CH-5-6 Highway Safety – Traffic and Highway Engineering : Book Guide

This Document Contains Chapters 5 to 6 Chapter 5 Highway Safety 5-1 Explain the difference between short-term and long-term driver expectancy, and provide an example of each. Expectancy is the ability of drivers to rely on past experience to assist with control, guidance, or navigation tasks. When driver expectancy is taking into consideration in the design of roadways, the negative effect of the driver’s limitations in processing information is reduced because drivers only need to process new information. Examples of long-term expectancies are: Freeway exists are located on the right-hand side of the road At an intersection of a major and minor road controlled by a stop sign, the stop sign is on the approaches of the road that appears to be the minor road At an intersection approach, a driver wishing to turn left will be in the left lane or on a through lane that allows left-turns Examples of short-term expectancies: A long section of roadway with gently winding characteristics is contiguous with a roadway section that has gentle curves A long section of roadway that allows for high speed driving is contiguous with a roadway sections that also allows for high speed driving Driving at a consistent speed along a well coordinated system of traffic signals on an arterial should not suddenly lead to an isolated intersection with a significantly different cycle length 5-2 Briefly explain the steps in the roadway safety management process as outlined in the Highway Safety Manual. The HSM recommends that the Roadway Safety Management process should involve the following steps: Network Screening – In this step, sites at which the potential for reducing average crash frequency exists are identified and ranked, through a review of the transportation network Diagnosis – In this step crash patterns are determined through the collection and analysis of historic data, and evaluation of site conditions Select Countermeasures – This step involves determining probable crash related factors at a site and selecting countermeasures that will be effective in reducing crash frequency. Economic Appraisal – This step involves conducting an appropriate economic analysis to identify specific projects that are economically justifiable. Prioritize Projects – In this step, the projects that are economically justifiable at specific sites and across multiple sites are prioritized with respect to their potential to achieving safety objectives within the available budget. Prioritization factors may include low cost, mobility enhancement potential or reduced negative environmental impact. Safety Effectiveness Evaluation – In this step the selected countermeasures at a site or multiple sites are evaluated to determine how effective they are or would be in reducing annual crash frequency or severity. 5-3 Describe the type of information on a collision diagram. Collision diagrams include symbols used to represent different types of maneuvers, types of accidents, and severity of accidents. The date and time the accident occurs is also included on the diagram. Because the diagrams provide the information in a pictorial format, the location of accidents is immediately known. 5-4 Identify and explain the system for identifying levels of crash severity as used in the Highway Safety Manual. Crash severity is the level of injury or property damage that is incurred as a result of a crash, where injury is defined as bodily harm to a person. There are several different methods to rank severity. However, the HSM uses the KABCO scale. There are five levels in this scale and are given as: K – Fatal Injury: an injury that results in death – Incapacitating Injury: An injury that does not result in death, but causes the injured to be incapable of walking, or driving or continue to perform in activities that he/she could before the occurrence of the injury. – Non-incapacitating evident injury: An injury that is neither a fatal or an incapacitating injury that is visible to observers at the site at which the crash occurred. – Possible injury: Any injury that is neither incapacitating nor a non-incapacitating evident, including a claim of injury that is not evident. O – No injury, i.e. Property Damage Only (PDO) 5-5 Determine the crash modification factor (using Table 5.14) that can be used to estimate the change in frequency of the crash type targeted by a proposed improvement that will widen a two-lane rural highway from 18 to 22 ft if the highway has an AADT of 1250 veh/day. The CMFexist can be calculated according to Table 5.14, for a rural two-lane roadway with 9ft lane width and 400 < AADT < 2000: = 1.05 + 2.81 × 10 ( − 400) = 1.05 + 2.81 × 10 (1250 − 400) = 1.28885 The CMFnew can be calculated according to Table 5.14, for a rural two-lane roadway with 11ft lane width and 400 5) = (30/40)6 = 0.178 for P(n<=5) = 1 – 0.178 = 0.822 Average waiting time per vehicle Using Equations 6.73 and 6.72, E(v) = 1/(Q – q) = 0.1 hr = 6 minutes wait time including queue time and service time E(w) = q/(Q(Q – q)) = 0.075 hr = 4.5 minutes wait time in the queue 6-27 An expressway off-ramp consisting of a single lane leads directly to a tollbooth. The rate of arrival of vehicles at the expressway can be considered to be Poisson with a mean of 45 veh/hr, and the rate of service to vehicles can be assumed to be exponentially distributed with a mean of 1 min. What is the average number of vehicles waiting to be served at the booth (that is, the number of vehicles in queue, not including the vehicle being served)? What is the length of the ramp required to provide storage for all exiting vehicles 90% of the time? Assume the average length of a vehicle is 18 ft and that there is an average space of 10 ft between consecutive vehicles waiting to be served. What is the average waiting time a driver waits before being served at the tollbooth (that is, the average waiting time in the queue)? Expected queue length Using Equation 6.71, E(m) = q2/[Q(Q – q)] = (45)2/[60(60 – 45)] = 2.25 vehicles Ramp length 1.00 – 0.90 = (q/Q)^(N+1) 0.10 = (45/60)^(N+1) 0.10 = (0.75)^N+1 ln(0.10) = ln(0.75)(N+1) 8 = N+1 N = 7 vehicles Ramp length = 7 veh (18 ft/veh) + 6 spaces (10 ft/space) = 186 feet c) Average waiting time per vehicle Using Equations 6.73 and 6.72, E(v) = 1/(Q – q) = 1/(60 – 45) = 0.067 hr = 4 minutes wait time including queue time and service time E(w) = q/(Q(Q – q)) = 45/(60(60 – 45)) = 0.05 hr = 3 minutes wait time in the queue Solution Manual for Traffic and Highway Engineering Nicholas J. Garber, Lester A. Hoel 9781133605157