Transportation

Student copy
Module 2

Transportation
Module 2
Geometric Design

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 Transportation
Student copy
Module 2
Problem 2-1
Two tangents with bearings as shown below, intersect.

E
15
O
30

O
015 E
N 69 3

If a circular curve of radius 500 ft is to be fitted between the tangents, the length of the curve in
feet is most nearly:

(A) 150
(B) 160
(C) 170
(D) 200

Solution

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 Transportation
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Module 2

Problem 2-2
An arc of length 781 feet 10.9 inches is subtended at its center by an angle of 40o. If the arc
forms a horizontal curve, the radius of the curve, in feet, is most nearly:
L = 781' 10.9"

14%

40o

(A) 900
(B) 1,000
(C) 1,100
(D) 1,200
Solution:

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 Transportation
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Module 2
Problem 2-3
Two tangents to a right hand circular curve intersect at an angle of 110. The long chord
connecting the tangent points PC and PT is 600 ft in length. Find the radius of the circular curve.
If the station of the first tangent point, PC is 150+00, the station of the mid point of the circular
curve is most nearly:
Stations PC PI

= 110o

R Required Station

=110o

/2

PT

(A) 150+50
(B) 153+00
(C) 153+50
(D) 155+00

Solution

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 Transportation
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Module 2
Problem 2-4
Given a highway circular curve with an intersection angle = 1230 , radius of curve =
580 feet, and a PI (Point of Intersection) station = 9 + 25.5, the station of the point of tangency
(PT) is most nearly:
(A) 9+50
(B) 10+00
(C) 10+50
(D) 11+00

Solution
Tangent distance

PC (Point of Curvature) station

Curve length L

PT(Point of Tangency) station

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 Transportation
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Module 2

Problem 2-5
A circular curve connects two tangents that deflect at an angle of 58o. The point of intersection
of the tangents is located at station (732 + 82.08). If the design speed of the highway is 60 mph,
determine both the point of the tangent and the deflection angles to whole stations for laying out
the curve. Assume that superelevation will be 0.08 and that the coefficient of side friction will
be 0.12.

PI
732 + 82.08
58

PC PT

1 2

Solution
u2
R=
15(e + f s )

R=

R=

The length of the tangent, T, is given by:

T = R tan(/2)

T=

T=

The length of the curve, L, is given by:

L = R/180 =

L/2 =

PC = 732+82.08

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 Transportation
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Module 2
PT =

The first station (from PC) is located at 727+00

1 l1
=
L

and C1 = 2R sin(1/2)

where 1 = first deflection angle

l1 = first arc

C1 = first chord

= intersection angle

1 / 58 =

1 =

and 1 /2 =

from C1 = 2R sin(1/2)

C1 =

C1 =

Hint: {if C1 >100 ft, there must be a mistake}

The first deflection angle to station 727 is 1 /2 =

The first station from PT is located at

l2 =

2 = l2 /L

2 =

2 =

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 Transportation
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Module 2
C2 = 2R sin (2/2) =

C2 =

For other deflection angles:

= 100 /L

= /2 =

C = 2R sin (2/2) =

Station Deflection Angle Chord Length

72616.91 0.000 0
72700 1.984 83.09
72800 4.371 100
72900 6.758 100
73000 9.146 100
73100 11.533 100
73200 13.920 100
73300 16.308 100
73400 18.695 100
73500 21.082 100
73600 23.470 100
73700 25.857 100
73800 28.244 100
73831.66 29.000 31.66

Note: there are 2 checks that you can make to ensure that youve reached the correct set of
angles and chords. First, the last deflection angle must be half the intersection angle of the two
tangents. Second, the sum of the chords must be close to the arc length.

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 Transportation
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Module 2
Problem 2-6
A transition curve is to be designed for a horizontal curve of radius 950 feet for a highway with a
design speed of 70 miles per hour. If the rate of change of centrifugal acceleration is 2.0 feet per
second, per second, the minimum length of the transition curve, in feet, is most nearly:
(A) 550
(B) 700
(C) 750
(D) 900
Solution:
3.15ud3
L=
RC
where
L= minimum length of spiral (ft)
ud = design speed (mph)
R = curve radius (ft)
C = rate of increase of centrifugal acceleration (ft/sec2).
L=

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 Transportation
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Module 2
Problem 2-7
The design speed of a 2-lane roadway with 11 foot lanes is 50 mph. The sag curve begins at
PVC at station 47+00 as shown below. If the gradients of the two tangents are -2.0% and
+3.0%, the stopping sight distance, in feet, is most nearly:

