Practical Class 2: Traffic Flow Fundamentals – Essaylink

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MONASH UNIVERSITY

DEPARTMENT OF CIVIL ENGINEERING

Transport and Traffic Engineering

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Practical Class 2: Traffic Flow Fundamentals and Traffic Flow Theory

The exercises in this practical class examine different types of fundamental diagrams in traffic flow. These exercises draw on the material presented in the lecture slides “Traffic Flow Fundamentals 2” and “Fundamental diagrams’’.

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 5

Part 2

Two aerial photographs were taken, 30 seconds apart, over one east-bound lane of a freeway. The following results were recorded.

Vehicle	Position in Photo 1 (m)	Position in Photo 2 (m)	Δx (m)	Speed (km/h)
1	609	897	288	34.56
2	640	814	174	20.88
3	579	792	213	25.56
4	457	732	275	33
5	152	497	345	41.4
6	0	305	305	36.6

Note: The vehicle positions (in metres) are taken from a single reference location.

Question 17a: Plot these trajectories on a sheet of paper (or in a spreadsheet) and compute the average flow, density and space-mean speed over the 900 metre length of the lane.

Space mean speed is the numerical average of the speeds of the six vehicles shown on (either of) the aerial photographs:

= AVERAGE(34.56, 20.88, 25.56, 33, 41.4, 36.6).

Note that at every fixed time (0-30s), the speed of vehicle i (i=1,…,6) is constant. The space mean speed is calculated by the observations obtained at a fixed time, so the mean formula is used.

The Space-mean speed is 32 kilometres per hour

We need to assume that these six vehicles are the only ones visible.
Over the length, L= 900 m, there are N=6 vehicles

Density, k = N / L = 6/900 = 0.00667 veh/m

The Density is 6.67 vehicles per kilometre

[Grab your reader’s attention with a great quote from the document or use this space to emphasize a key point. To place this text box anywhere on the page, just drag it.] Note: the density is observed at a fixed time over a specified length (900m). At any point of time, we see 6 vehicles within

900m road section.

From the Fundamental relationship, q = kv_s

Flow = Density × Space Mean Speed

Flow = 6.67 × 32 = 213 veh/h

The Average Flow is 213 vehicles per hour

Note: the flow is observed at a fixed point over a time period (30s)

Question 17b

Time (s)	1st car	2nd car	3rd car	4th car	5th car	6th car	7th car	vs (m/s) =mean (v1:v7)
1	11.9024	13.1155	12.5517	13.3015	13.716	13.6459	13.716	13.13557
3	11.6129	12.7437	13.079	13.1186	13.7099	13.7008	13.716	13.09727
5	10.8936	11.8476	13.271	12.5639	13.716	13.716	13.716	12.81773
7	10.5522	11.11	13.274	12.2133	13.6947	13.716	13.716	12.61089
9	10.6436	10.7594	13.018	12.189	13.3411	13.716	13.716	12.4833
11	10.671	10.6589	12.4176	12.192	12.9357	13.716	13.716	12.3296
13	10.6345	10.665	12.1128	12.1676	12.9479	13.716	13.6703	12.27344
15	10.2474	10.665	12.1707	12.0731	12.9875	13.6764	13.6459	12.20943
17	9.5494	10.6497	12.192	12.1097	12.9784	13.6246	13.7008	12.11494

EXERCISE 6

Data collected on a metropolitan freeway was used to find that speed (v in km/h) and density (k in veh/km) were related by the following expression:

v = 100 – 0.7k

Question 18: What is the free-flow speed on the freeway?

Free-flow speed (v_f) is the greatest possible speed, occurring when there are no other vehicles on the road (i.e. density, k = 0)

Substitute k = 0 into the given equation

v_f = 100 – 0.7× 0 = 100

The Free-flow Speed is 100 km/h

Question 19: What is the jam density?

The jam density (k_jam) is the greatest possible density, occurring when the road is completely congested and traffic is stationary (i.e. speed, v=0)

Substitute v = 0 into the given equation

0 = 100 – 0.7× k_jam

k_jam = 100 ÷ 7 = 142.8

The Jam Density is 143 vehicles per kilometre

Question 20: What is the jam spacing (average spacing between fronts of stationary vehicles)?

This Jam Density of 143 metres corresponds to a jam spacing of 1000 ÷ 143 = 7.0 m/veh)

The Jam Spacing is 7 metres per vehicle

Question 21: What is the formula for the fundamental relationship, expressing flow
(q in veh/h) in terms of density (k in veh/h)?

Flow = Density × Speed

Substitute the above relationship for speed as a function of density into this equation.

v = 100 – 0.7k

q = 100k – 0.7k²

This is the equation for a parabola

Question 22: What is the maximum flow that can be accommodated on the freeway?

The maximum flow, q_cap occurs at a value of density (k_cr) when flow does not change with changing density, i.e. dq/dk = 0

dq/dk = 100 – 0.7×2k

100 – 1.4k_cr = 0

k_cr = 71.43 veh/km

Substitute into the equation for the parabola

q_cap = 100k_cr – 0.7 k_cr ² = 100 × 71.43 – 0.55 × (71.43)² = 3571 veh/h

Alternatively, q_cap = v_f× k_jam ÷ 4 = 100 × 142.8 ÷ 4 = 3571

The Maximum Flow, q_cap, is 3571 vehicles per hour

Question 23: At what speed would vehicles be travelling to give that maximum flow?

