Unformatted text preview: The Pennsylvania State University The Pennsylvania State University Department of Civil and Environmental Engineering CE 321 Highway Engineering Class #10: Vehicle Braking – Part 2 Spring 2008 Role in Highway Engineering Role in Highway Engineering Braking behavior of vehicles is critical in the determination of:
– Stopping Sight Distance – Roadway Design Elements – Accident Avoidance Systems Ramps Intersections How do we move from braking How do we move from braking to stopping distance? Relate braking forces to stopping and highway design Stopping Sight Distance (SSD)
– PerceptionReaction (P/R) Time – Braking distance Perception / Reaction Perception / Reaction Distance Perception / Reaction time of a driver is a function of: – driver’s age – physical condition – emotional state – complexity of the situation – the strength of the stimuli requiring a stop Perception / Reaction Perception / Reaction Distance 2 Four distinct actions for a stimulus (e.g. a child running into the street) PIEV Time
– P = Perception – I = Identification – E = Emotion or Decision – V = Volition or Reaction PIEV Time PIEV Time P = Perception see object (i.e. impact retina) and perceive it as a threat I = Identification – identify its meaning E = Emotion or Decision generate alternative actions, sort through alternatives; decide action to take V = Volition or Reaction execute the selected decision What is the P/R Time for an What is the P/R Time for an Average Driver ?? What are the Limits??
– Fastest P/R Time is: – Slowest P/R Time is: P/R Time for Highway P/R Time for Highway Design Purposes
2.5 seconds Perception / Reaction Distance Perception / Reaction d r = V1 * t r V1 = initial vehicle speed (ft/sec) tr = the time (in seconds) required to perceive and react Once you react, how long Once you react, how long does it take to stop? Practical Stopping Distance Practical Stopping Distance “Adequate” driver sight distance to permit a safe stop… with different driver skills, vehicles, weather, braking efficiency… etc. From physics: V = V + 2ad
2 2 2 1
where: V2 = final vehicle speed (ft/sec) V1 = initial vehicle speed (ft/sec) a = acceleration, negative if deceleration (ft/sec2) d = deceleration distance (ft) Practical Stopping Distance 2 Practical Stopping Distance 2 Rearrange motion equation
V12 − V22 d= 2a If V2 = 0 (vehicle comes to a complete stop) V12 d= 2a Practical Stopping Distance 3 Practical Stopping Distance 3 V1 d= a 2g ± G g where: v is in ft/s G is in decimal form 2 Practical vs. Theoretical Stopping Distance Practical vs. Theoretical Stopping Distance Ignore aerodynamic resistance Vehicle comes to stop (V2 = 0) sin θ g = tan θ g = G Mass factor (γ b) & coefficient of rolling resistance (frl) are small ignore S= V 2 g (ηb µ ± G ) 2 1 All variables as defined previously Total Required Stopping Total Required Stopping Distance
Braking Distance + Distance Traveled during P/R Time Total Required Stopping Distance ds = d + dr where: ds = total stopping distance (ft) d = braking distance (ft) dr = P/R time distance traveled (ft) AASHTO Stopping Sight Distance AASHTO Stopping Sight Distance AASHTO 2001 Stopping AASHTO 2001 Stopping Sight Distance Criteria Driver’s eye height = 3.5 ft Object Height = 2.0 ft Wet Pavement Marginal Tires a = 11.2 ft/s2 Practical Stopping Distance Practical Stopping Distance AASHTO 2001 V d = 1.075 a 2 where: v is in mi/h a = 11.2 ft/s2 Practical Stopping Distance Practical Stopping Distance with Grade effects included AASHTO 2001
V2 d= a 30 ± G 32.2 where: v is in mi/h a = 11.2 ft/s2 G = %grade/100 Design Stopping Sight Design Stopping Sight Distance (Level Grade)
d = 1.47Vt + 1.075 V2/a P/R Time Distance Braking Distance Example 3 Example 3 A driver has a fatigued reaction time of 3.0 seconds and an alert reaction time of 1.0 seconds. Compare the distance traveled during perception/reaction if the driver is traveling 60 miles per hour. What is the practical stopping distance for the alert driver if the roadway grade is + 5%? Example 3 Example 3 Part 1 d = 1.47vt d1 = 1.47(60)(3.0) = 265 ft. d2 = 1.47(60)(1.0) = 88 ft. t1 = 3.0s t2 = 1.0s Part 2 Using eq. 2.47 d2=88+(1.075)(60)2/2(32.2)[11.2/32.2 + 0.05] = 88 +150.23 = 238.23 Example 4 (2.9 in text) Example 4 (2.9 in text) W = 2200lbs. CD = 0.25 Af = 21.5 ft.2 ηb = 1.00 Level road; poor wet pavement ρ = 0.00238 FIND: How inaccurate practical SSD is at 60mph Example 4 (2.9 in text) Example 4 (2.9 in text) Eq. 2.5 frl = 0.0064 Eq. 2.37 Ka = 0.0064 Eq. 2.42 S = 200.35 ft. If brake efficiency is 85% S = 234.11 ft. From Eq. 2.46 d = 345.71 ft. What can we conclude? Example 4 (2.9 in text) Example 4 (2.9 in text) Eq. 2.5 frl = 0.0064 Eq. 2.37 Ka = 0.0064 Eq. 2.42 S = 200.35 ft. If brake efficiency is 85% S = 234.11 ft. From Eq. 2.46 d = 345.71 ft. What can we conclude? Practical stopping sight distance is very conservative (assumes 0.35 g deceleration) Example 4 (2.9 in text) Example 4 (2.9 in text) Rearranging 2.46 to solve for a:
200.35 ft. 234.11 ft. 19.33 ft./s2 16.54 ft./s2 Values in range of observed drivers ...
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