vtu belagavi notes

1. PREPARED BY
SHIVKUMAR M H [SONOFGOD]
M.E(UVCE),AMIE
ASSISTANT PROFESSOR AND HIGHWAY ENGINEER
EWIT,BANGALORE

2. CHAPTER – 2
SYLLABUS
TRAFFIC CHARACTERISTICS: Road user characteristics, vehicular characteristics –
static and dynamic characteristics, power performance of vehicles, Resistance to the
motion of vehicles – Reaction time of driver – Problems on above. 6Hr
QUESTIONS
1. Mention the various factors that affect the road user characteristics explain briefly any
two of them
2.Explain with neat sketch, PIEV theory in analyzing driver reaction time and its significance.
Indicate how it is different from SSD and OSD calculations. Discus factors influencing
perception reaction time.
3.List and explain the various static and dynamic characteristic of vehicles in designing traffic
facilities.
4. List and explain with equations the various resistances considered for power calculation of
vehicles. Give the values of various constants used in equations.
5.Explain the various forces that offer resistance to motion of a vehicle.
6. Which are the design elements affected by length, speed, power and braking characteristics
of vehicles.
7.Explain physiological factors affecting driver performance.
8. Describe visual aspects of road users affecting highway design.

3. TRAFFIC CHARACTERISTICS
Road User Characteristics
 The human element is involved in all actions of the road users either as pedestrian, cyclist,
motorist.
 The physical, mental and emotional characteristics of human beings affect their ability to
operate motor vehicle safely or as a pedestrian.
 Hence it is important to the traffic engineer to study the characteristics and limitations of
the road users.
The various factors which affect human characteristics as road users may broadly be
classified under four heads:
(a) Physical
(b) Mental
(c) Psychological and
(d) Environmental
Physical characteristics
 The physical characteristics of the road users may be either permanent or temporary.
 The permanent characteristics are the vision, hearing, strength and the general reaction to
traffic situations.
 Vision plays the most important role of all these.
 These include the acuity of vision, peripheral vision, eye movement and depth judgment.
 Glare vision and glare recovery time of drivers play important role during night.
 Minimum standards for acuity of vision are often laid down by licensing authorities.

4.  Field of clearest and acute vision is within a cone whose angle is only 3-
degrees, though the vision is fairly satisfactory up to 10° in general and even
up to 20° in the horizontal plane.
 However in the vertical plane the field of clear vision may be about two
thirds of that in the horizontal plane.
 These factors are particularly taken care of while designing and installing
traffic control devices.
 The depth judgment is important for a driver in judging distance and speed
of vehicles and other objects ahead.
 Hearing helps drivers in a way, though it is more important for pedestrians
and cyclists.
 Though strength is not an important factor in "general, lack of strength
may make parking is difficult, particularly for heavy vehicles.
 The temporary physical characteristics of the road users affecting then-
efficiency are fatigue, alcohol or drugs and illness.
 All these reduce alertness and increase the reaction time and also do affect
the quality of judgment in some situations.

5. Mental factors
 Knowledge, skill, intelligence experience' and literacy can affect the road
user characteristics.
 Knowledge Of vehicle characteristics, traffic behaviour, driving practice,
rules of roads and psychology of road 'users will be quite useful for safe
traffic operations.
 Reactions to certain traffic situations become more spontaneous with
experience.
 Understanding the traffic regulation and special instruction to road users
and timely action depends on intelligence and literacy.
Psychological factors
 The emotional factors such as attentiveness, fear, anger, and superstition
impatience, general attitude towards traffic and regulations and maturity also
come under this category.
 Distractions by non-traffic events and worries reduce attentiveness to
traffic situations.
 Dangerous actions are likely due to impatience.
 Some road users do not pay due regard to the traffic regulations and do
not have the right attitude towards the traffic.

6. ENVIRONMENTAL FACTORS
 The various environmental conditions affecting the behaviour of road user are
traffic stream characteristics, roadside features, atmospheric conditions and the
locality.
 The traffic stream may consist of mixed traffic or heavy, traffic.
 The other environmental factors of importance are the weather, visibility and
other atmospheric conditions.
o The total reaction time or the "PIEV" time of the drivers varies considerably
from driver to driver BASED ON THE ABOVE ROAD USER
CHARACTERISTICS.
BEHAVIOUR OF DRIVERS AS ROAD USERS
 It is observed that in general, the behaviour of drivers vary depending on the
vehicle classes they drive.
For example
 The behaviour and attitude of drivers of heavy commercial vehicles or large public
transport buses are found to be different from those of smaller vehicles like cars or
two wheelers; one of the reasons may be relative difficulty in handling the
heavy/large vehicles along.
 The congested roads with mixed traffic flow.
 The traffic engineer has to take into consideration the variation in behaviour of
different types of drivers, before designing the roadway components and also
while implementing appropriate traffic management measures.