G1 = -2.0% G2 = +3.0%

PVC PVT
STA 47+00.00 STA 52+00.00
PVI
STA 49+50.00
ELEV 4100.00

HIGHWAY VERTICAL ALIGNMENT

NOT TO SCALE

(A) 300
(B) 460
(C) 530
(D) 650

Solution
The curve length is given by the station of PVT - station of PVC =

G1 = -2.0% G2 = +3.0%

PVC PVT
STA 47+00.00 STA 52+00.00
PVI
STA 49+50.00
ELEV 4100.00

HIGHWAY VERTICAL ALIGNMENT

NOT TO SCALE

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 Transportation
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Module 2
The algebraic difference in the grades for the sag curve is given by:
A = G2 G1 =

Curvature K = curve length/ algebraic difference of grades = L/A = 500/5 = 100

Refer to 2001 AASHTO A Policy on Geometric Design of Highways and Streets, exhibit 3-
76, p.274.

For 50-mph design speed, and K = 100, the stopping sight distance should fall within the range
of 425 to 495 feet. It interpolates to about 460 ft.

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 Transportation
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Module 2
Problem 2-8
A new vertical curve is needed. There are two options. Replace the sag curve:
A. Over the existing alignment, at least 22 feet higher at station 75+00.
B. Under the existing alignment, at least 20 feet lower at station 75+00.
g y g
over existing highway

Sta. 75 + 00
-2 % %
+3
22 clearance
20 clearance Existing highway alignment

El. 510.00
New highway alignment
Sta. 78 + 00

under existing highway

El. 482.00

P.V.I

Calculate the minimum length of curve above the existing and the maximum length of curve
under the existing.
Solution
A. Replacing the curve with a new curve Over Existing highway
The vertical offset y, at any point x, along a vertical (parabolic) curve, is given by:
q p 2
y= x
2L
where p and q are the approach and following slopes of the tangents. Now when x=L/2,
y=e, the vertical offset at the PVI.
q pL
2
L
Therefore, e = = (q p)
2L 4 8
e=
p=
Therefore, L =
B. Replacing the curve with a new curve Under Existing highway
e=
p=
Therefore, L =

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 Transportation
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Module 2

Problem 2-9

An ascending 5% grade and a descending 3% grade are joined by an 800-ft vertical curve. The
elevation at Station 24 + 00 on the 5% grade is 1776.39 feet, and the elevation at Station 36 + 00
on the 3 % grade is 1780.39 feet. The station of the start of the vertical curve, PVC, is most
nearly:

(A) 25+00
(B) 25+59
(C) 26+45
(D) 27+10

800'

P.V.I

-3%
+5%
P.V.C P.V.T

1776.39 ' 1780.39'

36 + 00
24 + 00
y

Solution
Consider the vertical alignment between Station 24 and Station 36:

8y = 40

P.V.I. Sta
P.V.C. Sta

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 Transportation
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Module 2
Problem 2-10
A gradient of -1% joins a -2% grade by means of a parabolic curve of length 700 feet.

-1%

Vertical offset

-2%

The vertical offset in inches, at the point of intersection of the tangents is most nearly:

(A) 6
(B) 10
(C) 12
(D) 18

Solution
Assume that the length of the curve is equal to its horizontal projection. The actual difference
between the two can be neglected for all practical purposes. Thus L
The vertical offset 'y' at any point 'x' along the curve is given by:

[G2 G1 ]x 2 Ax 2
y= =
2L 2L

when x = L/2, y = E, the vertical offset at the point of vertical intersection (P.V.I.)

AL
E=
8

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 Transportation
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Module 2
Problem 2-11
A sag vertical curve is to be designed to join a 3 percent grade to a +3 percent grade. The
design speed is 40 mph, the coefficient of skidding friction f, is 0.32, and the perceptionreaction
time is 2.5 seconds. The minimum length of curve in feet, that will satisfy all minimum criteria
is most nearly:

(A) 450
(B) 600
(C) 700
(D) 950

Solution
Find the stopping sight distance.
u2
SSD = 1.47ut +
30( f G )

Determine whether S < L or S > L for the headlight sight distance criterion. For S > L
(400 + 3.5S )
L = 2S =
A
(This condition is not appropriate since
For S< L, then,
AS 2
L= =
400 + 3.5S
(This condition applies.)
Determine minimum length for the comfort criterion.
Au 2 6* 402
L= = = 206.5 ft
46.5 46.5
Determine minimum length for the general appearance criterion.
L = 100 A = 100 x6 = 600 ft
The minimum length to satisfy all criteria is 600 ft.

