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This Document Contains Chapters 3 to 4 Chapter 3 Characteristics of the Driver, the Pedestrian, the Vehicle and the Road 3-1 Briefly describe the five characteristics of visual reception relevant to transportation engineering. Visual acuity: is the ability to see fine details of an object. It can be represented by the visual angle, which is the reciprocal of the smallest pattern detail in minutes of arc. The ability to resolve a pattern detail with a visual acuity of one minute of arc (1/60 of a degree) is considered the normal vision acuity (20/20 vision). Peripheral vision: is the ability of people to see objects beyond the cone of clearest vision. Although objects can be seen within this zone, details and color are not clear. Color vision: is the ability to differentiate one color from another. Deficiency in this ability, usually referred to as color blindness, is not of great significance in highway driving because other ways of recognizing traffic information devices (e.g., shape) can compensate for it. Glare vision and recovery: there are two types of glare vision: direct and specular. Direct glare occurs when relatively bright light appears in the individual’s field of vision and specular glare occurs when the image reflected by the relatively bright light appears in the field of vision. Both types of glare result in a decrease of visibility and cause discomfort to the eyes. Depth perception: it affects the ability of a person to estimate speed and distance. It is particularly important on two-lane highways during passing maneuvers, when head-on crashes may result from a lack of proper judgment of speed and distance. 3-2 Briefly describe the two types of visual acuity. The two types of visual acuity are static and dynamic. The ability of a driver to identify an object when both the object and the driver are stationary depends on one's static acuity. Some factors that affect static acuity include the background brightness, contrast, and time. The ability of a driver to clearly detect a moving object depends on the driver's dynamic visual acuity. 3-3 (a) What color combinations are used for regulatory signs (e.g. speed limit signs) and for general warning signs (e.g. advance railroad crossing signs) (b) Why are these combinations used? Regulatory signs use a color combination of black lettering on white background, and most advance warning signs use the color combination of black lettering on yellow background. These color combinations are used because they have been shown to be those to which the eye is most sensitive. 3-4 Briefly describe the changes that occur in the abilities of older drivers (aged 65 and over) that influence their driving capability. As one grows older, his or her sensory, cognitive, and physical functioning ability declines, which can result in older drivers being less safe than their younger counterparts, and with a higher probability of being injured when involved in a crash. Specific declining abilities of older drivers include reduced visual acuity, ability to see at night, and flexibility and motion range. This group also suffers from narrower visual fields, greater sensitivity to glare, higher reaction times, and reduced muscle strength, which may result in the older driver having a higher crash risk. 3-5 Determine your average walking speed. Compare your results with that of the recommended walking speed in the MUTCD. Which value is more conservative and why?

Pass # Intersection Width (ft) Walk Time (sec) Walking Speed (ft/sec)