7. BEHAVIOUR OF PEDESTRIANS AS ROAD USERS
 It is important for the traffic planners and engineers to understand the behaviour and
psychology of pedestrians as road users.
 It is desirable to plan and construct an attractive, even and firm surface for the pedestrians
to safely and comfortably walk along the road side comfortably;
 The longitudinal and transverse slopes should also be within acceptable limits.
 This will attract the pedestrians to make use of the specified side-walks, rather than
walking along the carriageway.
 Similarly safe pedestrian crossing facility has to be made available at identified locations
to enable the pedestrians to cross the roadway safely
VEHICULAR CHARACTERISTICS
 The various vehicular characteristics affecting the road design may be classified as static
and dynamic characteristics of the vehicles.
 Static characteristics' of vehicles affecting road design are the relevant dimensions,
maximum turning angle and the total weight.
 The height of driver seat affects the visibility distance and the height of head light affects
the head light sight distance at valley curves.
 The field of vision ahead for the driver also depends on the design of wind and the front
portion of the vehicle body.
 Approach, departure of the vehicle affect the design of vertical profile of drive ways,
humps and dips.
 Gross weight, axle and wheel loads of vehicle govern the structural design of pavements
cross drainage structures.

8. o Dynamic characteristics of vehicles that affect the road design are speed,
acceleration/deceleration and braking characteristics.
o The speed and acceleration depend upon the power of the engine and the
resistances to be overcome {including the road carried)
o The deceleration and braking characteristics guide safe vehicle
operation.
 The stability of vehicle and its safe movement on horizontal curves are
affected by the width of wheel base and the height of center of gravity.
 The riding comfort on vertical curves depends on the design of
suspension system of the vehicle.
 The impact characteristics on collision and the injuries to the occupants
depend on the design of the bumper and body of vehicle.
 Some of the important static and dynamic characteristics of road
vehicles to be considered in traffic engineering are: (i) vehicle dimensions
(ii) weight of loaded vehicles (iii) power of vehicle (iv) speed of vehicle (v)
braking characteristics and (vi) off tracking are presented below.

9. VEHICLE DIMENSIONS
 The dimensions of vehicles that significantly control the design of highway
geometric elements are the overall width, length and height of different
vehicles, particularly of the largest ones.
 The width of the vehicle affects the width Of the traffic lanes, shoulders and
parking facilities.
 If the width of the lanes is not adequate due to wider vehicles using the roads.
 The capacity of road will decrease significantly.
 The length of the vehicle and the length of the wheelbase are important
factors in the design of horizontal alignment as they affect the extra width of
pavement at horizontal curves, the minimum turning radius, safe
overtaking distance, and capacity of the road and parking facilities.
 Height of the vehicle affects the clearance to be provided under structures
such as over-bridges under-bridges, electric and other service lines

10. WEIGHT OF LOADED VEHICLES
 The maximum weight of loaded vehicle and the load on the wheels and axles
affect the design of pavement thickness and gradients.
 In fact the limiting gradients are governed by both the weight and the pulling
power of the heavy vehicles.
POWER OF VEHICLE
 The power of the heaviest vehicles and their loaded weights govern the
permissible and limiting values of gradient on roads.
 In this-regard the total resistances to traction consisting of inertia, rolling
resistance, air resistance and grade resistance are considered.
 From the total hauling capacity and the power required to over acceleration of
the vehicle which in turn are useful in planning, design of road geometries and
traffic regulation come the total tractive resistance, it is possible to determine the
speed and
Speed of vehicle
The vehicle speed affects design:
(i) sight distances (ii) super elevation, length of transition curve and limiting
radius on horizontal curves (iii) length of transition curves on vertical valley
curves and on humps (iv)width of each traffic lane/pavement and shoulders on
straight and on horizontal curves (v) design gradient (vi) capacity of traffic lane
(vii) design and control measures at intersections.
Thus the design speed controls most of the geometric elements of highways.