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 Transportation
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Module 2

Problem 2-12
A two lane horizontal circular curve with a design speed of 50mph is shown below. Assuming a
coefficient of side friction of 0.14 and superelevation of 0.08, a degree of curvature of 7.0O, and a
perception reaction time of 2.5 seconds, the required sight distance, in feet, between the
centerline of the road and the building is most nearly:

(A) 425
(B) 450
(C) 475
(D) 500

PI Sta 12 + 08

= 33.67

Roadway 30 ft
Centerline

Building

Solution
The curve radius corresponding to D = 7.0O is R = The required sight
distance is given by:
u2
S = 1.47ut +
30(a / g G )
u=

t=

a=

g=

assume G =

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 Transportation
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Module 2
S=

S=

Additional Calculation
The closest distance that an object can be obscuring the line of sight, m, is given by
28.65
m = R 1 cos S
R

25%m
R

20

28.65
where = S
R
28.65
So, m = 818.511 cos 423.3
818.51
m = 27.22 ft

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 Transportation
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Module 2
Problem 2-13
A two lane horizontal circular curve has a design speed of 50mph. Assuming a coefficient of side
friction of 0.14 and superelevation of 0.08, the maximum degree of curvature of the horizontal
curve is most nearly:

(A) 6.0
(B) 6.5
(C) 7.0
(D) 7.5

Solution

R = u2/15 (e + fs)

D =

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 Transportation
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Module 2

Problem 2-14
An accident analysis indicated that a car was traveling at 25 mph when it struck a pole. A site
investigation indicated that the car was traveling on a 3% downgrade. The skid marks from the
car are 400 feet long and testing of the road indicated a 0.30 coefficient of friction. If the
accident happened on a curve with a 60 mph design speed, and a radius of 1500 feet, the
appropriate super elevation, in percentage, for a safe design should be most nearly:

(A) 3
(B) 4
(C) 5
(D) 6

Solution
u2
Radius ( ft ) =
15(e + f s )
The radius of 1500 feet and the design speed of 60 mph is provided.
If a side friction factor (fs) is not given, then you should use the recommended values found in
Exhibit 3-14 on page 145 of the 2001 AASHTO A Policy on Geometric Design of Highways
and Streets , which is based on the design speed.

For a 60 mph design speed, the side friction factor is

Note that the side friction factor is different from the coefficient of friction used in the SSD
calculation. Be careful not to mix these factors because this may lead to a false solution. Also,
note that the SSD calculation is based on actual car speed or speed limit while the radius and
super elevation calculations should be based on Design Speed.

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 Transportation
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Module 2
Problem 2-15
A simple circular curve has an external angle of 100E. The design speed is 50 mph, the
corresponding value of fs is 0.14, and the maximum design value for e is 0.10. The maximum
degree of curvature, in degrees, is most nearly:
(A) 2
(B) 4
(C) 6
(D) 8

Solution
u2
e + fs =
15R
where R =

D=

u2
For a radius of 800 ft, Eq. ed es = f s for R>R min yields
gR
edes = 0.21 0.14 = 0.07 ft/ft

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 Transportation
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Module 2
Problem 2-16
The space mean speed of the curb lane on a freeway is 55 mph. An exit ramp is designed for a
safe speed of 25 mph. Assume a drive perception reaction time of 2.5 seconds and a deceleration
rate of 5 ft/sec2. The curb lane is on a downgrade of 1%.

Curb Lane EXIT

Exit Ramp
Safe Speed
Required
25 mph
Distance

Exit
Sign

The sign is to be located so that drivers will have enough time to see it, react, and exit the
freeway safely. Empirical studies show that drivers are able to read the sign at a distance of 180
ft. The required distance that the sign should be located measured from the exit ramp, in feet, is
most nearly:
(A) 400
(B) 585
(C) 600
(D) 750
Solution
u12 u22
S = 1.47u1t +
a
30 G
g
where
S = stopping (slowing) sight distance, feet

u1 =

t =

u2 =

a =

g =

G=

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 Transportation
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Module 2
S=

S=

Deduct for the distance drivers can read the sign:

Required distance of sign from exit =

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 Transportation
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Module 2

