CIV2282: Transport and Traffic Engineering

Group

Participating Group Members

MONASH UNIVERSITY

DEPARTMENT OF CIVIL ENGINEERING

CIV2282: Transport and Traffic Engineering

Semester 2/2020, Week 6

Practical Class 6: Review week 1-week 6

The exercises in this practical class draw on the material presented in the lecture slides from week 1 to week 6.

You will probably need a spreadsheet such as Excel to complete these exercises,
remember to bring a computer with spreadsheet software into the practical class.

Please complete this exercise sheet and hand it in at the end of the practice class. Remember to include the names of all group members who participated in solving this exercise.

QUESTION 1

The speeds of five vehicles at a fixed location on a long straight road section were measured as 80 km/h, 96 km/h, 75 km/h, 105 km/h and 69 km/h. What is the value below to represent most accurately the space-mean speed for this stream of traffic?

1. 83 km/h
2. 85 km/h
3. 87 km/h
4. 90 km/h

QUESTION 2

Regarding vehicle speeds, which statements below are false?

1. The spread of vehicle speeds on a road is measurable using the spot speeds
2. The space-mean speed cannot be larger than the time-mean speed
3. In general, there are four approaches to calculate the space-mean speeds
4. Time-mean speeds can be estimated directly from the fundamental diagram
5. The flow is increased when the space-mean speed is increased

QUESTION 3

Which factor(s) below can be incorporated into the Level of Service

1. Speed and travel time
2. Safety
3. Nature of interruptions
4. All of above

QUESTION 4

When shockwave occurs at a bottleneck, which direction it will travel

1. Downstream of the bottleneck
2. Upstream of the bottleneck
3. Either of the above

QUESTION 5

Which of the following variable(s) cannot be determined directly from vehicle trajectories on a time-space diagram?

1. Average density
2. Average flow
3. Space-mean speed
4. Space headway
5. Capacity

QUESTION 6

Figure 1

A one-lane, one-way road of 6 m wide, carrying traffic at 800 veh/hr, has a pedestrian crossing (as described in Figure 1). Assume walking speed is 3 km/h and the pedestrian perception time is 2 seconds. Assume that the vehicle headway follows a dichotomized distribution with minimum practical headway β=2 seconds and proportion of free vehicles α= 0.25.

1. What is the probability of safe gap for the pedestrian to cross the road

Calculate the required gap for crossing

t = the pedestrian perception time + time to cross the road = 2 + 6/(3/3.6) = 9.2 s

Calculate the probability of safe gap for the pedestrian to cross the road

q = 800/3600 = 0.222 vhe/s

beta = 2s

alpha = 0.25

Prob(h>=t) =1-= 0.25exp(…) (see the dichotomized distribution below) =0.122

The probability of safe gap for the pedestrian to cross the road is 0.122

Dichotomized headway distribution

2. What is the frequency of suitable gap for the pedestrian to cross the road safely?

The frequency of suitable gap = flow (veh/h) x the probability of safe gap = 800 x 0.122 = 97.35

The frequency of suitable gap for the pedestrian to cross the road safely is 97.35 (97 is acceptable) in one hour

QUESTION 7

The arrival times and spot speeds of 10 vehicles were recorded at a fixed location on a single lane of road, as shown below

Time	Speed (km/h)
7:15:52 PM	88.52
7:17:00 PM	103.06
7:17:24 PM	99.23
7:17:42 PM	65.63
7:18:00 PM	114.65
7:18:06 PM	89.09
7:19:17 PM	112.28
7:20:08 PM	82.75
7:21:49 PM	104.60
7:22:06 PM	96.19

1. What is the average time headway between these vehicles?

Once the arrival times are obtained, the individual headway can be calculated as the time difference between the consecutive arrival times. The average headway is the arithmetic mean of all the individual headways.

• h₁=t₂–t₁, h₂=t₃–t₂… average headway = mean(h_i) where i=1~n-1
• Or alternatively, average headway= (t_n–t₁)/(n-1)

Final answer: 41.46s

2. What is the volume recorded at the point on the roadway?

Volume=3600 ÷ average headway where average headway (in seconds) is obtained from a)

The volume is ~86.6 vehicles per hour.

3. What is the density of the recorded traffic?

Calculate the space mean speed

=93.27 km/h

Density = flow/space mean speed = 86.6/93.27 = 0.93 vhe/km

The density is ~0.93 vhe/km.

4. What is the average space headway (or spacing) between these vehicles?

Average space headway = average time headway x space mean speed or inverse of the density.

The average space headway is ~1075 m.