11. BRAKING CHARACTERISTICS
 The deceleration and braking characteristics of vehicles depend on the design
and type of braking system (such as mechanical, fluid or air brake) and its
efficiency.
 The safety of vehicle operation, stopping distance and the spacing between the
two consecutive vehicles in a traffic stream are affected by the efficiency of the
braking system.
 Thus the highway capacity and overtaking sight distance requirements also
indirectly get affected by the acceleration and deceleration characteristics as well
as the braking characteristics of vehicles.
BRAKING TEST
 It is possible to measure the skid resistance of pavement surface under the
prevailing conditions by conducting braking tests on the road at the desired
running speed.
 If the brakes are applied till the vehicle comes to stop, it may be assumed that
wheels are fully locked and the 'brake efficiency' is 100 percent.
 At least two of the following three measurements are needed during the braking
tests in order to determine the skid resistance of the pavement:
(a) Braking distance, L meter
(b) Initial speed, u m/sec
(c) Actual duration of brake application, t sec
 The method of calculating the average skid resistance of the pavement using two
of the above three values has been illustrated with the help of Examples 1, 2 and 3.

12. Example 1
In a braking test, a vehicle travelling at a speed of 30 kmph was stopped by applying
brakes fully and the skid marks were 5.8 in in length. Determine the average skid resistance of
the pavement surface. Solution
Given: Initial speed, u = 30 kmph,
Braking distance, L = 5.8 m
Initial speed, 30/3.6 = 8.33 m/sec
Braking distance, L = 5.8 m = u2/2gf
Average skid resistance, f = 8.332/ (2 x 9.8x 5.8) =0.61
Example 2
A vehicle travelling at 40 kmph was stopped within 1.8 seconds after the application of
the brakes. Determine the average skid resistance.
Solution
Given:
Initial speed, u = 40 kmph,
Duration of brake application, t = 1.8 sec,
Initial speed u = 40/ 3.6 = 11.11 m/sec;
Braking time t = 1.8 sec
Using the fundamental relation of motion for uniform acceleration/retardation,
v= u +at, v = 0, retardation, a= u/ t = 11.11/1.8 = 6.17 m / sec2
from the relation, friction force F = m a,
W* f = (W *a)/g
Average skid resistance, f = a/g = 6.17/ 9.88 = 0.63

13. Example 3
A vehicle was stopped in 1.4 sec by fully jamming the brakes and the skid marks measured
7.0 m. Determine the average skid resistance developed.
Solution
Given:
Braking duration, t = 1.4 sec,
Braking distance, L = 7.0 m
Using the fundamental relations of motion for uniform acceleration/retardation
v = u + at, as the final velocity v = 0, u = - at
v2 - u2 = 2as
s = - u2/ 2a= a2t2/ 2a and therefore, a = 2s/ t2
Substituting the values of s = L = 7.0 m and t = 1.4 sec, a = 2 x 7/1.42
Average skid resistance, f = a/ g = 2 x 7.0/ 9.8x1.42 = 0.729
Example 4
A vehicle moving of 40 kmph speed was stopped by applying the brake and the length of skid
mark was 12.2 m. If the average skid resistance of the pavement is known to be 0.70, determine the
brake efficiency of the test vehicle.
Solution
Given:
Speed, v = 40 kmph,
Braking distance L = 12.2 m, skid resistance,
f = 0.7
v = 40/ 3.6 = 11.11 m/sec
Average skid resistance developed
F1 = v2/ 2gL = 11.112/ (2 x 9.8 x 12.2) = 0.516
Braking efficiency = 100 F1/ f = 100 x 0.516 /0.70= 73.7%

14. OFF-TRACKING
 When a four wheeled vehicle such as car or a six wheeled vehicle such as truck or bus
negotiates a horizontal curve at relatively slow speed, the rear wheels do not follow the
path traced by the corresponding front wheels, as explained under 'Mechanical,
Widening'.
 At relatively slow speeds when the centrifugal force developed is lesser than the
counteracting forces due to the super elevation and transverse friction, the rear wheels-
follow paths on the inner side of the horizontal curve as compared with the path followed
by the corresponding set of front wheel.
 This difference in distance between the curved wheel paths of a particular set of front and
rear wheel (i.e., either the set of front and rear wheels on the outer side of the horizontal curve
or the set on the inner side) is called off-tracking or the mechanical widening for a vehicle
which is equal to 12/2R
Thus the off tracking depends on two factors
(a) The length of wheel base, l or the distance between the front and rear axles of the vehicle
(b) Turning angle or the mean radius of the horizontal curve traversed
Example:
A vehicle has a wheel base of 6.5 m. What is the off tracking while negotiating a curved
path with a mean radius 32 m?
Solution:
Given: Wheel base, 1= 6.5 m,
Radius, R = 32 m
Off tracking = l2/2R = 6.52/(2*32) = 0.66m

