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 Federal Highway Administration > Publications > Research Publications > 98108 > Chapter5.Cfm > Capacity Analysis of Pedestrian and Bicycle Facilities
 Publication Number: FHWA-RD-98-108 Date: February 1998

# Capacity Analysis of Pedestrian and Bicycle Facilities

Recommended Procedures for the "Bicycles" Chapter of the Highway Capacity Manual

## 5. EXAMPLE PROBLEMS

Example 1 - Uninterrupted, Exclusive Bicycle Path

For this example, the following is assumed:

• the bicycle path operates with two effective lanes;
• the path runs approximately north-south;
• peak-hour volume of 90 bicycles/h;
• peak-hour factor of 0.60; and
• a 70:30 directional split northbound:southbound.

The first step is to convert the peak-hour volume to a peak flow rate as follows:

1. Adjusted Peak-Hour Flow Rate = Peak-hour volume/peak-hour factor = 90/0.60 = 150 bikes/h
2. The total frequency of events and LOS for each direction is then computed using Equation  (which incorporates Equations  and ) and the computed peak flow rate of 150 bikes/h:

Ftotal= 0.5 [F meet] + [F pass]

= 0.5 [2{V bike-op}] + [0.188{V bike-sm}]

= 0.5 [2{(opp. dir. split)(peak flow rate)}] + [0.188{(same dir. split)(peak flow rate)}]

NORTHBOUND: F total= 0.5 [ 2{(0.30)(150)}] + [0.188{(0.70)(150)}] = 65 events/h

Using Table 1, this represents LOS Cfor the northbound direction.

SOUTHBOUND: F total= 0.5 [ 2{(0.70)(150)}] + [0.188{(0.30)(150)}] = 113 events/h

Using Table 1, this represents LOS Dfor the southbound direction.

Example 2 - Uninterrupted, Mixed-Use Path

For this example, the following is assumed:

• the bicycle path operates with three effective lanes;
• the path runs approximately east-west;
• adjusted peak-hour flow rate of 150 bicycles/h;
• adjusted peak-hour flow rate of 80 pedestrians/h;
• a 60:40 directional split of bicycles eastbound:westbound; and
• a 50:50 directional split of pedestrians eastbound:westbound.

The total frequency of events and LOS for each direction is then computed using Equation  which incorporates Equations  and ):

F total= 0.5 [F meet] + [F pass]

= 0.5 [5{V ped-op} + 2{V bike-op}] + [3{V ped-sm} + 0.188{V bike-sm}]

= 0.5 [5{(opp. dir. ped split)(ped peak flow rate)} + 2{(opp. dir. bike split)(bike peak flow rate)}] + [3{(same dir. ped split)(ped peak flow rate)}

+ 0.188{(same dir bike split)(bike peak flow rate)}]

EB: F total= 0.5 [5{(0.5)(80)} + 2{(0.4)(150)}] + [3{(0.5)(80)} + 0.188{(0.6)(150)}]

= 297 events/h

Interpolation between 100 and 200 bikes/h on Table 3 produces the same results. Using Table 4, this represents LOS Dfor the eastbound direction.

WB: F total= 0.5 [5{(0.5)(80)} + 2{(0.6)(150)}] + [3{(0.5)(80)} + 0.188{(0.4)(150)}]

= 321 events/h

Interpolation between 100 and 200 bikes/h on Table 3 produces the same results. Using Table 4, this represents LOS Efor the westbound direction.

Example 3 - On-Street Bicycle Lane

For this example, the following is assumed:

• a bicycle lane with allowance for passing;
• hourly bicycle volume of 150 bicycles/h, and a PHF of 0.75;
• heavy side friction characterized by large vehicle volume and high driveway density; and
• observed mean speed of 18 km/h and standard deviation of 4.5 km/h.

Since the standard deviation of speeds of 4.5 km/h is different from the default value of 3 km/h, equations  to  cannot be used. Table 5 must be used to predict the number of passing events and LOS.

First, the bicycle volume is converted to a peak flow rate as follows:

Bicycle flow rate = Hourly Volume/ PHF = 150/0.75= 200 bikes/h

Referring to Table 5, for 200 bikes/h, a mean speed of 18 km/h and standard deviation of 4.5 km/h, the predicted number of passing events is 113/h. This represents a LOS Don the facility.

For comparison purposes, if the default values were used (18 km/h, 3 km/h standard deviation), the predicted number of events would drop to 75 events/h. This would incorrectly represent LOS Con the facility.

Example 4 - Interrupted, Signalized Intersection For this example, the following is assumed:

• effective green time for movement in question = 20 s, cycle length = 50 s; and
• adjusted peak-hour flow rate is 120 bicycles/h for approach in question.

First, the capacity is computed using Equation , with an assumed bicycle saturation flow rate ( s bike) of 2000 bikes/h of green:

c bike= s bike(g/C) = 2000(20/50) = 800 bikes/h

The average signal delay is then computed using Equation :

d= 0.5C [1 - (g/C)] 2/ {1 - (g/C)[Min (V bike/ c bike, 1.0)]}

= 0.5(50) [ 1 - (20/50) ] 2/ { 1 - (20/50)[Min (120/800, 1.0)] } = 9.6 s

Using Table 6, this represents LOS B.

Example 5 - Combined Bicycle Facility

For this example, the following is assumed:

• the 2-km (1.2-mi) arterial contains four links and three signalized intersection nodes;
• the peak direction for this link during the peak hour is westbound;
• the four links are 0.5, 0.2, 1.0, and 0.3 km (0.31, 0.12, 0.62, and 0.19 mi) in length, respectively;
• the signalized intersections all have cycle lengths (C) of 100 s with g/C ratios of 0.3, 0.5 and 0.4 on the westbound approaches; and
• the adjusted peak-hour flow rate in the westbound direction is 600 bicycles/h.

The average delays for each of the intersections are computed using Equation  (which incorporates Equation ) and an assumed bicycle saturation flow rate ( s bike) of 2000 bikes/h of green:

d i = 0.5C [1 - (g/C)] 2/ {1 - (g/C)[Min (V bike/{ c bike}, 1.0)]}

= 0.5C [1 - (g/C)] 2/ {1 - (g/C)[Min (V bike/{( sbike)(g/C)}, 1.0)]}

d 1= 0.5 (100) [1 - (0.3) ] 2/ { 1 - (0.3) [Min (600/{(2000)(0.3)}, 1.0 ] } = 35.0 s

d 2= 0.5 (100) [1 - (0.5) ] 2/ { 1 - (0.5) [Min (600/{(2000)(0.5)}, 1.0 ] } = 17.9 s

d 3= 0.5 (100) [1 - (0.4) ] 2/ { 1 - (0.4) [Min (600/{(2000)(0.4)}, 1.0 ] } = 25.7 s

Using an average uninterrupted travel speed (as i) of 25 km/h (15.5 mi/h) for all links, the average travel speed for the arterial is computed using Equation 9: = 2 / [ { (0.5 + 0.2 + 1.0 + 0.3) / (25)} + { (35.0 + 17.9 + 25.7 ) / 3600 }] = 19.6 km/h

Using Table 7, this represents LOS B.