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Chapter 1

Traffic Studies and Analysis

13 teaching hours 16 marks Transportation Engineering II

Introduction to Traffic Studies

Course Objectives and Syllabus

Internal evaluation: Attendance (4), Assignments and traffic data collection analysis reporting, pavement design (10) Others case study group projects for pavement failure and traffic control and management (6), Midterms (20)

Traffic Studies

Need of Traffic Studies

Various Traffic Studies

  1. Traffic Volume Studies
  2. Traffic Speed and Delay Studies
  3. Origin and Destination Study
  4. Basic Traffic Flow Parameters
  5. Traffic Capacity and Level of Service
  6. Parking Studies
  7. Crash Studies

1.1 Traffic Volume Studies

Past Exam Questions on this Topic

  1. a) The traffic counts for an approach from 8:30 AM to 10:30 AM (in 15-minute intervals) are given in the table below. Determine (i) the design flow rate for the approach, (ii) the peak hourly volume, (iii) the peak hour and (iv) the peak hour factor. [4+4] b) Travel time data (in seconds) for 14 vehicles traversing a two kilometre road segment are: 85.7, 78.3, 80.9, 93.5, 76.6, 86.7, 80, 90, 96, 84.7, 72, 79.1, 120, 97.3. Estimate (i) the time mean speed, (ii) the space mean speed and (iii) the average density of the above traffic stream, if the mean headway is 8 sec. [2+6] (2080 Baishakh, Q2)
    Approach volume counts (15-min intervals)
    TimeCount (veh)
    8:30–8:45500
    8:45–9:00390
    9:00–9:15300
    9:15–9:30300
    9:30–9:45260
    9:45–10:00350
    10:00–10:15400
    10:15–10:30490
    Solution
    15-minute counts over two hours (veh): 500, 390, 300, 300, 260, 350, 400, 490 · travel-time run over L = 2 km with 14 vehicles · mean headway 8 s

    (a) Peak-hour analysis

    Slide a one-hour window over the 15-minute counts

    each window = 4 consecutive intervals 500+390+300+300 = 1490 390+300+300+260 = 1250 300+300+260+350 = 1210 300+260+350+400 = 1310 260+350+400+490 = 1500 ← maximum

    Peak hour and peak hourly volume

    peak hour = 09:30 – 10:30 peak hourly volume = 1500 veh/h

    Design flow rate from the busiest 15 minutes

    highest 15-min count = 490 veh design flow rate = 490 × 4 = 1960 veh/h

    Peak hour factor

    PHF = PHV / (4 × V15,max) = 1500 / (4 × 490) = 0.77

    (b) Speeds from the travel-time run

    Mean travel time

    Σt = 1220.8 s over 14 vehicles t̄ = 1220.8 / 14 = 87.2 s

    Space mean speed

    Vs = L / t̄ = 2000 / 87.2 = 22.9 m/s = 82.6 km/h

    Time mean speed

    vi = 7200 / ti (km/h for L = 2 km) Vt = mean of the spot speeds = ≈ 83.9 km/h

    Flow and density from the headway

    q = 3600 / h = 3600 / 8 = 450 veh/h k = q / Vs = 450 / 82.6 = 5.4 veh/km
    Peak hour 09:30–10:30 with PHV = 1500 veh/h, design flow rate 1960 veh/h, PHF = 0.77; space mean speed 82.6 km/h, time mean speed 83.9 km/h, density 5.4 veh/km.

    Vt > Vs always, which is the check on parts (b).

  2. A car driver was driving along an 800 m road. She counted the number of cars that she meets while driving southwards (Ns), the number of cars that overtake her while driving northward (No), and the number of cars that she overtakes while driving northward (Np), as shown in the table below. Find the average traffic flow and journey speed, given that she travels at a constant speed of 20 kmph. [8] (2082 Baishakh, Q4)
    Moving-observer counts
    RunCars met (driving south)Overtaking (driving north)Overtaken (driving north)
    11071074
    21132541
    330155
    479189
    Solution
    Moving-observer (floating-car) survey · test vehicle speed 20 km/h · section length L = 0.8 km · four runs, each recording vehicles met against the stream (x) and vehicles overtaking / overtaken with the stream

    Trip time for one pass of the section

    ta = tw = L / v = 0.8 / 20 = 0.04 h (each way)

    Wardrop moving-observer relations

    q = (x + y) / (ta + tw) t̄ = tw − y/q journey speed = L / t̄ x = vehicles MET while driving against the stream y = (overtaking − overtaken) while driving with the stream

    Pool the four runs

    Σx = 107 + 113 + 30 + 79 = 329 Σy = (10−74) + (25−41) + (15−5) + (18−9) = −61 Σta = Σtw = 4 × 0.04 = 0.16 h

    Flow

    q = (329 − 61) / (0.16 + 0.16) = 268 / 0.32 = ≈ 838 veh/h

    Mean travel time and journey speed

    mean y per run = −61/4 = −15.25 t̄ = 0.04 − (−15.25)/838 = 0.0582 h journey speed = 0.8 / 0.0582 = ≈ 13.7 km/h
    Stream flow ≈ 838 veh/h with a journey speed of ≈ 13.7 km/h.

    Trip times are taken equal because the test vehicle held a constant 20 km/h, and the four rows are treated as four test runs. Moving-observer method per Marsani.

  3. A bicycle racer practices every day in the morning. Her route includes a ride along an 800 m bikeway and back. Since she is a traffic engineer, she has made it a habit to count the number of cars in lane A that she meets while riding southward, the number of cars in lane A that overtake her while riding northward, and the number of cars in lane A that she overtakes while riding northward, as shown in the table below. Find the average traffic flow and travel time of lane A. [8] (2076 Ashwin, Q1)
    Moving-observer counts (lane A)
    RunAvg travel speed (km/hr)Vehicles metVehicles overtakenVehicles overtaking
    1321171074
    234932541
    33230155
    43370189
    Solution
    Moving-observer survey of lane A (northbound) on an L = 0.8 km bikeway · each run is one round trip, so the observer goes against lane A southward (counting cars met, M) and with it northward (counting cars overtaking her O and cars she overtakes P)

    Round-trip time of each run

    ta + tw = 2L / v = 1.6 / v (h)

    Tabulate each run

    Runv (km/h)MO − PM + (O − P)t = 1.6/v (h)
    13211774 − 10 = 641810.0500
    2349341 − 25 = 161090.0471
    332305 − 15 = −10200.0500
    433709 − 18 = −9610.0485
    Σ (pooled)3710.1955

    Flow of lane A

    qA = Σ[M + (O − P)] / Σ(ta + tw) = 371 / 0.1955 = ≈ 1898 veh/h

    Mean travel time of lane A

    one-way with-run time t̄w = ½(Σt)/4 = 0.02444 h mean net overtaking = 61/4 = 15.25 t̄ = t̄w − (Ō − P̄)/qA = 0.02444 − 15.25/1898 = 0.01640 h = ≈ 0.98 min (59 s)

    Mean speed of lane A

    V = L / t̄ = 0.8 / 0.01640 = ≈ 48.8 km/h
    Lane A carries ≈ 1898 veh/h at a mean speed of ≈ 48.8 km/h, a mean travel time of about 59 s over the 0.8 km.

