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 ftb. 399.1 ftc. 433.8 ftd. 444.0 ft

Crest Vertical Curves

h2

h1

S>L L = 2 S - 200(h1+h2)2 /|A|

SL L = |A| S2 / [200(h1+ h2)2]

SSD

L

h1 = 3.5 ft and h2 =2.0 ftS>L L = 2 S - 2158/|A| S L L = |A| S2/2158

A = G2 – G1 Note: Gi in grade

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 ftb. 579.2 ftc. 587.7 ftd. 594.1 ft

Sag Vertical Curves

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

HSSD

L

β

A = G2 – G1 Note: Gi in grade

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 ftb. 531.57 ftc. 533.33 ftd. 535.85 ft

Horizontal Curves

Superelevation 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 ftb. 7.4 ftc. 8.2 ftd. 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. 0.08d. 0.10

Vertical Curves

y = (g2 - g1)x2/2LYP= YPVC + g1 x + (g2 - g1)x2/2LE=a (L/2)2 = (g2 - g1)L/8Xm = g1 L/(g1-g2)

PVC

PVI

PVT

g1 g2

L/2 L/2x

yE

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. 300c. 333d. 367

Examples (2/3)

12. Given the diagram here, the length of the curve will be:

a. 1303 ftb. 1424 ftc. 1512 ftd. 1626 ft

-3% +1%

Sta

12+

75.0

0Ele

v.

44.8

5

Sta

20+

25.0

0Ele

v.

30.2

5

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 ftb. 2.0 ftc. 3.5 ftd. 4.5 ft

Horizontal Curves

L = π RI/180T = R tan(I/2)C = 2R sin(I/2)D = 5729.58/RStake out

d = 0.5(180/π)(x/R) x = 2 R sin(d)

PC

PI

PT

I

I

RR

D

100'

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 ftb. 479.1 ftc. 955.4 ftd. 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 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 isa. 27o 30’b. 41o 35’c. 69o 05’d. 110o 55’

Earthwork

Average end 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

kjqmax

qmax

kj

uf

q=kuqmax=kj uf/4

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/hrb. 1625 veh/hr c. 2167 veh/hrd. 3250 veh/hr

Signalized Intersections

vy = t +

(2a64.4 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 secb. 3.0 sec c. 3.7 secd. 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)

Runway 5-23Use 96%

5

23

Runway 23Use 43%

23