HCM UNSIGNALIZED SAMPLE PROBLEM

Highway Capacity Manual 2000

17-49 Chapter 17 - Unsignalized IntersectionsExample Problems

PART D. EXAMPLE PROBLEMS

Problem No. Description Application

1 TWSC Unsignalized Intersection Operational LOS2 TWSC Unsignalized Intersection Operational LOS3 TWSC Unsignalized Intersection Operational LOS4 AWSC Unsignalized Intersection Operational LOS5 AWSC Unsignalized Intersection Operational LOS6 Roundabout Capacity and v/c ratio


Chapter 17 - Unsignalized Intersections 17-50Example Problems

EXAMPLE PROBLEM 1

The Intersection A TWSC T-intersection with an exclusive westbound left-turn lane.

The Question What are the delay and level of service?

The Facts√ Two-lane major street, √ 10 percent HV,√ Two-lane minor street, √ No special intersection geometry, and√ Level grade, √ No pedestrians.√ Stop-controlled on minor street

approach,

Outline to Solution The steps below show the northbound approach calculations only.Calculations for other approaches are shown on the worksheets.

Steps1. Data input. Worksheets 1 and 2

2. Site characteristics. Worksheet 3 - lane designation, grade, right-turn channelization, and arrival type

3. tc and tf (use Equations 17-1 and17-2 and Exhibit 17-5 ).

tc,x = tc,base + tc,HV PHV + tc,G G – tc, T – t3,LT

tf,x = tf,base + tf,HV PHV

tc,4 = 4.1 + 1.0(0.10) + 0 – 0 – 0 = 4.200 s

tf,x = 2.2 + 0.9(0.10) = 2.290 s

4. Skip Worksheets 5a through 5e. No upstream signals within 0.4 km

5. Movement capacity cm,x accountingfor impedance (use Equation 17-4 ). vc,9 =

v2N + 0.5v3 + v14 + v15

vc,9 = 250 + 20 + 0 + 0 = 270 veh/h

cp,x = vc,xe−vc,xtc,x /3,600

1− e−vc,xtf,x /3,600

cp,9 = 270 *e−270*6.300/3,600

1− e−270*3.390/3,600 = 750 veh/h

cm,9 = cp,9 * Pp,9 = 750 (1) = 750 veh/h

P0,i = 1− vi

cm,i

P0,9 = 1− 120750

= 0.840

6. Skip Worksheets 7a and 7b. No two-stage gap acceptance

7. Shared-lane capacity (use Equation17-15).

Worksheet 8 - Movements 7 and 9 share thesame lane

cSH =vy

y∑

vy

cm,yy∑

cSH(NB) =40 +12040

274+ 120

750

=523 veh/h

8. Skip Worksheet 9. No flared minor-street approach



9. Control delay and LOS (useEquation 17-38 and Exhibit 17-2).

Worksheet 10

d = 3,600cm,x

+ 900T[... ] + 5

dNB =3,600523

+ 900 0.25( )[... ] + 5 = 14.9 s,

LOS B

10. Skip Worksheet 11. No Rank 1 vehicle delay

Example Problem 1

STOP

STOP

v1v2v3

v12

v7

v10

v9

v13

v14v15

v16v6

v4S2

S5

v11

v8

v5

TWSC - UNSIGNALIZED INTERSECTIONS WORKSHEET

Worksheet 1

Worksheet 2

Vehicle Volumes and Adjustments

Pedestrian Volumes and Adjustments


Movement 1 2 3 4 5 6 7 8 9 10 11 12

Volume (veh/h)

Peak-hour factor, PHF

Hourly flow rate (veh/h)

Proportion of heavy vehicles, PHV

Movement 13 14 15 16

Flow, Vx (ped/h)

Lane width, w (m)

Walking speed,1 Sp (m/s)

Percent blockage, fp (Equation 17-11)

1. Default walking speed = 1.2 m/s

General Information Site Information

Analyst ______________________ Intersection ________________________Agency or Company ______________________ Jurisdiction ________________________Date Performed ______________________ Analysis Year ________________________Analysis Time Period ______________________

Geometrics and Movements

Length of study period = _____ h

grade = ______

grade = ______

grade = ______

grade = ______

Show NorthStre

et

Street

KMN Jones/MarketCEI Latah County5/17/99 1999AM Peak

Jone

s

Market

25040

300150

40 120

250 40 150 300 40 120

1.00 1.00 1.00 1.00 1.00 1.00

250 40 150 300 40 120

0.10 0.10 0.10 0.10 0.10 0.10

0%

0.25

0%

0%

0 0 0 0




Worksheet 3

Movement 9 " Yes " No Storage space, n ________________(number of vehicles)


Type

Movements 7 and 8 " Yes _____ " No Storage space, m ________________(number of vehicles)


* Includes raised or striped median (RM), or two-way left-turn lane (TWLTL)

Lane Designation

Movements Lane 1 Lane 2 Lane 3 Grade, G Right TurnChannelized?