17. POWER PERFORMANCE OF VEHICLES
A knowledge of the power performance of a vehicle is necessary to
determine the vehicle running costs and the geometric design elements like grades.
Resistance to motion of a vehicle.
The power developed by the engine should be sufficient to overcome all
resistance to motion at the desired speed and to accelerate at any desired rate to
the desired speed.
The following forces have to be overcome for this purpose
1. Rolling resistance. 2. Air resistance. 3. Grade resistance.
4. Inertia forces during acceleration and deceleration.
5. Transmission losses .

18. ROLLING RESISTANCE
 When the vehicle wheels roll over the road surface; the irregularities and the
roughness of the surface cause of deformation of the tires,
 The road surface itself may undergo deformations.
 Shocks and impacts are caused by such a motion
 The rolling resistance varies with the type of surfacing.
The rolling resistance is given by P(f)= m*f *g
Where, m = mass of the vehicle in kg
f = coefficient of rolling resistance
P(f) = rolling resistance in N
g = acceleration due to gravity in m/sq (sec)
 The rolling resistance depends on the speed of the vehicles also .
 Though its value is approximately constant up to a speed of about 50 Kmph at
higher values of speed the 'coefficient increases in value.
 The following approximate equation accounts for this increase
f (v)= f(o) [1+ 0·01 (V-50)]
where f(v) = Coefficient of rolling resistance at speed V.
V = speed in Kmph
f (o) = coefficient of rolling resistance, assumed constant up to a speed of 50 Kmph

20. AIR RESISTANCE
When a vehicle is in motion, air resists it in the following ways
1. Since air has density, it exerts a reaction pressure against the front of the vehicle when it
moves at speed.
2. The friction of air against the sides vehicle body causes resistance
3.The air stream behind the vehicle, under body and around the wheels causes power loss.
4.The flow of air through the vehicle for ventilating and cooling causes resistance to motion.
The following formula can be used to determine the air resistance,
P(a) = C (a) *A* sq{v}
Where P(a) = air resistance in N
A = projected front area of vehicle in sq. meters on a plane at right angles to the direction of
motion
V = speed of the vehicle relative to air in m/sec.
C (a) = Coefficient of air resistance
g = Acceleration due to gravity, 9·81 m/sq(sec)

21. GRADE RESISTANCE
 When a vehicle which was moving on a level of stretch at a particular speed
has to move up an incline, additional work has to be done in keeping the vehicle
at the same speed as in the level stretch.
 The additional work is equal to the work that will be needed to lift the vehicle
through a height represented by the inclination.
 If the horizontal distance is unity (i.e. 1 meter) and the slope is i percent, the rise will
be ( i/100 * m).
 If the mass of the vehicle is m kg, the additional force to move the vehicle up
the incline,
P(i) = m*i*g/100
 It may be noted that if the slope becomes downward, i becomes -ve, and P(i) also
becomes -ve, representing a reduction in the force to move the vehicle.

22. INERTIA FORCES DURING ACCELERATION AND DE ACCELERATION
 When the speed of a moving vehicle needs to be increased some additional
power is needed to accelerate.
 Similarly if the vehicle has to gather a desired speed from a stopped position,
additional force is, needed to accelerate.
 The additional force P(j) is given by
Force = Mass x Acceleration
± P(j) = m*a = m * dv/dt
P(j) = Force to accelerate, N
m = Mass of the vehicle, kg
a = Average acceleration of the vehicle, m/sq(sec) = dv/dt .
The value of P(j) will be +ve if the vehicle is to accelerate
and -ve if the vehicle is to de-accelerate

23. TRANSMISSION LOSSES
 Losses in power occur to the mode of power transmission (clutch or
automatic fluid coupling) from the engine to the gear system
 The vehicle has a system of gears such that the speed of the vehicles can
be altered relative to the engine speed.
 At the start of the vehicle, high power is needed but at low speed.
 Similarly, a high engine power is needed while climbing uphill, which is
accomplished at a lower road speed than when driving at a level stretch.
 These manoeuvres are made at the lowest gears.
 For movement along a good road where the resistance to motion will be
small, a high gear will tend to be used.
 The highest forward gear will generally be 1 : 1, representing direct drive.
 A further gear reduction is made at the rear axle.
 The total effect of all the above is to consume about 10-15 percent of the
engine power, which maybe as high as 25% in case of trucks in their
lowest gear