1 36 7 5.1

2 36 8.2 4.4

3 36 9.5 3.8

4 36 7.6 4.7

5 36 8 4.5

Average 4.5

In this case, the MUTCD value, 3.5 ft/sec, is more conservative than the observed speeds. This value is more conservative because it is a slower speed and it will allow most slower people, such as elderly, individuals with small children, and handicapped individuals to traverse the intersection safely. 3-6 Describe the three types of vehicle characteristics. The three types of vehicle characteristics are static, kinematic, and dynamic. Static vehicle characteristics include the vehicle's weight and size. Kinematic characteristics involve the motion of the vehicle, and dynamic characteristics involve the force that causes the motion of the vehicle. 3-7 Determine the maximum allowable overall gross weight of the WB-67 Design Vehicle. From Table 3.2, the extreme distance between the axle groups is 43.4 – 45.4 ft (use 45.4 in this case). The number of axles in the group is 4. Use Eq. 3.2, W= 500 NLN−1+12N+36= 500× 45.44 1−×4 +12×4+36= 72267lb . The maximum allowable overall gross weight is 72267 lb, which, rounded to the nearest 500 lb, is 72,000 lb. 3-8 The design speed of a multilane highway is 60 mi/h. Determine (a) the minimum stopping sight distance that should be provided for a level roadway, and (b) the minimum stopping sight distance that should be provided for a roadway with a maximum grade of 7 percent. a Note: the term in the appropriate equation is typically rounded to 0.35 in g calculations. Assume perception reaction time = 2.5 sec. The minimum sight distance required in these cases is the stopping sight distance (SSD), given by Equation 3.27: S =1.47ut + u2 ; a 30( ±G) g where u = design speed (mi/h) t = perception-reaction time (sec) a = rate of deceleration (taken as 11.2 ft/sec2) g = gravitational acceleration (taken as 32.2 ft/sec2) G = grade Determine the minimum sight distance that should be provided for a level roadway. Since the roadway is level, G = 0. = (1.47)(60)(2.5) + (60)2 / (30)(0.35 + 0) S = 220.50 + 345.00 S = 565.50 feet Therefore, the minimum sight distance for this horizontal roadway is 565 feet. Determine the minimum sight distance that should be provided for a roadway with a maximum grade of 7 percent. Since this roadway has a maximum grade of 7 percent, G = –0.07. The downgrade (negative) case provides the most conservative (higher) value for design. = 1.47(60)(2.5) + (60)2 / 30(0.35 – 0.07) S = 220.50 + 428.57 S = 649.07 feet Therefore, the minimum sight distance for this roadway should be 650 feet. 3-9 The acceleration of a vehicle can be expressed as: du = 3.6 − 0.06u dt If the vehicle speed, u, is 30 ft/sec at time T0, Determine: Distance traveled when the vehicle has accelerated to 45 ft/sec. Time for vehicle to attain the speed of 45 ft/sec. Acceleration after 4 seconds. Determine the distance traveled by the vehicle when accelerated to 45 ft/sec. First, determine the time it took for the vehicle to accelerate to 45 ft/sec. Using a rearrangement of Equation 3.10: – βt = ln [(α– βut) / (α – βu0)] therefore; t = (–1/β) ln [(α– βut) / (α– βu0)] t = (–1/0.06) ln [(3.6 – 0.06(45)) / (3.6 – 0.06(30))] t = 11.55 seconds Next, determine the distance traveled during this time. From Equation 3.12: x = (α / β)t – (α / β2) (1 – e–βt) + (uo / β) (1 – e–βt) x = (3.6 / 0.06)11.55 – [(3.6 /(0.06)2)(1 – e–0.06(11.55))] + [(30 / 0.06)(1 – e–0.06(11.55))] x = 443.04 feet. Therefore, the vehicle traveled 443 feet when accelerating from 30 ft/sec to 45 ft/sec. Determine the time it takes for the vehicle to attain the speed of 45 ft/sec. This time was determined in Part (a) of this problem. Therefore, it took 11.55 seconds for the vehicle to attain the speed of 45 ft/sec. Determine the acceleration of the vehicle after 4 seconds. First, determine the velocity of the vehicle after 4 seconds. From Equation 3.11: ut = (α / β) (1 – e–βt) + uoe–βt ut = (3.6 /0.06)(1 – e–(0.06)(4)) + 30e–(0.06)(4) ut = 36.40 ft/sec. Since acceleration is: a = du/dt = 3.6 – 0.06u a = 3.6 – 0.06(36.40) a = 1.42 ft/sec2 Therefore, the acceleration of the vehicle after 4 seconds is 1.42 ft/sec2 3-10 The gap between two consecutive automobiles (distance between the back of a vehicle and the front of the following vehicle) is 65 ft. At a certain time the front vehicle is traveling at 40 mi/h and the following vehicle at 30 mi/h. If both vehicles start accelerating at the same time, determine the gap between the two vehicles after 10 sec if the acceleration of the vehicles can be assumed to take the following forms: du = −3.4 0.07ut (leading vehicle) dt du = −3.3 0.065ut (following vehicle) dt where ut is the vehicle speed in ft/sec. First, determine the distance each vehicle travels during the elapsed time (10 seconds) using Equation 3.12. For the leading vehicle: x = (α / β)t – (α / β2) (1 – e–βt) + (uo / β) (1 – e–βt) x = (3.4 / 0.07)10 – [(3.4 /(0.07)2)(1 – e–0.07(10))] + [((40)(1.47) / 0.07) (1 – e–0.07(10))] x = 559.27 feet. Similarly, for the following vehicle: x = (3.3 / 0.065)10 – [(3.3 /(0.065)2)(1 – e–0.065(10))] + [((30)(1.47) / 0.065) (1 – e–0.065(10))] x = 458.65 feet. Since the leading vehicle traveled further, the gap between vehicles increased by the difference in the distances, 559.27 – 458.65 = 100.62 ft. The initial gap was given as 65 ft, so after 10 sec, the gap is 65 + 100.62 = 165.62 ft. 3-11 The driver of a vehicle on a level road determined that she could increase her speed from rest to 50 mi/h in 12.8 sec and from rest to 65 mi/hr in 19.8 sec. If it can be assumed that the acceleration of the vehicle takes the form: du = −α βut dt determine the maximum acceleration of the vehicle First, convert miles/hour to feet/second. 50 mi/h = 73.33 ft/sec 65 mi/h = 95.33 ft/sec Next, use Equation 3.11 to develop the equation for each case as follows: Case 1 (50 mi/h) ut = (α / β) (1 – e–βt) + uoe–βt 73.33 = (α / β) (1 – e–β(12.8)) + 0e–β(12.8) 73.33 = (α / β) (1 – e–β(12.8)) Equation 1 Case 2 (65 mi/h) ut = (α / β) (1 – e–βt) + uoe–βt 95.33 = (α / β) (1 – e–β(19.8)) + 0e–β(19.8) 95.33 = (α / β) (1 – e–β(19.8)) Equation 2 Solve for α in equation 1 and substitute into equation 2. α = 73.33β / (1 – e–β(12.8)) Equation 3 Substitute this into Equation 2. 95.33 = [(73.33β / (1 – e–β(12.8))) / β) × (1 – e–β(19.8)) 95.33 = (73.33 / (1 – e–β(12.8))) × (1 – e–β(19.8)) (95.33 / 73.33) × (1 – e–β(12.8)) = (1 – e–β(19.8)) 1.30 × (1 – e–β(12.8)) = (1 – e–β(19.8)) 1.30 – 1.30e–β(12.8) = (1 – e–β(19.8)) 1.30e–β(12.8) – e–β(19.8) = 0.30 The above equation can be solved by assuming a value for β, evaluating the left hand side of the equation, and comparing the result to 0.3 as follows.

β 1.30 e –β (12.8) – e –β (19.8)