1, 2, 3

4, 5, 6

7, 8, 9

10, 11, 12

Flared Minor-Street Approach

Median Storage*

Computing Delay to Major-Street Vehicles

Upstream Signals

Data for Computing Effect of Delay to Major-Street Vehicles S2 Approach S5 Approach

Shared-lane volume, major-street through vehicles, vi1, blocked by LT

Shared-lane volume, major-street right-turn vehicles, vi2, blocked by LT

Saturation flow rate, major-street through vehicles, si1

Saturation flow rate, major-street right-turn vehicles, si2

Number of major-street through lanes

Length of study period, T (h)

Movements Distance to Prog Speed, Cycle Green Time, Arrival Saturation Flow Progressed Flow,Signal, D (m) Sprog (km/h) Length, C (s) geff (s) Type Rate, s (veh/h) Vprog (veh/h)

protected LT 3

TH

protected LT 3

TH

S2

S5

General Information

Project Description _________________________________________________________________________________

Example Problem 1

2, 3 0 N

4 5 0 N

7, 9 0 N

X

X

X

X

Example Problem 1



tf = tf,base + tf,HV PHV

tc = tc,base + tc,HV PHV + tc,G G - tc,T - t3,LT

Major LT Minor RT Minor TH Minor LT

Movement 1 4 9 12 8 11 7 10

tf,base (Exhibit 17-5)

tf,HV

PHV (from Worksheet 2)

tf (Equation 17-2)

Worksheet 5a

Time to Clear Standing Queue (Computation 1)

Movement 2 Movement 5vT,prog vL,prot vT,prog vL,prot

Effective green, geff (s)

Cycle length, C (s)

Saturation flow rate, s (veh/h)

Arrival type 3 3

vprog (veh/h)

Rp (from Chapter 16) 1.00 1.00

Proportion of vehicles arrivingon green, P (Equation 17-17)

gq1 (Equation 17-18)


gq (Equation 17-20)

Major LT Minor RT Minor TH Minor LTMovement 1 4 9 12 8 11 7 10

tc,base (Exhibit 17-5)

tc,HV


tc,G - - 0.1 0.1 0.2 0.2 0.2 0.2

G (from Worksheet 3)t3,LT

single stage

two stage

single stage

two stage

tc,T

tc (Equation 17-1)


Worksheet 4

Critical Gap and Follow-Up Time

General Information


Example Problem 1

4.1 6.2 7.11.0 1.0 1.00.10 0.10 0.10

0 0 00.0 0.0 0.70.0 0.0 0.0

4.200 6.300 6.500

2.2 3.3 3.50.9 0.9 0.90.10 0.10 0.102.290 3.390 3.590

Example Problem 1



7

7

7

10

10

10

0,4 0,1


Worksheet 6

Step 1: RT from Minor Street v9 v12

Conflicting flows (Exhibit 17-4) vc,9 = vc,12 =

Potential capacity (Equation 17-3 or 17-29) cp,9 = cp,12 =

Ped impedance factor (Equation 17-12) pp,9 = pp,12 =

Movement capacity (Equation 17-4) cm,9 = cp,9 pp,9 = cm,12 = cp,12 pp,12 =

Prob of queue-free state (Equation 17-5) p0,9 = p0,12 =

Step 2: LT from Major Street v4 v1




Movement capacity (Equation (17-4) cm,4 = cp,4 pp,4 = cm,1 = cp,1 pp,1 =


Major left shared lane prob of queue-free state (Equation 17-16) p* = p* =

Step 3: TH from Minor Street (4-leg intersections only) v8 v11




Capacity adjustment factor due to impeding movement f8 = p0,4 p0,1 pp,8 = f11 = p0,4 p0,1 pp,11 =(shared lane use p*) (Equation 17-13)