    The mean speed exceeds the cyclist's, which is consistent with a positive net overtaking count — a useful sanity check.

    Wardrop moving-observer relations (Marsani): q = (M + O − P)/(ta + tw), t̄ = tw − (O − P)/q.

Traffic Volume

Traffic Volume Study

Objectives or Uses of Traffic Volume Studies

To evaluate and analyze existing facilities, and plan and prioritize improvements

Traffic Count Periods

Methods of Traffic Count

Manual Method

Automatic Method

Pneumatic Tubes
Piezoelectric Strips
Magnetic/Inductive Loops

Ultrasound

Infrared (https://www.swarco.com/)

Microwave Radar

Video Camera

TDC1 Traffic Detector

Video Taping Method

Photographic Method

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Moving Observer (Floating Car) Technique20822076

Traffic Volumes and Their Uses

Daily Volumes and Use

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Traffic Volume

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Hourly Volume and Use

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Sub-Hourly Volume and Use2080

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Traffic Volume

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Cordon Volume Count and Intersection Count

Presentation of Traffic Volume Data

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Poisson Distribution20812076

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Numerical Example: The hourly flow rate in a road section is 120 veh/hr. Use Poisson distribution to model vehicle arrival at this road for 1 min time interval.

Tasks

1.2 Traffic Speed and Delay Studies

Past Exam Questions on this Topic

  1. Define spot speed and explain the methods of conducting spot speed study. Write down the advantages and limitations of rotary intersection: [8] (2081 Bhadra, Q2)
    Solution

    Spot speed = instantaneous speed of a vehicle at a specified point on the road.

    Methods of conducting a spot-speed study:

    • Manual: stopwatch + short measured base length, or enoscope (mirror box).
    • Automatic: pneumatic road tubes, inductive loop detectors, Doppler radar gun, LIDAR speed gun, and video / image-processing systems.

    Advantages of a rotary intersection: continuous (stop-free) movement; all crossing manoeuvres converted to merging, weaving and diverging; self-regulating (no signal); eliminates severe right-angle and head-on conflicts; orderly and aesthetic.

    Limitations: requires large land area; unsuitable for very high volumes or heavy pedestrian movement; causes delay under heavy flow; not suited to high-speed traffic or steep gradients; costly where land is expensive.

  2. In a field survey of spot speed, the following twenty observations were taken (in km/h): 50, 40, 60, 54, 45, 31, 72, 18, 43, 52, 46, 56, 43, 65, 33, 69, 34, 51, 47, 41. Find the time mean speed and space mean speed. Also, assuming these vehicle speeds are fixed over a half km segment, calculate the corresponding travel times and show that the space mean speed calculated using travel times is equal to the point estimate. (2075 Chaitra, Q2)
    Solution
    20 spot speeds (km/h): 50, 40, 60, 54, 45, 31, 72, 18, 43, 52, 46, 56, 43, 65, 33, 69, 34, 51, 47, 41

    Summations

    n = 20 Σv = 950 Σ(1/v) = 0.4643

    Time mean speed — arithmetic mean

    Vt = Σv / n = 950 / 20 = 47.5 km/h

    Space mean speed — harmonic mean

    Vs = n / Σ(1/v) = 20 / 0.4643 = 43.1 km/h

    Check against travel times over a 0.5 km segment

    ti = 0.5 / vi Vs = total distance / total time = (n × 0.5) / Σti = (n × 0.5) / (0.5 × Σ(1/vi)) = n / Σ(1/vi) = 43.1 km/h ✓
    Time mean speed 47.5 km/h, space mean speed 43.1 km/h.

    Vs < Vt always — the harmonic mean is pulled down by the slow vehicles, which occupy the road for longer.

  3. Spot speed observations at a particular link are given in the table below. From the data, calculate the maximum speed limit, minimum speed limit, design speed and modal speed for regulation of traffic. (The frequencies of the lowest two speed classes were not legible in the scanned paper.) (2074 Ashwin, Q2)
    Spot-speed frequency distribution
    Speed range (km/h)Frequency
    6–10–
    10–14–
    14–187
    18–2220
    22–2644
    26–3080
    30–3482
    34–3879
    38–4249
    42–4636
    46–5026
    50–549
    54–5810
    58–623
    Solution
    Spot-speed frequency distribution in 4 km/h classes, N = 445 legible observations

    The two lowest classes (6–10 and 10–14 km/h) are illegible in the paper, but they carry negligible weight for the percentiles required here.

    Build the cumulative distribution

    Upper bound (km/h)Cum. freq.Cum. %
    1871.6
    22276.1
    267116.0
    3015133.9
    3423352.4
    3831270.1
    4236181.1
    4639789.2
    5042395.1
    5443297.1
    5844299.3
    62445100

    Minimum speed limit — 15th percentile

    lies between 22 km/h (6.1%) and 26 km/h (16.0%) V15 = 22 + (15 − 6.1)/(16.0 − 6.1) × 4 = ≈ 25.6 km/h

    Median speed — 50th percentile

    V50 = 30 + (50 − 33.9)/(52.4 − 33.9) × 4 = ≈ 33.5 km/h

    Safe / maximum speed limit — 85th percentile

    V85 = 42 + (85 − 81.1)/(89.2 − 81.1) × 4 = ≈ 43.9 km/h

    Design speed — 98th percentile

    V98 = 54 + (98 − 97.1)/(99.3 − 97.1) × 4 = ≈ 55.6 km/h

    Modal speed — peak of the frequency histogram

    peak class 30–34 km/h with f = 82 Vmode = 30 + (82 − 80)/[(82 − 80) + (82 − 79)] × 4 = ≈ 31.6 km/h
    Minimum speed limit 25.6 km/h · median 33.5 km/h · safe (85th percentile) speed limit 43.9 km/h · design speed 55.6 km/h · modal speed 31.6 km/h.

    Method per Marsani (speed studies): plot cumulative % against the upper class boundary and read the 15th, 50th, 85th and 98th percentiles; the modal speed is the peak of the frequency histogram.