0.010 0.32

0.005 0.31

0.001 0.30

As can be seen, the solution is: β = 0.001 Substituting into Equation 3 gives: α = 73.33 × 0.001 / (1 – e (–0.001 × 12.8)) = 5.77 ft/sec2 Since the maximum acceleration is achieved when the velocity of the vehicle is 0, the value for α determined above is the maximum acceleration. 3-12 If the vehicle in Problem 3-11 is traveling at a speed of 40 mi/h, how long will it take after the driver starts accelerating for the vehicle to achieve a speed of 45 mi/h? The acceleration model found in Problem 3-11 was du/dt = 5.77 – 0.001 ut The values α =5.77 and β=0.001 can then be substituted in Equation 3.10: t = (–1/β) ln [(α – βut) / (α – βu0)] t = (–1/0.001) ln [(5.77 – ((0.001)(40)(1.47)) / (5.77 – ((0.001)(45)(1.47))] t = 1.3 seconds 3-13 Determine the horsepower developed by a passenger car traveling at a speed of 50 mi/h on an upgrade of 5% with a smooth pavement. The weight of the car is 3500 lb and the cross-sectional area of the car is 40 ft2. First, determine all of the resistive forces, air, rolling, and grade, acting on the vehicle, and then determine its horsepower requirement. The air resistance is (using Equation 3.13): Ra = 0.5 × [(2.15pCDAu2) / g] Ra = 0.5 × [(2.15(0.0766)(0.4)(40)(50)2 / 32.2] Ra = 102.3 lb The rolling resistance is (using Equation 3.14): Rr = (Crs + 2.15Crvu2) × W Rr = (0.012 + 2.15(0.65 × 10–6)(50)2) × 3500 Rr = 54.2 lb The grade resistance is: RG = WG RG = (3500)(0.05) RG = 175.0 lb Since these are the only forces acting on the vehicle, one can now determine the horsepower requirement. Use Equation 3.17. P = 1.47Ru / 550 P = [(1.47(102.3 + 54.2 + 175.0)(50)) / 550] P = 44.3 hp 3-14 Repeat Problem 3-13 for a 24,000-lb truck with a cross-sectional area of 100 ft2 and coefficient of drag of 0.5 traveling at 50 mi/hr. First, determine all of the resistive forces, air, rolling, and grade, acting on the vehicle, and then determine its horsepower requirement. The air resistance is (using Equation 3.13): Ra = 0.5 × [(2.15pCDAu2) / g] Ra = 0.5 × [(2.15(0.0766)(0.5)(100)(50)2 / 32.2] Ra = 319.7 lb The rolling resistance is (using Equation 3.15): Rr = (Ca + 1.47Cbu) × W Rr = (0.02445 + 1.47(0.00044)(50)) × 24,000 Rr = 1363 lb The grade resistance is: RG = WG RG = (24,000)(0.05) RG = 1200 lb Since these are the only forces acting on the vehicle, one can now determine the horsepower requirement. Use Equation 3.17. P = 1.47Ru / 550 P = [(1.47(319.7 + 1361 + 1200)(50)) / 550] P = 385 hp 3-15 A 2500-lb passenger vehicle originally traveling on a straight and level road gets onto a section of the road with a horizontal curve of radius=850 ft. If the vehicle was originally traveling at 55 mi/h, determine (a) the additional horsepower on the curve the vehicle must produce to maintain the original speed, (b) the total resistance force on the vehicle as it traverses the horizontal curve, and the total horsepower. Assume that the vehicle is traveling at sea level and has a front cross-sectional area of 30 ft2. First, determine all of the resistive forces, air and rolling, acting on the vehicle while it is traveling straight and then determine its horsepower requirement. The air resistance is (using Equation 3.13): Ra = 0.5 × [(2.15pCDAu2) / g] Ra = 0.5 × [(2.15(0.0766)(0.4)(30)(55)2 / 32.2] Ra = 92.74 lb The rolling resistance is (using Equation 3.14): Rr = (Crs + 2.15Crvu2) × W Rr = (0.012 + 2.15(0.65 × 10–6)(55)2) × 2500 Rr = 40.57 lb Since these are the only forces acting on the vehicle, one can now determine the horsepower requirement on the straight segment. Use Equation 3.17. P = 1.47Ru / 550 P = [(1.47(92.74 + 40.57)(55)) / 550] P = 19.60 hp The curve resistance is (using Equation 3.16): Rc = 0.5 × [(2.15u2W) / gR] Rc = 0.5 × [(2.15(55)2(2500) / (32.2)(850)] Rc = 297.03 lb Determine the total additional horsepower for the curve section of roadway. P = 1.47Ru / 550 P = [(1.47(297.03)(55)) / 550] P = 43.66 hp Determine the additional required for the vehicle to maintain its original speed hprequired = hpcurve + hpstraight hprequired = 43.66 + 19.60 hprequired = 63.26 hp Therefore, the vehicle will need to produce 63.26 horsepower to traverse the curve at its original velocity. The total resistive force acting on the vehicle while in the curve is: Rtotal = Ra + Rr + Rc Rtotal = 92.7 + 40.6 + 297.0 Rtotal = 430.3 lb Therefore, the total resistive force acting on the vehicle in the curve is 430 pounds. 3-16 A horizontal curve is to be designed for a section of a highway having a design speed of 50 mi/h. If the physical conditions restrict the radius of the curve to 400 ft, what value is required for the superelevation at this curve? Is this a good design? Determine required superelevation First, determine the coefficient of side friction, fs, from Table 3.3. fs = 0.14 Next, use equation 3.34 and solve for the superelevation value. R = u2 / 15(e + fs) e = [u2 / 15R] – fs e = [(50)2 / 15(400)] – 0.14 e = 0.28 Is this a good design? The superelevation for this curve would be 0.28. Since e = 0.28 > 0.10 (allowable maximum superelevation, this would NOT be a good design. 3-17 Determine the minimum radius of a horizontal curve required for a highway if the design speed is 60 mi/h and the superelevation rate is 0.08. First, determine the coefficient of side friction, fs, from Table 3.3. fs = 0.12 Next, use Equation 3.34 to solve for R. R = u2 / 15(e + fs) R = [(60)2 / 15(0.08 + 0.12) R = 1,200 feet. The minimum radius for this curved section of roadway was found to be 1200 ft. 3-18 The existing posted speed limit on a section of highway is 55 mi/h and studies have shown that that the current 85th percentile speed is 65 mi/h. If the posted speed limit is to be increased to the current 85th percentile speed, what should be the increase in the radius of a curve that is just adequate for the existing posted speed limit? Assume a superelevation rate of 0.08 for the existing curve and for the redesigned curve. For the existing curve, use Equation 3.34 to determine the radius. R = u2/15(e+fs) R = (55)2/15(0.08+0.13) R = 960 ft Similarly, determine the radius for the curve to be redesigned. R = (65)2/15(0.08+0.11) R = 1482 ft The increase should then be 1482 – 960 = 522 ft 3-19 The radius of a horizontal curve on an existing highway was field-measured to be 810 ft. The pavement on this two-lane highway is 22 ft wide, and the elevation difference between the inside and outside of the curve is 1.43 ft. The posted speed limit on the road is 60 mi/h. Is this a hazardous location? If so, why? What action will you recommend to correct the situation? Assume that the design speed of this section of roadway is 70 mi/h (10 mi/h above the posted speed limit. Next, determine the coefficient of side friction, fs, from Table 3.3. fs = 0.10 Next, determine the rate of superelevation, e: e = 1.43/22 e = 0.065 Next, determine the maximum permissible speed on this existing curve by using Equation 3.34. R = u2 / 15(e + fs) u2 = 15(R)(e + fs) u = [15(810)(0.065 + 0.10)]1/2 u = 44.77 mi/h This curve is hazardous since the speed limit is posted at 60 mi/h yet the maximum safe speed in the curve is approximately 45 mi/h. One low cost measure to increase the safety of this curve would be to reduce the speed limit to 45 mi/h, or to post a curve warning sign with an advisory (maximum safe) speed of 45 mi/h. A long-term solution to improve safety would be to increase the radius of curvature to permit safe operation at the speed limit. This can be accomplished by using the above equation (equation 3.34): R = (60)2 / 15(0.065 + 0.10) R = 1,454.54 ft. Therefore, to permit safe travel at the maximum speed limit, the