Movement capacity (Equation 17-7) cm,8 = cp,8 f8 = cm,11 = cp,11 f11 =

Prob of queue-free state p0,8 = p0,11 =

Step 4: LT from Minor Street (4-leg intersections only) v7 v10




Major left, minor through impedance factor p'' = p0,11 f11 = p'' = p0,8 f8 =

Major left, minor through adjusted impedance factor p' = p'

=(Equation 17-8)

Capacity adjustment factor due to impeding movements f7 = p' p0,12 pp,7 = f10 = p' p0,9 pp,10 =

(Equation 17-14)

Movement capacity (Equation 17-10) cm,7 = f7 cp,7 = cm,10 = f10 cp,10 =

Step 5: LT from Minor Street (T-intersections only) v7 v10






Impedance and Capacity Calculation

General Information


Notes

1. For 4-leg intersections use Steps 1, 2, 3, and 4.2. For T-intersections use Steps 1, 2, and 5.

2707501.000

7500.840 1.000

29012271.000

12270.878 1.000

8703121.000

0.878

274

Example Problem 1

Example Problem 1



Example Problem 1


Worksheet 8

Movement v (veh/h) cm (veh/h) cSH (veh/h)

7

8

9

10

11

12

Worksheet 9

Movement 7 Movement 8 Movement 9 Movement 10 Movement 11 Movement 12

csep (from Worksheet 6 or 7)

Volume (from Worksheet 2)

Delay (Equation 17-38)

Qsep (Equation 17-34)

Qsep + 1

Round (Qsep + 1)

nmax (Equation 17-35)

cSH

csep

n

cact (Equation 17-36)

Effect of Flared Minor-Street Approaches

Shared-Lane Capacity

General Information


(Equation 17-15)cSH =

∑vy

∑

y

yvy

cm,y( )

40 274

523

120 750

Example Problem 1



Example Problem 1

*


Worksheet 10

Delay to Rank 1 Vehicles

Worksheet 11

Control Delay, Queue Length, Level of Service

S2 Approach S5 Approach

p0,j (Equation 17-5)

vi1, volume for Stream 2 or 5


si1, saturation flow rate for Stream 2 or 5



dmajor left, delay for Stream 1 or 4

N, number of major-street through lanes

dRank 1, delay for Stream 2 or 5 (Equation 17-39)

Lane v (veh/h) cm (veh/h) v/c Queue Length Control Delay LOS Delay and LOS(Equation 17-37) (Equation 17-38) (Exhibit 17-2)

1

2

3

1

2

3

Movement v (veh/h) cm (veh/h) v/c Queue Length Control Delay LOS(Equation 17-37) (Equation 17-38) (Exhibit 17-2)

1

4

General Information


121110

121110

121110

987

987

987

p0,1 = p0,4 =

p* = p* =0, 1 0, 4

Example Problem 1

160 523 0.306 < 2 14.9 B 14.9B

X X X

150 1227 0.12 < 1 8.3 A



EXAMPLE PROBLEM 2

The Intersection A TWSC intersection with upstream signals. The major street isWalnut St. (EB/WB) and the minor street is Elm St. (NB/SB).

The Question What are the delay and level of service of the minor-street approaches?

The Facts√ Four-lane major street, √ 10 percent HV,√ Two-lane minor street, √ Upstream signals in both major-street directions, and√ Level grade, √ No pedestrians.


Steps1. Data input. Worksheets 1 and 22. Site characteristics. Worksheet 3 - lane designation, grades, right-turn

channelization, and upstream signals3. tc and tf (s) (use

Equations 17-1 and 17-2 and Exhibit 17-5).

tc,x = tc,base + tc,HV PHV + tc,G G – tc, T – t3,LT

tc,1 = 4.1 + 2.0(0.10) + 0 – 0 – 0 = 4.300 s

tf,x = tf,base + tf,HV PHV

tf,x = 2.2 + 1.0(0.10) = 2.300 s

4. Effect of upstreamsignals

Worksheets 5a through 5e

Proportion of vehiclesarriving during green, P(use Equation 17-17).

P = Rp

geffC

P = 0.33 3080 = 0.124

Discharge time (useEquations 17-18, 17-19, and 17-20 ).

