Characteristics of Driver, Vehicle, and Roadway

Chapter 2 includes practical problems and questions related to the characteristics of driver, vehicle, and roadway. The characteristics of driver (road user), vehicle (moving object on the road), and the road itself are the three major components that compose the whole traffic system. Hence, safety and efficiency of the system is dependent on these three components. The ideal characteristics of these modules will lead to idea conditions and operation of the traffic system. However, variability, inconsistency, and unpredictability commonly prevail in any traffic system, which result in some complications. Consequently, there is a need to understand these characteristics and to compute motion parameters and travel measures in order to predict the operation of the system. The following practical problems and questions will focus on these aspects of traffic engineering. 2.1

The design speed on a highway is 100 km/h (62.1 mph), and the highway has an upgrade of 3%. In this case and by assuming the perception–reaction time = 2.5 s and the average deceleration rate of vehicles is 3.4 m/s² (11.2 ft/s²), estimate the minimum stopping sight-distance on this highway.

Solution:

The following two formulas are used; the first one is in SI units and the second one is in US customary units:

2.1 SSD = 0.278   u t + u 2 254 ( a g ± G ) ( SI   units ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0001.tif"/>

2.2 SSD = 1.47   u t + u 2 30 ( a g ± G ) ( US   customary   units ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0002.tif"/>

Since the highway has an upgrade of 3%, a +ve sign will be used for the G in the denominator of the formula:

SSD = 0.278 ( 100 ) ( 2.5 ) + ( 100 ) 2 254 ( 3.4 9.81 + 0.03 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0003.tif"/>

SSD = 174.0   m   ( 570.3   ft ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0004.tif"/>

The MS Excel worksheet used for rapid and efficient solution is shown in Figure 2.1.6

2.2

In a study to verify the design speed on a roadway with a downgrade of 4%, the estimated stopping sight distance for vehicles is 200 m (656.2 ft), what would the recommended design speed be on this roadway assuming that the average perception–reaction time of drivers is 2.5 s and the average deceleration rate of vehicles is 3.4 m/s² (11.2 ft/s²).

Solution:

The same two formulas as above will be used:

SSD = 0.278   u t + u 2 254 ( a g ± G ) ( SI   units ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0005.tif"/>

SSD = 1.47   u t + u 2 30 ( a g ± G ) ( US   customary   units ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0006.tif"/>

Since the highway has a downgrade of 4%, a −ve sign will be used for the G in the denominator of the formula:

200 = 0.278 ( 2.5 )   u + u 2 254 ( 3.4 9.81 − 0.04 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0007.tif"/>

By simplifying and reformulating the above equation, the following equation is obtained:

0.012791   u 2 + 0.695   u − 200 = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0008.tif"/> 7

Following the solution of a quadratic equation, the speed can be obtained as follows:

u = − 0.695 ∓ ( 0.695 ) 2 − 4 ( 0.012791 ) ( − 200 ) 2 ( 0.012791 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0009.tif"/>

Two values for the speed are obtained; one is negative (not possible) and the other one is positive (practical).

u = 100.6   kph   ( 62.6   mph ) . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0010.tif"/>

The MS Excel Solver tool can be used to solve this problem as well.

Excel Solver Tool Procedure

If you do not have it: in MS Office 2003-2000, from “File” in the top menu, select “Options”, “Add-Ins”, “Solver Add-In”, and hit “Go”, Select “Solver Add-In” and hit “OK”. Now, you will find it under “Data” in the top menu of the Excel worksheet.

Set up the Solver worksheet before using the tool:

The left side of the equation in this case = 200.

The right side is a function of the speed and the other known variables. Start with an initial value for the speed, the right side will have a computed value now based on the speed initial value and the other known parameters. Define the error, which basically represents the deviation between the left side and the right side of the equation. It is defined as the square of the difference between the two.

Now, go to “Data”, “Solver”, you will see a window like the one below.8

Use “Set Objective” to set the error to a value of zero or minimum; choose the cell of the error in the Excel worksheet, and use “By Changing Variable Cells” to select the cell of the “speed” value. Afterwards, hit “Solve”. The Excel Solver will keep doing iterations until the error is close to zero by changing the speed value. The results obtained using the Excel Solver optimization are shown in Figure 2.2.

The MS Excel worksheet is shown in Figure 2.3.