radius of the curve should be increased to 1,455 feet. 3-20 A section of highway has a superelevation of 0.05 and a curve with a radius of only 300 ft. What is the maximum safe speed at this section of the highway? A trial value for u must be assumed and the corresponding fs found and then checked for safety. Using Equation 3.34, solve for the value of fs associated with u = 35 mi/h R = u2 / 15(e + fs) 300 = 352 / 15(0.05 + fs) fs = 0.22 Interpolating in Table 3.3, for u =35 mi/h, fs = 0.18 should be assumed. Therefore, try a lower speed of u = 30 mi/h. Using Equation 3.34, fs = 0.15, which is less than the assumed to be provided value of 0.20. Therefore, the maximum safe speed at this section of the highway should be 30 mi/h. 3-21 What is the distance required to stop an average passenger car when brakes are applied on a 3.3% downgrade if that vehicle was originally traveling at 35 mi/h? Use equation 3.25 to determine the braking distance. u2 Db = a 30( ± G) g Note that a is taken as 11.2 ft/sec2; therefore, a/g is equal to 0.35. Db = 352 / 30(0.35 – 0.033) Db = 128.81 feet The braking distance required to stop the vehicle is 129 feet. 3-22 A driver on a level two-lane highway observes a truck completely blocking the highway. The driver was able to stop her vehicle only 30 ft from the truck. If the driver was driving at 55 mi/h, how far was she from the truck when she first observed it (assume perception-reaction time is 1.5 seconds)? How far was she from the truck at the moment the brakes were applied (use a/g = 0.35)? Use equation 3.35 to determine the stopping sight distance used. SSD =1.47ut + u2 =1.47(55)(1.5) + (55)2 = 409.37 ft a ± G) 30(0.35 + 0) 30( g Therefore, the distance from the point at which the driver observed the stopped truck to the truck is 30 + 409.37 = 439.37 ft. Next, use equation 3.25 to determine braking distance. Db = u2 = (55)2 = 288.09 ft a ± G) 30(0.35 + 0) 30( g Therefore, the distance from the point at which the driver applied the brakes to the truck is 30 + 288 = 318 ft. 3-23 A temporary diversion has been constructed on a highway of +4% grade due to major repairs that are being undertaken on a bridge. The maximum speed allowed on the diversion is 20 mi/h. Determine the minimum distance from the diversion that a road sign should be located informing drivers of the temporary change on the highway. Assume that a driver can read a road sign within his or her area of vision at a distance of 30 ft for each inch of letter height. Speed limit on highway = 65 mi/h Letter height of road sign = 8 inches Perception-reaction time = 2.5 sec Use equation 3.35 to determine the stopping sight distance. u2 SSD =1.47ut + a 30( ± G) g While the first term in this equation is simply the distance traveled during the perception-reaction time, second term is the distance traveled during braking. a/g is taken as 0.35. Since the vehicle is not stopping (the final speed is not equal to zero), the equation needs to be modified to take this into consideration. The second term of equation 3.35 is replaced by equation 3.26 as follows (in which u1 is the initial velocity and u2 is the final velocity): =1.47ut + u12 −u22 SSD a 30( ±G) g SSD = 1.47(65)(2.5) + (652 – 202) / (30 (0.35 + 0.04)) SSD = 238.87 + 326.92 SSD = 565.79 feet. Next, determine the readability of the roadway sign. Readability = (Letter height in inches) × 30 feet / inch of letter height Readability = 8 inches × 30 feet / inch Readability = 240 feet The sign can be read at a distance of 240 feet. Next, determine the distance from the diversion the sign should be placed. x = SSD – readability distance x = 565.79 – 240.00 x = 325.79 feet The sign should be placed approximately 325 feet prior to the diversion to alert drivers of the change on the highway. 3-24 Repeat Problem 3-23 for a highway with a downgrade of –3.5% and the speed allowed on the diversion is 15 mph. Assume that a driver can read a road sign within his or her area of vision at a distance of 40 ft for each inch of letter height. Use equation 3.35 to determine the stopping sight distance. u2 SSD =1.47ut + a 30( ± G) g While the first term in this equation is simply the distance traveled during the perception-reaction time, second term is the distance traveled during braking. a/g is taken as 0.35. Since the vehicle is not stopping (the final speed is not equal to zero), the equation needs to be modified to take this into consideration. The second term of equation 3.35 is replaced by equation 3.26 as follows (in which u1 is the initial velocity and u2 is the final velocity): =1.47ut + u12 −u22 SSD a 30( ±G) g SSD = 1.47(65)(2.5) + (652 – 152) / (30 (0.35 – 0.035)) SSD = 238.87 + 423.28 SSD = 662.15 ft Next, determine the readability of the roadway sign. Readability = (Letter height in inches) × 40 feet / inch of letter height Readability = 8 inches × 40 feet / inch Readability = 320 feet The sign can be read at a distance of 320 feet. Next, determine the distance from the diversion the sign should be placed. x = SSD – readability distance x = 662.15 – 320 x = 342.15 ft The sign should be placed approximately 345 feet prior to the diversion to alert drivers of the change on the highway. 3-25 An elevated expressway goes through an urban area, and crosses a local street as shown in Figure 3.9. The partial cloverleaf exit ramp is on a 2% downgrade and all vehicles leaving the expressway must stop at the intersection with the local street. Determine (a) minimum ramp radius and (b) length of the ramp for the following conditions: Maximum speed on expressway = 60 mi/h Distance between exit sign and exit ramp = 260 ft Letter height of road sign = 3” Perception-reaction time = 2.5 sec Maximum superelevation = 0.08 Expressway grade = 0% Assume that a driver can read a road sign within his or her area of vision at a distance of 50 ft for each inch of letter height, and the driver sees the stop sign immediately on entering the ramp. First, determine the readability of the roadway sign. Readability = (Letter height in inches) × 50 feet / inch of height Readability = 3 inches × 50 feet / inch Readability = 150 feet Next, determine the speed of the vehicle just prior to it entering the exit ramp (uexit). Modify equation 3.27 by replacing the second term (braking distance) with equation 3.26. a/g is taken as 0.35. SSD =1.47ut + u12 −u22 – readability distance a 30( ±G) g 260 = 1.47(60)(2.5) + [(602 – u2exit) / 30 (0.35 + 0)] – 150 410 = 220 + (602 – u2exit) / 10.5 602 – u2exit = 1995 uexit = 40.06 mi/h Next, determine the minimum radius for this exit ramp, using Equation 3.34. For the exit speed of 45 mi/h, the new coefficient of side friction should be fs = 0.14. = u2exit / 15(e + fs) R = (40.06)2 / 15(0.08 + 0.14) R = 486.35 feet. Next, determine the length required for this exit ramp. = 1.47uexitt + u2exit / 30 ((a/g) ± G) S = 1.47(40.06)(2.5) + (40.06)2 / 30(0.35 - 0.02) S = 309.33 feet Therefore, the minimum radius for this exit ramp is approximately 490 feet and the minimum length of the exit ramp was found to be approximately 310 feet. Chapter 4 Traffic Engineering Studies 4-1 What are the advantages and disadvantages of machine vision (video image detection) when compared with other forms of detection? The greatest advantage of video image detection is that it is non-intrusive; detectors need not be installed in the roadway (as with inductive loops or magnetometers). Systems such as the Autoscope can detect traffic in many locations within the camera’s field of view. Because of this, a single camera can replace many inductive loop detectors, thereby providing a wide-area detection system. Unlike inductive loops, with video image processing traffic need not be disrupted to install the device and the detection configuration can be changed either manually or by using a software routine. Disadvantages of this method include higher initial costs and that cameras must be dedicated to this purpose and detection zones must be placed precisely for accurate detection. 