6.1
Determine Problem the hourly volume. BFFS := 65 fLW := 1.9 fLC := 0.80 fN := 0.0 fID := 0.0 FFS := BFFS - fLW - fLC - fN - fID vp := 1616 PT := 0.10 ET := 2.5 1 fHV := 1 + P T E T - 1 FFS = 62.3 (given) (Table 6.3) (Table 6.4) (Table 6.5) (Table 6.6)
(Eq. 6.2)
(Table 6.1, by interpolation) (given) (Table 6.8)
(
)
fHV = 0.87
(Eq. 6.5)
PHF := 0.90 V := vp PHF N fHV fp
fp := 1.0
N := 3 V = 3794
(given)
veh/h
(Eq. 6.3)
Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
Problem 6.3
Determine the maximum number of large trucks and buses. first, determine the heavy vehicle factor assume urban freeway BFFS := 70 fLW := 0.0 fLC := 0.0 fN := 4.5 fID := 0.0 FFS := BFFS - fLW - fLC - fN - fID PHF := 1800 700 4 N := 2 1800 PHF N fHV fp 1400 fHV PHF = 0.6429 FFS = 65.5 (given) (Table 6.3) (Table 6.4) (Table 6.5) (Table 6.6) (Eq. 6.2)
(Eq. 6.4) (given)
fp := 1.0 vp
(Eq. 6.3)
interpolate from Table 6.1 to find vp
vp := 1680 + ( FFS - 65) 1400 vp
1770 - 1680 70 - 65
vp = 1689
fHV :=
fHV = 0.8289
now, determine the number of trucks ET := 2.5 fHV
(rolling terrain)
(Table 6.7) 1 1 1 + 1.5PT
1
1 + P T E T - 1
(
)
1 + PT ( 2.5 - 1)
(Eq. 6.5)
1 + 1.5 PT
1 fHV
PT = 0.1376 1800 PT = 247.7143 therefore 247 trucks and buses
Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
Problem 6.5
Determine the driver population factor. PHF := 0.80 Solve for f HV ET := 2.0 ER := 3.0 Table 6.7 PT := 0.08 PR := 0.06 N := 3 V := 3900 (given)
1 fHV := 1 + P T E T - 1 + P R E R - 1
(
)
(
)
fHV = 0.833 vp := 2250
(Eq. 6.5)
Freeway is operating at capacity, so for FFS = 55 mi/h
(Table 6.1)
vp
V PHF fHV N fp
fp = 0.867
Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
Problem 6.6
Determine the new level of service. BFFS := 60 fLW := 1.9 fLC := 0.6 (Table 6.3) (Table 6.4) fN := 0.0 fID := 0.0 (Table 6.5) (Table 6.6) FFS = 57.5 mi/h (Eq. 6.2)
FFS := BFFS - fLW - fLC - fN - fID calculate heavy vehicle adjustment PT := 0.12 ET := 4.5 PR := 0.06 ER := 4.0
(Table 6.7)
1 fHV := 1 + P T E T - 1 + P R E R - 1
(
)
(
)
fHV = 0.625
(Eq. 6.5)
calculate LOS before and after lane addition N1 := 2 N2 := 3 = 2275 fp := 0.90 PHF := 0.88 vp := 2275 V := vp PHF fHV N1 fp
2300 + 2250 2 vp V
(Table 6.1 by interpolation)
PHF fHV N1 fp
(Eq. 6.3)
V = 2252.25 V vp2 := PHF fHV N2 fp vp2 = 1516.667 max service flow: 1495 pc/h/ln for LOS C 1965 pc/h/ln for LOS D Calculate FFS after lane addition fLC := 0.4 For vp of 1517, FFS := BFFS - fLW - fLC - fN - fID S := FFS FFS = 57.7 pc/h/ln (Eq. 6.3)
LOS D (Table 6.1, by interpolation)
D :=
vp2 S
D = 26.285
pc/mi/ln
LOS D - 26 < 26.3 < 35
(Eq. 6.6)
Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
Problem 6.7
Determine the LOS. find the analysis flow rate ( 1500 0.03) + ( 1000 0.04) 2500 2500 5280 = 0.473 mi = 0.034 (average grade)