Problem 2-17
An alert driver (with a reaction time 0.5 second) is driving downhill on a 4% grade at 35 mph on
a dry pavement when suddenly a person steps from behind a parked car in the path of the driver,
at a distance of 125 ft.
(a) Can the driver stop in time with emergency braking assuming a deceleration rate of 14.8
ft/s2?
(b) Can the driver stop in time on a rainy day with comfortable braking assuming a
deceleration rate of 11.2ft/s2?
Solution

(a) S = + 1.47 35 0.5 = ft

0.04

(b) S = + 1.47 35 0.5 = ft

0.04

Discussion
The driver's reaction time, the condition of the road pavement, vehicle braking system, and
the prevailing weather all play a significant role in this problem.

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 Transportation
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Module 2
Problem 2-18

A deceleration lane is being provided on a freeway for traffic at an exit ramp. Assuming a 65-
mph initial speed and a 2% downgrade, in order to provide sufficient space to reduce a vehicle
speed to 25-mph, the braking deceleration lane length, in feet, is most nearly:

(A) 305
(B) 370
(C) 444
(D) 516

Solution
The question asks how long the braking deceleration lane should be so in the stopping sight
distance equation, only the braking portion is used for the solution. The PIEV distance is not
necessary and if you include it you will arrive at an incorrect solution.

If a coefficient of friction is not given, then you can derive it from a/g where a is the assumed
deceleration rate and g is gravity. For the recommended braking rate, refer to the AASHTO
section on Braking Distance and Effect of Grade on Stopping p.111-114, for determining
stopping sight distances: 11.2 ft/sec2.
a
AASHTO represents f as .
g
The equation for braking distance is
(u12 u2 2 ) (u12 u2 2 )
Db = =
30( f G ) 30(a / g G )
Where:
f = a/g
a=
g=
G=
u1 =
u2 =
(u12 - u2 2 )
Braking Distance = Db = =
30( f G )
So the braking deceleration length is

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 Transportation
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Module 2
Problem 2-19
A car approaches a railroad crossing at 50 miles per hour. The wet pavement coefficient of
friction is 0.30 and the highway grade is 3.0%. The stop line is 20 feet from the track. For the
vehicle to stop at the stop line, the distance a driver must see the train from the rail, in feet, is
most nearly:

(A) 310
(B) 480
(C) 515
(D) 535

Solution
SSD in feet = reaction distance plus braking distance.
Assume tp = 2.5 sec
Given:
tp = 2.5 sec, f = .G= , u1 = mph, u2 = mph
u12 u22
SSD = t p (1.47)(u1 ) +
30( f G )
SSD =

SSD =
Distance from Rail = SSD + stop line from rail =

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 Transportation
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Module 2
Problem 2-20
An accident analysis indicated that a car was traveling at 25 miles per hour when it struck a
telegraph pole. A site investigation indicated that the car was traveling on a 3% downgrade. The
skid marks from the car were 400 feet long and testing of the road surface indicated a 0.30
coefficient of friction. The original speed of the car, in miles per hour, is most nearly:

(A) 56
(B) 62
(C) 68
(D) 74

Solution
The 400 foot long skid marks are related to the braking distance portion of the SSD. Using the
braking portion of the SSD equation, solve for the initial speed (u1).
u12 u22
Braking Distance (Db-) =
30( f G )
Db =
f =
G =
u1 =
u2 =
Although speed and friction coefficient are related, do not assume you can simply look up the
coefficient of friction and find the speed. In this case, the friction was found from testing.
u 2 u22
Braking Distance (Db) = = 1 =
30( f G )
Solving for u1 =

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 Transportation
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Module 2
Problem 2-21
A horizontal curve has a 440-foot radius with 3% super elevation. The design speed of the
curve, in miles per hour, is most nearly:

(A) 30
(B) 35
(C) 40
(D) 45

Solution
u2
Radius ( ft ) =
15(e + f s )
Given the radius of 440 feet and super elevation of 3%, two unknowns remain: the design speed
and the side friction factor. Since the two are related, the solution needs a trial and error
approach using the side friction factor and corresponding design speed found in Exhibit 3-14 on
page 145 of the 2001 AASHTO A Policy on Geometric Design of Highways and Streets.

By trial and error,

Radius at mph (with fs = )=
Radius at mph (with fs = )=
Radius at mph (with fs = ) =
Radius at mph (with fs = )=
So, from the trials, when fs = the radius is . Therefore, the associated speed of
mph is the design speed.

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