For Movement 2gq = gq1 + gq2

gq1 =vprogC(1−P)

s= 250(80)(1− 0.124)

3,600= 4.867

gq2 =vprogCPgq1

sgeff − vprogCP= 250(80)(0.124)(4.867)

3,600(30) − 250(80)(0.124)= 0.114

gq = 4.867 + 0.114 = 4.981

Maximum platoon flowrate, vc,max (useEquation 17-21).

vc,max = sf[1 – (1 – F)gq]

vc,max = 3,600(0.751)[1 – (1 – 0.253)4.981] = 2,071 veh/h

Minimum platoon flowrate, vc,min.

vc,min = 1,000(2) = 2,000

Duration of blockedperiod, tp,i (useEquation 17-22).

tp,i = gq −

ln 1−vc,min

sf

vc,max − vprogRpf

vc,min − vprogRpf

ln(1−F)

tp,2 = 4.981−ln 1− 2,000

3,600(0.751)

2,071−250(0.33)(0.751)2,000 −250(0.33)(0.751)

ln(1−0.253)=

0.489

Proportion of timeblocked (use Equation17-23).

pi =tP,T + tP,L

C

p2 = 0.489

80 = 0.006



4. (continued) Dominantand subordinateplatoons (useEquations 17-24 and17-25).

pdom = max(p2, p5)

psubo = min(p2, p5)

pdom = P2 = 0.006

psubo = P5 = 0.000

Conflicting flow duringunblocked period (useEquation 17-28).

vc,u,x =vc,x − s(1− px )

px

vc,u,1 =400 − 3,600(1−1.000)

1.000=400 veh/h

Capacity duringunblocked period, cr,x(use Equation 17-3).

cr,x = vc,u,xe−vc,u,xtc,x /3,600

1− e−vc,u,xtf,x /3,600

cr,1 = 400e−400(4.300)/3,600

1− e−400(2.300)/3,600 =1,100 veh/h

Potential capacity,cplat,x, accounting forplatooning (useEquation 17-29).

cplat,x = px cr,x

cplat,1 = 1.000(1,100) = 1,100 veh/h

5. Movement capacity,cm,x, accounting forimpedance (useEquation 17-4, 17-7, or17-10).

cm,x = cplat,x fxcm,1 = 1,100(1.000) = 1,100 veh/h

6. Skip Worksheets 7aand 7b.

No two-stage gap acceptance

7. Shared-lane capacityWorksheet 8 (useEquation 17-15). cSH =

vyy∑

vy

cm,yy∑

cSH(NB) =44 +132 + 55

44202

+ 132254

+ 55867

=288 veh/h

8. Skip Worksheet 9. No flared minor-street approaches9. Control delay and LOS

(use Equation 17-38and Exhibit 17-2).

d = 3,600cm,x

+ 900T[... ] + 5

dNB =3,600288

+ 900 0.25( )[... ] + 5 = 53.5 s , LOS F

10. Skip Worksheet 11. No Rank 1 vehicle delay



Example Problem 2

STOP

STOP

v1v2v3

v12

v7

v10

v9

v13

v14v15

v16v6

v4S2

S5

v11

v8

v5


Worksheet 1

Worksheet 2




Movement 1 2 3 4 5 6 7 8 9 10 11 12

Volume (veh/h)





Flow, Vx (ped/h)

Lane width, w (m)







Length of study period = _____ h grade = ______

grade = ______

grade = ______

grade = ______

Show NorthStre

et

Street

JME Walnut/ElmCEI Latah County5/7/99 1999AM Peak

28 110 11

100

300

66

44 132 55

33

250

50

33 250 50 66 300 100 44 132 55 11 110 28

1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000

33 250 50 66 300 100 44 132 55 11 110 28

0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10

0 0 0 0

Walnut

Elm

0.25

0%

0%

0%

0%



Example Problem 2


Worksheet 3



Type




Lane Designation


1, 2, 3

4, 5, 6

7, 8, 9

10, 11, 12


Median Storage*


Upstream Signals









protected LT 3

TH

protected LT 3

TH

S2

S5

General Information


1 2 2, 3 0 N

4 5 5, 6 0 N

7, 8, 9 0 N

10, 11, 12 0 N

X

X

X

X

135 55 80.0 30.0 1 3600 250

200 50 70.0 20.0 1 3600 250

Example Problem 2



Example Problem 2


Worksheet 4





Movement 1 4 9 12 8 11 7 10


tf,HV


tf (Equation 17-2)

tc,T

tc (Equation 17-1)

Worksheet 5a




Cycle length, C (s)