24. POWER REQUIREMENTS OF THE VEHICLE
 The mechanical power developed by the engine is transmitted to the driving
wheels by the transmission system.
 The torque developed at the flywheel is converted to a torque at the rear axle and
the following equation holds good
Rear axle torque, T(a) = k*T(c)*G(t)*G(a)
where T (a) = rear-axle torque
k =efficiency of the transmission system which takes into account
the loss of power due to overcoming the resistance of all mechanism between the
engine and the driving wheels, and can be taken to be about 0·85 to 0·90
T(c) = engine torque at the fly-wheel
G(t) = transmission gear ratio
G(a) = rear-axle gear ratio.
The rear axle torque imparts a tractive force P(p) at the contact of the wheel and
the road.
This Tractive force also known as tyrerim pull, is given by the following
equation

25. P (p) = Rear axle torque/ Radius of the rolling drive tire = k*T(e)*G(t)*G(a) / r(w)
r(w) is related to the radius of the tyre r(o) by the following formula,
r(w) = λ*r(0)
λ = tyre deformation factor value of 0·945-0·950 for high pressure air tyres
λ = tyre deformation factor value of 0·930-0·935 for low-pressure tyres, on hard
surfaces
The horse power corresponding to the tractive effort P(p) when the vehicle moves at a
speed of ‘v’ m/sec is
Power output = P(p)* v
But v = V/ 3·6
Power output = [P (p ) x V]/ 3.6
V = 2*π*r(w)*n *3.6 / 60* G(t)*G(a)
where n is the engine speed in R.P.M.
V = 0·377 r(w)*n / G(t)*G(a)
Power output = P(p) x 0·377* r(w)*n / G(t)*G(a)
Engine power (in Watts) = {P(p) x 0·377 x r(w) x n} /{G(t), x G(a) x k }
where k = transmission efficiency
Engine horse-power (metric) = Engine power (in watts)/735
It may be noted that the tractive effort developed at the' wheels
P (p) = Rolling resistance + Air resistance + Grade resistance + Inertia force's due to
acceleration and deceleration = p(f) + or- p(a) + or - p(i)+ p(j)
The following example will illustrate the use of the formulae developed above,

26. Problem – 1
A passanger car weighing 2 tonnes (2000 kg) is required to accelerate at a rate of 3 m/sec2 in the first gear
from a speed of 10 kmph . The gradient is + 1 percent and the road has black topped surface. The frontal
projection area of the car is 2.15 m2 . The car tyres have radius of 0.33 m . The rear axle gear ratio is
3.82 : 1 and the first gear ratio is 2.78 : 1. Calculate the engine horse power needed and the speed of
engine. Make the suitable assumptions
Solution :
Tractive force needed = p(p) = p(f) ± p(a) ± p (i) ± p (j)
P(f) = m * g * f
f = 0.02
p(f) = 2000 * 9.81 * 0.02 = 392.4 N
c(a) = 0.39 from table (2.3)
V = 10 kmph initially and increased to 20 kmph. So assume average speed = (10 + 20)/2 = 15 kmph
p(a) = c(a) * A * v2 = 0.39 * 2.15 * (15/3.6)2 = 14.5 N
p(i) = m * g * i = 2000 * 9.81 * (1/100) = 196.2 N
p(j) = m * a = 2000 * 3 = 6000 N
P(p) = 392.4 + 14.5 + 196.2 + 6000 = 6603.1 N
Note : V = 10 kmph initially
Power output = P(p) * v = p(p) * (V/3.6) = 6603.1 * (10/3.6)
= 18341.9 Watt = (18341.9/735) = 24.95 Hp
Assume a transmission efficiency of (0.85 to 0.90) - 0.90 ,
the engine horse power = (24.95/0.90) = 27.72 Hp
Assuming λ = 0.935 ( for low tyre pressure with hard surface),
r (w) = λ * r (o) = 0.935 * 0.33 = 0.308 m
V = [0.377 * r (w) * n ]/ G(t) * G (a)
n = [G(t) * G (a) * V ] / [ 0.377 * r (w)] = [ 2.78 * 3.82 * 10 ] / [0.377 * 0.308 ] = 915 RPM

