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Publication Number: FHWARD98108
Date: February 1998 
Capacity Analysis of Pedestrian and Bicycle FacilitiesRecommended Procedures for the "Bicycles" Chapter of the Highway Capacity Manual
5. EXAMPLE PROBLEMSExample 1  Uninterrupted, Exclusive Bicycle Path For this example, the following is assumed:
The first step is to convert the peakhour volume to a peak flow rate as follows:
F_{total}= 0.5 [F _{meet}] + [F _{pass}] = 0.5 [2{V _{bikeop}}] + [0.188{V _{bikesm}}] = 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, MixedUse Path For this example, the following is assumed:
The total frequency of events and LOS for each direction is then computed using Equation [6] which incorporates Equations [4] and [5]): F _{total}= 0.5 [F _{meet}] + [F _{pass}] = 0.5 [5{V _{pedop}} + 2{V _{bikeop}}] + [3{V _{pedsm}} + 0.188{V _{bikesm}}] = 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  OnStreet Bicycle Lane For this example, the following is assumed:
Since the standard deviation of speeds of 4.5 km/h is different from the default value of 3 km/h, equations [1] to [3] 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:
First, the capacity is computed using Equation [7], 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 [8]: 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 average delays for each of the intersections are computed using Equation [8] (which incorporates Equation [7]) 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}/{( s_{bike})(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. 