  4. Assuming a linear speed-density relationship, the mean free speed near zero density is 80 km/h and the corresponding jam density is 130 veh/km. Assume the average length of vehicles is 6 m. i) Write down the speed-density and flow-density equations. ii) Compute the speed and density corresponding to a flow of 1000 veh/hr. iii) Compute the average headways, spacing, clearance and gaps when the flow is maximum. Second part: The following data were collected for a section of road 2.5 km long during a floating-car method study. Assuming equivalency factors of 1, 2 and 3 for car, bus and truck respectively, calculate the directional flow, the journey speed and the running speed of the traffic stream. (The complete floating-car data table - journey time, stopped delay, number of vehicles met from the opposite direction, and overtaking/overtaken counts for each N-S and S-N run - is given in the scanned exam paper on the Past Questions page.) (2072 Chaitra, Q3)
    Solution
    Greenshields model with vf = 80 km/h and kj = 130 veh/km · vehicle length 6 m · plus a floating-car survey (data table on the scanned paper)

    Model equations

    v = 80(1 − k/130) = 80 − 0.615 k q = kv = 80k − 0.615 k²

    Capacity and the critical point

    k0 = kj/2 = 65 veh/km v0 = vf/2 = 40 km/h qmax = vfkj/4 = 80 × 130/4 = 2600 veh/h
    Greenshields linear modeldensity kspeed vvfkj(k₀, v₀)density kflow qk₀qmaxkjflow qspeed vv₀vfuncongestedcongested
    The three Greenshields relationships. Speed–density is a straight line; flow–density and speed–flow are parabolas whose peak is the capacity, at k₀ = kj/2 and v₀ = vf/2.

    Speed and density at q = 1000 veh/h

    0.615k² − 80k + 1000 = 0 k = 14.0 veh/km → v = 1000/14.0 = 71.4 km/h (free flow) k = 116.0 veh/km → v = 1000/116.0 = 8.6 km/h (congested)

    Spacing, headway, clearance and gap at maximum flow

    spacing = 1000/k0 = 1000/65 = 15.4 m headway = 3600/qmax = 3600/2600 = 1.38 s clearance = spacing − vehicle length = 15.4 − 6 = 9.4 m gap = clearance / v0 = 9.4 / (40/3.6) = 0.85 s

    Floating-car study — the data table on the paper

    PCU factors 1 (car), 2 (bus), 3 (truck). The paper prints the section as "25.5 km", which would give journey speeds over 340 km/h; the only sensible reading is L = 2.5 km. For each stream the with-runs are its own rows and the against-runs are the opposite ones.

    q = (Σna + Σnw) / (Σta + Σtw) t̄ = t̄w − n̄w/q journey speed = L / t̄ running speed = L / (t̄ − mean stopped delay)
    StreamΣna (PCU)Σta (s)ΣnwΣtw (s)mean stop (s)
    N–S21415616157780.5
    S–N22215775156187.5

    North–south stream

    q = (214 + 6)/(1561 + 1577) = 0.07011 veh/s = 252 PCU/h t̄ = 262.8 − 1.000/0.07011 = 248.6 s journey speed = 2.5/248.6 × 3600 = 36.2 km/h running time = 248.6 − 80.5 = 168.1 s running speed = 2.5/168.1 × 3600 = 53.5 km/h

    South–north stream

    q = (222 + 5)/(1577 + 1561) = 0.07234 veh/s = 260 PCU/h t̄ = 260.2 − 0.833/0.07234 = 248.6 s journey speed = 2.5/248.6 × 3600 = 36.2 km/h running time = 248.6 − 87.5 = 161.1 s running speed = 2.5/161.1 × 3600 = 55.8 km/h
    Greenshields part: capacity 2600 veh/h at 65 veh/km and 40 km/h; at 1000 veh/h the speed is 71.4 or 8.6 km/h; at capacity spacing 15.4 m, headway 1.38 s, clearance 9.4 m, gap 0.85 s.
    Floating-car part: flow 252 PCU/h north–south and 260 PCU/h south–north, journey speed 36.2 km/h both ways, running speed 53.5 and 55.8 km/h.

    Running speed exceeds journey speed because roughly a third of the journey time is stopped delay — the gap between the two measures the congestion on the section.

    The free speed vf = 80 km/h is taken from the sibling exam papers (the value did not OCR). The floating-car table was transcribed from a render of the scanned paper, whose text layer is unreadable.

  5. Average trip time for office is 30 minutes with a standard deviation of 5 min. Assuming a normal distribution of trip time, calculate the following: a) Probability of trip time being at least 35 minutes. b) If the working hour starts at 10:00 AM and the trip starts at 9:40 AM, what is the probability of being late? (2071 Chaitra, Q3)
    Solution
    Trip time is normally distributed with mean μ = 30 min and standard deviation σ = 5 min · the traveller leaves at 9:40 and must arrive by 10:00

    (a) Probability the trip takes 35 minutes or more

    z = (35 − 30) / 5 = 1.00 P(T ≥ 35) = 1 − Φ(1.00) = 1 − 0.8413 = 0.159 (15.9%)

    (b) Probability of being late

    time available = 10:00 − 9:40 = 20 min late if T > 20 min z = (20 − 30) / 5 = −2.00 P(T > 20) = P(z > −2) = Φ(2) = 0.977 (97.7%)
    There is a 15.9% chance the trip takes 35 min or more, and a 97.7% chance of being late.

    Leaving at 9:40 for a 10:00 appointment allows only 20 min against a mean trip time of 30 min, so being late is almost certain — the traveller needs to leave about 30 + 2σ = 40 min early to be ~98% sure of arriving on time.

Review

Speed Studies

Types of Speed

Spot Speed

Average Speed

Time Mean Speed (TMS)

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Space Mean Speed (SMS)20752066

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Relationship Between SMS &TMS

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SMS & TMS

Numerical Example: Travel time (in second) measured for vehicles traversing 300m segment of a feeder road are 20.6, 21.7, 19.8, 20.3, 22.5, 18.5, 19.0, and 21.4. Determine the time mean speed and space mean speed of these vehicles.

Running Speed

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Journey (Travel) Speed

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Methods of Spot Speed Measurement2081

Location

Sample Selection

Stop Watch (Direct) Method

Enoscope Method

Pneumatic Tube Detector

Inductive Loop Detector

Radar Meter

Photographic Method

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Video graphic Method

Presentation and Analysis of Spot Speed Data

Tabular Presentation

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Statistical Analysis

Central Tendency

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Dispersion

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Sample Size

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Graphical Presentation

Histogram

Frequency Curve

Cumulative Frequency Curve2074

Spot Speed

Numerical Example The data collected in kmph on 86 vehicles in a rural road during spot speed study is shown in table below. Develop the frequency histogram, frequency distribution, and

cumulative frequency of data and determine: i) Mean speed and space mean speed ii) Standard deviation iii) Median speed iv) Modal speed v) Pace vi) Speed limit for regulation vii) Speed to check geometric design viii) Lower speed limit

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Normal Distribution to Spot Speed2071

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Numerical Example: A spot speed at particular location are normally distributed with a mean of 51.7 kmph, standard deviation of 8.3kmph. What is the probability that the speed exceeds a) 65kmph. b) Speed lies between 40-70kmph. c) What is the 85th percentile speed?

