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MONASH UNIVERSITY
DEPARTMENT OF CIVIL ENGINEERING
Transport and Traffic Engineering
(With Answers)
Practical Class week 1: Traffic Flow Fundamentals
The exercises in this practical class examine the variables used to characterise traffic flow and the basic relationships between those variables. These exercises draw on the material presented in the lecture slides “Traffic Flow Fundamentals 1”.
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.
EXERCISE 1
An observer at the side of a road conducted a traffic survey using a stop watch to record the times when vehicles passed. The recorded values are presented below:
| Vehicle Number | Observation Time (secs) | Time (s) | Headway (s) |
| 0 | 08:00:15 | 15 | |
| 1 | 08:00:21 | 21 | 6 |
| 2 | 08:00:24 | 24 | 3 |
| 3 | 08:00:29 | 29 | 5 |
| 4 | 08:00:35 | 35 | 6 |
| 5 | 08:00:42 | 42 | 7 |
| 6 | 08:00:54 | 54 | 12 |
| 7 | 08:01:06 | 66 | 12 |
| 8 | 08:01:13 | 73 | 7 |
| 9 | 08:01:22 | 82 | 9 |
| 10 | 08:01:35 | 95 | 13 |
Question 1: What is the average headway?
There are a total of 10 headways (between 11 vehicles). These are shown in the table as the difference between observation times. The average of these 10 numbers is 8.0 seconds
The average headway is 8.0 seconds
Question 2a: Based on that average headway, what is the volume (flow in vehicles per hour) on the road?
Flow is the reciprocal of average headway, which we calculated to be 8.0 seconds.
Flow = 1 ÷ 8.0 = 0.125 veh/s, or (since 3600 seconds in 1 hour) = 3600 ÷ 8.0 = 450 veh/h
Alternatively,
Flow = (number of vehicles – 1) ÷ (time of last vehicle – time of first vehicle)
Flow = (11 – 1) ÷ (95 – 15) = 0.125 veh/s
Volume = 3600 × 0.125 = 450 veh/h
The hourly flow (volume) is 450 vehicles per hour
Note that we cannot simply use number of vehicles ÷ total time because we don’t know the total time. The time recording may have started before 8:00:15, and may have finished after 8:01:35. Thus an answer of volume = 495 veh/h would be wrong.
Question 2b and 2c: This data type is not giving enough information to calculate either speed or density. See live workshop lecture.
EXERCISE 2
A circular test track is 2 km in length. A traffic survey conducted beside the track measured a traffic flow of 180 vehicles/hour and the space mean speed of the traffic was be 60 km/h.
Question 3: How many vehicles are travelling around the track?
Flow = Density × Space Mean Speed
180 veh/h = Density × 60 km/h
Density = 3 veh/km
The track is 2 kilometres in length, so there are 2 km × 3 veh/km = 6 veh
There are 6 vehicles travelling around the track
EXERCISE 3
Five vehicles are observed travelling along a long straight section of rural highway over a one hour period. A radar gun was used to record the speeds of the vehicles as they pass. Two of the vehicles are travelling at 100 km/h, one is travelling at 90 km/h, one is travelling at 80 km/h and the slowest is travelling at 75 km/h. We wish to determine the average speed of the traffic.
Question 4: What is the average of the spot speeds?
Average of the spot speeds = (100 + 100 + 90 + 80 + 75) ÷ 5 = 445 ÷ 5 = 89.0
The average of the spot speeds is 89.0 kilometres per hour
Question 5: Is this value the time-mean speed or the space-mean speed?
Since this survey was taken at one particular location over a period of time (i.e. the distances travelled in a fixed time), the average of the spot speeds is the time-mean speed
At the same time, two traffic surveyors position themselves one kilometre apart along the highway. They record the times at which these five vehicles are seen, and subtract those times to give the travel time of each vehicle over the one-kilometre distance. Assuming the above five vehicles are travelling at constant speeds, the surveyors determined how long it had taken each of the five vehicles to cover the one kilometre section of road.
| Vehicle | Spot Speed (km/h) | Time to travel 1 km (s) |
| 1 | 100 | 3600/100 = 36 |
| 2 | 100 | 3600/100 = 36 |
| 3 | 90 | 3600/90 = 40 |
| 4 | 80 | 3600/80 = 45 |
| 5 | 75 | 3600/75 = 48 |
Question 6: Based on the average time taken to travel the one kilometre section, what is the average speed?
Average time to travel 1 km = (36 + 36 + 40 + 45 + 48) ÷ 5 = 41 seconds
Average speed = 1 km / 41 s × 3600 s/h = 87.8 km/h
The average speed based on times to travel 1 km is 87.8 kilometres per hour
Question 7: Is this value the time-mean speed or the space-mean speed?
Since this is a measure of times taken to travel a fixed distance, this is the space-mean speed
Question 8: Are the relative magnitudes of the two average speeds as you would expect?
