13-45)-11-41.45)
"
q=MsK
Ms
= 12.43 Mls
Ms =^ 12.43/3.6)
sample
Problem
Ms
kph
Mk
-451--11-4-
MK
MK
MK =^ 44.
kph
b.)
Mk
= Ms
t 0
'
Ms
44.89=44.751-
O
'
/ variance^ ) = 6.
3.) During peak hours, 3800 vehicles pass through
a certain highway from 9:00 am to 11:00 am,
with space mean speed of 20 kph. What is the
average space headway per vehicle?
FAHAD SINSUAT
Rea'd^
: sit
= ?
Given :^ n
-3800rem
= 2hr
Given
:
'
M= (^20) Kph 9=
Veh (^) / hv
soln
:
Ms
__
kph
Required
:S#
= ? 2420Wh
= 52km
9=38OzOn%h_
=
1900 Veh
/hrs
q=Msk
q=Msk (^) 2340= /
k
)
/
°¥T
/K )
(^52) km/hr
1<=45 Veh / km
k= 95 Veh
/ km
sit
=s*=
1*11%1-
Sit
= I^
45rem
K SH
'
22.72in
Sit =^ 1km
95rem |Y¥
Sit
= 10.53M / Veh
FREE FL OW SPEED AN D JAM DENSIT Y
speed
no. Of
vehicles
Monday /
4am) Fast^ 10W
High
Monday
18am) "0W
symbolize
as Sam
Density
Free Flow (^) speed
if
the jam density
is zero
M=Mf
; K=O
N
= speed
Mf
= free-flow speed
= 0 j K^
= K
,
G
man,
= maximum flow
g-
Msmax
=
¥
kmax
=
combine
:
9ma×=M÷
.
k{
Cfmax
=
Mfkj
9ma×=Mf¥
ylk
)
49ma?
( (^0). Kj)
y
y
= At Bx
y=
At
BM
UFO )
✗ (^) (m)
Formula :
Ms
= Mf /
l
K
)
ks
The Southern Luzon Expressway was
designed to carry a free-flow speed of 50
kph and a capacity of 4000 vehicles per
interaction of a road leading to Laguna
there were 3600 vehicles counted, what
is the space mean speed of these
vehicles?
Sample
Problem 2.
Given :
Mf
= (^50) Kph
qmax
= 4000 Veh
1hr
g-
Msk
of
= 3600 ✓^ eh (^) / hr
qmai-dyll.sk
Required
:
Ms
= ?
wht
solution
:
4000=
Mnr ) 19
* Ms =
Mf
/
I
¥ ;)
Kj
= (^320)
VM
/ try
qmax
=
4- (Mt)
( Kj)
q=msk
Ms
Mf (^) /
I
¥ )
Msk Ms
= 50 (I^
K
=
→
eq
. I
Ms
9 max
=
4-
( Mt)^ /^
K) )
4000%1=14-
KIT
) Kj
Kj
= 320 Ved
km
substitution eq
l to
formula
Ms
=
50/
)
Ms =
50/
320ms
)
Ms
=
32.91K¥
A portion of the extension of SLEX has a
free flow speed of 60 km/hr and a capacity
of 3,600 veh/hr. If in a given hour, 3,
vehiclea were countered at a specified
point along this highway, determine the
traffic density (vehicle/km) of these 3344
vehicles.
A car moves along florida at maximum speed
of 64 miles/hour. If the maximum flow is 1849
vehicles/hr, determine the density at which
the flow is maximum in veh/km.
V
Sample
Problem 3
Sample problem
Given
:Ms=64mi÷ (^) /
1-6091'm
)
_- 103
Kph
1 mile
Given
:
Mf
Kph qmax
=
1849 Ven
qmax
vehlhr Tv
9=3,
vehlhr (^) Required
:
K
=
Required
: K=
? Solution :
Sol 'n^
:
=
Mf Kj
Ms
÷
)
g=
Msk
g-
K=
Kj
3344M¥
=μsK
Ms
= Mf
Ms
→ eg.^ y
k
103 =^ Mf
9 max^
=
Mf
Kj
Nf
--
Kph
3600M¥
=
4-
160km (^) / nr) Kj
1849=14-1206)^
/ Kj)
Kj
= 240 Veh^ / km
Kj
=
Veh
Tam
substitute eq
-1 to^
formula
:
K
=
3344=60/
¥- )
k
19=17.95 Veh
11=88 Veh Tam
Tam
A freeway has three lanes in each sirection and has a maximum flow of 100 vehicles/min. It
is operating at 50 vehicles/min. A collision occurs, blocking the two lanes and restricting the
flow of the third lane to 25 vehicles/min. The freeway has a constant speed of 60 vehicles/hr
and its three-lane jam density is 60 vehicles/min. The incident is completely cleared in 30
minutes and traffic returns to normal as soon this happened.