QUESTION 8

Consider a 10-km three-lane freeway, with a triangular fundamental diagram. The free flow speed is 100 km/h, the capacity is 2400 veh/h/lane and the jam density is 150 veh/km/lane. The demand is constant at 2500 veh/h. From t=0.5h to t=1.5h, an incident occurs at x=5km, limiting the capacity to 1200 veh/h.

1. Calculate the traffic states and the shock waves in different traffic conditions during the period t=0 to t=2h and draw them in the fundamental diagram.

State A: Free-flow: q=2500 veh/h, k = 25 veh/km

State B: Upstream the incident location (t = 0.5h -1.5h): q = 1200 veh/h, k = 387 veh/km (in congested state)

State C: Downstream of the incident location: q = 1200 veh/h, k =12 veh/km (in free-flow state, this is where the traffic is flowing out of the bottleneck freely)

State D: Back to capacity after the incident clear (t=1.5h-2h)

Shockwave AB = (1200-2500)/(387-25) = -3.6 km/h, and moves upstream

Shockwave BD = (1200-7200)/(387-72) = -19 km/h, and moves upstream

Wave AC = 100km/h (not shockwave since A and C are in free-flow condition, also called free-flow wave, in this case travels with free-flow speed)

2. Draw the space-time diagram and several vehicle trajectories.

3. Determine the queue tail

Queue tail is the furthest the queue propagates backwards, and determined as the height of the triangular (bounded by the slope of the shockwaves, which is -3.6km/h and -19km/h)

After t=1.5h, the shock AB and BD still travel upstream for t hour, which defines the queue tail. Then it is straightforward to show that19t = 3.6 (t+1) which leads to t=0.23h. Then:

The queue tail is located at 3.6*(0.23+1) = 4.4 km (or 19*0.23) upstream of the incident location

QUESTION 9

A road is carrying traffic at a speed of 100 km/h at uniform headway of 2 seconds.

The road changes from level to having an uphill gradient for a length of 3 kilometres.

This uphill gradient causes a heavy truck to slow down to a speed of 25 kilometres per hour.

The road does not have any overtaking opportunities, and traffic has to follow the truck up the hill at this reduced speed and a spacing of 20 metres between vehicles.

1. What are the volumes and densities of the traffic streams in the following two regions?

1. Region A: vehicles arriving behind the back of the queue of slow-moving vehicles following the truck up the hill.
2. Region B: vehicles within the slow-moving queue following the truck up the hill.

Flow, q_A = 3600 / h_A = 3600 / 2 = 1800 veh/h

Density, k_A = q_A / v_A = 1800 / 100 = 18 veh/km

Density, k_B = 1000 / s_B = 1000 / 20 = 50 veh/km

Flow, q_B = v_B × k_B = 25 × 50 = 1250 veh/h

2. Sketch a flow-density curve, showing these regions A and B, and a third region (C) corresponding to the empty space in front of the truck; and the waves between those three regions.

3. What are the speeds of those waves between the three regions?

w_AC = (1800-0) / (18-0) = 100 km/h

w_AB = (1250-1800) / (50-18) = -17.2 km/h

w_BC = (1250-0) / (50-0) = 25 km/h

4. Classify each of those waves into either of the following: shockwave, moving forwards or backwards

Shock Wave AB is “Backward Forming” – moving backwards

Wave AC is “Forward Recovery” – moving forwards (NOTE: This is not a shock-wave as A and C belong to the same free-flow regime)

Shock Wave BC is “Forward Forming” – moving forwards, forming increased congestion

5. Sketch a time-space diagram showing the three regions, the shock waves between them, and some representative trajectories of vehicles travelling up the hill.

6. How much time does the truck take to climb the hill (in minutes)?

Hill is 3 km in length (given)

Truck Speed is 25 km/h (given)

Travel Time is 3 / 25 = 0.12 hours = 7.2 minutes

7. How long is the queue of slow-moving vehicles behind the truck at the time it reaches the top of the hill?

In 0.12 h, the back-of-queue has moved backwards a distance of 0.12 w_AB = 0.12 × 17.2 = 2.06 km

In the same time, the truck has moved forwards 3 km,
so the queue length is 3 + 2.06 = 5.06 km

The queue of slow-moving vehicles behind the truck is 5.06 km in length

8. How many vehicles are in the slow-moving queue behind the truck at the time it reaches the top of the hill?

Queue length = 5.06 km,

Density in Region B is k_B = 50 veh / km,

so the number of vehicles is 5.06 × 50 = 253

There are 253 vehicles in the slow-moving queue behind the truck