PT := 0.05 ET := 2.0 1 fHV := 1 + P T E T - 1
(given) (Table 6.8)
(
)
fHV = 0.952 fp := 1.0 vp = 1166.667 V := 2000
(Eq. 6.5)
N := 2.0
PHF := 0.90
(given)
V vp := PHF N fHV fp determine FFS BFFS := 65 fLW := 0.0 fLC := 1.8 fN := 4.5 fID := 0.0
pc/h/ln
(Eq. 6.3)
(given) (Table 6.3) (Table 6.4) (Table 6.5) (Table 6.6) FFS = 58.7 (Eq. 6.2)
FFS := BFFS - fLW - fLC - fN - fID find level of service for vp = 1167, S = FFS D := vp FFS D = 19.88
pc/mi/ln therefore LOS is C
(Eq. 6.6)
Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
Determine density and level of service before and after the ban. Before: PT := 0.06 PB := 0.05 PTB = 0.11
Problem 6.9
(given)
PTB := PT + PB ETB := 2.5
(Table 6.7)
1 fHVTB := 1 + PTB ETB - 1
(
)
fHVTB = 0.858
(Eq. 6.5)
PHF := 0.95
fp := 1.0
N := 4 vp = 1655.526
V := 5400
(given)
V vp := PHF fHVTB fp N BFFS := 70 fLW := 1.9 fLC := 0.4 fN := 1.5 fID := 3.7
(Eq. 6.3)
(given) (Table 6.3) (Table 6.4) (Table 6.5) (Table 6.6) FFS = 62.5 (Eq. 6.2) (Figure 6.2)
FFS := BFFS - fLW - fLC - fN - fID S := 62.5 D := vp S D = 26.5
pc/mi/ln
LOS D
(Eq. 6.6)
After: Vnew := V 1 - PT
(
)
Vnew = 5076 NumBuses = 270
NumBuses := V PB
Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
PBnew := EB := 2.5
NumBuses Vnew PBnew = 0.053 (given)
1 fHVB := 1 + PBnew EB - 1
(
)
fHVB = 0.926
(Table 6.7) (Eq. 6.5)
vp := PHF fHVB fp N S := 62.5 D := vp S D = 23.1
Vnew
vp = 1442.368 (Eq. 6.3)
pc/mi/ln
LOS C
(Figure 6.2) (Eq. 6.6)
Solutions Manual to accompany of Principles Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
Problem 6.10
Determine the density, v/c ratio, and LOS before and after the strike. calculate free-flow speed BFFS := 65 fLW := 0.0 fLC := 0.0 fN := 3.0 fID := 5.0 FFS := BFFS - fLW - fLC - fN - fID calculate volume after bus strike V1 := 3800 PT := 0.02 PB := 0.04 PHF := 0.90 V2 = 4560 PTB := PT + PB (given) FFS = 57 mi/h (given) (Table 6.3) (Table 6.4) (Table 6.5) (Table 6.6) (Eq. 6.7)
V2 := V1 - PB V1 + 6 PB V1
(
)
calculate heavy vehicle adjustments before and after bus strike ETB := 4.0 ET := 5.0 (Table 6.8)
1 fHVTB := 1 + PTB ETB - 1
(
)
fHVTB = 0.847 V1 PT V2
(Eq. 6.5)
Calculate new P T 1 fHVT := 1 + P T E T - 1
PT :=
PT = 0.017
(
)
fHVT = 0.938
N := 3
fp := 1.0 vp1 = 1660.741
(given)
V1 vp1 := PHF N fHVTB fp
(Eq. 6.3)
Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
vp1 c
= 0.732
D :=
vp1 FFS LOS D
(Eq. 6.6)
D = 29.14
pc/mi/ln
V2 vp2 := PHF N fHVT fp vp2 c = 0.794 D := vp2 FFS
vp2 = 1801.481
(Eq. 6.3)
D = 31.6
pc/mi/ln
LOS D
(Eq. 6.6)
Solutions Manual to accompany Principles of Highway Engineering and Traffic Analysis, 4e, by Fred L. Mannering, Scott S. Washburn, and Walter P. Kilareski. Copyright 2008, by John Wiley & Sons, Inc. All rights reserved.
