Traffic Density: What Does That Mean?
publictransit.us Special Report No. 7.2
 
 
Leroy W. Demery, Jr. • Michael D. Setty• September 27, 2007
 
Copyright 2005-2007 Publictransit.us
 
1) Introduction
The concept of “Traffic Density,” or simply “Density,” appears to be relatively unfamiliar in the U.S. but is very simple to grasp. Traffic density refers to the number of people who travel over each length of line or guideway, on average, during some interval.
For example, if an average of 5,000 people travel over each mile of a bus line during the average weekday, then that line serves (or carries) a “weekday traffic density” of 5,000.
Technically, a “weekday traffic density of 5,000” refers to “5,000 passenger-miles per route-mile per weekday.” Perhaps someday, someone will adopt a shorter name for this clumsy-sounding unit, but until then we’ll refer to “weekday traffic density” and “passenger-miles per mile of route per weekday” interchangeably. (We use “passenger-miles per mile of route” simply because it has a smoother sound.) Also, we note that “weekday traffic density” refers to the annual average, but constant reminders of this fact should not be necessary.
One may refer to  “one-way” as well as “two-way” traffic density. Either makes perfect sense, but we’ll use “two-way” density as the “standard.” This makes sense unless passenger travel patterns show a strong imbalance by direction.
One may also refer to traffic density with reference to some other interval of time besides the “weekday,” e.g. “daily traffic density” or “annual traffic density.” The “annual” statistic is useful for comparison among systems located in different countries, owing to differences in the temporal distribution of urban travel and lack of data regarding “weekday” ridership levels. We note, for example, that Monday-Friday ridership in Japan is not identical to Saturday and Sunday ridership, but “weekday” ridership data are scarce.
The authors note that traffic density has an apparent cachet of mystery in the U.S. We note in addition that the U.S. remains the world’s only industrialized nation where the metric system is not in common use. The two, we suspect, are not coincidental. (We will add that the major European industrial powers had all “gone metric” prior to World War I):
and that the English-speaking countries did not begin large scale metrification until 1965.) We believe that greater “metric literacy” would facilitate understanding of concepts such as traffic density. But we admit that traffic density is a remarkably friendly concept to the metrically challenged. In short, 5,000 passenger-miles per mile of route per weekday, converted to metric measures, gives the same number: 5,000 passenger-kilometers per kilometer of route  per weekday. The number remains the same given any unit of linear measure: e.g. 5,000 passenger-cubits per cubit of route.
The importance of the traffic density concept stems from what is being measured – work. A “passenger-mile” or “passenger-kilometer” (or “passenger-cubit”) is a unit of work, as “work” is defined in physics (“a force, moving through a distance”).
One could convert pass-mi or pass-km to “foot-pounds” or “joules” (the metric unit of “work”) if one wished – simply estimate the average “mass” per passenger. Using this estimate, one could even convert to the “freight” units: ton-miles or tonne-kilometers (here we use the spelling “tonne” to emphasize the reference is to metric tons).
It is worth remembering: “mass” and “weight” are not the same concept, a fact that we metrically-challenged Americans sometimes overlook. “Mass” is independent of the force of gravity, but “weight” arises because of the force of gravity. One would weigh about one-sixth as much on the moon as on earth, but one’s mass would not change.
“Traffic density” may therefore be described as a measure of work, performed over a given distance, within a given time interval.
Obviously, it takes more “work,” in the form of energy input, to move 5,000 passengers for an average distance of 5 miles than for an average distance of 0.5 mile. This very important point tends to get overlooked during early planning. In addition, other important planning issues arise as the result of traffic density.
 
2) Calculation
Calculation of traffic density requires three statistics: passenger count, line length and average travel distance. These are used in the following straightforward formula:
Traffic Density = (Passenger Count * Average Travel Distance) / Line Length.
The “passenger count” is the number of boardings during the interval of interest (e.g. annual, daily or weekday traffic density).
Obviously, “Passenger Count * Average Travel Distance” gives passenger-miles (or passenger-km). Therefore, one may simply divide the passenger-mile statistic by the line length over which this travel occurs.
Determining the number of passengers-miles (or pass-km) carried by a transit service is quite simple. All one needs to know is 1.) the number of passengers on board each vehicle, between each stop, and 2.) the distance between each stop. A little bit of math, will give the passenger-mile statistic. It is not necessary to keep track of where individual passengers board and alight.
Some transit operators can easily determine the number of passengers traveling between each pair of stations from fare gate data. Others use statistical sampling methods established by the Federal Transit Administration. Precision is very good given adherence to FTA guidelines and proper statistical methods.
Lack of data for passenger-miles or average travel distance is much less of a problem than one might think. Average travel distance can be estimated from other transit statistics. Reynolds (1971) demonstrates one technique for doing this, using data for total passengers, total revenue, and total revenue vehicle-miles. This was used to estimate an average bus transit trip length of 4.2 miles for the nine largest Canadian cities. The following examples use the statistics derived by Reynolds.
Average fare paid per passenger = total revenue / total passengers = $0.19.
Average revenue per bus-mile = total revenue / total revenue vehicle miles = $0.82.
Let:
Average bus occupancy (passenger-mile / vehicle-mile) = O.
Average fare per passenger-mile ($ / passenger-mile) = C.
Average travel distance (passenger-mile / passenger) = D.
Equation 1.): C * O = Average revenue per bus-mile.
($ / passenger-mile) * (passenger-mile / vehicle-mile) = $ / vehicle-mile.
Equation 2.): C * D = Average fare paid per passenger.
($ / passenger-mile) * (passenger-mile / passenger) = $ / passenger.
Based on the example presented by Reynolds (1971):
C * O = 0.82.
C * D = 0.19.
Strictly speaking, it is not possible to solve a system of three unknowns in two equations such as that above. However, the domains of the three variables are limited, as follows:
--Average bus occupancy O cannot be less than zero nor greater than the maximum number of people likely to travel on any single bus. As a practical matter, buses are of limited size and occupancy but vehicle occupancy cannot be “too small.”
--Average travel distance D must be greater than zero and less than the line length. In practical terms, people are more likely to walk for distance up to a mile than use a bus. However, cities and towns have finite size, and so average travel distance cannot be “too large.”
--Average fare per passenger-mile C must be greater than zero, and less than the average fare paid per passenger (because an average travel distance less than one mile is unlikely).
It is therefore possible to select the “most likely” values for average travel distance (D) and average bus occupancy (O). This process is illustrated in Table 1, below:
 