Yes, as expected, time mean speed is greater than space mean speed (89.0 > 87.8)
EXERCISE 4
Three vehicles are travelling round a circular track which is 1 km in length. One vehicle is travelling at 40 km/h, another at 50 km/h and the third is travelling at 60 km/h.
An observer records speeds of all vehicles passing a fixed point over a period of 6 minutes.
Question 9: What is the density of the traffic on the test track?
There are three vehicles on the track, and the track is one kilometre in length
Density = number of vehicles per unit length = 3 vehicles / 1 km
The Density is 3 vehicles per kilometre
Question 10: What volume (flow in vehicles per hour) does the observer record?
In the time period of 6 minutes (1/10 of an hour), the 40 km/h vehicle would pass the point 4 times, the 50 km/h vehicle would pass the point 5 times and the 60 km/h vehicle would pass the point 6 times. That is, a total of 4+5+6 = 15 vehicle passes would be observed.
Flow = number of vehicles per time = 15 vehicles / 6 minutes = 2.5 vehicles per minute
= 2.5 × 60 = 150 vehicle per hour
The Hourly Volume is 150 vehicles per hour
Question 11: Based on the values from questions 9 and 10, what is the space-mean speed?
Flow = Density × Space-Mean Speed
150 veh/h = 3 veh/km × Space Mean Speed (km/h)
Space Mean Speed = 150 ÷ 3
The Space Mean Speed is 50 kilometres per hour
Question 12: What is the space-mean speed, calculated as the harmonic mean of the spot speeds observed at the fixed point?
(Hints: Refer to Lecture: Traffic Flow Fundamentals (2), slide 42. How many spot speeds would there be in the 6 minute period?)
Harmonic mean means the inverse of the average of the inverses.
In the time period of 6 minutes (1/10 of an hour), the 40 km/h vehicle would pass the point 4 times, the 50 km/h vehicle would pass the point 5 times and the 60 km/h vehicle would pass the point 6 times. That is, a total of 4+5+6 = 15 vehicle passes would be observed.
Thus, n = 15.
The space mean speed calculated as the harmonic mean of spot speeds is 50 kilometres per hour
Question 13: What is the time-mean speed?
In the time period of 6 minutes (1/10 of an hour), the 40 km/h vehicle would pass the point 4 times, the 50 km/h vehicle would pass the point 5 times and the 60 km/h vehicle would pass the point 6 times. The time mean speed is the numerical average of the speeds of these 15 vehicles.
The time mean speed calculated as the numerical mean of spot speeds is 51.3 kilometres per hour
Question 14: Are the relative magnitudes of the space-mean speed and the time-mean speed as would be expected?
Yes, as expected, time mean speed is greater than space mean speed (51.3 > 50)
Question 15: Calculate the time-mean speed using the formula that uses space-mean speed and the variance of the spot speeds from the space-mean speed.
The time mean speed can be calculated from the space mean speed and the variance of the spot speeds from the space mean speed using the following formula:
The time mean speed calculated as the numerical mean of spot speeds is 51.3 kilometres per hour
EXERCISE 5
Part 1
A platoon of six vehicles was observed over a distance of 300 m on a single lane of Normanby Road, entering at Point A and departing at Point B.
| Vehicle | Time at Point A (sec) | Time at Point B (sec) | Δt (sec) |
| 1 | 0 | 35 | 35 |
| 2 | 2 | 37 | 35 |
| 3 | 3 | 39 | 36 |
| 4 | 5 | 42 | 37 |
| 5 | 6 | 44 | 38 |
| 6 | 8 | 48 | 40 |
Question 16: Calculate the average volume, density, and mean speed for the six vehicles observed, considering the first 10 seconds at Point A as the period of observation.
In the first ten seconds at A, six (6) vehicles arrive.
Flow = 6 ÷ 10 = 0.6 veh/s. Volume = flow expressed in vehicles/hour = 0.6 veh/s × 3600 s/h
The Volume is 2160 vehicles per hour
Since we are measuring the times taken to travel a fixed distance, a simple average will give us the space-mean speed.
Space-mean speed is calculated from the average of the times to travel over the fixed distance of 300 m
Average time = (35+35+36+37+38+40) ÷ 6 = 36.83 s
Space-mean speed = 300 ÷ 36.83 = 8.144 m/s
Space-mean speed = 8.144 m/s × 3600 s/h ÷ 1000 m/km = 29.3 km/h
The Space-Mean Speed is 29.3 kilometres per hour
From the Fundamental relationship, q=kvs
Flow = Density × Space Mean Speed
Density = 2160 veh/h ÷ 29.3 km/h = 72.5
The Density is 73.7 vehicles per kilometre
(This corresponds to an average spacing of 1000 ÷ 72.5 = 13.6 m/veh)