characteristics 1. drivers' characteristics · 2018. 12. 2. · 3. vehicle characteristics...

Civil Engineering Department TRAFFIC ENGINEERING Third Year 2018

Lecturer: Dr. Mohammed Bally Mahdi 14

CHARACTERISTICS

1. Drivers' characteristics

a) Driving task: القیادة مھمة

By keeping the vehicle at a desired speed and position with a lane, interaction

with other traffic, and reading guide signs.

b) Vision: الرؤیة

1. Visual acuity البصر حدة

2. Peripheral vision المحیطیة الرؤیة

3. Colour vision اللون رؤیة

4. Depth perception ابعاد بثلاثة الاجسام ادراك على البصریة القدرة( العمیق الادراك )

5. Hearing perception

c) Perception-reaction time (P.I.E.V): الفعل رد زمن – الادراك

1. Perception (seeing the stimuli) المحفزات رؤیة( الادراك )

2. Interpretation (understanding the stimuli) المحفزات فھم – ترجمة

3. Evaluation of appropriate response (i.e., decision) المناسبة الاستجابة تقییم

4. Volition or response (i.e., apply the reaction) الاستجابة او القرار



P.I.E.V. Refers to the time taken to detect the target, identify the target, decide on

response and initiate the response. Perception-reaction time does not include the

time to execute the decision (e.g. stop by applying a brake). The perception-

reaction time is not fixed for all drivers and also changes for a same driver depend

on the situations.

Factors affecting perception-reaction time

1. Age العمر

2. Fatigue الاجھاد

3. Complexity of Cues الدلائل( الاشارات تعقید )

4. Presence of Drugs or Alcohol الكحول او المخدرات تعاطي

5. Expectation التنبأ( التوقع )

Complexity of Cues

2. Pedestrian characteristics

Same characteristics of drive with addition of others witch influent the design and

location of the pedestrian control devices such as:

1. Special pedestrian signals.

2. Safety zones and islands at intersections.



3. Pedestrian underpasses.

4. Elevated walkways.

5. Crosswalks.

all-red ال طور تصمیم ان مثلا, السیطرة وسائل تصمیم في مباشر تاثیر لھا العبور في المشاة ممیزات

سرعة معرفة یتطلب حیث للیسارات الكثیف المرور خلال التقاطعات في الاشخاص لمرور یسمح والذي

.النفسیة والحالة التعلم ودرجة والذكاء والجنس العمر حسب تختلف قد والتي للاشخاص المشي

3. Vehicle characteristics

Criteria for the geometric design of highways are partly based on:

A. Static characteristics

Height, Width, Length and Minimum and Maximum Turning Radii as they

control of payment design

B. Kinematic characteristics

Involve the movement of the vehicle without considering the forces that causes

the motion such as acceleration capability of the vehicle, braking, turning,…….

C. Dynamic characteristics

Involve the forces that causes the motion of the vehicle:

1. Air resistance

2. Grade resistance

3. Rolling resistance

4. Curve resistance

5. Friction resistance



Vehicle performance

Vehicle performance is defined by how well a vehicle can accelerate, decelerate,

brake and maneuver.

Vehicle types

In general, there are three common types of vehicles, these are:

1. Passenger cars

Passenger cars are two-axle, four-tires, generally with seating for two to six

passengers.

2. Trucks

Vehicles with at least two-axle and six tires, and have a cargo area. Trucks are

used for commercial goods transportation.

3. Buses

Buses are designed to carry passengers and have more than four tires.

4. Road characteristics

The characteristics that affects on :

Stopping sight distance للوقوف الرؤیة مجال

Passing sight distance للتجاوز الرؤیة مجال

These characteristics are:

a) Gradient

b) Superelevation

c) Geometric design of the road



Examples:

Vehicle time headways and spacing were measured at a point along a highway,

from a single lane, for an hour. The average values were calculated as 2 .5 s for

headway and 61 m for spacing . Calculate the average speed of the traffic.

Average speed = spacing / headway

= (61/1000) / (2.5/3600)

= 87.84 km/h

Or

q = 3600/ha

= 3600/ 2.5= 1440 veh

Density = 1000 / 61 = 16.39

Average speed = q/D

= 1440 / 16.39 = 87.85 km / h



Examples 1:

An observer counts 360 veh/h at a specific highway location. Assuming that the

arrival of vehicles at this highway location is Poisson distributed, estimate the

probabilities of having 0, 1, 2, 3, 4, and 5 or more vehicles arriving over a 20-

second time interval .

SOLUTION

The average arrival rate, 20, is 360 veh/h, or 0.1 vehicles per second (veh/s) .