4-2 Select and describe the method and equipment you will recommend be used for traffic data collection for each of the road sections given below. Give reasons for your recommendations. A private road leading to an industrial development A residential street A rural collector road A section of an Interstate highway For a private road leading to an industrial development, which would likely be low volume and low speed, pneumatic tubes would be an appropriate choice. Concerns about impact of data collection method on driver behavior would likely be minimal on such a facility. For a residential street, which would likely be low volume and low speed, radar would be an appropriate choice as concerns may exist about the conspicuity of pneumatic tubes affecting driver behavior. For a rural collector road, where speeds and volumes may be somewhat high, radar would be appropriate. If expense were not substantial, machine vision would be worth considering. For a section of Interstate highway, given the high speed and high volume nature of the facility, it would be preferable to not have observers close the roadway. Therefore, methods such as machine vision or inductive loops (if an opportunity to temporarily close a lane is available) would be preferable. If observers can safely be placed, radar would also be acceptable. 4-3 Speed data collected on an urban roadway yielded a standard deviation in speeds of ±5.6 mi/h. If an engineer wishes to estimate the average speed on the roadway at a 95% confidence level so that the estimate is within ±2 mi/h of the true average, how many spot speed observations should be collected? If the estimate of the average must be within ±1 mi/h, what should the sample size be? (a) Using Equation 4.5, N = [(z σ)/d]2 N = [1.96(5.6) / 2]2 N = 30.1 => 31 spot speed observations Note: z =1.96 for 95% confidence interval (b) For speeds within ±1 mi/h: N = [1.96(5.6) / 1]2 N = 120.5 => 121 spot speed observations 4-4 An engineer wishing to obtain the speed characteristics on a bypass around her city at a confidence level of 95%, and an acceptable error limit of ± 2.0 mi/h, collected a total of 104 spot speed observations and determined that the standard deviation is 4.8 mi/h. Has the engineer met with all of the requirements of the study? This can be answered using Equation 4.5 to determine whether minimum sample size requirements are met. For the 95% confidence level, z = 1.96 The acceptable margin of error, d =2 The standard deviation σ = 4.8 N = Zσ2 = (1.96)(4.8)2 = 22.1 d 2 The minimum sample size is 23 observations. Since 104 were collected, the data collection requirements are met. Fewer observations would decrease time spent in the field and associated costs. 4-5 A spot speed study was conducted on a freeway on which the mean speed and standard deviation were found to be 72.4 mi/h and 5.4 mi/h respectively. What is the range of values that would be expected to contain the middle 95% of speeds? If it can be assumed that the true mean of the speeds in a section of highway is μ and the true standard deviation is σ, it can be concluded that all vehicle speeds will be between (μ – Zσ) and (μ + Zσ). For this problem, it is known that: μ = 72.4 mi/h σ = 5.4 mi/h Z = 1.96 for a 95% confidence level Therefore, the range of values that would be expected to contain the middle 95% of speeds will be: (μ – Zσ), (μ + Zσ) = (72.4 – 1.96 × 5.4), (72.4 + 1.96 × 5.4) = (61.8, 83.0). 95% of all vehicle speeds will be between 61.8 and 83.0 mi/h. 4-6 For a spot speed study in which 100 observations were obtained, the mean speed was 52 mi/h and the standard deviation was 4.6 mi/h. What is the margin of error on the estimate of true mean speed obtained from this sample? What is a reasonable estimate of the 85th percentile speed based on the information available? Using Equation 4.5, N = Zσ2 d = Zσ 1.96× 4.6 = 0.9mi h/ d N For a 95% confidence level, usually used for speed data, the margin of error on the estimate of true mean speed is ± 0.9 mi/h. Based on the information available, the 85th percentile speed must be estimated solely from the mean speed and standard deviation. From Table 4.1, it can be seen that z=1.00 for a confidence interval of 68.3% centered about the mean. Half of this range (about 34.2% of the entire range) falls above the mean. The 85th percentile speed is the value at which 85% of observations fall below; according to the properties of the normal distribution, which is the entire range below the mean plus 35% that falls immediately above the mean. Since z=1.00 for the 34.2% of the entire range that falls immediately above the mean, then the 85th percentile can be approximately estimated as (μ + Zσ) when z=1. Therefore, a reasonable estimate of the 85th percentile speed is the sum of the mean and the standard deviation; 52 + 4.6 = 56.6 mi/h. 4-7 An engineer wishing to determine whether there is a statistically significant difference between the average speed of passenger cars and that of large trucks on a section of highway, collected the data shown below. Determine whether the engineer can conclude that the average speed of large trucks is the same as for passenger cars. Trucks Passenger Cars Average Speed (mi/h) 70 73 Standard deviation of speed ± mi/h 5.9 7.3 Sample size 150 300 To determine whether the difference in mean speeds was statistically significant, first, the pooled standard deviation must be determined, using Equation 4.6. 2 2 2 Sd = S1 + S2 = (5.9) + (7.3) = 0.64 n1 n2 150 300 Then, compare the absolute value of the difference of the sample means with the product of the appropriate z-statistic and the pooled standard deviation. Absolute difference in means = 73 – 70 = 3. ZSd = (1.96)(0.64) > 1.25 Therefore, a statistically significant difference exists between the two data sets, and it cannot be concluded that the average speed of large trucks is the same as that of passenger cars. 4-8 Assume that the data shown in Table 4.2 were collected on a rural road in your state and consideration is being made to set the speed limit on the road. Speed limits of 50, 55, 60, and 65 mi/h are being considered. Plot the expected non-compliance percentages vs. the associated speed limit on a graph and recommend the speed for the road. Give reasons for your selection. Examination of the data in Table 4.2 shows that the number of observations exceeding 50, 55, 60, and 65 mi/h are 38, 17, 5, and zero, respectively. In percentages, these are 44.2%, 19.7%, 5.8%, and 0%, respectively. The plot appears below. The 85th percentile speed, speed above which 15 percent of vehicles travel, is usually recommended for the road. Therefore, 55 mi/h will be recommended as the speed for the road, as 19.7% of vehicles travel above it. 4-9 The accompanying data show spot speeds collected at a section of highway located in a residential area before and after an increase in speed enforcement activities. Using the student’s t test, determine whether there was a statistically significant difference in the average speeds at a significance level of α=0.05 (the 95% confidence level). Also report, for both the before and after cases, the mean speed, standard deviation, 85th percentile speed, and percentage of traffic exceeding the posted speed limit of 30 mi/h.