Arrival type 3 3

vprog (veh/h)





gq (Equation 17-20)

General Information




tc,HV


tc,G - - 0.1 0.1 0.2 0.2 0.2 0.2


single stage

two stage

single stage

two stage

4.1 4.1 6.9 6.9 6.5 6.5 7.5 7.52.0 2.0 2.0 2.0 2.0 2.0 2.0 2.00.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10

0 0 0 0 0 0 0 00.0 0.0 0.0 0.0 0.0 0.0 0.0 0.00.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0- - - - - - - -4.300 4.300 7.100 7.100 6.700 6.700 7.700 7.700- - - - - - - -

2.2 2.2 3.3 3.3 4.0 4.0 3.5 3.51.0 1.0 1.0 1.0 1.0 1.0 1.0 1.00.10 0.10 0.10 0.10 0.10 0.10 0.10 0.102.300 2.300 3.400 3.400 4.100 4.100 3.600 3.600

30.0 20.080.0 70.03600 36001 1

250 2500.33 0.330.124 0.094

4.867 4.4040.114 0.103

4.981 4.507

Example Problem 2



Example Problem 2


Proportion of Time TWSC Intersection Is Blocked (Computation 2)

Worksheet 5b

Movement 2 Movement 5

vT,prog vL,prot vT,prog vL,prot

α (Exhibit 17-13)

β = (1 + α)-1

ta = D/Sprog (s)

F = (1 + αβta)-1

f = vprog/Vc ≥ 0

vc,Max (Equation 17-21)

vc,Min = 1000N

tp (Equation 17-22)

p (Equation 17-23)

Worksheet 5c

Platoon Event Periods (Computation 3)

General Information


p2 (from Worksheet 5b)

p5 (from Worksheet 5b)

pdom (Equation 17-24)

psubo (Equation 17-25)

Constrained or unconstrained (Equation 17-26, 17-27)

Proportion for Minor Movements, px

Single-Stage Two-Stage

(Exhibit 17-16) Stage I Stage II

p1

p4

p7 1 - p2 1 - p5

p8 1 - p2 1 - p5

p9

p10 1 - p5 1 - p2

p11 1 - p5 1 - p2

p12

0.50 0.50

0.667 0.667

8.836 14.400

0.253 0.172

0.751 0.536

2071 1105

2000 2000

0.489 0.000

0.006 0.000

0.006

0.000

0.006

0.000

Unconstrained

1.000

0.994

0.994

0.994

0.994

0.994

0.994

1.000

Example Problem 2




Worksheet 5d

Worksheet 5e

Conflicting Flows During Unblocked Period (Computation 4)

Two-Stage

Movements 7 8 10 11

Stage I Stage II Stage I Stage II Stage I Stage II Stage I Stage II

px (from Worksheet 5c)

cr,x (Equation 17-3)

cplat,x (Equation 17-29)

Single-Stage

Movements 1 4 7 8 9 10 11 12

vc,x (Exhibit 17-4)

s (veh/h)


vc,u,x (Equation 17-28)

Single-Stage

Movements 1 4 7 8 9 10 11 12


cr,x (Equation 17-3)

cplat,x (Equation 17-29)

Capacity During Unblocked Period (Computation 5)

Two-Stage

Movements 7 8 10 11

Stage I Stage II Stage I Stage II Stage I Stage II Stage I Stage II

vc,x (Exhibit 17-4)

s (veh/h)


vc,u,x (Equation 17-28)

General Information


Example Problem 2

400 300 678 873 150 739 848 200

3600 3600 3600 3600 3600 3600 3600 3600

1.000 0.994 0.994 0.994 0.994 0.994 0.994 1.000

400 280 660 857 129 722 831 200

1.000 0.994 0.994 0.994 0.994 0.994 0.994 1.000

1100 1223 333 279 872 300 289 783

1100 1216 331 277 867 298 287 783

Example Problem 2



7

7

7

10

10

10

0,4 0,1


Worksheet 6



























=(Equation 17-8)


(Equation 17-14)









General Information


Notes


150 200867 7831.000 1.000

867 783

0.937 0.964

300 400

1216 11001.000 1.000

1216 1100

0.946 0.970

873 848277 2871.000 1.000

0.917 0.917

254 2630.480 0.582

678 739331 298

1.000 1.0000.534 0.440

0.634 0.556

0.611 0.521

202 155

Example Problem 2

Example Problem 2



(Equation 17-15)vy

cm,y

cSH =

( )