Delay

Speed and Delay Study

Methods of Speed and Delay Study

License Plate Observations

Observation from Vantage Point

Photographic and Video Survey Method

Interview Method

Moving Observer (Floating Car) Technique2072

details:

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Numerical Example: The consolidated data collected from speed and delay studies by floating car method on a stretch of urban road of length 3.5 km running North-South are given below. Determine the average values of volume, journey speed and running speed of the traffic stream along either direction. Trip no. Direction of trip Journey time Total stopped delay No. of vehicles Overtaking Overtaken From opposite direction N-S S-N N-S S-N N-S S-N N-S S-N min sec min sec

Tasks

1.3 Origin and Destination Study

Past Exam Questions on this Topic

  1. What are the uses of origin and destination study? Briefly explain the methods of conducting this study (2071 Chaitra, Q2)
    Solution

    Uses of Origin-Destination (O-D) study: planning road networks, bypasses and new links; locating terminals, parking areas and expressways; planning mass-transit routes; establishing desire lines and travel-demand (trip) patterns; phasing road improvements; and land-use / zoning decisions.

    Methods of conducting an O-D study:

    • Road-side interview method
    • License-plate (registration-number) method
    • Return post-card method
    • Tag-on-vehicle method
    • Home-interview method

Review

Origin and Destination (O-D) Study

Uses of O-D Study2071

O-D Data

Methods of O-D Survey

Road Side Interview Method

Home Interview Method

Work Spot Interview Method

License Plate Method

Post Card Method

Tag on Vehicle Method

Commercial Vehicle Survey

Presentation of O-D Data

O-D Table

Desire Lines

Presentation of O-D Data

JICA (2024) Kathmandu

Valley Urban Transport System Master Plan:

JICA (2024) Kathmandu Valley Urban Transport System Master Plan

Shopping increased from 4 to 7.6%, Leisure/recreation increased from 1.8 to 5.5% JICA (2024) Kathmandu Valley Urban Transport System Master Plan

Tasks

1.4 Basic Traffic Flow Parameters

Past Exam Questions on this Topic

  1. Assuming a linear speed-density relationship, the mean free speed is observed to be 80 kmph near zero density and the corresponding jam density is 120 veh/km. [8] a) Write down the speed-density and flow-density equations. b) Draw the v-k, v-q and q-v diagrams indicating the critical values. c) Compute the speed and density corresponding to a flow of 800 veh/hr. d) Compute the average headways and spacing when the flow is maximum. (2081 Bhadra, Q3)
    Solution
    Greenshields linear model · free speed vf = 80 km/h · jam density kj = 120 veh/km

    (a) Speed-density and flow-density equations

    v = vf(1 − k/kj) = 80(1 − k/120) = 80 − 0.667 k q = k · v = 80k − 0.667 k²

    (b) Critical values at maximum flow

    k0 = kj/2 = 120/2 = 60 veh/km v0 = vf/2 = 80/2 = 40 km/h qmax = vfkj/4 = 80 × 120/4 = 2400 veh/h

    v–k is a straight line falling from vf to 0 at kj; q–k and v–q are parabolas peaking at the critical values above.

    Greenshields linear modeldensity kspeed vvfkj(k₀, v₀)density kflow qk₀qmaxkjflow qspeed vv₀vfuncongestedcongested
    The three Greenshields relationships. Speed–density is a straight line; flow–density and speed–flow are parabolas whose peak is the capacity, at k₀ = kj/2 and v₀ = vf/2.

    (c) Speed and density at q = 800 veh/h

    800 = 80k − 0.667k² 0.667k² − 80k + 800 = 0 k = 11.0 veh/km → v = q/k = 800/11.0 = 72.7 km/h (free flow) k = 109.0 veh/km → v = q/k = 800/109.0 = 7.3 km/h (congested)

    (d) Spacing and headway at maximum flow

    spacing = 1000/k0 = 1000/60 = 16.7 m headway = 3600/qmax = 3600/2400 = 1.5 s
    v = 80 − 0.667k, q = 80k − 0.667k²; capacity 2400 veh/h at 60 veh/km and 40 km/h. At 800 veh/h the stream runs either at 72.7 km/h or 7.3 km/h; at capacity the spacing is 16.7 m and the headway 1.5 s.
  2. Assuming a linear speed-density relationship of V = 60 - 0.43K: a) Draw the V-K, V-Q and Q-K diagrams showing the critical values. b) Find the saturation flow. c) Find the speed and density at a flow of 1000 veh/hr. (2074 Ashwin, Q4)
    Solution
    Linear speed-density relation v = 60 − 0.43 k

    Read the model parameters off the relation

    at k = 0 : vf = 60 km/h at v = 0 : kj = 60/0.43 = 139.5 veh/km

    v–k is the given straight line; v–q and q–k are parabolas. Mark vf, kj and the peak on each sketch.

    Greenshields linear modeldensity kspeed vvfkj(k₀, v₀)density kflow qk₀qmaxkjflow qspeed vv₀vfuncongestedcongested
    The three Greenshields relationships. Speed–density is a straight line; flow–density and speed–flow are parabolas whose peak is the capacity, at k₀ = kj/2 and v₀ = vf/2.

    (b) Saturation (maximum) flow

    k0 = kj/2 = 69.8 veh/km v0 = vf/2 = 30 km/h qmax = vfkj/4 = 60 × 139.5/4 = ≈ 2093 veh/h

    (c) Speed and density at q = 1000 veh/h

    q = kv = k(60 − 0.43k) 0.43k² − 60k + 1000 = 0 k = 19.4 veh/km → v = 1000/19.4 = 51.7 km/h (free flow) k = 120.2 veh/km → v = 1000/120.2 = 8.3 km/h (congested)
    Capacity ≈ 2093 veh/h at 69.8 veh/km and 30 km/h. A flow of 1000 veh/h can occur at 51.7 km/h (uncongested) or 8.3 km/h (congested).
  3. Assuming a linear speed-density relationship, the mean free speed near zero density is 80 km/h and the corresponding jam density is 130 veh/km. Assume the average length of vehicles is 6 m. i) Write down the speed-density and flow-density equations. ii) Compute the speed and density corresponding to a flow of 1000 veh/hr. iii) Compute the average headways, spacing, clearance and gaps when the flow is maximum. Second part: The following data were collected for a section of road 2.5 km long during a floating-car method study. Assuming equivalency factors of 1, 2 and 3 for car, bus and truck respectively, calculate the directional flow, the journey speed and the running speed of the traffic stream. (The complete floating-car data table - journey time, stopped delay, number of vehicles met from the opposite direction, and overtaking/overtaken counts for each N-S and S-N run - is given in the scanned exam paper on the Past Questions page.) (2072 Chaitra, Q3)
    Solution
    Greenshields model with vf = 80 km/h and kj = 130 veh/km · vehicle length 6 m · plus a floating-car survey (data table on the scanned paper)

    Model equations

    v = 80(1 − k/130) = 80 − 0.615 k q = kv = 80k − 0.615 k²

    Capacity and the critical point

    k0 = kj/2 = 65 veh/km v0 = vf/2 = 40 km/h qmax = vfkj/4 = 80 × 130/4 = 2600 veh/h
    Greenshields linear modeldensity kspeed vvfkj(k₀, v₀)density kflow qk₀qmaxkjflow qspeed vv₀vfuncongestedcongested
    The three Greenshields relationships. Speed–density is a straight line; flow–density and speed–flow are parabolas whose peak is the capacity, at k₀ = kj/2 and v₀ = vf/2.