Sample Problem^
a.) Determine the
length of qnen
20 minutes
after
the collision
b.) Determine the
longest
vehicle qnen
c.) In^
how
many
minutes will^ the^ auen
dissipate
d.) How^ many vehicles were^ affected by
the accident.
e.) Compute the^
total delay
due to accident
f)
What (^) is the
average
Given :^
qma×=
woven (^) / min /_^ three^
lanes
b.)
max
=
✗ =^50 vehlmin
)
y >^ μ^
i. (^) Q
^
μ =^25
Veh
/ min^
<
.
to -
-1=30 mins
loot
= (^100)
to -30)
solution :
50T
'
i.
→
Qmax
✗ =^50 Veh^ /
Min est^!
I 1- I 1 I 1 1 )
M =^25
Veh / min
to (^) to 30 4050 60 W
Me
= 100 Veh /^ min
M =^ 25-1 ①
30=50/3)^
no.
Of
vehicles
Me
= 100T
Qzo
= (^750) Veh
I
I
c.) 0=0. :X =M
if do^ to^ do^ to^
t^ / mins)
50-1=100 /to -30) 1-^750
v v
Ven Ven
a) Qzo
= (^)?
to =^ 45min
veh
^
d.) Affected / always
arrival rate
@
intersect
✗
4g
= 50T
got ;
0120 ;
)
01020
30 4050 60 70
✗ 45=
Veh
Qzo =^ A^ M
Qzo =^ 50T -25T (^) e.)
Qzo=
(^50120) ) -25120)^ n
zo
= 500 Veh
i.
i.
.
!
i ,
Qmaxi
TOTAL DELAY
, ,
É
, É^ l s
£
(Qmax^ ) ( to)
(^10 20 3040 )
1-D=
12-(7507/30)
t
2-
f- (7507/45)
= (^16875) Veh_
1-D=^16875 Veh / min min
f.)
Dave
= ?
Dave
= TD
N
= 16875 Veh^. min
2250 Veh
Dave
= 7- b- mins
A ramp meter operates during the morning peak period. Ramp meter cycles
vary with time as shown in the table below. The metering scheme allows one
vehicle per cycle to pass the signal. The table below gives the number of
vehicles demanding service on the ramp during the particular time intervals, the
cumulative demand for the ramp for the morning peak and the ramp meter cycle
for each interval.
sample Problem
'
Time Period^ 15min volume cumulative volume^ Meter
cycle
6 : 30 - 6 : 45 75 75
7 : 00
7 : 15
7 : 30
7 : 45
a.) What are^ the service^ rates^ for
meter
cycle 12s^?
spy
,
=
15min /^60 )
= 754th /15min
12s
Iveh SRZ
12s
=
-15min (^160) )
b.) What are the^
service rates
for
meter cycle
10s
90rem
SRO =
10
=
15min,
c.) What^ are the^ service^
rates
cycle
les
= (^) -
F-
= 150
Veh
15 mins
d.) Determine^ the^
time qneu
on the (^) ramp
begins
and ends^
.
6 : 45 and 8 :00am
e.)
Determine the^
longest qn.eu
Given
from
an observation^
along
EDSA
during
rush hour
.
Mean free (^) speed
=
64 Kph
Jam Density
= (^120) Veh
1km
a.) Determine^ the^ maximum^ flow of (^) traffic
9 max
=
Mf Kj
9 max
(64*14)/
no
Y÷m (^) )
9 max^
=
1920M¥
b.)
Determine at which (^) the flow (^) of
traffic
is maximum^
.