∑vy

∑

y

y


Worksheet 8


7

8

9

10

11

12

Worksheet 9






Qsep + 1

Round (Qsep + 1)


cSH

csep

n




General Information


Example Problem 2

44 202

132 254 288

55 867

11 155

110 263 284

28 783

Example Problem 2




Worksheet 10


Worksheet 11













1

2

3

1

2

3


1

4

General Information


Example Problem 2

121110

121110

121110

987

987

987

p0,1 = p0,4 =

p* = p* =0, 1 0, 4*

Example Problem 2

231 288 0.802 6 53.5 F 53.5F

149 284 0.525 3 30.9 D 30.9D

X X X

33 1100 0.030 - 8.4 A

66 1216 0.054 - 8.1 A

X X X



EXAMPLE PROBLEM 3

The Intersection A TWSC intersection with flared approaches and a median storage.The major street is Walnut St. (EB/WB), and the minor street is Elm St. (NB/SB).

The Question What are the delay and level of service of the minor-street approaches?

The Facts√ Four-lane major street, √ 10 percent HV,√ Two-lane minor street, √ Flared approach, storage for one vehicle, and√ Level grade, √ Median storage, storage for two vehicles.√ No pedestrians,


Steps1. Data input. Worksheets 1 and 22. Site characteristics. Worksheet 3 - lane distribution, grades, right-turn

channelization, flared minor-street approach, andmedian storage

3. tc and tf (use Equations 17-1 and17-2).

Worksheet 4 - Movement 1tc = tc,base + tc,HV PHV +...

tc = 4.1 + 2.0(0.10) + 0.0 – 0.0 – 0.0 = 4.300 s


tf = 2.2 + 1.0(0.10) = 2.300 s

4. Skip Worksheets 5a through 5e. No upstream signals within 0.4 km5. Movement capacity, cm,x, for

minor RT and major LTmovements accounting forimpedance (use Equations 17-3,17-4, 17-5, and 17-12).

Worksheet 6 - Movement 9

vc,9 = v2N + 0.5v3 + v14 + v15

vc,9 = 2502 + 0.5(50) + 0 + 0 = 150 veh/h

cp,9 = vc,9e−vc,9tc,9 /3,600

1− e−vc,9tf,9 /3,600

cp,9 = 150e−150(7.1)/3,600

1− e−150(3.4)/3,600 =845 veh/h

pp,9 = 1.0 – fp = 1.0 – 0.000 = 1.000

cm,9 = cp,9 pp,9 = (845)(1.000) = 845 veh/h

p0,9 = 1 – v9

cm,9 = 1 –

55845 = 0.935

6. Movement capacities, cm,x, forminor TH and minor LTmovements, accounting forimpedance and two-stage gapprocess (use Equations 17-32,17-33, or 17-7 and Exhibit 17-4).

Worksheets 7a and 7bvc,I,8 = 341 veh/h

vc,II,8 = 532 veh/h

vc,8 = 873 veh/h

Single Stagecm,8 = cp,8 f8 = 273(0.917) = 250 veh/h

Two-Stage

cT =α

ym+1 −1y(ym −1)...[ ]

cT =0.949

1.8082+1 −11.808(1.8082 −1)...[ ] =390 veh/h



7. Shared-lane capacity (useEquation 17-15).

Worksheet 8

cSH =vy

y∑

vy

cmy∑

cSH (NB) = 44 +132 + 5544370

+ 132390

+ 55845

=442 veh/h

8. Flared approach capacity (useEquations 17-36 and 17-38).

Worksheet 9 - Movement 7

d = 3,600cm,x

+ 900T ...[ ] + 5

d = 3,600369

+ 900 0.25( ) ...[ ] + 5 = 16.070 s

Qsep =d7v7

3,600=

16.0( ) 44( )3,600

= 0.196

cact (NB) = csepi∑ − cSH

nnMax

+ cSH

cact = (1,605 – 442) 12 + 442 = 1,024 veh/h

9. Control delay and LOS (useEquation 17-38 and Exhibit17-2).

Worksheet 10

dNB = 3,600cm,x

+ 900T[... ] + 5

dNB = 3,6001,024

+ 900 0.05( )[... ] + 5 = 9.5 s

LOS A10. Skip Worksheet 11. No Rank 1 vehicle delay



Example Problem 3

STOP

STOP

v1v2v3

v12

v7

v10

v9

v13

v14v15

v16v6

v4S2

S5

v11

v8

v5


Worksheet 1

Worksheet 2




Movement 1 2 3 4 5 6 7 8 9 10 11 12

Volume (veh/h)





