Crash Rates for Roadway Segments

Transportation · FE Reference Handbook section

Transportation

6 formulas

10 exam-style examples

~57 min

All Transportation lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Crash Rates for Roadway Segments within Transportation. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what crash rates for roadway segments describes physically and when it applies.
State every one of the 6 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: grades as decimals in curve formulas, percent in the stem.

Lecture

Why this section exists. Crash Rates for Roadway Segments is the part of Transportation that lets you connect a vertical or horizontal alignment, or a traffic stream to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a curve geometry element or a capacity/flow relationship. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. grades as decimals in curve formulas, percent in the stem. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Dense peak-hour traffic queued on an urban arterial at dusk. — Photo 1. Where this shows up in practice: crash rates for roadway segments.
Wikimedia Commons, CC BY 2.0

Transportation — Crash Rates for Roadway Segments: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a vertical or horizontal alignment, or a traffic stream. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 6 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

RMVM	Quantity produced by "RMVM = VMT" — read its definition and unit from the handbook line directly above the equation.
A	Quantity produced by "A = number of crashes, total or by type at the study location, during a given period" — read its definition and unit from the handbook line directly above the equation.
VMT	Quantity produced by "VMT = vehicle miles of travel during the given period;" — read its definition and unit from the handbook line directly above the equation.
ADT	Quantity produced by "ADT = average daily traffic on the roadway segment" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

A # 1, 000, 000
where

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments

A 2.0 mile segment with an ADT of 31,354 vehicles per day experienced 20 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 25% crash reduction factor.

Given

Crashes = 20 in 3 yr
ADT = 31,354 veh/day
Length = 2.0 mi
CRF = 0.25

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 31354 \times 365 \times 3 \times 2.0 = 6.867 e + 7 v e h - mi$
Substituting
$R = (20 \times 1 0^{8}) /6.867 e + 7 = 29.13 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$20 (1 - 0.25) = 15.00 cr a s h ese x p ec t e d$
Prevented
$20 - 15.00 = 5.00 cr a s h eso v er 3 y e a r s$

Answer: Rate = 29.1 crashes/100 MVM; 5.0 crashes prevented

Why the other options are there

10.00 (crashes per mile, no exposure)
15.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Example 2

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments (2)

A 5.0 mile segment with an ADT of 15,775 vehicles per day experienced 26 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 40% crash reduction factor.

Given

Crashes = 26 in 5 yr
ADT = 15,775 veh/day
Length = 5.0 mi
CRF = 0.40

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 15775 \times 365 \times 5 \times 5.0 = 1.439 e + 8 v e h - mi$
Substituting
$R = (26 \times 1 0^{8}) /1.439 e + 8 = 18.06 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$26 (1 - 0.40) = 15.60 cr a s h ese x p ec t e d$
Prevented
$26 - 15.60 = 10.40 cr a s h eso v er 5 y e a r s$

Answer: Rate = 18.1 crashes/100 MVM; 10.4 crashes prevented

Why the other options are there

5.20 (crashes per mile, no exposure)
52.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Example 3

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments (3)

A 4.5 mile segment with an ADT of 35,816 vehicles per day experienced 24 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 30% crash reduction factor.

Given

Crashes = 24 in 3 yr
ADT = 35,816 veh/day
Length = 4.5 mi
CRF = 0.30

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 35816 \times 365 \times 3 \times 4.5 = 1.765 e + 8 v e h - mi$
Substituting
$R = (24 \times 1 0^{8}) /1.765 e + 8 = 13.60 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$24 (1 - 0.30) = 16.80 cr a s h ese x p ec t e d$
Prevented
$24 - 16.80 = 7.20 cr a s h eso v er 3 y e a r s$

Answer: Rate = 13.6 crashes/100 MVM; 7.2 crashes prevented

Why the other options are there

5.33 (crashes per mile, no exposure)
21.6 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Example 4

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments (4)

A 1.5 mile segment with an ADT of 9,555 vehicles per day experienced 40 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 25% crash reduction factor.

Given

Crashes = 40 in 3 yr
ADT = 9,555 veh/day
Length = 1.5 mi
CRF = 0.25

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 9555 \times 365 \times 3 \times 1.5 = 1.569 e + 7 v e h - mi$
Substituting
$R = (40 \times 1 0^{8}) /1.569 e + 7 = 254.9 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$40 (1 - 0.25) = 30.00 cr a s h ese x p ec t e d$
Prevented
$40 - 30.00 = 10.00 cr a s h eso v er 3 y e a r s$

Answer: Rate = 254.9 crashes/100 MVM; 10.0 crashes prevented

Why the other options are there

26.67 (crashes per mile, no exposure)
30.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Example 5

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments (5)

A 5.5 mile segment with an ADT of 31,028 vehicles per day experienced 35 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 20% crash reduction factor.

Given

Crashes = 35 in 5 yr
ADT = 31,028 veh/day
Length = 5.5 mi
CRF = 0.20

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 31028 \times 365 \times 5 \times 5.5 = 3.114 e + 8 v e h - mi$
Substituting
$R = (35 \times 1 0^{8}) /3.114 e + 8 = 11.24 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$35 (1 - 0.20) = 28.00 cr a s h ese x p ec t e d$
Prevented
$35 - 28.00 = 7.00 cr a s h eso v er 5 y e a r s$

Answer: Rate = 11.2 crashes/100 MVM; 7.0 crashes prevented

Why the other options are there

6.36 (crashes per mile, no exposure)
35.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Example 6

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments (6)

A 2.0 mile segment with an ADT of 37,491 vehicles per day experienced 58 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 35% crash reduction factor.