V=
21Mt
V =
12-
V
= (^32) kph
c.) Determine^ the^
density
at which the^
flow of traffic
is maximum
K=
kik
=
lZ0vzehlkm_
K =^60 Veh
1hr
1.) The^ mean^ flow speed^
at a portion of
the
highway
is 62.
m¥es_
& a jam density (^) of
110 vent mile
Given
:
Nf
mif÷ /
= 101-
Yf
Kj
= (^110) Veh I
mile/Hain )
Kj
=
' 37
Vk÷
a.) Density^
when the
flow
is maximum
=
& Mf^
Kj
Km/hr Veh /km
=
f-
(101-045)/6837)
9
¥ 1
of
=
Ms
K
÷h
= Us K
Ms
=MY
Ms
= Mf (^) /
I
÷ )
1M¥
= (1010451/1-6%37) K=^
34.22k¥
6.)
Determine the^
service rate
of
one toll booth
every
10min
a.) (^) se
10k¥
SR
=
10 min
b.)
maximum
qneu
0=300 Vehicles
c.) longest delay
longest
delay
-40min
Peak Hour Factor^ (^
PHF
the ratio
of
total
hourly
volume to^ the maximum^ rate
of (^) flow within^
the hour.
V > total Volume in 1 hour
PHF =
4 /Vig
)
Peak 15min flow
1.) (^) Time a.) Peak^ hour
8 : 00 - 8 : 15 32 b.)
Peak Hour Volume
8 : 15
8 : 30 37 c.) Peak Hour^ Factor
8 : 30
8 : 45
9 : 15
a.) 8 :^00
321-371-291-33=131 Veh
8 : 15
321-371-297331-37=136 Veh
8 : 30
37 1- 291-331-371-19=^
Ven
b.) Peak^
Hour Volume
PHV= 136 Veh
c.) Peak^ Hour Factor
PHF =^
I
= 136
= (^) 0.
2.) For^
the given
data (^) shown
time V
6 : 15
6 : 30
a.) Peak Hour
3751-3801-4121-390=1557 Ven
b.) PHF =L
c) (^) Design Hourly Volume
DHV
=PHV_
= 1557 PHF
=I557_
=
= 1656.^
Ven
2.) The number^
of
accident
for
6
years
recorded
in (^) a certain
intersection
of
a highway
is 4877.
If the^ average
is
647. What^ is the accident^ rate^ per million^
.
R =^ A (1×104)
ADT (N)/ 3651
=4877/1×-
(647)
= 3442
3.) The^ number^ of
accidents
for
b-
years
recorded in^ a 10 mile
long (^) highway
is 2345. If the
average daily
traffic
is
502 , what^ accident^ rate^
per
hundred million^
entering
ven ?
¥¥¥:÷
(100×106)
(365×10)
13=25596.
R =^25597
4.) Data on^ traffic
accident
for b-^ years
recorded on a certain sketch
of
two highway
is shown below. Determine the SR.
Year Property (^) Damage injury
fatal
(^2002 )
I
SR =^ fatal t injury
fatal t^
injury
1- property damange
SR = 15 + 314
SR
=
seat work
g.) A^ freeway
is to be designed
as passenger
Only (^) facility for
an AADT
of
35000 Veh /
day.
is estimated that the
freeway
will have a
free flow^ speed^ of
mph
. The
design
will (^) be
for
commuters
and the^ peak
hour factor traffic^ travelling
in the
peak
direction . The K -
factor
for
this
freeway
is
0.148 , determine the directional
design hourly^
volume
.
Given
M =^70 mph
AADT =^ 35.
D= 0-
Required
:
DDHV =^?
So /
'
n
:
DDHV =/D) ( AADT )
K)
= (0-65)/35000)(0-148)
DDHV = 3367 Veh
day
Given
: N =^ 0.4×2-36×+
✗ =^20 yrs
N
=
501h
:
N
= 0.4/2072-361207+
N =^
440 accidents
Given :
A- =
ADT =^476
N =^6
yrs
Required
: R :=
solution
R=
A
(1×106)
ADT (^) (N (^) ) / 365 )
R= (^) 5432/1× 106 )
(^476) / 6) (^1365) )