2.3

In a traffic safety study to reduce pedestrian–vehicle accidents on a level road segment, it was estimated by a video camera that vehicles moved during deceleration an average distance of 60 m (590.6 ft) at the moment right before hitting pedestrians. If the average speed of vehicles on this segment is 80 kph (49.7 mph), compute the maximum safest speed of vehicles on this segment assuming the deceleration rate of vehicles is 3.4 m/s²:

Solution:

The following two formulas are used; the first one is in SI units and the second one is in US customary units:

Braking   Distance = u 2 254 ( a g ± G ) ( SI   units ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0011.tif"/>

2.5 Braking   Distance = u 2 30 ( a g ± G ) ( US   customary   units ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0012.tif"/>

Since the road segment is level, G = 0 in the denominator of the formula:9

60 = u 2 254 ( 3.5 9.81 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0013.tif"/>

⇒

u = 60 ( 254 ) ( 3.4 9.81 ) u = 72.7   kph   ( 45.2   mph) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0014.tif"/>

The MS Excel worksheet is shown in Figure 2.4.

2.4

The average speed of vehicles on a road with a downgrade of 4% is 90 kph. A traffic study was conducted to improve the braking distance (reduce the braking distance) due to a high risk of accident occurrence on this road during the night. If the existing braking distance is 120 m, determine what would be the recommended speed limit based on a reduction in braking distance of 20%. If the grade of the road is to be reduced to achieve this reduction in braking distance, what would the new grade of the road be?

(Assume the deceleration rate of vehicles is 3.4 m/s²):

Solution:

The recommended speed is computed using the formula shown below, and at an improved braking distance equal to 80 m:10

Braking   Distance = u 2 254 ( a g ± G ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0015.tif"/>

96 = u 2 254 ( 3.5 9.81 − 0.04 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0016.tif"/>

⇒

u = 96 ( 254 ) ( 3.4 9.81 − 0.04 ) u = 86.5   kph   ( 53.8   mph) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0017.tif"/>

In other words: to reduce the braking distance on this road segment from 120 m to 96 m, the recommended speed of vehicles will have to be 86.5 kph (53.8 mph).

In the second part of the problem: if the grade is to be changed for this road and the speed is to be kept the same, the same formula will be used to solve for the grade (G) and using u = 90 kph. Hence:

96 = ( 90 ) 2 254 ( 3.4 9.81 − G ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0018.tif"/>

⇒

G = ( 3.4 9.81 − ( 90 ) 2 96 × 254 ) G = 0.014   or   1.4 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0019.tif"/>

In conclusion, if the braking distance is to be reduced by 20% from 120 m to 96 m and the speed is kept the same (90 kph), the grade has to be changed from 4% to 1.4% on this road segment.

An image of the MS Excel worksheet used to perform the computations of this problem is shown in Figures 2.5 and 2.6.11 12

2.5

A vehicle is traveling at 80 kph. If the acceleration of the vehicle is given by the equation:

a = d u d t = 4 − 0.06   u https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0020.tif"/>

compute the velocity of the vehicle after 5 seconds:

Solution:

If

a = d u d t = 4 − 0.06   u https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0021.tif"/>

then:

by integration, the velocity is obtained as shown in the equation below:

u = α β ( 1 − e − β t ) + u 0 e − β t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0022.tif"/>

Therefore,

u = 33.7   m/s   or   110.7   ft/s https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0023.tif"/>

The problem is solved using the MS Excel worksheet shown in Figure 2.7.13

2.6

In Problem 2.5 above, compute the acceleration of the vehicle after 8 seconds.

Solution:

After 8 seconds, the vehicle will have a velocity of:

u = α β ( 1 − e − β t ) + u 0 e − β t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0024.tif"/>

u = 39.2   m/s   or   128.5   ft/s https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0025.tif"/>

And therefore, the acceleration is computed using the formula given in Problem 2.5:

u = α β ( 1 − e − β t ) + u 0 e − β t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0026.tif"/>

a = d u d t = 4 − 0.06   u https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0027.tif"/>

a = 1.65   m/s 2   or   5.41   ft/s 2 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0028.tif"/>

The problem is solved using the MS Excel worksheet shown in Figure 2.8.14

2.7

In the same Problem, 2.5, what is the distance travelled by the vehicle after 10 seconds?

Solution:

By integrating the velocity given in the formula in Problem 2.5, the following expression is obtained for the distance:

x = α β t − α β 2 ( 1 − e − β t ) + u 0 β ( 1 − e − β t ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0029.tif"/>

And hence,

x = 332.5   m   or   1090.7   ft . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0030.tif"/>

The problem is solved using the MS Excel worksheet shown in Figure 2.9.