Given

Crashes = 58 in 5 yr
ADT = 37,491 veh/day
Length = 2.0 mi
CRF = 0.35

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 37491 \times 365 \times 5 \times 2.0 = 1.368 e + 8 v e h - mi$
Substituting
$R = (58 \times 1 0^{8}) /1.368 e + 8 = 42.38 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$58 (1 - 0.35) = 37.70 cr a s h ese x p ec t e d$
Prevented
$58 - 37.70 = 20.30 cr a s h eso v er 5 y e a r s$

Answer: Rate = 42.4 crashes/100 MVM; 20.3 crashes prevented

Why the other options are there

29.00 (crashes per mile, no exposure)
101.5 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Example 7

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments (7)

A 3.5 mile segment with an ADT of 33,511 vehicles per day experienced 59 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 45% crash reduction factor.

Given

Crashes = 59 in 3 yr
ADT = 33,511 veh/day
Length = 3.5 mi
CRF = 0.45

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 33511 \times 365 \times 3 \times 3.5 = 1.284 e + 8 v e h - mi$
Substituting
$R = (59 \times 1 0^{8}) /1.284 e + 8 = 45.94 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$59 (1 - 0.45) = 32.45 cr a s h ese x p ec t e d$
Prevented
$59 - 32.45 = 26.55 cr a s h eso v er 3 y e a r s$

Answer: Rate = 45.9 crashes/100 MVM; 26.5 crashes prevented

Why the other options are there

16.86 (crashes per mile, no exposure)
79.7 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Example 8

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments (8)

A 3.5 mile segment with an ADT of 36,887 vehicles per day experienced 21 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 35% crash reduction factor.

Given

Crashes = 21 in 5 yr
ADT = 36,887 veh/day
Length = 3.5 mi
CRF = 0.35

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 36887 \times 365 \times 5 \times 3.5 = 2.356 e + 8 v e h - mi$
Substituting
$R = (21 \times 1 0^{8}) /2.356 e + 8 = 8.91 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$21 (1 - 0.35) = 13.65 cr a s h ese x p ec t e d$
Prevented
$21 - 13.65 = 7.35 cr a s h eso v er 5 y e a r s$

Answer: Rate = 8.9 crashes/100 MVM; 7.4 crashes prevented

Why the other options are there

6.00 (crashes per mile, no exposure)
36.8 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Example 9

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments (9)

A 5.5 mile segment with an ADT of 37,570 vehicles per day experienced 56 crashes in 5 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 30% crash reduction factor.

Given

Crashes = 56 in 5 yr
ADT = 37,570 veh/day
Length = 5.5 mi
CRF = 0.30

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 37570 \times 365 \times 5 \times 5.5 = 3.771 e + 8 v e h - mi$
Substituting
$R = (56 \times 1 0^{8}) /3.771 e + 8 = 14.85 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$56 (1 - 0.30) = 39.20 cr a s h ese x p ec t e d$
Prevented
$56 - 39.20 = 16.80 cr a s h eso v er 5 y e a r s$

Answer: Rate = 14.8 crashes/100 MVM; 16.8 crashes prevented

Why the other options are there

10.18 (crashes per mile, no exposure)
84.0 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Example 10

Crash rate for a roadway segment and the benefit of a countermeasure — Crash Rates for Roadway Segments (10)

A 3.0 mile segment with an ADT of 17,701 vehicles per day experienced 48 crashes in 3 years. Compute the crash rate per hundred million vehicle-miles, then estimate the crashes prevented by a countermeasure with a 35% crash reduction factor.

Given

Crashes = 48 in 3 yr
ADT = 17,701 veh/day
Length = 3.0 mi
CRF = 0.35

Find

Crash rate (per 100 MVM) and the crashes prevented

Start with the thinking

Segment crash rates are normalised by exposure: vehicle-miles of travel, not just by length.
A crash reduction factor multiplies the expected crashes at the treated location.

Step-by-step solution

Formula
$R = (A \times 100, 000, 000) / (A D T \times 365 \times N \times L)$
Exposure
$A D T \times 365 \times N \times L = 17701 \times 365 \times 3 \times 3.0 = 5.815 e + 7 v e h - mi$
Substituting
$R = (48 \times 1 0^{8}) /5.815 e + 7 = 82.55 cr a s h es p er 100 M V M$
Formula
$cr a s h es a f t er = cr a s h es \times (1 - C R F)$
Substituting
$48 (1 - 0.35) = 31.20 cr a s h ese x p ec t e d$
Prevented
$48 - 31.20 = 16.80 cr a s h eso v er 3 y e a r s$

Answer: Rate = 82.5 crashes/100 MVM; 16.8 crashes prevented

Why the other options are there

16.00 (crashes per mile, no exposure)
50.4 (years double-counted)

Reference: FE Reference Handbook — Transportation → Crash Rates for Roadway Segments

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a vertical or horizontal alignment, or a traffic stream, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Crash Rates for Roadway Segments contains 6 relations; you must be able to find this page in under 15 seconds.
Exam style: a curve geometry element or a capacity/flow relationship.
Unit rule: grades as decimals in curve formulas, percent in the stem.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

grades as decimals in curve formulas, percent in the stem
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.