Stopping Sight Distance V2 S = 1.47 V t + (30[(a/32.2)"G)]) Note: G in percentage (decimal)

Examples (1/2) 2. For a roadway with a 55 mph design speed, a minimum required stopping sight distance of 500 ft, a deceleration rate of 8 f/s2, and driver reaction time of 2 second, the grade will be: a. 3.0% b. 4.0% c. 5.0% d. 6.0%

Examples (2/2) 3. The minimum stopping sight distance for a vertical curve connecting a +3% to a -2% with a deceleration rate of 10 ft/s2, a design speed of 50 mph, and driver reaction time of 2 seconds, will be: a. 391.7 ft b. 399.1 ft c. 433.8 ft d. 444.0 ft

Crest Vertical Curves h2 h1 L S>L L = 2 S - 200(qh1+qh2)2 /|A| A = G2 – G1 Note: Gi in grade SSD h2 h1 L S>L L = 2 S - 200(qh1+qh2)2 /|A| S#L L = |A| S2 / [200(qh1+ qh2)2] h1 = 3.5 ft and h2 =2.0 ft S>L L = 2 S - 2158/|A| S# L L = |A| S2/2158

Examples 4. The minimum length of a vertical curve with a required 500 ft stopping sight distance, object height of 2 ft, driver’s eye height of 3.5 ft, and connecting a +2% to a -3% grade will be: a. 568.4 ft b. 579.2 ft c. 587.7 ft d. 594.1 ft

Sag Vertical Curves S>L L = 2 S - (400 + 3.5S)/|A|] H SSD L β A = G2 – G1 Note: Gi in grade S>L L = 2 S - (400 + 3.5S)/|A|] S #L L = |A| S2/(400 + 3.5 S) Comfort criteria L = |A|V2/46.5

Examples 6. The stopping sight distance for a 500-ft vertical curve connecting a -3% to a +1% grade and with object height of 2.0 ft and driver’s eye height of 3.5 ft will be: a. 528.76 ft b. 531.57 ft c. 533.33 ft d. 535.85 ft

Horizontal Curves Superelevation Clearance from roadside obstruction 0.01e + f = V2/15R Clearance from roadside obstruction HSO = R [1 - cos (28.65 S/R)]

Examples (1/2) 7. The required clear distance for a horizontal curve with a radius of 700 ft and stopping sight distance of 215 ft will be: a. 6.3 ft b. 7.4 ft c. 8.2 ft d. 9.1 ft

Examples (2/2) 9. For a 900-foot radius horizontal curve designed for 50 mph and superlevation of 8%, the required side friction should be: a. 0.04 b. 0.05 c. 0.08 d. 0.10

Vertical Curves g1 g2 y x PVI E PVC PVT L/2 y = (g2 - g1)x2/2L YP= YPVC + g1 x + (g2 - g1)x2/2L E=a (L/2)2 = (g2 - g1)L/8 Xm = g1 L/(g1-g2)

Examples (1/3) 11. The highest point of 500-foot curve connecting a +4% grade to a -2% will be located feet from the PVC. a. 267 b. 300 c. 333 d. 367

Examples (2/3) 12. Given the diagram here, the length of the curve will be: a. 1303 ft b. 1424 ft c. 1512 ft d. 1626 ft -3% +1% Sta 12+75.00 Elev. 44.85 Sta 20+25.00 Elev. 30.25

Examples (3/3) 13. For a point 150 ft to the left of the PVT on a 450-foot curve connecting a -3% to -1% grade, the tangent offset will be: a. 0.5 ft b. 2.0 ft c. 3.5 ft d. 4.5 ft

Horizontal Curves L = π RI/180 T = R tan(I/2) C = 2R sin(I/2) D = 5729.58/R Stake out d = 0.5(180/π)(x/R) x = 2 R sin(d)

Examples (1/3) 18. The radius for a horizontal curve with a 650-foot long cord and intersection angle of 85o 45’ will be: a. 477.7 ft b. 479.1 ft c. 955.4 ft d. 958.2 ft

Examples (2/3) 20. The deflection angle for a the first whole station of a horizontal curve with the PC station at 34+47.50 and a radius of 350 ft will be: a. 3.9o b. 4.3o c. 7.8o d. 8.6o

Examples (3/3) 21. A horizontal curve with an external distance of 5.2 ft and a radius of 800 ft requires for the intersection angle between the tangents. a. 13o b. 20o c. 26o d. 33ο

Angle Measurements Azimuths Deflection angles Bearings from N 48o or S 228o Deflection angles 48o R (sometimes left) Bearings N 48o E N 48o

Example 27.The first sighting(PC-PI) for a horizontal curve is N 62o 30’ E and the next is S 48o 25’ E. The intersection angle for the curve is a. 27o 30’ b. 41o 35’ c. 69o 05’ d. 110o 55’

Earthwork Average end Prismoidal Pyramid V = L (A1+A2)/2 V = L (A1/2 + A2 + A3+…+An-1 + An/2) Prismoidal V = L(A1 + 4Am+A2)/6 Pyramid V= L A/3

q-k-u Relationships uf uf qmax kj qmax q=ku qmax=kj uf/4 kj

Example 27. The capacity of a roadway with free flow speed of 50 mph and maximum density of 130 vehicles per mile will be: a. 812 veh/hr b. 1625 veh/hr c. 2167 veh/hr d. 3250 veh/hr

Signalized Intersections v y = t + (2a"64.4 G) r = (W+l)/v Note: G in percentage (decimal)

Example 29. The required clearance interval for a signalized intersection approach with speeds of 35 mph, approach grade 1%, deceleration rate of 10f/s2, driver reaction time of 2 sec, average vehicle length of 20 ft, and a side street with three 12-foot lanes will be: a. 1.1 sec b. 3.0 sec c. 3.7 sec d. 4.8 sec

Elevation Measurements Differential leveling Δh = ΣBS- ΣFS hi= hi-1 +BSi-1 - FSi

Wind Analysis Determine max crosswind

Wind Analysis (cont’d) 23 23 5 Runway 5-23 Use 96% Runway 23 Use 43%