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CE 463 Lecture 07 Car following models _1_

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					CE 463 Lecture 07 Car Following Models (1)

CE 643 Highway Transportation Characteristics Civil Engineering Purdue University

Outline
Introduction Steady-state model - Model derivation - Application to safety consideration Linear car-following model - Model derivation - Example application to simulation

Driving sub-tasks in the car-following situation
• Perception – observation of the leading car motion in relation to the drivers’ car (vehicle speeds, acceleration, inter-vehicle spacing, relative speed, etc.) and interpretation of the situation Decision making – selection of a proper reaction (acceleration vs. deceleration, magnitude of reaction)

•

•

Reaction – change in speed (acceleration, deceleration)

Driving sub-tasks in the car-following situation
One lane is considered No lane changes and passing Driver’s behavior is consistent over time

Notation:

x(t )

dx   v(t )  x(t ) dt

dv  a(t )  (t ) x dt

Steady-State Model

 x1 (t )  b1 (t )  x2 (t )    x2 (t )  b2 (t )  s

 Let x(t )  v and b(t )  b
x1 (t )  b1  x2 (t )    v  b2  s x1 (t )  x2 (t )    v  (b2  b1 )  s
x    v  (b2  b1 )  s

Exercise
A driver has been following another car for a quite long time. The lead car has been maintaining a constant speed of 73.3 ft/s. The driver in the following car attempts to maintain a safe distance based on the following assumptions and preferences: – – – – – His reaction time is above average and equals 2 s Preceding car may rapidly decelerate at the rate of 10 ft/s2 Driver of the second car prefers deceleration of 8 ft/s2 20-foot clear distance between stopped vehicles is desirable Preceding car is 18 feet long

1. Calculate the inter-vehicle distance maintained by the second driver. 2. The first car hits an object on the roadway. The average deceleration of the lead car during the collision is 90 ft/s2. Has the second driver avoided the collision if his actual reaction time was 0.9 s and the wheels were locked during the breaking? The coefficient of friction was 0.45 and the road was level.

Safety Consideration

Deceleration Deceleration Case of leading of following vehicle vehicle a ∞ normal bn

b
c d

emergency be
∞

normal bn
emergency be

equal

e

no braking

Steady-State Model
x    v  (b2  b1 )  s Let b1  b2

x    v  s
If τ = 1.5 s and s = 25 ft, than Δx for v = 70 mph is Δx=1.5x(70x5280/3600) + 25 = 179 ft The time separation is Δx/v = (179/5280)/70x3600 = 1.74 s Number of vehicles per hour per lane = 3600/1.74 = 2070 veh/h/lane

Linear Car-Following Model
   x2 (t )  b2 (t )  b1 (t )  s  x1 (t )  x2 (t )

Let b1 (t )  b2 (t )
   x2 (t )  s  x1 (t )  x2 (t ) d dt

    2 (t )  x1 (t )  x2 (t ) x
2 (t )  x 1   x1 (t )  x2 (t )



Linear car-following model
2 (t )  x 1



  x1 (t )  x2 (t )

Due to the time lag in the reaction of the second driver, the following is more appropriate:
2 (t   )  x 1



  x1 (t )  x2 (t )

response = sensitivity x stimulus A general form of the linear model:
2 (t   )  x C



  x1 (t )  x2 (t )

Example Simulation
2 (t   )  x C



  x1 (t )  x2 (t )

2 (0.0  1.0)  x
2 (1.3  1.0)  x

0.5 0.00  0.00 ft/s2 1.0

0.5  1.50  0.75 ft/s2 1.0

  x2 (2.3)  x2 (2.2)  t  2 (2.2)  x  x2 (2.3)  73.28  0.1 (0.50)  73.23 ft/s
1 2 (2.2)  t 2  x2 (2.2)  t  x 2 1 x2 (2.3)  161 .26   (0.50 )  0.12  73 .28  0.1 2 x2 (2.3)  168 .59 ft x2 (2.3)  x2 (t ) 

Example Results

Example Results


				
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