Flow, Vx (ped/h)

Lane width, w (m)







Length of study period = _____ h grade = ______

grade = ______

grade = ______

grade = ______

Show NorthStre

et

Street

JME Walnut/ElmCEI Latah County5/7/99 1999AM Peak

28 110

132 55

33

250

50

33 250 50 66 300 100 44 132 55 11 110 28

1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00

33 250 50 66 300 100 44 132 55 11 110 28

0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10

0 0 0 0

Walnut

Elm

0%

0%

0%

0%0.25

100300

6644

11




Worksheet 3



Type




Lane Designation


1, 2, 3

4, 5, 6

7, 8, 9

10, 11, 12


Median Storage*


Upstream Signals









protected LT 3

TH

protected LT 3

TH

S2

S5

General Information


Example Problem 3

1 2 2, 3 0 N

4 5 5, 6 0 N

7, 8, 9 0 N

10, 11, 12 0 N

X

X

X

X

1

1

2

2

RM

RM

Example Problem 3

2.2 2.2 3.3 3.3 4.0 4.0 3.5 3.51.0 1.0 1.0 1.0 1.0 1.0 1.0 1.00.10 0.10 0.10 0.10 0.10 0.10 0.10 0.102.300 2.300 3.400 3.400 4.100 4.100 3.600 3.600




Worksheet 4




tc,T

tc (Equation 17-1)

Worksheet 5a




Cycle length, C (s)


Arrival type 3 3

vprog (veh/h)





gq (Equation 17-20)

General Information




tc,HV


tc,G - - 0.1 0.1 0.2 0.2 0.2 0.2


single stage

two stage

single stage

two stage

Example Problem 3


Movement 1 4 9 12 8 11 7 10


tf,HV


tf (Equation 17-2)

4.1 4.1 6.9 6.9 6.5 6.5 7.5 7.52.0 2.0 2.0 2.0 2.0 2.0 2.0 2.00.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10

0 0 0 0 0 0 0 00.0 0.0 0.0 0.0 0.0 0.0 0.0 0.00.0 0.0 0.0 0.0 0.0 0.0 0.0 0 .0- - - - 1.0 1.0 1.0 1.04.300 4.300 7.100 7.100 6.700 6.700 7.700 7.700- - - - 5.700 5.700 6.700 6.700

Example Problem 3



7

7

7

10

10

10

0,4 0,1


Worksheet 6



























=(Equation 17-8)


(Equation 17-14)









General Information


Notes


Example Problem 3

150 200845 7831.000 1.000

845 7830.935 0.964

300 4001202 11001.000 1.000

1202 11000.945 0.970

Example Problem 3




Worksheet 7a

Effect of Two-Stage Gap Acceptance

Step 3: TH from Minor Street v8 v11

Part I - First Stage

Conflicting flows (Exhibit 17-4) vc,I,8 = vc,I,11 =

Potential capacity (Equation 17-3 or 17-29) cp,I,8 = cp,I,11 =

Ped impedance factor (Equation 17-12) pp,I,8 = pp,I,11 =

Capacity adjustment factor due to impeding movement fI,8 = p0,1 pp,I,8 = fI,11 = p0,4 pp,I,11 = (shared lane use p*) (Equation 17-6 or 17-13)

Movement capacity (Equation 17-7) cm,I,8 = cp,I,8 fI,8 = cm,I,11 = cp,I,11 fI,11 =

Prob of queue-free state (Equation 17-5) p0,I,8 = p0,I,11 =

Part II - Second Stage

Conflicting flows (Exhibit 17-4) vc,II,8 = vc,II,11 =

Potential capacity (Equation 17-3 or 17-29) cp,II,8 = cp,II,11 =

Ped impedance factor (Equation 17-12) pp,II,8 = pp,II,11 =

Capacity adjustment factor due to impeding movement fII,8 = p0,4 pp,II,8 = fII,11 = p0,1 pp,II,11 = (shared lane use p*) (Equation 17-6 or 17-13)

Movement capacity (Equation 17-7) cm,II,8 = cp,II,8 fII,8 = cm,II,11 = cp,II,11 fII,11 =

