Traffic Engineering – Intersection Capacity

July 5, 2019 by

Intersection Capacity is more than the volume of vehicles the intersection can accommodate in a given time interval. First, as transportation engineers we must give consideration to pedestrians, mass transit, and other forms of travel. Second, we must design for the time required for an intersection to fill up and then empty given the various speeds of pedestrians, private vehicles, buses, light rail, etc. And third, we must be able to coordinate each intersection with the overall traffic configuration of other intersections, traffic flow patterns, access points, speed limits, etc.

The study of intersection capacity utilization is filled with opportunities for creativity and problem-solving.

Intersection Capacity

Using information from Traffic Engineering – Street Segment Interrupted Flow and based on the traffic counts for the intersection shown (assume up is north):

Intersection Capacity

1. Calculate the East-West Critical Lane Volumes:

  • East bound traffic turning left + (West bound traffic + West bound traffic turning right) = 105 + (250 + 130) = 105 + 380 = 485 vph
  • West bound traffic turning left + (East bound traffic + East bound traffic turning right) = 70 + (215 + 60) = 70 + 275 = 345 vph

2. Calculate the North-South Critical Lane Volumes:

  • North bound traffic turning left + (South bound traffic + South bound traffic turning right) = 75 + (245 + 65) = 75 + 310 = 385 vph
  • South bound traffic turning left + (North bound traffic + North bound traffic turning right) = 110 + (225 + 125) = 110 + 350 = 460 vph

3. Calculate the Critical Volume for Phase (Vci):

  • Vci = 485 vph (from calculations in steps 1 and 2)

4. Calculate the Total Critical Volume (Vc):

  • Vc = sum of Critical Lane Volumes = 380 + 275 + 310 + 350 = 1315 vph

5. Calculate the Saturation Flow Rate (given: headway (h) = 2.3 sec/veh):

  • Saturation Flow Rate (s) = 3600/h = 3600/2.3 = 1565 veh/hr

6. Calculate the Cycle Length (C) if the number of phases (N) = 4, the total lost time per phase (tL) = 4 sec, the Peak Hour Factor (PHF) = 1.0, and the desired volume/capacity (v/c) ratio = 0.97:

  • C = (N x tL) ÷ [1 – (Vc ÷ (PHF x (v/c) x s))]
  • C = (4 x 4) ÷ [1 – (1315 ÷ (1.0 x 0.97 x 1565))] = 120 sec

7. Calculate the Effective Green Time (gi):

  • gi = (Vci ÷ Vc) x (C – L), where Vci = Critical Volume for Phase, Vc = Total Critical Volume, and L = Total Lost Time = N x tL
  • gi = (485 ÷ 1315) x (120 – (4 x 4)) = 38.4 sec

8. Calculate the capacity of one approach lane:

  • c = 3600gi ÷ hC = (3600 x 38.4) ÷ (2.3 x 120) = 500vph
Related Content:

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Jeff Setzer, PE
Fire Protection PE Exam Prep Course

Intersection Capacity

NCEES

Filed Under: Civil Engineering Transportation PE Exam Tagged With: , , , , , , , , ,

Traffic Engineering – Uninterrupted Flow

July 5, 2019 by

Traffic Engineering Uninterrupted Flow is defined and regulated by:

  • Vehicle-to-Vehicle interactions:
    • Speed
    • Proximity
    • Volume
  • Vehicles-to-Roadway interactions:
    • Weather conditions
    • Surface condition
    • Site distance

Uninterrupted Traffic Flow
Examples of Uninterrupted Traffic Flow are:

  • Interstate highways
  • Expressways
  • Rural highways and roads
Concepts and Equations
  • Flow (q)

The rate at which vehicles pass a given point on a roadway in a given time interval. q = n/t (vehicles per hour) and q = Space Mean Speed x Density.

  • Volume (V)

The number of vehicles passing a given point on a roadway in a specified period of time.