    Speed and density at q = 1000 veh/h

    0.615k² − 80k + 1000 = 0 k = 14.0 veh/km → v = 1000/14.0 = 71.4 km/h (free flow) k = 116.0 veh/km → v = 1000/116.0 = 8.6 km/h (congested)

    Spacing, headway, clearance and gap at maximum flow

    spacing = 1000/k0 = 1000/65 = 15.4 m headway = 3600/qmax = 3600/2600 = 1.38 s clearance = spacing − vehicle length = 15.4 − 6 = 9.4 m gap = clearance / v0 = 9.4 / (40/3.6) = 0.85 s

    Floating-car study — the data table on the paper

    PCU factors 1 (car), 2 (bus), 3 (truck). The paper prints the section as "25.5 km", which would give journey speeds over 340 km/h; the only sensible reading is L = 2.5 km. For each stream the with-runs are its own rows and the against-runs are the opposite ones.

    q = (Σna + Σnw) / (Σta + Σtw) t̄ = t̄w − n̄w/q journey speed = L / t̄ running speed = L / (t̄ − mean stopped delay)
    StreamΣna (PCU)Σta (s)ΣnwΣtw (s)mean stop (s)
    N–S21415616157780.5
    S–N22215775156187.5

    North–south stream

    q = (214 + 6)/(1561 + 1577) = 0.07011 veh/s = 252 PCU/h t̄ = 262.8 − 1.000/0.07011 = 248.6 s journey speed = 2.5/248.6 × 3600 = 36.2 km/h running time = 248.6 − 80.5 = 168.1 s running speed = 2.5/168.1 × 3600 = 53.5 km/h

    South–north stream

    q = (222 + 5)/(1577 + 1561) = 0.07234 veh/s = 260 PCU/h t̄ = 260.2 − 0.833/0.07234 = 248.6 s journey speed = 2.5/248.6 × 3600 = 36.2 km/h running time = 248.6 − 87.5 = 161.1 s running speed = 2.5/161.1 × 3600 = 55.8 km/h
    Greenshields part: capacity 2600 veh/h at 65 veh/km and 40 km/h; at 1000 veh/h the speed is 71.4 or 8.6 km/h; at capacity spacing 15.4 m, headway 1.38 s, clearance 9.4 m, gap 0.85 s.
    Floating-car part: flow 252 PCU/h north–south and 260 PCU/h south–north, journey speed 36.2 km/h both ways, running speed 53.5 and 55.8 km/h.

    Running speed exceeds journey speed because roughly a third of the journey time is stopped delay — the gap between the two measures the congestion on the section.

    The free speed vf = 80 km/h is taken from the sibling exam papers (the value did not OCR). The floating-car table was transcribed from a render of the scanned paper, whose text layer is unreadable.

Review

Basic Traffic Flow Parameters

  1. Flow (q): Number of vehicles/volume passing a fixed roadside point per unit time (vph)
  2. Speed (v): Space mean speed of vehicles (kmph)
  3. Density (k): Number of vehicles located in a given length of lane or roadway at a particular instant in time (vpkm)

The relation between speed, density, and flow is: 𝑞= 𝑘𝑉 (Fundamental relationship in traffic flow theory)

Other Traffic Flow Parameters

S1

S2 𝑘= 1 𝑠

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Lecture slide

Time Space Diagram

Basic Traffic Flow

Parameters

Traffic Flow Conditions

Analysis of Speed, Flow and Density Relationship20812074

Uninterrupted traffic flow:

Speed Density Relationship20812072

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𝑣𝑓 :

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Flow Density Relationship

Lecture slide

Speed Flow Relationship

Lecture slide

Basic Traffic Flow Parameters

Example: A study of freeway flow at a particular site has resulted in a calibrated speed density relationship as follows: 𝑉 = 57.5(1 − 0.008𝐾). For this relationship, determine, (a) free flow speed (b) jam density (c) capacity and d) plot speed flow relationship and e) flow density relationship. What will be the speed and density value for highest level of service possible if the flow is 80% of the capacity flow? Calculate spacing, headway, gap and clearance when there is capacity flow if the length of vehicle is 6.1m.

1.5 Traffic Capacity and Level of Service

Past Exam Questions on this Topic

  1. Define the capacity of a road and list out its types. Two vehicles A and B approaching at right angles, A from west and B from south, collide with each other. After the collision, vehicle A skids in a direction 45° north of west and vehicle B 50° east of north. The initial skid distances of vehicles A and B are 35 m and 25 m respectively before collision. The skid distances after collision are 30 m and 10 m respectively. If the weights of vehicles A and B are 3.5 and 6 tonnes respectively, calculate the original speeds of the vehicles. The average skid resistance of the pavement is found to be 0.55. [2+6] (2080 Bhadra, Q2)
    Solution

    Capacity of a road = the maximum hourly rate at which vehicles can reasonably be expected to traverse a point or uniform section of a lane/roadway under prevailing traffic and roadway conditions. Types: basic (theoretical) capacity, possible capacity, and practical (design) capacity.

    Accident reconstruction

    g = 9.81 m/s² · skid coefficient f = 0.55 · vehicle A 3.5 t travelling W→E, vehicle B 6 t travelling S→N · post-collision skid 30 m (A) and 10 m (B) · pre-collision skid 35 m (A) and 25 m (B) · after impact A deflects 45° toward N and B 50° E of N

    Speeds just after the collision, from the post-collision skid

    v = √(2 g f d) vA' = √(2 × 9.81 × 0.55 × 30) = 17.99 m/s vB' = √(2 × 9.81 × 0.55 × 10) = 10.39 m/s

    Conservation of momentum at impact

    take +x east (A's approach) and +y north (B's approach): x: mAuA = mAvA' cos45° + mBvB' sin50° y: mBuB = mAvA' sin45° + mBvB' cos50° with mA = 3.5 t and mB = 6 t: uA ≈ 26.4 m/s , uB ≈ 14.1 m/s

    Original speeds, adding the energy lost in the pre-collision skid

    u0 = √(uimpact² + 2 g f dpre) A (dpre = 35 m): u0A = √(26.4² + 2×9.81×0.55×35) = 32.7 m/s = ≈ 118 km/h B (dpre = 25 m): u0B = √(14.1² + 2×9.81×0.55×25) = 21.6 m/s = ≈ 78 km/h
    Vehicle A was travelling at about 118 km/h and vehicle B at about 78 km/h before braking — both far above any urban speed limit.