2.8

The acceleration of a vehicle is given by the equation below:

d u d t = 2 − 0.04   u https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0031.tif"/>

Compute the time when the acceleration of the vehicle will be 1.0 m/s² if the vehicle was initially travelling at 100 kph:

Solution:

The acceleration of the vehicle is given by:

a = d u d t = 2 − 0.04   u https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0032.tif"/>

When a = 1.0 m/s², u = 25.0 m/s.15

By integrating the acceleration, the velocity is obtained as below:

u = α β ( 1 − e − β t ) + u 0 e − β t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0033.tif"/>

The time is obtained at u = 25.0 m/s, and given the parameters α = 2 and β = 0.04 and the initial speed, u₀ = 80 kph = 22.2 m/s:

25 = 2 0.04 ( 1 − e − 0.04   t ) + 22.2 e − 0.04   t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0034.tif"/>

Solving the above equation for t provides:

The Excel Solver tool is used to obtain t:

t = 2.635   seconds ≅ 2.64   seconds . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0035.tif"/>

The MS Excel worksheet along with the Excel Solver tool are used to obtain the time in this problem as shown in Figure 2.10.16

The False-Position numerical method or the Newton–Rapshon method ( Numerical Methods for Engineers by Chapra and Canale, 2015) can also be used to solve the above equation as shown below:

2.11 f ( t ) = 2 0.04 ( 1 − e − 0.04   t ) + 22.2 e − 0.04   t − 25 = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0036.tif"/>

According to the False-Position numerical method, the root (time) at any iteration is obtained using the formula:

2.12 Root = UL − f ( UL ) ( LL − UL ) f ( LL ) − f ( UL ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0037.tif"/>

Where:

UL = upper limit of the selected interval in the previous iteration

LL = lower interval of the selected interval in the previous iteration

f(LL) = the function value at the selected upper limit, and

f(LL) = the function value at the selected lower limit

Starting with an initial interval having lower and upper limits of (0, 5), and after 5 iterations, a good estimate of the time (root of f(t)) with a percent approximate relative error of 0.011% is obtained:

t = 2.63   seconds . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0038.tif"/>

A screen image of the MS Excel worksheet that is used to perform the iterative approach in the False-Position numerical method is shown in Figure 2.11.17 18

Using the Newton–Raphson method, the root (time) at any iteration is obtained using the formula:

2.13 t i + 1 = t i − f ( t i ) f ′ ( t i ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0039.tif"/>

Where:

t_i+1 = the root after iteration i+1

t_i = the root after iteration i

f(t_i) = the function value at t_i, and

f′(t_i) = the derivative value at t_i

Using an initial value t₀ = 0, the estimate of the root after five iterations is equal to 2.654 with a percent relative error of 0.000%.

Therefore,

t = 2.65   seconds . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH002_eqn_0040.tif"/>

A screen image of the MS Excel worksheet that is used to perform the iterative approach in the Newton–Raphson numerical method is shown in Figure 2.12.

Figure 2.1 An image of the MS Excel worksheet used for the computations of Problem 2.1. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_1_B.tif"/> Figure 2.2 An image of the MS Excel Solver used for determining the design speed for Problem 2.2. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_2_B.tif"/> Figure 2.3 An image of the MS Excel worksheet used for the computations of Problem 2.2. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_3_B.tif"/> Figure 2.4 An image of the MS Excel worksheet used for the computations of Problem 2.3. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_4_B.tif"/> Figure 2.5 An image of the MS Excel worksheet used for the computations of Problem 2.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_5_B.tif"/> Figure 2.6 An image of the MS Excel worksheet used for the computations of Problem 2.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_6_B.tif"/> Figure 2.7 An image of the MS Excel worksheet used for the computations of Problem 2.5. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_7_B.tif"/> Figure 2.8 An image of the MS Excel worksheet used for the computations of Problem 2.6. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_8_B.tif"/> Figure 2.9 An image of the MS Excel worksheet used for the computations of Problem 2.7. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_9_B.tif"/> Figure 2.10 An image of the MS Excel worksheet used for the computations of Problem 2.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_10_B.tif"/> Figure 2.11 An image of the MS Excel worksheet used for performing the False-Position numerical method for Problem 2.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_11_B.tif"/> Figure 2.12 An image of the MS Excel worksheet used for performing the Newton–Raphson numerical method for Problem 2.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig2_12_B.tif"/>