CEE 320 Fall 2010
Homework 3 Additional Problems
Prof. Goodchild
Homework 3 Solution Additional Problems:
1. Estimate the FFS from the loop detector data. Explain your method, any assumptions you made, and any limitations and advantages to calculating FFS this way.
FFS can be obtained directly from loop detector data; however, several approaches will be accepted: The average speed of vehicles when flow is less than 1300 vehicles per hour per lane (64 miles per hour). The free flow speed obtained in part 5 of homework 1, which estimates the intercept for the speed occupancy curve (64.2 miles per hour). Free flow speed should be close to 65 miles per hour. If another method was used it will need to be fully justified.
2. Calculate the Level of Service. Use the maximum hourly volume from this data (you identified this in Homework 1) and calculate the PHF for this hour from the available 20second volume data. Assume 10% of the traffic stream is trucks and busses, and the freeway runs over level terrain. The area is heavily used by commuters, and RVs can be neglected.
Calculate the analysis flow rate using the expression:
V is the peak hour volume obtained from the data provided. This occurs between 6:41 and 7:40 am. Total volume for the two lanes is 3800 vehicles, or 1900 vehicles per hour per lane. To determine peak hour factor, identify the 15 minute period within the peak hour with the highest volume. This occurs between 7:07 and 7:22 am, when volume is 519.5 vehicles per lane. PHF = 1900/(519.5*4)=0.914 fp=1, PT=0.1, PRV=0 Analysis flow rate is 2188 passenger cars per hour per lane. With FFS = 65 mph and vp=2188 pcphpl, Table 6.1 or Figure 6.2 can be used to identify LOS E. Average speed (s) under these conditions is approximately 57 miles per hour. Density which is vp/s is 2188/57 or approximately 38.39 passenger cars per mile per lane.
3. If traffic volume is expected to grow at a rate of 4% per year for the next 5 years, what will LOS be in 5 years (assuming no other changes)? The following equation can be used for estimating traffic volume after n years with an average yearly growth rate of i: Vn
= V (1 + i )
n
We do not expect FFS to change, so this will remain approximately 65 miles per hour. However, the analysis flow rate will change. N does not change, there are still 2 lanes in each direction. We also have no information that makes us believe fHV or fp will change. PHF does not change, as the volume in the numerator, and the denominator, will both grow at the same rate:
CEE 320 Fall 2010
Homework 3 Additional Problems
Prof. Goodchild
So we need only adjust for growth in the volume expected over the next 5 years. 3800*(1+0.04)^5 = 4624 vehicles/hour
So the analysis flow rate is now 2662 passenger cars per hour per lane. Referring to Table 6.1 or Figure 6.2, we see that this is clearly in LOS F.
4. If the AADT for this stretch of roadway is currently 35,000 vehicles per day, how many lanes should be added to ensure LOS A in 5 years? Use Figure 6.7 in the text to estimate K. Assume the FFS is the same you identified in Part (1).
Again, we assume FFS is fixed at 65 mph. There is no reason that this value would change. We know the data is for two lanes in the same direction, so D=1. A reasonable K value is 0.12, corresponding to the 30th highest hour used for planning purposes. We can use the growth rate equation as above to calculate what the current AADT is 35,000*(1+0.04)^5 = 42,583 veh/day. Using equation 6.16: DDHV = KxDxAADT = (0.12)(1)(42,583)=5109.94 vehicles per hour To achieve LOS A with FFS is 65 mph, the maximum service flow rate is 710 passenger cars per hour per lane (Table 6.1). So N=8 (round up, we can't build 7.97 lanes!). In addition to the two lanes that already exist, if traffic grows at an annual rate of 5 percent, and we'd like to guarantee freely flowing traffic, then we'd have to build an additional 6 lanes westbound on 520!