Prob of queue-free state (Equation 17-5) p0,II,8 = p0,II,11 =

Part III - Single Stage




Capacity adjustment factor due to impeding movement f8 = p0,4 p0,1 pp,8 = f11 = p0,4 p0,1 pp,11 =(shared lane use p*) (Equation 17-13 or 17-16)


Result for Two-Stage Process

a (Equation 17-30) a = a =

y (Equation 17-31) y = y =

cT (Equation 17-32 or 17-33) cT = cT =


General Information


Example Problem 3

341 482

618 532

1.000 1.000

0.970 0.945

599 503

0.780 0.781

532 366

504 601

1 .000 1.000

0.945 0.970

476 583

0.723 0.811

873 848

273 283

1.000 1.000

0.917 0.917

250 260

0.949 0.949

1.808 0.946

390 405

0.662 0.728

Example Problem 3



7

7

7

10

10

10


Worksheet 7b

Effect of Two-Stage Gap Acceptance

Step 4: LT from Minor Street v7 v10

Part I - First Stage

Conflicting flows (Exhibit 17-4) vc,I,7 = vc,I,10 =

Potential capacity (Equation 17-3 or 17-29) cp,I,7 = cp,I,10 =

Ped impedance factor (Equation 17-12) pp,I,7 = pp,I,10 =

Capacity adjustment factor due to impeding movements fI,7 = p0,1 pp,I,7 = fI,10 = p0,4 pp,I,10 =

Movement capacity (Equation 17-7) cm,I,7 = fI,7 cp,I,7 = cm,I,10 = fI,10 cp,I,10 =

Part II - Second Stage

Conflicting flows (Exhibit 17-4) vc,II,7 = vc,II,10 =

Potential capacity (Equation 17-3 or 17-29) cp,II,7 = cp,II,10 =

Ped impedance factor (Equation 17-12) pp,II,7 = pp,II,10 =

Capacity adjustment factor due to impeding movements fII,7 = p0,4 p0,I,11 p0,12 pp,II,7 = fII,10 = p0,1 p0,I,8 p0,9 pp,II,10 =

Movement capacity (Equation 17-7) cm,II,7 = fII,7 cp,II,7 = cm,II,10 = fII,10 cp,II,10 =

Part III - Single-Stage




Major left, minor through impedance factor p" = p0,11 f11 = p" = p0,8 f8 =

Major left, minor through adjusted impedance factor p' = p' =(Equation 17-8)

Capacity adjustment factor due to impeding movements f7 = p' p0,12 pp,7 = f10 = p' p0,9 pp,10 =(Equation 17-9 or 17-14)


Result for Two-Stage Process

a (Equation 17-30) a = a =

y (Equation 17-31) y = y =

cT (Equation 17-32 or 17-33) cT = cT =

General Information


Example Problem 3

341 482

626 514

1.000 1.000

0.970 0.945

607 486

337 257

629 703

1.000 1.000

0.711 0.707

447 497

678 739

323 291

1.000 1.000

0.668 0.607

0.742 0.694

0.715 0.649

231 189

0.949 0.949

2.055 1.227

369 347

Example Problem 3



(Equation 17-15)vy

cm,y

cSH =

( )

∑vy

∑

y

y


Worksheet 8


7

8

9

10

11

12

Worksheet 9






Qsep + 1

Round (Qsep + 1)


cSH

csep

n




General Information


Example Problem 3

44 369

132 390 442

55 845

11 347

110 405 439

28 783

369 390 845 347 405 783

44 132 55 11 110 28

16.070 18.881 9.557 15.714 17.171 9.768

0.196 0.692 0.146 0.048 0.525 0.076

1.196 1.692 1.146 1.048 1.525 1.076

1 2 1 1 2 1

2 2

442 439

1604 1535

1 1

1023 987

Example Problem 3




Worksheet 10


Worksheet 11













1

2

3

1

2

3


1

4

General Information


Example Problem 3

*

121110

121110

121110

987

987

987

p0,1 = p0,4 =

p* = p* =0, 1 0, 4

Example Problem 3

231 1024 0.226 < 1 9.5 A 9.5A

149 987 0.151 < 1 9.3 A 9.3A

X X X

33 1100 0.030 < 1 8.4 A

66 1216 0.055 < 1 8.2 A

X X X

Date post:	16-Nov-2014
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