Before After Before After

40 23 38 25

35 33 35 21

38 25 30 35

37 36 30 30

33 37 38 33

30 34 39 21

28 23 35 28

35 28 36 23

35 24 34 24

40 31 33 27

33 24 31 20

35 20 36 20

36 21 35 30

36 28 33 32

40 35 39 33

First calculate average and standard deviation for both sets of data. The mean speeds for before and after cases will be: Before: = 35.1 mi/h After: = 27.5 mi/h And the standard deviation will be: sb sa2 = 5.4442 Next, calculate the pooled standard deviation for both sets of data. sd = [(sb2/ nb) + (sa2/ na)]½ = [(3.2092 / 30) + (5.4442 / 30)]½ = 1.154 Then compare the difference in average speed with the product of the z-statistic and standard deviation. |ub-ua | = 7.6 Zsd = 1.96(1.154) = 2.262 Since 7.6 > 2.262 therefore the speeds are significantly different. The percentage of traffic exceeding the posted speed limit of 30 mi/h is: Before = = 86.7% After = = 33.3% The 85th percentile speed can be determined from the table below:

Before

Speed (mi/h) Frequency Percentage of Observations Cumulative Percentage of Observations

28 1 0.03 0.03

29 0 0.00 0.03

30 3 0.10 0.13

31 1 0.03 0.17

32 0 0.00 0.17

33 4 0.13 0.30

34 1 0.03 0.33

35 7 0.23 0.57

36 4 0.13 0.70

37 1 0.03 0.73

38 3 0.10 0.83

39 2 0.07 0.90

40 3 0.10 1.00

After

Speed (mi/h) Frequency Percentage of Observations Cumulative Percentage of Observations

