1. 20/01/2013 1

2. General Equation of Parabola
Equal Tangent Vertical Curves
y  a x2  b x  c
y  a x 2  g1 x  (elevation of BVC )
A
2a  r
L
r 2
y x  g1 x  (elevation of BVC )
2
r is the rate of change of grade per station.
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3. Design of a Vertical Curve
A vertical curve design convex/concave means determining;
• Radius of curvature R,
• Central angle subtended by circular curve , which is the intersection angle of the
two lines of alignments 1 and 2.
• Length of the circular curve L.
Curves Radii R (m)
Design Speed Road
Km/hr Classification
Concave Convex
120 Highway 5,000 15,000
100 1st Order 3,000 10,000
80 2nd Order 2,000 5,000
60 3rd Order 1,500 3,000
40 4th Order 1,200 2,000
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4. Design of a Vertical Curve
Problem:
Design a concave vertical curve appropriate for connecting two lines of alignment 1
and 2 of 1st order road, their grades are g1 = - 2% and g2 = +3%. The elevation of
intersection point V is 390.55 m.
1 R
2

V
Solution:
As the road is of 1st order, the design speed = 100 km/hr match R = 3,000.00 m.
See previous Table.
As the grades of the two lines of alignment 1 and 2 are very small,
 = (g2 – g1)% = (+ 3 – (- 2))% = 0.05 rad.
The curve length, L = R  = 150.00 m.
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5. Passing a Vertical Curve
Through a Fixed Point
Problem:
Using the grades, intersection station and elevation given in Figure below, it is
required to the vertical curve to pass through station 18 + 25 at an elevation of
881.20 ft. What should be the new length of the curve?
18 + 00 V
g1 = + 1.25% 886.10 ft
g2 = - 2.75%
15 + 00
BVC 21 + 00
300 ft
EVC
600 ft
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6. Passing a Vertical Curve
Through a Fixed Point
Solution:
The curve must be lengthened. The new value of L must be computed and a new
value for r results.
r 2
y x  g1 x  (elevation of BVC )
2
r A g  g1  2.75  1.25
  2 
2 2L 2L 2L
L
x   0.25
2
gL
elevation of BVC  886.10  1
2
r 2
y x  g1 x  (elevation of BVC )
2
 2.75  1.25 L L 1.25L
881.21  [ (  0.25) 2 ]  [1.25(  0.25)]  [886.10  ]
2L 2 2 2
L2  9.425L  0.25  0
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7. Passing a Vertical Curve Through a Fixed Point
Solution Continued:
L2 – 9.425L + 0.25 = 0
Solving for L gives:
L = 9.3984 stations and L/2 = 4.6992 stations.
The elevation of BVC is:
886.10 – 1.25  4.6992 = 880.23 ft.
The station of BVC is:
18 – 4.6992 = 13.3008 stations.
The value x is:
18 – 13.3008 = 4.9492
r/2 = (- 2.75 – 1.25)/(2  9.3984) = - 0.2128%/station.
Then:
y = - 0.2128  4.94922 + 1.25  4.9492 + 880.23 = 881.20 ft.
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8. Unequal Tangent Vertical Curve
In Figure below;
l1 l2 a line v1v2 is drawn parallel with AB
Av1 = v1V and Vv2 = v2B
The curve from A to K is
L
consisted from two equal
parabolic vertical curves.
l1 l2
The curve from K to B is
EVC consisted from two equal
C parabolic vertical curves.
BVC (B)
K
g v2
(A) v1
V
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9. Unequal Tangent Vertical Curve
In Figure below;
The elevation of v1 is the average of that of A and V.
The elevation of v2 is the average of that of V and B.
L
CK = KV
l1 l2
CV = l1l2/L(g2 – g1)
CK = l1l2/2L(g2 – g1)
EVC
C l1, l2 and L are distances in
BVC (B)
K stations and g2 and g1 are
g v2 percent grades.
(A) v1
V
The grade g of v1v2 is the same as AB; g = (elev. B – elev. A)/L
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10. Unequal Tangent Vertical Curve
Problem:
In Figure below, g1 = - 4%, g2 = + 3.5%, l1 = 250 ft, l2 = 400 ft, the elevation
of V is 450.00 ft. Point V is at station 55 + 00. Compute the elevations of
each half station between A and B.
Solution:
L The elevation of A is:
450.00 + 2.50  4 = 460.00 ft.
The elevation of B is:
l1 l2 450.00 + 4.00  3.5 = 464.00 ft.
g = (464.00 – 460.00)/6.50 = + 0.615%
EVC
C Elevation of K:
BVC (B) 460.00 - 1.25  4 + 1.25  0.615 = 455.77 ft.
K
g v2
(A) v1 Elevation of B:
V
455.77+2.00  0.615+2.00  3.5 = 464.00 ft.
(check).
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11. Unequal Tangent Vertical Curve
Solution Continued:
For the vertical curve from A to K:
r = (0.615 – (- 4.00))/2.50 = 1.846%/station
Elevation of BVC at A = 460.00 ft.
