Traffic And Highway Engineering
Traffic And Highway Engineering
5th Edition
ISBN: 9781337631020
Author: Garber, Nicholas J.
Publisher: Cengage,
Question
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Chapter 9, Problem 18P
To determine

The LOS, v/c, veh-mi in the peak 15min and peak hr, and total travel time in the peak 15min.

Expert Solution & Answer
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Answer to Problem 18P

The level of service is E, volume to capacity ratio is 0.35, the total number of vehicle miles during the peak 15min period is 1276veh/mi, the total number of vehicle miles during the peak period is 4950veh/mi and the total travel time in peak 15min period is 33.5veh/h.

Explanation of Solution

Given:

Base free flow speed is 55mi/h.

Lane width is 11ft.

Shoulder width is 3ft.

Access points per mile is 15.

Formula used:

Write the expression to calculate the value of fHV.

fHV=11+PT(ET1)+PR(ER1)   ...... (I)

Here, PT is decimal portion of trucks in the traffic stream, PR is decimal portion of RV's in the traffic stream, ET is passenger car equivalent for trucks and ER is passenger car equivalent for RV's.

Write the expression for peak 15min hourly passenger car equivalent.

vp=V(PHF)(fG)(f HV)   ...... (II)

Here, V is demand volume for the entire peak hour, PHF is the peak hour factor, fG is grade adjustment factor for level or rolling terrain and fHV is adjustment factor to account for heavy vehicles in the traffic stream.

Write the expression to calculate the base percent time spent following.

BPTSF=100(1e0.000879vp)   ...... (III)

Here, BPTSF is the base time percent spent following and vp is the peak 15min hourly passenger car equivalent.

Write the expression to calculate the percent time following.

PTSF=BPTSF+fdnp   ...... (IV)

Here, fdnp is the adjustment in PTSF to account for the combined effect.

Write the expression to calculate the free flow speed.

FFS=BFFSfLSfA   ...... (V)

Here, FFS is the free flow speed, BFFS is the base free flow speed, fLN is adjustment for lane and shoulder width and adjustment for number of access point per mile.

Write the expression to calculate the average travel speed.

ATS=FFS0.00776vpfnp   ...... (VI)

Here, ATS is the average travel speed and fnp is adjustment for the percentage of no passing zones.

Write the expression to calculate the volume to capacity ratio.

vc=vpc   ...... (VII)

Here, v is the volume and c is the capacity.

Write the expression to calculate the total number of vehicle miles during 15min peak.

VMT=0.25(VPHF)Lt   ...... (VIII)

Here, Lt is the total length of the analysis segment.

Write the expression to calculate the number of vehicle miles during the peak hour.

VMT60=VLt   ...... (IX)

Write the expression to calculate the total travel time in peak 15min period.

TT15=VMT60ATS   ...... (X)

Calculation:

Refer table 9.5, "Passenger car equivalents for trucks and RV's to determine the percent time spent following on two way and directional segments.

The value of ET for level terrain is 1.1.

The value of ER for level terrain is 1.0.

Substitute 10% for PT, 1.1 for ET, 7% for PR and 1.0 for ER in equation (I).

fHV=11+10%( 1.11)+7%( 11)=11+0.10( 0.1)=0.99

Substitute 1100veh/h for V, 0.97 for PHF, 1 for fG and 0.99 for fHV in equation (II).

vp=1100veh/hr( 0.97)(1)( 0.99)=1146pc/h

Substitute 1146pc/h for vp in equation (III).

BPTSF=100(1e 0.00879×1146)=63.48%

Refer table 9.3, "Adjustment for combined effect of directional distribution of traffic and percentage of no passing zones on percent time spent following on two way segments.

The value of fdnp is 7.47%.

Substitute 63.48% for PTSF and 7.47% for fdnp in equation (IV).

PTSF=63.48%+7.47%=70.95%

Substitute 55mi/h for BFFS, 3mi/h for fLS and 3.75mi/h for fA in equation (V).

FFS=55mi/h3mi/h3.75mi/h=48.25mi/h

Refer table 9.6, "Adjustment for effect of no passing zones on average travel speed on two way segments".

The value of fnp is 1.284mi/h.

Substitute 48.25mi/h for FFS, 1146pc/h for vp and 1.284mi/h for fnp in equation (VI).

ATS=48.25mi/h(0.00776×1146pc/h)1.284=38.07mi/h

Refer table 9.1, "Level of service criteria for two lane highways in class I.

The value of PTSF is 70.95% and ATS IS 38.07mi/h, therefore the level of service is E.

Substitute 1146pc/h for vp and 3200pc/h for c in equation (VII).

vc=1146pc/h3200pc/h=0.35

Substitute 1100veh/h for V, 0.45mi for Lt and 0.97 for PHF in equation (VIII).

VMT=0.25( 1100 veh/ hr 0.97)×4.5mile=1276veh/mi

Substitute 1100veh/h for V and 4.5mile for Lt in equation (IX).

VMT60=1100veh/h×4.5mile=4950veh/mile

Substitute 1276veh/mi for VMT and 38.07mi/h for TT15 in equation (X).

TT15=1276veh/mi38.07veh/h=33.5veh/h

Conclusion:

Therefore, the level of service is E, volume to capacity ratio is 0.35, the total number of vehicle miles during the peak 15min period is 1276veh/mi, the total number of vehicle miles during the peak period is 4950veh/mi and the total travel time in peak 15min period is 33.5veh/h.

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