TABLE 1: Estimation of Average Travel Distance
Equation 1.): C * O = 0.82
Equation 2.): C * D = 0.19
Average fare per passenger-mile (C)
Average bus occupancy (O)
Average fare per passenger-mile (C)
Average travel distance (D)
$0.03
27.3
$0.03
6.3
$0.035
23.4
$0.035
5.4
$0.04
20.5
$0.04
4.7
$0.045
18.2
$0.045
4.2
$0.05
16.4
$0.05
3.8
$0.055
14.9
$0.055
3.5
$0.06
13.7
$0.06
3.2
$0.065
12.6
$0.065
2.9
An average fare per passenger-mile (C) of $0.05 implies 16.4 passenger-miles per vehicle-mile, which is low given the dominance of Montreal and Toronto. An average of $0.04 implies 20.5 passenger-miles per vehicle-mile, but the associated average travel distance, 4.7 miles, is high. The fare that implies the most likely values of O and D appears to be $0.045. These “most likely” values are 18 pass-mi per veh-mi and 4 miles, respectively.
The authors addressed uncertainty of estimation by rounding the “most likely” values to the nearest unit.
Reynolds’ technique provides a valuable tool for analysis of historic data.
 
3) Transit Capacity
Transit “capacity” is a function of traffic density, rather than “boardings” or “passengers.” In other words, the shorter the average travel distance, the greater the number of passengers that can be carried, all else equal. This fact was well known to a previous generation of transit professionals:
. . . allowance must be made for the difference in length of haul [i.e. average travel distance], as with a shorter haul more passengers per mile can be carried (Taylor 1913).
The amount of work that can be performed over a given length of line, during a given interval, is finite. There are well-defined limits to the number of vehicles that can be operated safely, or efficiently, over a given length of line during a given interval. Although less well defined, the number of people who will choose to occupy an enclosed space such as a transit vehicle is also limited. Therefore, the number of passenger-miles that a transit service can carry is limited; the number of passengers, at least in theory, is not. As illustrated below, the shorter the average travel distance (ATD), the greater the number of passengers that can be carried, all else equal.
In the hypothetical examples above, a 100-passenger vehicle operates between line terminals spaced one mile apart. We have attempted to keep things simple by considering only a single one-way trip, within a fixed interval of unspecified duration.
In Case A, 100 passengers board the vehicle and alight after one mile. The vehicle carries 100 passengers, and 100 passenger-miles; the ATD is one mile.
In Case B, 100 passengers board the vehicle and alight at an intermediate stop. An additional 100 passengers then board, and alight at the end of the line. The vehicle carries 200 passengers. However, the passenger-mile count remains unchanged because the average number of persons aboard between terminals does not change. In other words, 200 passengers travel aboard the vehicle for an average distance of 0.5 mile, rather than 1.0 mile as in Case A.
In Case C, 100 passengers board the vehicle and alight at the first of two intermediate stops. Another 100 customers then board, and alight at the second intermediate stop. A third group of 100 passengers then boards, and rides the end of the line. The vehicle carries 300 passengers, but the passenger-mile count again remains unchanged. The ATD is 1/3 (0.33) mile.
One could continue this exercise by adding stations. Although the passenger-mile count would not change, the number of passengers could in theory become very large as ATD became very small. An important inference from Taylor (1913) is that corridors with relatively higher ATD require more service than one with relatively lower ATD. This may be demonstrated mathematically as follows:
A.) Traffic density = Service effectiveness * Service density
where:
ATD = average travel distance (e.g. miles or kilometers), with route length expressed in the same unit.
Traffic density = (passenger boardings * ATD) / route length, relative to some interval.
Service effectiveness = passenger-miles [or km] / vehicle-miles [or km].
Service density = vehicle-miles / route length, relative to the same interval as traffic density.
“Service effectiveness” may also be interpreted as “vehicle occupancy,” the average number of passengers aboard each vehicle when in service. Equation A.) above implies that a higher ATD will lead to a larger number of people aboard each vehicle, all else equal – including the number of boarding passengers.
A transit facility that carries a large number of boarding passengers can use relatively small vehicles if ATD is sufficiently low (and boardings are distributed along the line rather than concentrated in a few locations). For example, given a very low ATD, a rail line could use vehicles as small as the “Birney Safety Car,” which was typically 26 feet (8 meters) in length. However, the same line, with the same number of weekday boardings might require 90-foot (27-meter) articulated vehicles given a sufficiently high ATD.
 
References
Reynolds, D. J. 1971. Research Monograph No. 3: The Urban Transport Problem in Canada, 1970-2000. (Prepared for the Honourable R. K. Andras, Minister Responsible for Housing, Government of Canada.) Ottawa.
Taylor, Abraham Merritt. 1913. Report of Transit Commissioner, Volume 1. Philadelphia: City of Philadelphia.