The critical speed, v_cr is the speed at which flow is a maximum (3571 veh/h) and density is the critical density (71.43 veh/km).

Speed = flow / density = 3571 / 71.43 = 50

The Critical Speed, v_cr, is 50 kilometres per hour

Alternatively, we can substitute k=k_cr into the original equation relating speed and density

v_cr = 100 – 0.7× 71.43 = 50

Question 24: Plot the relationship for pace (t_1km in seconds to travel 1 kilometre) in terms of density (k in veh/km)

Speed / Density (v/k) relationship

Flow/ Density (q/k) relationship – the Fundamental Diagram

Rate of change of Flow vs Density – This is used to find maximum flow and critical density

Speed / Flow (v/q) relationship

The time taken to travel 1 kilometre is the inverse of the speed

EXERCISE 7

Data collected on a metropolitan freeway was used to find that speed (v in km/h) and density (k in veh/km) were related by the following expression:

Question 25: What is the free-flow speed on the freeway?

Free-flow speed (v_f) is the greatest possible speed, occurring when there are no other vehicles on the road (i.e. density, k = 0)

Substitute k = 0 into the given equation

v_f = 100 × (2 + 0.008 × 0 – e^0.008×0) = 100 × (2 + 0 – 1) = 100

The Free-flow Speed is 100 km/h

Question 26: What is the jam density?

(Hint: You may need to use a numerical solver such as Solver in Excel)

The jam density (k_jam) is the greatest possible density, occurring when the road is completely congested and traffic is stationary (i.e. speed, v=0)

Substitute v = 0 into the given equation

We need to solve this numerically, for example using Solver in Microsoft Excel, to give:

k_jam = 143.3 veh/km

The Jam Density is 143.3 vehicles per kilometre

Question 27: What is the jam spacing (average spacing between fronts of stationary vehicles)?

This Jam Density of 143.3 m corresponds to a jam spacing of 1000 ÷ 143.3 = 6.97 m/veh)

The Jam Spacing is 6.97 metres per vehicle

Question 28: What is the formula for the fundamental relationship, expressing flow
(q in veh/h) in terms of density (k in veh/h)?

Flow = Density × Speed

Substitute the above relationship for speed as a function of density into this equation.

Question 29: What is the maximum flow that can be accommodated on the freeway?

(Hint: You will need to use a numerical solver such as ‘Solver’ in Excel)

The maximum flow, q_cap occurs at a value of density (k_cr) when flow does not change with changing density, i.e. dq/dk = 0

We need to solve this numerically, for example using ‘Solver’ in Microsoft Excel, to give:

k_cr = 86.64 veh/km

Substitute this into the equation for flow as a function of density:

q_cap = 6005 veh/h

Alternatively, we could one again use a numerical method such Solver in Microsoft Excel to find the maximum value of flow, q, by varying the density, k.

The Maximum Flow, q_cap, is 6005 vehicles per hour

Note that the formula q_cap = v_fk_jam/4 no longer is valid because we do not have a parabola.

Question 30: At what speed would vehicles be travelling to give that maximum flow?

The critical speed, v_cr is the speed at which flow is a maximum (6005 veh/h) and density is the critical density (86.64 veh/km).

Speed = flow / density = 6005 / 86.64 = 69.3 km/h

The Critical Speed, v_cr, is 69.3 kilometres per hour

Alternatively, we can substitute k=k_cr into the original equation relating speed and density

v_cr = 69.3

Question 31: Plot the relationship for pace (t_1km in seconds to travel 1 kilometre) in terms of density (k in veh/km)

Speed / Density (v/k) Relationship – this is no longer a straight line

Flow / Density (q/k) Relationship – this is no longer a parabola

Rate of change of Flow vs Density – This is used to find the maximum flow and the critical density. It is no longer straight line and must be solved numerically

Speed / Flow (v/q) Relationship – this is no longer a sideways parabola

The time taken to travel 1 kilometre is the inverse of the speed

EXERCISE 8

A section of highway is known to have a free-flow speed of 75 km/h and a capacity of 3000 veh/h. In a given hour, 2000 vehicles were counted at a specified point along this highway section.

Question 32: Given that the linear speed-density relationship applies, what would you estimate the space-mean speed of these vehicles to be?

The linear relationship between speed and density is:

We are given v_f = 75 km/h and q_cap = 3000 veh/h

For this linear speed-density relationship, we can use

So k_jam = 4 × 3000 ÷ 75 = 160 veh/km

Inverting the linear speed density equation gives

Multiplying by the fundamental equation q = kv gives

Substituting in the given value of v_f= 75 and the calculated value of k_jam = 160 gives

This is a quadratic equation that can be solved for q = 2000 veh/h

Solve for v = 15.85 or v = 59.15

Check: when v = 15.85, k = 126.2, q = kv = 2000

Or, when v = 59.15, k = 33.8, q = kv = 2000

The space-mean speed of these vehicles is either 15.85 km/h, or 59.15 km/h