# Practical Class 2: Traffic Flow Fundamentals – Essaylink

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**Transport and Traffic Engineering**

**(With Answers)**

**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 – 0.7*k*

**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 – 0.7*k*

*q* = 100*k* – 0.7*k*^{2}

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 – 0.7×2*k*

100 – 1.4*k*_{cr} = 0

*k*_{cr} = 71.43 veh/km

Substitute into the equation for the parabola

*q*_{cap} = 100*k*_{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*_{f}*k*_{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**

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