  • Time-Mean Speed (vt)

Arithmetic mean of speeds of vehicles passing a point (mph).

The Time-Mean Speed for six vehicles travelling 55 mph, 53 mph, 50 mph, 47 mph, 45 mph, and 44 mph is:

vt = (55 + 53 + 50 + 47 + 45 + 44) ÷ 6 = 49 mph

  • Space-Mean Speed (vs)

The average speeds over a length of roadway (mph).

The Space-Mean Speed for the six vehicles is:

vs = 6 ÷ (1/55 + 1/53 + 1/50 + 1/47 + 1/45 + 1/44) = 48.7 mph

  • Free Flow Speed (vf)

The average speed that a motorist would travel if there were no congestion or other adverse conditions (i.e. bad weather).

  • Headway (h)

The time interval or distance between two vehicles traveling in the same direction over the same route. The time elapsed between the arrival of one vehicle and a following vehicle at a designated test point. The inverse of headway is flow (h = 1/q).

  • Time Headway (ht)

Time difference between the front of a vehicle crossing a point on a roadway and the front of the next vehicle crossing the same point (measured in seconds).

  • Space Headway (hs)

Distance between the front of a vehicle and the front of the next vehicle (ft).

  • Density (k)

If n = number of vehicles and ℓ = length in miles, then k = n/ℓ.

  • Critical Density (kc)

The maximum density achievable under free flow.

  • Jam Density (kj)

The maximum density achieved during congestion.

  • Spacing (s)

The center-to-center distance between two vehicles. Spacing is the inverse of density (s = 1/k).

  • Gap (g)

The time elapsed between the rear bumper of the first vehicle passing a point and the front bumper of the second vehicle passing the same point

  • Clearance (c)

The distance between the rear bumper of a vehicle and the front bumper of the second vehicle. Clearance is also equal to the spacing minus the length of the first vehicle.

Greenshield’s Model of Uninterrupted Traffic Flow

Predicts and explains trends observed in real traffic flows:

v = A – B x k, where v = speed (mph)

where A and B are constants determined from field observations, and k = density (vehicles/mile)

Also, vs = vf x (vf/kj)k

Example 1

The Space-Mean Speed for a given freeway is vs = 43.9(1.7 – 0.015k).

Calculate:

  1. Free-flow speed
  2. Jam Density
  3. Maximum flow
  4. Speed at maximum flow

Solution
Using Greenshield’s Model:

vs = vf – (vf/kj)k = 43.9(1.7 – 0.015k) = 74.63 – 0.6585k

(1) Free-flow speed: (vf) = 74.63 mph

(2) Jam density: vf/kj = 0.6585 or kj = vf/0.6585 = 74.63/0.6585 = 113 veh/mile

(3) Maximum flow: (Definition of Flow)

q = Space Mean Speed x Density = vs x k
q = (43.9(1.7 – 0.015k))k = 74.63k – 0.6585k2

to determine maximum flow: dq/dk = 74.63 – 0.6585(2)k = 0

k = 56.7 vehicles per mile = density at maximum flow
q = 74.63k – 0.6585k2 = 74.63(56.7) – 0.6585(56.7)2
q = 2114.5 veh/hr

(4) Speed at maximum flow:

vmax = 74.63 – 0.6585k = 74.63 – 0.6585(56.7) = 37.3 mph

Related Content:

If there is anything we can do to help you prepare for the Civil Engineering Transportation PE Exam, please do not hesitate to Contact Us.

For additional PE Exam resources, go to EngineeringDesignResources.com.

To Your Success …

Jeff Setzer, PE
Fire Protection PE Exam Prep Course

Uninterrupted Flow

NCEES

Filed Under: Civil Engineering Transportation PE Exam Tagged With: , , , , , , , , , ,

Quantity Take-Off Methods

August 2, 2016 by

Quantity Take-Off Methods

Quantity Take-Off Methods

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