27. OTHER VEHICLE CHARACTERISTICS
The other vehicle characteristics that are of concern to the traffic engineer are:
(i) Dimensions and weight
(ii) Turning capability
(iii) Braking system
(iv) Acceleration and deceleration.
(v) Vehicle lighting system
(vi) Features of the vehicle body.
(vii) Tyres etc.
 The dimensions and weight of road vehicles and their turning capability have
an important influence on the geometric design..
 The braking system of a vehicle is important from the point of view of safety.
 The safe stopping distance is composed of the distance travelled by the vehicle
during the perception and brake reaction time and the distance required to
stop the vehicle after the brakes are applied,
 If f is the coefficient of friction between the tyre and the pavement, the braking
distance is given by the formula:
d = V2/ 254*f
d = braking distance, meters
V = speed-in Kmph
f= coefficient of friction between the tyre and the pavement,
 The braking capability of a vehicle influences the skidding that can take
place where brakes are applied suddenly, because of its close connection with
road accidents,

28.  Acceleration characteristics of a vehicle need to be understood when designing
the intersection elements (acceleration lane etc.) and overtaking sight distance.
 The deceleration rate that a driver utilizes is generally less than that which a
vehicle is capable of developing.
 Only in an emergency does the driver attempt to fully utilize the maximum
deceleration.
 The maximum deceleration is related to the coefficient Of friction between the tyre
and the pavement in the following way:
The force required to decelerate a vehicle is given by F= m*d
Where F= force required to decelerate, N
m = mass of the vehicle in Kg
d = deceleration in m/sq(sec)
Rearranging the terms, d=F/m
but F /m*g = f
f – is being coefficient of friction
d = f*g = 9.81*f
List gives observed deceleration rates
1) Comfortable to passenger, preferred by driver deceleration = 2.62, f = 0.27
2) Undesirable but not alarming to passengers: driver would rather not use deceleration
= 3.39, f = 0.34
3) Severe and uncomfortable to passengers: slides objects of seats- driver classifies as
emergency stop deceleration = 4.26, f = 0.43

29. When the vehicle which is travelling at particular speed is suddenly allowed to coast
by switching off the engine and putting the gear to neutral, deceleration is caused.
From the force of equation
P(p) = p(f) + p(a)+ or – p(i) + or – p(j)
p(p) = 0 and i (gradient) = 0
p(j) = p(a) + p (f)
m * dv/dt = c(a) * A * v2 + m* f *g
dv/dt = [c(v) * A * v2]/m + f * g

30. Problem – 2
An ambassador car travelling at a speed of 60 kmph on a level W.B.M road in good
condition is suddenly allowed to coast by switching off the engine and putting the gear in
neutral. What is the deceleration caused ?
Solution :
dv/dt = [ C(a) * A * v2 ]/m + f * g
for an ambassador car ( from the table 2.3 )
C(a) = 0.39 kg/m3
A = 2.15 m3
m = 1365 kg
for W. B.M road f= 0.025 (from table 2.2)
v = V/3.6 = 60/3.6 = 16.66
dv/dt = [0.39 * 2.15 * (16.66)2 ]/1365+ (0.025 * 9.81)
= 0.17 + 0.25 = 0.42 m/sec2

31. TOTAL REACTION TIME OF DRIVER
 Reaction time of driver is time taken from instant the object is visible to driver to instant
the brakes are effectively applied.
 The total reaction time may be explained with help of PIEV theory
• According to PIEV theory , the total reaction time of driver is split into four parts , time
taken by the driver for (i) perception (ii) intellection (iii) emotion (iv) volition
• Perception time is the time required for the sensations received by the eyes or ears of the
driver to be transmitted to the brain through nervous system and spinal chord .
 Intellection time is the time required for the driver to understand the situation
 It is also time required for comparing the different thoughts, regrouping and registering
new sensations
 Emotion time is time elapsed during emotional sensations and other mental disturbance
such as fear, anger or any emotional feelings like superstition etc. with reference to the
situation
 Volition time is time taken by the driver for final action, such as brake application
 The PIEV theory process has been illustrated in figure 4.12
 It is possible that the driver may apply the brakes or take any other avoiding action like
turning , by the reflex action, without the normal thinking process, which is probably the
minimum time for taking a preventive action like brake application
 In both SSD and OSD, total reaction time of driver very important and inter related
 Some authorities call perception and reaction time as PIEV time assumes a perception
time of 1.5 sec and total perception and brake reaction time of 2.5 sec for designing sight
distance.
 The safe stopping distance is composed of the distance travelled by the vehicle during
the perception and brake reaction time and the distance required to stop the
vehicle after the brakes are applied,