20 3 0.10 0.10

21 3 0.10 0.20

22 0 0.00 0.20

23 3 0.10 0.30

24 3 0.10 0.40

25 2 0.07 0.47

26 0 0.00 0.47

27 1 0.03 0.50

28 3 0.10 0.60

29 0 0.00 0.60

30 2 0.07 0.67

31 1 0.03 0.70

32 1 0.03 0.73

33 3 0.10 0.83

34 1 0.03 0.87

35 2 0.07 0.93

36 1 0.03 0.97

37 1 0.03 1.00

85th percentile speed before: between 38 and 39 mi/h 85th percentile speed after: between 33 and 34 mi/h 4-10 Using the data furnished in Problem 4-9, draw the histogram frequency distribution and cumulative percentage distribution for each set of data and determine (a) average speed, (b) 85th-percentile speed, (c) 15th-percentile speed, (d) mode, (e) median, and (f) pace. average speed = Σui / Σfi “before” average = 1053/30 = 35.1 mi/h “after” average = 824/30 = 27.5 mi/h 85%-ile speeds from cumulative distribution plots before = 38.4 mi/h after = 33.6 mi/h 15%-ile speeds from cumulative distribution plots before = 30.8 mi/h after = 20.5 mi/h mode from histograms before = 35 mi/h after = 21 mi/h median from cumulative distribution plots before = 34.5 mi/h after = 31 mi/h pace from histograms before = 30 – 40 mi/h after = 21 – 31 mi/h Before data set: After data set: 4-11 Define the following terms and cite examples of how they are used. Average annual daily traffic (AADT) Average daily traffic (ADT) Vehicle-miles of travel (VMT) Peak hour volume (PHV) Average annual daily traffic (AADT) is the average of 24-hour traffic counts collected every day in the year. These counts are used to estimate highway user revenues, compute accident rates, and establish traffic volume trends. Average daily traffic (ADT) is the average of 24-hour traffic counts collected over a number of days greater than one but less than a year. These counts are used for planning of highway activities, measuring current traffic demand, and evaluating existing traffic flow. Vehicle miles of travel (VMT) is a measure of travel usage along a section of road. It is the product of the volume (ADT) and the length of roadway in miles to which the volume is applicable. This measure is used mainly as a base for allocating resources for maintenance and improvement of highways and to establish highway system usage trends. Peak hour volume (PHV) is the maximum number of vehicles that pass a point on a highway during a period of sixty consecutive minutes. This volume is used for functional classification of highways, geometric design standards selection, capacity analysis, development of operational programs, and development of parking regulations. 4-12 Describe the different traffic count programs carried out in your state. What data are collected in each program? The Traffic Engineering Division of the Virginia Department of Transportation conducts an "Interstate, Arterial and Primary Traffic Count Program." Vehicle classifications consist of passenger cars, 2-axle 4-tire trucks, 2-axle 6-tire trucks, 3-axle 6-10 tire trucks, tractor-trailers, twin trailers, and buses. There are a total of 1,345 counting stations for this program. Counts are taken either 2, 4, or 9 times per year at each location. The data collected is then published in a manual entitled “Average Daily Traffic Volumes on Interstate, Arterial and Primary Routes” which includes 24-hour average daily traffic (ADT) by section of roadway, 24-hour vehicle miles traveled (VMT) by route and county, and 24-hour VMT statewide. In addition, counts are taken on secondary, and unpaved roads on a less frequent basis. Intersection turning movement counts are taken at specific locations when needed for detailed traffic operational studies. 4-13 A traffic engineer, wishing to determine a representative value of the ADT on 250 highway links having similar volume characteristics, conducted a preliminary study from which the following estimates were made: Mean volume = 45,750 veh/day, Standard deviation = 3750 veh/day. Determine the minimum number of stations for which the engineer should obtain 24-h volume counts for a 95-5 precision level. Use Equation 4.9: n = [t2α/2,N–1(s2/d2)]/ [1+ (1/N)(t2α/2,N–1)(s2/d2)] α = 100 – 95 = 5 S = 3750 = 45,750 therefore d = (0.1)(45,750) = 4575 V = N – 1 = 250 –1 = 249 (tα/2,N–1) = 1.96 = [(1.96)2(37502/45752)]/ [1+(1/250)(1.96)2(37502/45752)] n = 2.57 Use 3 count stations. 4-14 Describe the following types of traffic counts and when they are used. screen line counts cordon counts intersection counts control counts Screen line counts involve dividing the study area into large sections by drawing imaginary lines (screen lines) across the study area. Counts are then taken at each place a road crosses this line. This data is then used to detect variations in traffic volumes and flow direction attributable to changes in land-use patterns in the area. Cordon counts are similar to screen line counts with the imaginary line completely surrounding an area, for example, a CBD. Counts are taken at each place a road crosses the line, giving information about vehicle accumulations within the area. The information from cordon counts can be used to plan for parking facilities, evaluate traffic operational techniques, and long range infrastructure planning. Intersection counts are counts of vehicular turning movements and classifications at an intersection. The data gathered is used to develop signal timing and phasing plans for signalized intersections and for geometric design improvements. Control counts are taken at strategically located locations chosen to obtain representative samples of traffic volumes for specific types of highways or streets. There are two types of control counts; major and minor. Major control counts are taken monthly with 24-hour directional counts being taken at least three days during the week. 4-15 A traffic count taken on a rural highway between 7:00 a.m. and 2:00 p.m. on a rural highway found hourly volumes as follows:

Time Hourly volume

7:00-8:00 a.m. 310

8:00-9:00 a.m. 289

9:00-10:00 a.m. 241

10:00-11:00 a.m. 251

11:00 a.m.-12:00 p.m. 267

12:00-1:00 p.m. 264

1:00-2:00 p.m. 243

If these data were collected on a Tuesday in March, estimate the AADT on this section of highway. Assume the expansion factors given in Tables 4.6, 4.7, and 4.8 apply. (a) Estimate the 24-h volume using the factors given in Table 4.6: 24-h volume = 5,447 vehicles per day (b) Adjust the 24-h volume for Tuesday to an average volume for the week using Table 4.7: 7 = 5,447 × 7.727 = 42,089 ℎ 24ℎ = = 6,013 ℎ (c) To obtain the AADT, use the factor for March shown in Table 4.8: = 6,013 × 1.635 = 9,832 ℎ 4-16 How are travel time and delay studies used? Describe one method for collecting travel time and delay data at a section of a highway. Explain how to obtain the following information from the data collected: (a) travel time, (b) operational delay, (c) stopped time delay, (d) fixed delay, and (e) travel time delay. Travel time and delay studies are used to aid the traffic engineer in identifying problem locations, which may require special attention in order to improve the overall flow of traffic on the route. Data from these studies may be used to determine the efficiency of a route with respect to its ability to carry traffic, identify bottleneck locations with relatively high delays and the causes for those delays, perform before and after studies to evaluate the effectiveness of traffic operations improvements, determine the relative efficiency of a route by developing congestion indices, determine travel times on specific links for use in assignment models, and perform economic studies in the evaluation of traffic operation alternatives that reduce travel time. There are several methods for collecting travel time and delay data including the floating car method, the average car method, and the moving vehicle technique. In the average car method, the test car is driven along the length of the test section at a speed that, in the opinion of the driver, is the average speed of the traffic stream. The time to traverse the section is noted and the test run is repeated for a minimum number of times, with the average time then being recorded as the travel time. An additional stop watch would be used to measure the amount of time the test vehicle is delayed by impedance of other traffic such as vehicles parking, or by a reduction in the capacity of the roadway such as a work zone. This value is recorded as the operational delay. During the test runs, the amount of time the vehicle is stopped would also be recorded and the average from all runs would be recorded as the stopped-time delay. The fixed delay would be measured as the time spent waiting for a traffic signal along the route to turn green. This delay is independent of other traffic. The travel time delay would be determined by subtracting the travel time for a vehicle to traverse the study section under uncongested conditions from the actual travel time. 4-17 Table 4.11 shows data obtained in a travel time study on a section of highway using the moving-vehicle technique. Estimate (a) the travel time and (b) the volume in each direction at this section of the highway. To determine volume using the moving vehicle technique, use Equation 4.9: VN = (N O PS + N + N )60 T TS + N

