Chapter 3
Characteristics of the Driver, the
Pedestrian, the Vehicle and the Road
3-1
Briefly describe the two types of visual acuity.
The two types of visual acuity are static and dynamic. The ability of a driver to
identify an object when both the object and the driver are stationary depends on
one's static acuity. Some factors that affect static acuity include the background
brightness, contrast, and time. The ability of a driver to clearly detect a moving
object depends on the driver's dynamic visual acuity.
3-2
(a) What color combinations are used for regulatory signs (e.g. speed limit signs)
and for general warning signs (e.g. advance railroad crossing signs)
(b) Why are these combinations used?
Regulatory signs use a color combination of black lettering on white
background, and advance warning signs use the color combination of black
lettering on yellow background. These color combinations are used because they
have been shown to be those to which the eye is most sensitive.
3-3
Determine your average walking speed.
Compare your results with that of the suggested walking speed in the MUTCD.
Which value is more conservative and why?
Pass # Intersection Width (ft) Walk Time (sec) Walking Speed (ft/sec)
1 36 7 5.1
2 36 8.2 4.4
3 36 9.5 3.8
4 36 7.6 4.7
5 36 8 4.5
Average 4.5
, In this case, the MUTCD value, 4.0 ft/sec, is more conservative than the observed
speeds. This value is more conservative because it is a slower speed and it will
allow most slower people, such as elderly, individuals with small children, and
handicapped individuals to traverse the intersection safely.
3-4
Describe the three types of vehicle characteristics.
The three types of vehicle characteristics are static, kinematic, and dynamic.
Static vehicle characteristics include the vehicle's weight and size. Kinematic
characteristics involve the motion of the vehicle, and dynamic characteristics
involve the force that causes the motion of the vehicle.
3-5
Determine the maximum allowable overall gross weight of WB-20 Design Vehicle.
(The WB-20 is same as WB-65 and WB-67.)
From Table 3.2, the extreme distance between the axle groups is 43.4 – 45.4 ft
(use 45.4 in this case). The number of axles in the group is 4.
Use Eq. 3.2,LN
45.4 4
W 500 12N 36 500 12 4 36 72267lb .
N 1 4 1
The maximum allowable overall gross weight is 72267 lb.
3-6
The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping
sight distance that should be provided on the road if (a) the road is level and (b) the
road has a maximum grade of 4%? Assume the perception-reaction time = 2.5 sec.
The minimum sight distance required in these cases is the stopping sight distance
(SSD), given by Equation 3.27:
u2
S 1.47ut ;
a
30( G)
g
where u = design speed (mi/h)
t = perception-reaction time (sec)
a = rate of deceleration (taken as 11.2 ft/sec2)
, g = gravitational acceleration (taken as 32.2 ft/sec2)
G = grade
a
Note: the term is typically rounded to 0.35 in calculations.
g
(a) Determine the minimum sight distance that should be provided for a
level roadway.
Since the roadway is level, G = 0.
S = (1.47)(60)(2.5) + (60)2 / (30)(0.35+0)
S = 220.50 + 345.00
S = 565.50 feet
Therefore, the minimum sight distance for this horizontal roadway is 565 feet.
(b) Determine the minimum sight distance that should be provided for a
roadway with a maximum grade of 4 percent.
Since this roadway has a maximum grade of 4 percent, G = -0.04. The
downgrade (negative) case provides the most conservative (higher) value for
design.
S = 1.47(60)(2.5) + (60)(0.35 – 0.04)
S = 220.50 + 387.10
S = 607.60 feet
Therefore, the minimum sight distance for this roadway should be 608 feet.
3-7
The acceleration of a vehicle can be expressed as:
du
3.6 0.06u
dt
If the vehicle speed, u, is 45 ft/sec at time T0, Determine:
(a) Distance traveled when the vehicle has accelerated to 55 ft/sec.
(b) Time for vehicle to attain the speed of 55 ft/sec.
(c) Acceleration after 3 seconds.
(a) Determine the distance traveled by the vehicle when accelerated to 55
ft/sec.
First, determine the time it took for the vehicle to accelerate to 55 ft/sec.
Using a rearrangement of Equation 3.10:
-t = ln [( - ut) / ( - u0)] therefore;
t = (-1/) ln [( - ut) / ( - u0)]