    The compass directions in the source paper are ambiguous; resolve the momentum components along the axes of your own sketch. Method: Marsani, Crash / Accident studies (skid-to-speed plus conservation of momentum).

  2. Define PIEV theory. Categorize LOS and discuss the factors need to be considered to evaluate LOS . [2+6] (2080 Bhadra, Q4)
    Solution

    PIEV theory describes the total reaction (perception-reaction) process of a driver in four stages: Perception, Intellection, Emotion and Volition. The total PIEV time (about 2.5 s for design) is used in stopping-sight-distance calculations.

    Levels of service (LOS) are graded A to F (A = free flow, best; F = forced/breakdown flow, worst), based mainly on the volume-to-capacity ratio and the resulting speed, density and delay.

    Factors to evaluate LOS: speed and travel time, traffic density/volume-to-capacity ratio, freedom to manoeuvre, traffic interruptions/delay, comfort and convenience, and safety.

  3. Describe highway capacity. Explain the factors which affect capacity and level of service. (2074 Ashwin, Q3)
    Solution

    Highway capacity = the maximum number of vehicles that can pass a given point on a lane or roadway during one hour under prevailing road and traffic conditions.

    Factors affecting capacity:

    • Roadway: lane width, lateral clearance, shoulder width, gradient, horizontal/vertical alignment, surface condition.
    • Traffic: traffic composition (trucks/buses), directional distribution, lane distribution, speed, driver behaviour.
    • Control/environment: traffic signals and control devices, weather, parking and side friction.

Traffic Capacity2080

Basic Capacity (Theoretical Capacity)

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Possible and Practical Capacities

Possible Capacity

Traffic Capacity

Example: Calculate the theoretical capacity of a traffic lane with one way traffic flow at a stream speed of 65 kmph. Assume coefficient of friction as 0.38, total reaction time as 0.68 sec and average length of vehicle as 6.1 m.

Capacity and Level of Service

Level of Service2080

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LOS A: LOS B: Free flow, free maneuver, average spacing 161m (528 ft), v/c ratio ≈ 0.3 Stable flow, reasonably free flow, maneuvering still easy, average spacing 100m (330 ft), v/c ratio ≈ 0.45

LOS C: LOS D: Stable flow, maneuverability noticeably restricted, satisfactory operating speed still maintained, average spacing 67m (220 ft), v/c ratio ≈ 0.7 Unstable flow, speed decline, maneuvering difficult and driver looks for gap, average spacing 50m (165 ft), v/c ratio ≈ 0.9

LOS E: LOS F: Unstable flow, operating at capacity, speed reduced, few gap for maneuvering, driver need to search gap, average spacing 33.5m (110 ft), v/c ratio ≈ 0.98 Forced or breakdown flow and low speed, demand exceed capacity, road in constant traffic jam.

Factors Affecting Capacity and LOS2074

Passenger Car Unit (PCU)

Factors Affecting PCU

Factors for Simple Analysis of PCU

Tasks

1.6 Parking Studies

Past Exam Questions on this Topic

  1. List out the importance of parking and ill effect of illegal parking: Explain about different types of On-Street and Off-street parking facilities with their pros and cons [8] (2076 Ashwin, Q3)
    Solution

    Importance of parking studies: assess parking demand vs supply, plan parking facilities, frame parking policy and pricing, reduce congestion from cruising/illegal parking, and improve safety.

    Ill effects of illegal parking: reduced carriageway width and capacity, congestion, obstruction to moving traffic and pedestrians, increased accidents, and hindrance to emergency vehicles.

    On-street parking (kerb parking): parallel, 30°/45°/60° angle, and 90° (right-angle) parking. Pros: cheap, convenient, no separate land. Cons: reduces road capacity, causes congestion and accidents.

    Off-street parking: surface lots and multi-storey / underground / mechanical garages. Pros: does not obstruct traffic, large capacity. Cons: needs land and high capital cost.

  2. What is the importance of parking studies? Describe the different types of parking. (2073 Shrawan, Q3)
    Solution

    Importance of parking studies: to assess parking demand versus supply, plan and locate parking facilities, frame parking policy and pricing (fees, time limits), reduce congestion caused by cruising and illegal parking, and improve road safety and capacity.

    Types of parking:

    • On-street (kerb) parking: parallel parking, and angle parking at 30°, 45°, 60° or 90° (right-angle) to the kerb.
    • Off-street parking: surface car parks and multi-storey / underground / mechanical (automated) garages.

Review

Parking Study2073

Issues of Parking

Parking Management as Solution

Parking Policies

Types of Parking Facilities20822076

  1. On Street/Kerbside Parking
  2. Off Street Parking

On Street Parking

Types:

Off Street Parking

Off Street Parking: Surface Car Parks

Off Street Parking: Multi Story Car Parks

Off Street Parking: Roof Parks

Off Street Parking: Mechanical Car Parks

Off Street Parking: Underground Parks

Parking Study

Parking Characteristics Study

Parking Inventory or Supply Study

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Parking Demand Study or Parking Usage Survey

Parking Demand Study: Cordon Count

Parking Demand Study: Interview Method

Parking Demand Study: In-Out Survey

Parking Demand Study: Fixed Period Survey

Parking Demand Study: License Plate Survey

Parking Demand Analysis

Parking Attributes

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Parking Study

Example: From an in-out survey conducted for a parking area consisting of 40 bays, the initial count was found to be 25. The number of vehicles coming in and out of the parking lot for a time interval of 5 minutes is as shown in the table. Find the accumulation, total parking load, average occupancy and efficiency of the parking lot. Time In Out

Tasks

1.7 Crash Studies

Past Exam Questions on this Topic

  1. Two vehicles A and B approaching at right angles, A from North and B from West, collide with each other. After the collision, vehicle A skids in a direction 47° East of North and vehicle B skids 56° South of East. The initial skid distances of vehicles A and B are 32 and 18 m respectively before collision. The skid distances after collision are 14 and 33 m respectively. If the weights of vehicles A and B are 4 and 6 tonnes respectively, calculate the original speeds of the vehicles. The average skid resistance of the pavement is found to be 0.52. [8] (2082 Bhadra, Q2)
    Solution
    g = 9.81 m/s² · skid coefficient f = 0.52 · vehicle A 4 t travelling N→S, vehicle B 6 t travelling W→E · post-collision skid 14 m (A) and 33 m (B) · pre-collision skid 32 m (A) and 18 m (B) · after impact A moves 47° E of N and B 56° S of E

    Speeds just after the collision

    v = √(2 g f d) vA' = √(2 × 9.81 × 0.52 × 14) = 11.95 m/s vB' = √(2 × 9.81 × 0.52 × 33) = 18.35 m/s