NN = 104.2 NS = 93.9 VN = (93.9 + 1.6 – 1.1)60 / (5.25 + 4.90)

ON = 1.6 OS = 1.1 VN = 558 veh

PN = 1.1 PS = 1.0 VS = (104.2 + 1.1 – 1.0)60 / (5.25 + 4.90)

TN = 5.25 TS = 4.90 VS = 617 veh

To determine travel time using the moving vehicle technique, use Equation 4.10:

TN = TN – [60(ON – PN)/VN] TS = TS – [60(OS – PS)/VS]

TN = 5.25 – [60(1.6–1.1)/558] TS = 4.90 – [60(1.1–1.0)/617]

TN = 5.20 min TS = 4.89 min

4-18 An engineer, wishing to determine the travel time and average speed along a section of an urban highway as part of an annual trend analysis on traffic operations, conducted a travel time study using the floating-car technique. He carried out 10 runs and obtained a standard deviation of ±3.4 mi/h in the speeds obtained. If a 5% significance level is assumed, is the number of test runs adequate? Use Equation 4.10: N = (tα( )σ )2 d For trend analysis, assume ±3 mi/h acceptable error. N = [1.833(3.4)/3]2 N = 4.32 runs Therefore, 10 runs is adequate 4-19 Briefly describe the tasks you would include in a comprehensive parking study for your college campus, indicating how you would perform each task and the way you would present the data collected. Parking studies, in general, are used to determine the demand and supply of parking facilities in an area, the projection of the demand into the future, and the collection of the views of various interest groups on how best to solve any problems. A comprehensive parking study usually includes an inventory of existing parking facilities, collection of data on parking accumulation, parking turnover, and parking duration data, identification of parking generators, and obtaining information on parking demand. On a college campus, this would include faculty, staff, and student lots, as well as visitor parking facilities. The college administration would also have to be interviewed to determine their policy on parking as it relates to the provision of on-campus parking. Parking accumulation data would be obtained by checking the amount of parking at 2-hour intervals from 7:00 am to 5:00 pm on weekdays, when demand is highest. Parking accumulation would then be plotted against time of day for each lot. While collecting accumulation data, license plates of vehicles in selected spaces should be recorded to determine duration and turnover. When the data is analyzed, the average length of time an individual vehicle occupies a space can be estimated. The college itself is the parking generator and therefore this step can be omitted as parking facilities of varying capacity exist throughout campus; however, the location of the larger parking facilities could be identified on a map. Parking demand would be collected by interviewing drivers as they enter the parking facility. The interviewer should attempt to identify the driver’s trip origin, purpose of their trip and the driver’s destination after parking. This information could indicate a need for a parking facility in a new location or the enlargement of an existing facility. 4-20 Select a parking lot on your campus. For several hours, conduct a study of the lot using the methods described in this chapter. From the data collected, determine the turnover and duration. Draw a parking accumulation curve for the lot. Identify the capacity of the parking lot, the peak demand observed, when the peak demand occurred, and occupancy of the lot at that time. To conduct a parking accumulation study of a parking lot, detailed data on space usage and vehicle turnover must be collected. The number of spaces in use, as well as license plate numbers of those vehicles and their entry and exit times, must be collected to monitor turnover and duration. Turnover is calculated as the ratio of the total number of different vehicles parked to the number of available spaces. Turnover for the entire study period can be calculated by dividing the number of vehicles parked during the study period by the number of spaces available. Duration is the average amount of time a vehicle spends in a parking space, which is a surrogate measure for the availability of parking spaces. By recording the entry and exit times for each vehicle during the study period, durations for each of these vehicles can be calculated and then averaged to determine average duration. The parking accumulation curve can be drawn to display percent of spaces used (on the y-axis) as a function of time-of-day (on the x-axis). By noting the number of spaces that are occupied at discrete intervals (e.g. every hour), the data for drawing the accumulation curve are obtained. The capacity will be the number of parking spaces at the parking facility. The demand may be obtained by noting the times of arrival and departure of each vehicle. It is recommended that demand information should be obtained in a weekday, from 8:00 a.m. to 10:00 p.m.; however, on campus, data collection from 7:00 a.m. to 6:00 p.m. should be sufficient to identify when the peak demand occurs. Occupancy is the percentage of spaces used in the parking facility at a point in time. 4-21 Data collected at a parking lot indicate that a total of 300 cars park between 8 a.m. and 5 p.m. 5% of these cars are parked for an average of 2 hr, 20% for an average of 3 hr, 10% for an average of 4 hr, and the remaining cars are parked for an average of 9 hr. Determine the space-hours of demand at the lot. Use Equation 4.12: D = (0.05)(300)(2) + (0.20)(300)(3) + (0.10)(300)(4) + (0.65)(300)(9) D = 2,085 space-hours 4-22 If 10% of the parking bays are vacant on average (between 8 a.m. and 5 p.m.) at the parking lot of Problem 4-21, determine the number of parking bays in the parking lot. Assume an efficiency factor of 0.90. Use Equation 4.15: 2,085 + (2,085)(0.10) = 2,294 space-hours (assuming 10% vacancy) Use Equation 4.16: 2,294 = (0.90)(9)(N) N = 284 spaces 4-23 The owner of the parking lot of Problems 4-21 and 4-22 is planning an expansion of her lot to provide adequate demand for the following 10 years. If she has estimated that parking demand for all categories will increase by 3% a year, determine the number of additional parking bays that will be required. Find space-hours of demand in 10 years: (1+0.03)10 (2,294) = 3,083 space hours Additional space hours = 3,083 – 2,294 = 789 Find number of spaces: (0.90)(9)(N) = 789 N = 97.41; use N = 98 spaces Solution Manual for Traffic and Highway Engineering Nicholas J. Garber, Lester A. Hoel 9781133605157