Using t = 20 seconds, the probabilities of having exactly 0, 1, 2, 3, and 4

vehicles arrive are:

For five or more vehicles,

P(5)=1—P(n<5 )

=1— 0.135 — 0.271 — 0.271 — 0.180 — 0.090

= 0 .053



A histogram of these probabilities is shown in Figure

Number of vehicles arriving in a 20-s time interval

(probabilities in parentheses)

Example 2:

Traffic data are collected in 60-second intervals at a specific highway location

as shown in Table 1 . Assuming the traffic arrivals are Poisson distributed and

continue at the same rate as that observed in the 15 time periods shown, what is

the probability that six or more vehicles will arrive in each of the next three 60-

second time intervals (12 :15 P .M . to 12 :16 P.M., 12 :16 P.M. to 12 :17 P.M.,

and 12 :17 P.M. to 12 :18 P.M.)?



Table 1 shows that a total of 101 vehicles arrive in the 15-minute period from

12 :00 P .M . to 12 :15 P .M . Thus the average arrival rate, λ, is 0.112 veh/s

(101/900) . Then find the probabilities of exactly 0, 1, 2, 3, 4, and 5 vehicles

arriving. Applying Poisson Eq, with λ =0.112 veh/s and t = 60 seconds, the

probabilities of having 0, 1, 2, 3, 4, and 5 vehicles arriving in a 60-second time

interval are (using AAR = 6.733) { λ t =0.112*60} 0r {101/15}

Table 1 Observed Traffic Data for Example

No Time period Observed number of vehicles

1 12 :00 P .M . to 12 :01 P .M . 3

2 12 :01 P.M. to 12 :02 P.M. 5

3 12 :02 P .M . to 12 :03 P .M . 4

4 12 :03 P .M . to 12 :04 P.M. 10

5 12 :04 P.M. to 12 :05 P.M. 7

6 12 :05 P.M. to 12 :06 P.M. 4

7 12 :06 P.M. to 12 :07 P.M. 8

8 12 :07 P.M. to 12 :08 P.M. 11

9 12:08 P .M . to 12 :09 P.M. 9

10 12:09 P .M . to 12 :10 P.M. 5

11 12:10 P .M . to 12 :11 P.M. 3

12 12:11 P .M . to 12 :12 P .M . 10

13 12:12 P .M . to 12 :13 P.M. 9

14 12:13 P .M . to 12 :14 P.M. 7

15 12:14 P .M . to 12 :15 P.M. 6

Total 15 Minutes 101



The summation of these probabilities is the probability that 0 to 5 vehicles will

arrive in any given 60-second time interval, which is: = 0.0012+0.008+0.027+0.0606+0.102+0.137 = 0 .3358

So 1 minus P(n≤5) is the probability that 6 or more vehicles will arrive in any

60-second time interval, which is

P(n≥6)=1- P(n≤5)

=1-0.3358

= 0 .6642

The probability that 6 or more vehicles will arrive in three successive time

intervals (t1 , t2 , and t3) is simply the product of probabilities, which is:

P(n≥6) for three successive time periods = (0.6642)^3

= 0.293



Example 3: Five vehicles, as shown in the figure below, are traveling at

constant speeds on section of 230m length. Assuming that all vehicles have a

same length of 4m and if speeds and clear spacing between vehicles are as

shown in the figure, estimate the following:

1. Traffic density

2. Average time headway arriving a section A-A

3. Average clear spacing

Traffic density D = (veh/km) = (푁표.표푓 푣푒ℎ푖l푒푠 표푛 푎 푠푒푔푚푒푛푡/ 푙푒푛푔푡ℎ 표푓 푡ℎ푒 푠푒푔푚푒푛푡 (푘푚)) = 5/0.23 = 21.7 veh/km Estimation of average time headway: Arrival of vehicle 2 (t2)=(30+4)/(76*1000/3600)=1.61 sec Headway of vehicle 2 (h2)=1.61 Arrival of vehicle 3 (t3) =(80+8)/(75*1000/3600)=4.224 Headway of vehicle 3 (h3)=4.224-1.61=2.614 sec Arrival of vehicle 4 (t4)= (140+12)/(80*1000/3600)=6.84sec Headway of vehicle 4 (h4)=6.84-4.224=2.616sec Arrival of vehicle 5 (t5)= (175+16)/(75*1000/3600)=9.168 Headway of vehicle 5 (h5)=9.168-6.84=2.232sec Average time headway=(1.61+2.614+2.616+2.232)/4 = 2.268sec Average clear spacing=(30+50+60+35)/4=43.75m



Example 4:

The number of traffic accidents that occurs on a particular stretch of road during

a month follows a Poisson distribution with a mean of 9.4. Find the probability

that less than two accidents will occur on this stretch of road during a randomly

selected month.

푃(푋 = 0) = e . ( . )

! = 0.0000827

푃(푋 = 1) = e . ( . )!

= 0.00078

P(x < 2) = P(x = 0) + P(x = 1) = 0.000860

푃(푋 = x) = e−푚 (푚)푥

푥!

characteristics 1. drivers' characteristics · 2018. 12. 2. · 3. vehicle characteristics...

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