    Conservation of momentum at impact

    x (east): mBuB = mAvA' sin47° + mBvB' cos56° → uB = 16.1 m/s y (north): mAuA = mBvB' sin56° − mAvA' cos47° → uA = 14.7 m/s

    Original speeds, adding the pre-collision skid

    u0 = √(uimpact² + 2 g f dpre) u0A = √(14.7² + 2×9.81×0.52×32) = 23.3 m/s = ≈ 84 km/h u0B = √(16.1² + 2×9.81×0.52×18) = 21.0 m/s = ≈ 76 km/h
    Vehicle A was travelling at about 84 km/h and vehicle B at about 76 km/h before the brakes were applied.
  2. A passenger car of mass 430 kg moving east collides with a truck of mass 1299 kg that is moving south. Initial skid distances of the car and the truck are 15 m and 10 m respectively. After collision they lock together and skid through a distance equal to 8 m in a direction 65° south of east before stopping. Compute the initial speeds of the car and the truck. Assume average coefficient of friction as 0.5. [8] (2081 Baishakh, Q2)
    Solution
    Car mc = 430 kg travelling east · truck mt = 1299 kg travelling south · skid coefficient f = 0.5 · after the collision the two lock together and skid 8 m in a direction 65° south of east

    Because the vehicles lock together the impact is perfectly plastic, so there is a single common velocity immediately after it.

    Common speed just after the collision

    v' = √(2 g f d) = √(2 × 9.81 × 0.5 × 8) = 8.86 m/s , directed 65° S of E

    Combined mass

    mc + mt = 430 + 1299 = 1729 kg

    Momentum in the east direction gives the car's speed

    mcuc = (mc + mt) v' cos65° uc = (1729 × 8.86 × 0.4226) / 430 = 15.05 m/s = ≈ 54.2 km/h

    Momentum in the south direction gives the truck's speed

    mtut = (mc + mt) v' sin65° ut = (1729 × 8.86 × 0.9063) / 1299 = 10.69 m/s = ≈ 38.5 km/h
    The car was doing about 54.2 km/h and the truck about 38.5 km/h at the moment of impact.

    No pre-collision skid marks are given, so these are the speeds at impact rather than the speeds before braking.

  3. Define the capacity of a road and list out its types. Two vehicles A and B approaching at right angles, A from west and B from south, collide with each other. After the collision, vehicle A skids in a direction 45° north of west and vehicle B 50° east of north. The initial skid distances of vehicles A and B are 35 m and 25 m respectively before collision. The skid distances after collision are 30 m and 10 m respectively. If the weights of vehicles A and B are 3.5 and 6 tonnes respectively, calculate the original speeds of the vehicles. The average skid resistance of the pavement is found to be 0.55. [2+6] (2080 Bhadra, Q2)
    Solution

    Capacity of a road = the maximum hourly rate at which vehicles can reasonably be expected to traverse a point or uniform section of a lane/roadway under prevailing traffic and roadway conditions. Types: basic (theoretical) capacity, possible capacity, and practical (design) capacity.

    Accident reconstruction

    g = 9.81 m/s² · skid coefficient f = 0.55 · vehicle A 3.5 t travelling W→E, vehicle B 6 t travelling S→N · post-collision skid 30 m (A) and 10 m (B) · pre-collision skid 35 m (A) and 25 m (B) · after impact A deflects 45° toward N and B 50° E of N

    Speeds just after the collision, from the post-collision skid

    v = √(2 g f d) vA' = √(2 × 9.81 × 0.55 × 30) = 17.99 m/s vB' = √(2 × 9.81 × 0.55 × 10) = 10.39 m/s

    Conservation of momentum at impact

    take +x east (A's approach) and +y north (B's approach): x: mAuA = mAvA' cos45° + mBvB' sin50° y: mBuB = mAvA' sin45° + mBvB' cos50° with mA = 3.5 t and mB = 6 t: uA ≈ 26.4 m/s , uB ≈ 14.1 m/s

    Original speeds, adding the energy lost in the pre-collision skid

    u0 = √(uimpact² + 2 g f dpre) A (dpre = 35 m): u0A = √(26.4² + 2×9.81×0.55×35) = 32.7 m/s = ≈ 118 km/h B (dpre = 25 m): u0B = √(14.1² + 2×9.81×0.55×25) = 21.6 m/s = ≈ 78 km/h
    Vehicle A was travelling at about 118 km/h and vehicle B at about 78 km/h before braking — both far above any urban speed limit.

    The compass directions in the source paper are ambiguous; resolve the momentum components along the axes of your own sketch. Method: Marsani, Crash / Accident studies (skid-to-speed plus conservation of momentum).

  4. What are the basic requirements of intersection at grade? Mention the importance of street lighting. Explain the different types of traffic islands. How is accident study carried out? Two vehicles A and B approaching at right angles, vehicle A from West and vehicle B from south, collide with each other. After the collision, vehicle A skids in a direction 49° N of W and vehicle B skids 27° E of N. The initial skid distances of vehicles A and B are 37 m and 19 m respectively before collision. If the weight of vehicle A is 4 tonne and vehicle B is 6 tonne, and the skid distances after collision for vehicle A is 15 m and for vehicle B is 36 m, calculate the initial speeds of the vehicles if the average skid resistance of the pavement is found to be 0.55. (2075 Ashwin, Q1)
    Solution

    Basic requirements of an at-grade intersection: adequate sight distance, minimum conflict points, channelization to guide traffic, adequate capacity, low relative speed, clear and simple layout, provision for pedestrians and proper drainage/lighting.

    Importance of street lighting: improves night visibility & safety, reduces night accidents, aids pedestrian security, improves traffic flow and comfort, and enhances aesthetics.

    Types of traffic islands: divisional islands, channelizing islands, pedestrian refuge islands, and rotary (central) islands.

    Accident study is carried out by collecting accident data (spot maps, collision & condition diagrams), analysing causes (road user, vehicle, road, environment), and applying remedial 3-E measures (Engineering, Enforcement, Education).