The equation of the curve is:
y = 0.923 x2 – 4x + 460.00
For the vertical curve from K to B:
r = (3.5 – 0.615)/4.00 = + 0.721%/station
Elevation of BVC at K = 455.77 ft.
The equation of the curve is:
y = 0.361 x2 + 0.615x + 455.77
The computations for the two curves are shown in Table
below:
* The low point on this curve is at station 54 + 66.68 at
which the elevation is 455.67.
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12. Unequal Tangent Vertical Curve
Solution Continued:
Station x x2 (r/2)x2 g1x Elev. BVC Elev. Curve
52 + 50 0 0 0 0 460.00 460.00
53 + 00 0.5 0.25 + 0.23 - 2.00 460.00 458.23
53 + 50 1 1 + 0.92 - 4.00 460.00 456.92
54 + 00 1.5 2.25 + 2.08 - 6.00 460.00 456.08
54 + 50 2 4 + 3.69 - 8.00 460.00 455.69
55 + 00 2.5 6.25 + 5.77 - 10.00 460.00 455.77
55 + 50 0.5 0.25 + 0.09 + 0.31 455.77 456.17
56 + 00 1 1 + 0.36 + 0.62 455.77 456.75
56 + 50 1.5 2.25 + 0.81 + 0.92 455.77 457.50
57 + 00 2 4 + 1.44 + 1.23 455.77 458.44
57 + 50 2.5 6.25 + 2.26 + 1.54 455.77 459.57
58 + 00 3 9 + 3.25 + 1.85 455.77 460.87
58 + 50 3.5 12.25 + 4.42 + 2.15 455.77 462.34
59 + 00 4 16 + 5.78 + 2.46 455.77 464.00
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13. Minimum Length of Vertical Curve
The length of a vertical curve on a highway should be ample to provide a clear
sight that is sufficiently long to prevent accidents.
The American Association of State Highway and Transportation Officials,
AASHTO has developed criteria for the,
safe passing sight distance, Ssp and
safe stopping sight distance, Snp
Assumption:
• The eyes of the driver of a vehicle are about 3.75 ft above the pavement.
• The top of an oncoming vehicle is about 4.50 ft above the pavement.
• An obstruction ahead of the vehicle is 0.50 ft.
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14. Minimum Length of Vertical Curve
The values of length of Ssp and Snp are given in the following Table:
AASHTO Sight-Distance Recommendations
Design Speed (mph) Ssp (ft) Snp (ft)
30 1,100 200
40 1,500 275
50 1,800 350
60 2,100 475
70 2,500 600
80 2,700 750
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15. Minimum Length of Vertical Curve
The values of minimum length of a vertical curve are given here
without proof:
a ) If , S sp  L
S sp ( g1  g 2 )
2
L
33
b) If , S sp  L
33
L  2 S sp 
g1  g 2
a ) If , S np  L
S np ( g1  g 2 )
2
L
14.0
b) If , S np  L
14.0
L  2 S np 
g1  g 2
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16. Minimum Length of Vertical Curve
Problem:
The grades at a crest are g1 = + 2% and g2 = - 3% and the design speed is 60
mph. Compute the lengths of the vertical curves required for the safe passing
sight distance, Ssp and the safe stopping sight distance, Snp recommended by
the AASHTO.
Solution:
From previous Table, the safe passing sight distance, Ssp = 2100 ft and the
safe stopping sight distance, Snp = 475 ft.
If , S sp  L
S sp ( g1  g 2 )
2
2100 2  0.05
L   6682 ft
33 33
If , S np  L
S np ( g1  g 2 )
2
475 2  0.05
L   806 ft
14.0 14.0
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17. Minimum Length of Vertical Curve
Continued Solution:
Since the safe passing sight distance, Ssp = 2100 ft  L = 6682 ft and the safe
stopping sight distance, Snp = 475 ft  L = 806 ft, it would not be necessary to
apply the cases b) of Ssp and Snp longer than the curve.
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18. Crowns & Superelevations
x
0.015 – 0.020 ft/ft y1 y
W/2
Pavement Cross-Section Showing
W
Pavement Crown
Horizontal Curves, 8% superelevation
W = 24 ft, e = 0.08 (24) = 1.92 m
The curve to be banked
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19. Crowns & Superelevations
• Design Speed of the highway
• Side Friction
Climate and Area Classification (Urban/Rural)
Snow (7% - 8%)
10% - 12% the highest for gravel roads with cross drainage
Urban streets (4% - 6%)
Runoff, is the transition from the normal crown cross section on a
tangent to the fully superelevated cross section.
The length of the runoff transition may range from (500 ft – 600 ft) or
(100 ft– 250 ft).
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20. Crowns & Superelevations
4
3
Circular curve 2 1
3
2
1
Normal 3
3 4 2 1
2
1
Convex
Transition section
Normal crowned section
Section 1 - 1
Level Convex
Maximum superelevation
Transition section
Section 2 - 2
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