    Collision numerical

    skid coefficient f = 0.55 · vehicle A 4 t, vehicle B 6 t · post-collision skid 15 m (A) and 36 m (B) · pre-collision skid 37 m (A) and 19 m (B) · deflections A 49° N of W and B 27° E of N

    Speeds just after the collision

    v = √(2 g f d) vA' = √(2 × 9.81 × 0.55 × 15) = 12.72 m/s vB' = √(2 × 9.81 × 0.55 × 36) = 19.71 m/s

    Momentum conservation to recover the impact speeds

    The paper does not state which way each vehicle was heading. The deflections given — A to 49° N of W and B to 27° E of N — are only consistent with A travelling east and B travelling north; any other pairing makes one impact speed come out negative. Measuring all directions anticlockwise from east, A leaves at 131° and B at 63°.

    x (east): mAuA = mAvA' cos131° + mBvB' cos63° 4 uA = 50.89(−0.656) + 118.26(0.454) 4 uA = −33.38 + 53.69 = 20.30 uA = 5.08 m/s y (north): mBuB = mAvA' sin131° + mBvB' sin63° 6 uB = 50.89(0.755) + 118.26(0.891) 6 uB = 38.41 + 105.37 = 143.78 uB = 23.96 m/s

    Add the pre-collision skid to get the original speeds

    u0 = √(uimpact² + 2 g f dpre) A (dpre = 37 m): u0A = √(5.08² + 2×9.81×0.55×37) = √425.1 = 20.62 m/s = ≈ 74 km/h B (dpre = 19 m): u0B = √(23.96² + 2×9.81×0.55×19) = √779.6 = 27.92 m/s = ≈ 100 km/h
    Vehicle A was travelling at about 74 km/h and vehicle B at about 100 km/h before braking.

    The weights (A = 4 t, B = 6 t) enter only as the mass ratio in the momentum equations, so they need not be converted to newtons. Note that A leaves the impact faster than it entered it (12.72 vs 5.08 m/s) — the heavier, faster vehicle B drove it forward, which is exactly why the wreck slid 15 m.

  5. What are the causes of accident and how can accidents be prevented? Describe briefly the factors influencing street light design. (2075 Chaitra, Q3)
    Solution

    Causes of accidents (grouped by the 4 elements):

    • Road user: over-speeding, drunk/fatigued driving, disobeying rules, wrong overtaking, carelessness of pedestrians.
    • Vehicle: brake/steering/tyre failure, poor lights, overloading.
    • Road: poor geometry, inadequate sight distance, slippery/poor surface, poor markings.
    • Environment: rain, fog, poor lighting, glare.

    Prevention - the 3 E's: Engineering (better geometry, signs, lighting, road furniture), Enforcement (speed & traffic-law enforcement) and Education (road-safety awareness and training).

  6. A vehicle hits a bridge abutment at a speed estimated by investigations as 20 kmph. Approaching the abutment, the vehicle left skid marks of 60 m on the gravel shoulder (f = 0.50) followed by skid marks of 30 m on the pavement (f = 0.35). What was the initial speed of the vehicle? (2073 Shrawan, Q4)
    Solution
    A vehicle skids 60 m on the gravel shoulder (f = 0.50), then 30 m on the pavement (f = 0.35), and still strikes the abutment at 20 km/h · g = 9.81 m/s²

    Two surfaces means two skid phases. Work backwards from the impact, using vbefore² = vafter² + 2 g f d for each phase in turn.

    Speed at impact

    vimpact = 20 km/h = 20/3.6 = 5.56 m/s

    Back through the pavement skid (30 m, f = 0.35)

    v² = 5.56² + 2 × 9.81 × 0.35 × 30 = 30.9 + 206.0 = 236.9 v = 15.39 m/s (speed entering the pavement)

    Back through the gravel skid (60 m, f = 0.50)

    v0² = 15.39² + 2 × 9.81 × 0.50 × 60 = 236.9 + 588.6 = 825.5 v0 = 28.73 m/s = ≈ 103.4 km/h
    The vehicle was travelling at about 103 km/h when the driver first braked.

    Friction values as given in the paper: gravel shoulder f = 0.50, pavement f = 0.35, impact speed 20 km/h. Method: Marsani, Crash studies (skid-to-speed).

  7. a) Define rotary intersection. What are the advantages and disadvantages of rotary intersection? [8] b) A vehicle skids through a distance of 40m before colliding with another parked vehicle; the weight of which is 75 percent of the former After collision both the vehicles skid through 14m before stopping: Compute the initial speed of moving vehicle. Assume coefficient of friction of 0.62. [8] (2068 Baishakh, Q2)
    Solution

    Rotary intersection = a channelized at-grade intersection with a central island around which all traffic moves one-way (anticlockwise here), converting crossing conflicts into merging, weaving and diverging.

    Advantages: continuous stop-free movement, no severe right-angle/head-on conflicts, self-regulating, orderly, aesthetic. Disadvantages: large land area, poor for high volumes/heavy pedestrian flow, delay under heavy traffic, unsuitable at high speeds or steep grades.

    Numerical

    skid coefficient f = 0.62 · g = 9.81 m/s² · a moving vehicle skids 40 m, strikes a parked vehicle weighing 0.75 × its own weight, and the two then skid 14 m together

    Common speed of the pair just after impact

    v' = √(2 g f d) = √(2 × 9.81 × 0.62 × 14) = 13.06 m/s

    Speed of the moving vehicle just before impact

    plastic collision, the parked vehicle starts at rest: m · uimpact = (m + 0.75m) v' uimpact = 1.75 × 13.06 = 22.86 m/s

    Initial speed, adding the 40 m pre-impact skid

    u0 = √(uimpact² + 2 g f d) = √(22.86² + 2 × 9.81 × 0.62 × 40) = √(522.6 + 486.6) = √1009.2 = 31.77 m/s = ≈ 114 km/h
    The moving vehicle was travelling at about 114 km/h before it began to skid.

Review

Road Crashes

Crash Study20802070

Objectives of Crash Study:

Causes of Crashes2075

Causes of Crash: Road User Factors

Causes of Crash: Vehicles

Causes of Crash: Road and Its Condition

Causes of Crash: Other Environmental Factors

Crash Data Collection

Crash Reporting

Crash Records

Location Files

Spot Maps

Condition Diagram

Collision Diagram

outline map of the location and symbols showing the direction of movement of vehicles/pedestrians/object s involved in crashes.

Crash Study

Steps:

  1. Obtaining crash data
  2. Select high crash frequency locations
  1. Conduct site analysis
  1. Summarize the facts (major collision types, potential causes)
  2. Supplement with field observations
  3. Analyze summarized facts and field data and suggest remedial measures

Crash Prevention

Measures:

Engineering Measures

Enforcement Measures

Educational Measures

Crash Analysis2073

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Collision of Moving Vehicle with Parked Vehicle2068

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Crash Study

Example 1: A vehicle of weight 2 tonne skids through a distance equal to 40m before colliding with another parked vehicle of weight 1tonne. After collision both the vehicles kid through a distance equal to 12m before stopping. Compute the initial speed of the moving vehicle. Assume coefficient of friction as 0.5.

Collision of Vehicles Approaching from Right Angle2082208120802075

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Crash Study

Example 2: Two vehicles A and B approaching at right angles, A from West and B from South collide each other. After the collision vehicle A skids in North direction 500 North of West and vehicle B 600 East of North. The initial skid distances of vehicle A and B are 38m and 20m respectively before collision. The skid distances after collision are 15m and 36m respectively. If the weights of vehicles B and A are 6 tonnes and 4 tonnes. Calculate the original speeds of the vehicles. The average skid resistance of the pavement is found to be 0.55.

Tasks